Experiments With MatLab Moled 2011
144
Experiments with MATLAB Cleve Moler October 4, 2011 ii Copyright 2011 Cleve Moler Electronic edition published by MathWorks, Inc. http://www.mathworks.com/moler
Transcript of Experiments With MatLab Moled 2011
Experiments
withMATLAB
Cleve
Moler
October
4,2011
ii
Cop
yrigh
t2011
Cleve
Moler
Electronic
editionpublished
byMathWorks,
Inc.
http://www.mathworks.com/moler
Contents
Pre
face
iii
1Itera
tion
1
2FibonacciNumbers
17
3Calendars
and
Clock
s33
4M
atrices
45
5LinearEquations
63
6Fra
cta
lFern
75
7Google
PageRank
83
8ExponentialFunction
97
9T
Puzzle
113
10
Magic
Square
s123
11
TicTacToeM
agic
141
12
GameofLife
151
13
Mandelbro
tSet
163
14
Sudoku
183
15
Ord
inary
Differe
ntialEquations
199
16
Pre
dato
r-Pre
yM
odel
213
17
Orb
its
221
18
Shallow
WaterEquations
241
iii
ivCon
tents
19
MorseCode
247
20
Music
263
Experiments
withMATLAB
R ©
Cleve
Moler
Copyrightc ©
2011CleveMoler.
Allrights
reserved.Nopart
ofthis
e-bookmay
bereproduced,stored,ortrans-
mitted
inan
ymanner
withoutthewritten
permission
oftheau
thor.
For
more
inform
ation,contact
Theprogramsdescribed
inthis
e-bookhavebeenincluded
fortheirinstructional
value.
Theseprogramshavebeentested
withcare
butare
notguaranteed
foran
yparticularpurpose.
Theau
thordoes
notoffer
anywarranties
orrepresentations,
nordoes
heacceptanyliabilitieswithrespectto
theuse
oftheprograms.
These
programsshould
notberelied
onas
thesolebasisto
solveaproblem
whose
incorrect
solutioncould
result
ininjury
topersonorproperty.
Matlab
R ©is
aregisteredtradem
ark
ofMathWorks,
Inc.
TM.
For
more
inform
ationaboutrelevantMathWorkspolicies,see:
http://www.mathworks.com/company/aboutus/policies_statements
October
4,2011
iiCon
tents
Preface
Figure
1.exmguiprovides
astartingpointforsomeoftheexperimen
ts.
Welcometo
Experim
ents
withMATLAB.This
isnotaconventional
book
.It
iscu
rrentlyavailable
only
via
theInternet,atnocharge,
from
http://www.mathworks.com/moler
Theremay
eventuallybeahardcopyedition,butnotrightaw
ay.
AlthoughMatlabis
now
afullfledged
TechnicalComputingEnvironment,
itstarted
inthelate
1970sasasimple
“Matrix
Laboratory”.Wewantto
build
onthislaboratory
traditionbydescribingaseries
ofexperim
ents
involvingapplied
mathem
atics,technicalcomputing,andMatlabprogramming.
iii
ivPreface
Weexpectthatyou
alreadyknow
somethingab
outhighschoollevel
material
ingeometry,algebra,an
dtrigonometry.
Wewillintroduce
ideasfrom
calculus,
matrix
theory,an
dordinary
differentialequations,
butwedonotassumethatyou
havealreadytaken
courses
inthesubjects.
Infact,theseexperim
ents
areuseful
supplements
tosuch
courses.
Wealsoexpectthatyou
havesomeexperience
withcomputers,perhapswith
wordprocessors
orspreadsheets.If
you
know
somethingaboutprogrammingin
languages
likeC
orJava,thatwillbehelpful,butnotrequired
.Wewillintroduce
Matlabbyway
ofexamples.
Manyoftheexperim
ents
involveunderstandingand
modifyingMatlabscripts
andfunctionsthatwehavealreadywritten
.You
should
haveaccess
toMatlab
andto
ourexmtoolbox
,thecollection
ofprogramsan
ddata
thatare
described
inExperim
ents
withMATLAB.Wehop
eyou
willnoton
lyuse
theseprograms,butwillreadthem
,understandthem
,modify
them
,an
dim
prove
them
.Theexmtoolbox
istheapparatusin
our“Laboratory”.
Youwillwantto
haveMatlabhandy.
Forinform
ationab
outtheStudent
Version,see
http://www.mathworks.com/academia/student_version
For
anintroductionto
themechanicsofusingMatlab,seethevideosat
http://www.mathworks.com/academia/student_version/start.html
For
documentation,including“GettingStarted”,see
http://www.mathworks.com/access/helpdesk/help/techdoc/matlab.html
For
usercontributedprograms,programmingcontests,andlinksinto
theworld-w
ide
Matlabcommunity,
checkout
http://www.mathworks.com/matlabcentral
Toget
started,dow
nloadtheexmtoolbox
,use
pathtoolto
addexmto
the
Matlabpath,andrunexmguito
generate
figure
1.Youcanclickon
theiconsto
preview
someof
theexperim
ents.
Youwillwantto
makefrequentuse
oftheMatlabhelpanddocumentation
facilities.Toquickly
learn
how
touse
thecommandorfunctionnamed
xxx,enter
helpxxx
For
more
extensiveinform
ationab
outxxx,use
docxxx
Wehopeyou
willfindtheexperim
ents
interesting,andthatyouwilllearn
how
touse
Matlabalongtheway.Each
chapterconcludes
witha“Recap”section
thatis
actuallyanexecutable
Matlabprogram.Forexample,you
canreview
the
Magic
Squareschapterbyentering
magic_recap
Preface
v
Betteryet,enter
editmagic_recap
andruntheprogram
cell-by-cellbysimultaneouslypressingtheCtrl-Shift-Enter
keys.A
fairly
new
Matlabfacility
isthepublishcommand.You
canget
anicely
form
atted
web
pageaboutmagic_recapwith
publishmagic_recap
Ifyou
wantto
concentrate
on
learningMatlab,makesure
you
read,run,and
understandtherecaps.
CleveMoler
October
4,2011
viPreface
Chapter1
Iteration
Iterationis
akeyelem
entin
much
oftechnicalcomputation.Examplesinvolvingthe
Golden
Ratiointroduce
theMatlabassignmen
tstatemen
t,forandwhileloops,
andtheplotfunction.
Start
bypickinganumber,an
ynumber.Enteritinto
Matlabbytyping
x=yournumber
This
isaMatlabassignmen
tstatemen
t.Thenumber
youchose
isstoredin
the
variablexforlateruse.For
example,ifyoustart
with
x=
3
Matlabrespondswith
x=
3
Next,enterthis
statement
x=
sqrt(1
+x)
Theabbreviationsqrtis
theMatlab
nameforthesquare
rootfunction.
The
quantity
ontheright,√ 1
+x,iscomputedandtheresultstoredback
inthevariable
x,overridingthepreviousvalueof
x.
Somew
hereon
yourcomputerkeyboard,probably
inthelower
rightcorner,
you
should
beableto
findfourarrow
keys.
Theseare
thecommandlineeditingkeys.
Theup-arrow
key
allow
syou
torecallearliercommands,includingcommandsfrom
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
1
2Chapter1.
Iteration
previoussessions,
andtheother
arrowskeysallow
youto
revisethesecommands.
Use
theup-arrow
key,follow
edbytheenteror
return
key,to
iterate,or
repeatedly
execute,this
statement:
x=
sqrt(1
+x)
Hereis
whatyou
getwhen
you
start
withx=3.
x=
3
x=
2
x=
1.7321
x=
1.6529
x=
1.6288
x=
1.6213
x=
1.6191
x=
1.6184
x=
1.6181
x=
1.6181
x=
1.6180
x=
1.6180
Thesevalues
are
3,√ 1
+3,√ 1
+√ 1
+3,
√ 1+
√ 1+
√ 1+3,andso
on.After
10steps,
thevalueprintedremainsconstantat1.6180.Try
severalother
starting
values.Try
iton
acalculatorifyou
haveon
e.Youshould
findthatnomatter
where
you
start,youwillalwaysreach
1.6180in
abouttensteps.
(May
beafew
more
will
berequired
ifyouhaveaverylargestartingvalue.)
Matlabis
doingthesecomputationsto
accuracy
ofabout16
decim
aldigits,
butis
displayingonly
five.
Youcanseemore
digitsbyfirstentering
formatlong
andrepeatingtheexperim
ent.
Hereare
thebeginningandendof30stepsstarting
atx=
3.
x=
3
3
x=
2
x=
1.732050807568877
x=
1.652891650281070
....
x=
1.618033988749897
x=
1.618033988749895
x=
1.618033988749895
After
aboutthirty
orso
steps,
thevaluethatis
printeddoesn’t
changeanymore.
You
havecomputed
oneof
themost
famousnumbersin
mathem
atics,φ,the
Golden
Ratio.
InMatlab,an
dmost
other
programminglanguages,theequals
signis
the
assignmentoperator.
Itsayscompute
thevalueon
therightandstore
itin
the
variable
ontheleft.So,thestatement
x=
sqrt(1
+x)
takes
thecurrentvalueofx,computessqrt(1+
x),andstorestheresult
back
inx.
Inmathem
atics,theequals
signhasadifferentmeaning.
x=
√ 1+
x
isanequation.A
solutionto
such
anequationisknow
nas
afixedpoint.
(Becareful
notto
confuse
themathem
aticalusageoffixedpointwiththecomputerarithmetic
usageoffixedpoint.)
Thefunctionf(x)=
√ 1+
xhasexactly
onefixed
point.
Thebestway
tofindthevalueofthefixed
pointis
toavoid
computers
alltogether
andsolvethe
equationusingthequadraticform
ula.Take
alookatthehandcalculationshow
nin
figure
1.1.Thepositiverootofthequadraticequationis
theGolden
Ratio.
φ=
1+
√ 5
2.
YoucanhaveMatlabcompute
φdirectlyusingthestatement
phi=
(1
+sqrt(5))/2
Withformatlong,thisproducesthesamevalueweobtained
withthefixed
point
iteration,
phi=
1.618033988749895
4Chapter1.
Iteration
Figure
1.1.Compute
thefixedpointby
hand.
−10
12
34
−1
−0.50
0.51
1.52
2.53
3.54
Figure
1.2.A
fixedpointatφ=
1.6180.
Figure
1.2isourfirstexample
ofMatlabgraphics.
Itshow
stheintersection
ofthegraphsofy=
xandy=
√ 1+
x.Thestatement
x=
-1:.02:4;
generatesavectorxcontainingthenumbersfrom
-1to
4in
stepsof.02.
The
statements
y1
=x;
y2
=sqrt(1+x);
plot(x,y1,’-’,x,y2,’-’,phi,phi,’o’)
5
produce
afigure
thathasthreecomponents.Thefirsttw
ocomponents
are
graphs
ofxand√ 1
+x.The’-’argumenttellstheplotfunctionto
drawsolidlines.The
last
componentin
theplotis
asingle
pointwithbothcoordinatesequalto
φ.The
’o’tellstheplotfunctionto
draw
acircle.
TheMatlabplotfunctionhasmanyvariations,
includingspecifyingother
colors
andlinetypes.You
canseesomeofthepossibilitieswith
helpplot
φ
φ −
1
1
1
Figure
1.3.Thegolden
rectangle.
TheGolden
Ratioshow
supin
manyplacesin
mathem
atics;we’llseeseveral
inthis
book.TheGolden
Ratiogets
itsnamefrom
thegolden
rectangle,show
nin
figure
1.3.Thegolden
rectangle
hastheproperty
thatremov
ingasquare
leaves
asm
aller
rectangle
withthesameshape.
Equatingtheaspectratiosoftherectangles
gives
adefiningequationforφ:
1 φ=
φ−
1
1.
Multiplyingboth
sides
ofthisequationbyφproducesthesamequadraticpolynomial
equationthatweobtained
from
ourfixed
pointiteration.
φ2−φ−1=
0.
Theup-arrow
key
isaconvenientway
torepeatedly
execute
asinglestatement,
orseveralstatements,separatedbycommasorsemicolons,
onasingle
line.
Two
more
pow
erfulconstructsare
theforloopandthewhileloop.Aforloopexecutes
ablock
ofcodeaprescribed
number
oftimes.
x=
3
fork
=1:31
x=
sqrt(1
+x)
end
6Chapter1.
Iteration
produces32lines
ofoutput,on
efrom
theinitialstatementan
donemore
each
time
throughtheloop.
Awhileloopexecutesablock
ofcodean
unknow
nnumber
oftimes.Termi-
nationiscontrolled
byalogicalexpression,whichevaluatesto
trueor
false.Here
isthesimplest
whileloopforou
rfixed
pointiteration.
x=
3
whilex
~=
sqrt(1+x)
x=
sqrt(1+x)
end
This
producesthesame32
lines
ofoutputastheforloop.How
ever,this
codeis
open
tocriticism
fortw
oreasons.
Thefirstpossible
criticism
involves
theterm
i-nationcondition.Theexpressionx
~=
sqrt(1+x)is
theMatlabway
ofwriting
x�=
√ 1+
x.W
ithexact
arithmetic,xwou
ldnever
beexactly
equalto
sqrt(1+x),
theconditionwou
ldalway
sbetrue,
andtheloopwou
ldrunforever.How
ever,like
most
technicalcomputingenvironments,Matlabdoes
notdoarithmetic
exactly.
Inorder
toeconomizeonbothcomputertimeandcomputermem
ory,Matlabuses
floatingpointarithmetic.Eventuallyourprogram
producesavalueofxforwhich
thefloatingpointnumbersxandsqrt(1+x)are
exactly
equalandtheloopterm
i-nates.
Expectingexact
equality
oftw
ofloatingpointnumbersisadelicate
matter.
ItworksOK
inthisparticularsituation,butmay
notworkwithmore
complicated
computations.
Thesecondpossiblecriticism
ofoursimplewhileloopisthatitisineffi
cient.
Itevaluatessqrt(1+x)tw
iceeach
timethroughtheloop
.Hereisamore
complicated
version
ofthewhileloopthatavoidsbothcriticisms.
x=
3
y=
0;
whileabs(x-y)>
eps(x)
y=
x;
x=
sqrt(1+x)
end
Thesemicolonsat
theendsoftheassignmentstatements
involvingyindicate
that
noprintedoutputshould
result.Thequantity
eps(x),isthespacingof
thefloating
pointnumbersnearx.Mathem
atically,
theGreek
letter
ε,orepsilon,often
rep-
resents
a“small”quantity.This
versionoftheloop
requires
only
onesquare
root
calculationper
iteration,butthatis
overshadow
edbythead
ded
complexityof
the
code.
Both
whileloopsrequireaboutthesameexecutiontime.
Inthissituation,I
preferthefirstwhileloopbecause
itis
easier
toreadan
dunderstand.
HelpandDoc
Matlabhasextensiveon-linedocumentation.Statements
like
helpsqrt
helpfor
7
providebrief
descriptionsof
commandsandfunctions.
Statements
like
docsqrt
docfor
providemore
extensivedocumentationin
aseparate
window
.Oneob
scure,butveryim
portant,helpentryisab
outthevariouspunctuation
marksandspecialcharactersusedbyMatlab.Takealooknow
at
helppunct
docpunct
You
willprobably
wan
tto
return
tothis
inform
ation
asyou
learn
more
abou
tMatlab.
Numbers
Numbersareform
edfrom
thedigits0
through9,an
optionaldecim
alpoint,
aleading+or-sign,an
optionalefollow
edbyaninteger
forapow
erof10
scaling,
andan
optionalior
jfortheim
aginary
part
ofacomplexnumber.Matlabalso
know
sthevalueof
π.Herearesomeexamplesofnumbers.
42
9.6397238
6.0221415e23
-3+4i
pi
Assignmen
tstatemen
tsandnames
Asimple
assignmentstatementconsistsofaname,
an=
sign,andanumber.The
names
ofvariables,functionsan
dcommandsare
form
edbyaletter,follow
edbyan
ynumber
ofupper
andlower
case
letters,
digitsandunderscores.
Single
character
names,likexandN,an
danglicizedGreek
letters,
likepiandphi,are
often
used
toreflectunderlyingmathem
aticalnotation.Non-m
athem
aticalprogramsusually
employlongvariable
names.Underscoresan
daconventionknow
nascamel
casing
are
usedto
create
variable
names
outofseveralwords.
x=
42
phi=
(1+sqrt(5))/2
Avogadros_constant=
6.0221415e23
camelCaseComplexNumber=
-3+4i
Exp
ressions
Pow
eris
denoted
by^and
hasprecedence
over
all
other
arithmetic
operations.
Multiplicationanddivisionareden
otedby*,/,and\andhaveprecedence
over
additionandsubtraction,Additionandsubtractionare
denotedby+and-and
8Chapter1.
Iteration
havelowestprecedence.
Operationswith
equalprecedence
are
evaluated
left
toright.
Parentheses
delineate
subexpressionsthatareevaluatedfirst.
Blankshelp
readability,
buthavenoeff
ectonprecedence.
All
ofthefollow
ingexpressionshavethesamevalue.
Ifyou
don’t
already
recognizethis
value,
youcanaskGoogle
abou
titsim
portance
inpopularculture.
3*4+
5*6
3*
4+5
*6
2*(3+
4)*3
-2^4+
10*29/5
3\126
52-8-2
Rec
ap
%%
Iteration
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Iteration
chapter
of
"Experiments
in
MATLAB".
%You
can
run
it
by
entering
the
command
% %iteration_recap
% %Better
yet,
enter
% %edit
iteration_recap
% %and
run
the
program
cell-by-cell
by
simultaneously
%pressing
the
Ctrl-Shift-Enter
keys.
% %Enter
% %publish
iteration_recap
% %to
see
aformatted
report.
%%
Help
and
Documentation
%help
punct
%doc
punct
%%
Format
formatshort
100/81
formatlong
100/81
formatshort
9
formatcompact
%%Names
and
assignment
statements
x=
42
phi=
(1+sqrt(5))/2
Avogadros_constant=
6.0221415e23
camelCaseComplexNumber=
-3+4i
%%Expressions
3*4+
5*6
3*
4+5
*6
2*(3+
4)*3
-2^4+
10*29/5
3\126
52-8-2
%%Iteration
%Use
the
up-arrow
key
to
repeatedly
execute
x=
sqrt(1+x)
x=
sqrt(1+x)
x=
sqrt(1+x)
x=
sqrt(1+x)
%%For
loop
x=
42
fork
=1:12
x=
sqrt(1+x);
disp(x)
end
%%While
loop
x=
42;
k=
1;
whileabs(x-sqrt(1+x))>
5e-5
x=
sqrt(1+x);
k=
k+1;
end
k
%%Vector
and
colon
operator
k=
1:12
x=
(0.0:
0.1:
1.00)’
%%Plot
x=
-pi:
pi/256:
pi;
y=
tan(sin(x))
-sin(tan(x));
10Chapter1.
Iteration
z=
1+
tan(1);
plot(x,y,’-’,pi/2,z,’ro’)
xlabel(’x’)
ylabel(’y’)
title(’tan(sin(x))-
sin(tan(x))’)
%%
Golden
Spiral
golden_spiral(4)
Exe
rcises
1.1
Expressions.
Use
Matlabto
evaluate
each
ofthesemathem
aticalexpressions.
432
−34
sin1
4(32)
(−3)
4sin1◦
(43)2
4√ −3sin
π 34√ 32
−2−4/3
(arcsin1)/π
You
canget
started
with
help^
helpsin
1.2
Tem
perature
conversion.
(a)Write
aMatlabstatementthatconvertstemperature
inFah
renheit,f,to
Cel-
sius,c. c=somethinginvolvingf
(b)Write
aMatlabstatementthatconverts
temperature
inCelsius,c,to
Fah
ren-
heit,f. f=somethinginvolvingc
1.3
Barn-m
egaparsec.A
barn
isaunit
ofareaem
ployedbyhighenergyphysicists.
Nuclearscatteringexperim
ents
tryto
“hit
thesideof
abarn”.A
parsec
isaunit
oflength
employed
byastronomers.
Astarat
adistance
ofoneparsec
exhibits
atrigonometricparallaxof
onearcsecondastheEarthorbitstheSun.
Abarn-
megaparsec
istherefore
aunit
ofvolume–averylongskinnyvolume.
Abarn
is10
−28square
meters.
Amegaparsec
is106parsecs.
Aparsec
is3.262light-years.
Alight-yearis
9.461
·101
5meters.
Acubic
meter
is106milliliters.
Amilliliteris
1 5teaspoon.
11
Express
onebarn-m
egaparsec
inteaspoons.
InMatlab,theletter
ecanbeused
todenote
apow
erof10expon
ent,so
9.461·101
5canbewritten
9.461e15.
1.4
Complexnumbers.W
hathappensifyoustart
withalargenegativevalueofx
andrepeatedly
iterate
x=
sqrt(1
+x)
1.5
Comparison.W
hichis
larger,πφor
φπ?
1.6
Solvingequations.
Thebestway
tosolve
x=
√ 1+
x
or
x2=
1+
x
isto
avoid
computers
alltogether
andjust
doit
yourselfbyhand.But,
ofcourse,
Matlabandmost
other
mathem
aticalsoftwaresystem
scaneasily
solvesuch
equa-
tions.
Hereare
severalpossible
waysto
doitwithMatlab.Start
with
formatlong
phi=
(1
+sqrt(5))/2
Then,foreach
method,explain
whatis
goingonandhow
theresultingxdiffers
from
phiandtheother
x’s.
%roots
helproots
x1
=roots([1
-1
-1])
%fsolve
helpfsolve
f=
@(x)
x-sqrt(1+x)
p=
@(x)
x^2-x-1
x2
=fsolve(f,
1)
x3
=fsolve(f,
-1)
x4
=fsolve(p,
1)
x5
=fsolve(p,
-1)
%solve
(requires
Symbolic
Toolbox
or
Student
Version)
helpsolve
helpsyms
symsx
x6
=solve(’x-sqrt(1+x)=0’)
x7
=solve(x^2-x-1)
12Chapter1.
Iteration
1.7
Sym
bolicsolution.Ifyou
havetheSymbolicToolbox
orStudentVersion,explain
whatthefollow
ingprogram
does.
x=
sym(’x’)
length(char(x))
fork
=1:10
x=
sqrt(1+x)
length(char(x))
end
1.8
Fixed
points.VerifythattheGolden
Ratioisafixed
pointofeach
ofthefollow
ing
equations.
φ=
1
φ−
1
φ=
1 φ+1
Use
each
oftheequationsas
thebasisforafixed
pointiterationto
compute
φ.Do
theiterationsconverge?
1.9
Another
iteration.Before
you
runthefollow
ingprogram,predictwhatit
will
do.Then
runit.
x=
3
k=
1
formatlong
whilex
~=
sqrt(1+x^2)
x=
sqrt(1+x^2)
k=
k+1
end
1.10Another
fixedpoint.
Solvethis
equationbyhand.
x=
1√ 1
+x2
How
manyiterationsdoes
thefollow
ingprogram
require?
How
isthefinalvalueof
xrelatedto
theGolden
Ratioφ?
x=
3
k=
1
formatlong
whilex
~=
1/sqrt(1+x^2)
x=
1/sqrt(1+x^2)
k=
k+1
end
13
1.11cos(x).Findthenumericalsolutionof
theequation
x=
cosx
intheinterval[0,π 2],show
nin
figure
1.4.
00.
51
1.5
0
0.51
1.5
Figure
1.4.Fixed
pointofx=
cos(x).
−6−4
−20
24
6−6−4−20246
Figure
1.5.Threefixedpoints
ofx=
tan(x)
1.12tan(x).
Figure
1.5
show
sthreeofthemanysolutionsto
theequation
x=
tanx
Oneof
thesolutionsis
x=
0.Theother
twoin
theplotare
nearx=
±4.5.If
wedid
aplotover
alargerange,
wewou
ldseesolutionsin
each
oftheintervals
[(n−
1 2)π,(n+
1 2)π]forinteger
n.
(a)Does
this
compute
afixed
point?
x=
4.5
14Chapter1.
Iteration
fork
=1:30
x=
tan(x)
end
(b)Does
this
compute
afixed
point?
Whyis
the“+
pi”necessary?
x=
pi
whileabs(x-
tan(x))
>eps(x)
x=
atan(x)
+pi
end
1.13Summation.Write
amathem
aticalexpressionforthequantity
approxim
ated
bythis
program.
s=
0;
t=
Inf;
n=
0;
whiles
~=
t
n=
n+1;
t=
s;
s=
s+
1/n^4;
end
s
1.14Why.
ThefirstversionofMatlab
written
inthelate
1970’s,hadwho,what,
which,an
dwherecommands.
Soitseem
ednaturalto
addawhycommand.Check
outtoday
’swhycommandwith
why
helpwhy
fork
=1:40,
why,
end
typewhy
editwhy
Asthehelpentrysays,
please
embellish
ormodifythewhyfunctionto
suit
your
owntastes.
1.15Wiggles.Aglimpse
atMatlabplottingcapabilitiesisprovided
bythefunction
f=
@(x)
tan(sin(x))
-sin(tan(x))
Thisusesthe’@’signto
introduce
asimple
function.Youcanlearn
more
aboutthe
’@’signwithhelpfunction_handle.
Figure
1.6show
stheoutputfrom
thestatement
ezplot(f,[-pi,pi])
15
−3−2
−10
12
3
−2.5−2
−1.5−1
−0.50
0.51
1.52
2.5
x
tan(
sin(
x))−
sin(
tan(
x))
Figure
1.6.A
wigglyfunction.
(Thefunctionnameezplotis
intended
tobepronounced“Easy
Plot”.This
pun
doesn’t
workif
you
learned
topronounce
“z”
as“zed”.)
You
can
seethatthe
functionis
veryflatnearx=
0,oscillatesinfinitelyoften
nearx=
±π/2andis
nearlylinearnearx=
±π.
You
canget
more
controlover
theplotwithcodelikethis.
x=
-pi:pi/256:pi;
y=
f(x);
plot(x,y)
xlabel(’x’)
ylabel(’y’)
title(’Awigglyfunction’)
axis([-pipi
-2.8
2.8])
set(gca,’xtick’,pi*(-3:1/2:3))
(a)W
hatis
theeff
ectofvariousvalues
ofnin
thefollow
ingcode?
x=
pi*(-2:1/n:2);
comet(x,f(x))
(b)This
functionis
bou
nded
.A
numeric
valuenearitsmaxim
um
canbefound
with
max(y)
Whatisitsanalyticmaxim
um?(T
obeprecise,Ishould
ask
”W
hatisthefunction’s
supremum?”)
1.16Graphics.
Weuse
alotofcomputergraphicsin
thisbook,butstudyingMat-
labgraphicsprogrammingisnotou
rprimary
goal.How
ever,ifyouare
curious,the
16Chapter1.
Iteration
scriptthatproducesfigure
1.3
isgoldrect.m.Modifythis
program
toproduce
agraphic
thatcomparestheGolden
Rectangle
withTV
screenshav
ingaspectratios
4:3
and16:9.
1.17Golden
Spiral
Figure
1.7.A
spiralform
edfrom
golden
rectangles
andinscribedquarter
circles.
Ourprogram
golden_spiraldisplaysanever-expandingsequence
ofgolden
rectangleswithinscribed
quarter
circles.
Checkitout.
Chapter2
FibonacciNumbers
Fibonaccinumbers
introduce
vectors,functionsandrecursion.
LeonardoPisanoFibonacciwasborn
around1170anddiedaround1250in
Pisain
whatis
now
Italy.Hetraveled
extensivelyin
EuropeandNorthernAfrica.
Hewrote
severalmathem
aticaltexts
that,
amongother
things,
introducedEurope
totheHindu-A
rabic
notationfornumbers.
Even
thoughhisbookshadto
betran-
scribed
byhand,they
werewidelycirculated.In
hisbestknow
nbook
,Liber
Abaci,
published
in1202,heposed
thefollow
ingproblem:
Amanputs
apair
ofrabbitsin
aplace
surrounded
onallsides
byawall.
How
manypairsofrabbitscanbe
producedfrom
thatpair
inayearifit
issupposedthateverymonth
each
pair
begets
anew
pair
whichfrom
the
secondmonth
onbecomes
productive?
Today
thesolutionto
this
problem
isknow
nastheFibonaccisequen
ce,or
Fibonaccinumbers.
Thereis
asm
allmathem
aticalindustry
basedonFibonacci
numbers.
AsearchoftheInternet
for“Fibonacci”
willfinddozensof
Web
sitesan
dhundredsofpages
ofmaterial.Thereiseven
aFibonacciAssociationthatpublishes
ascholarlyjournal,theFibonacciQuarterly.
Asimulationof
Fibonacci’sproblem
isprovided
byourexmprogram
rabbits.
Just
execute
thecommand
rabbits
andclickon
thepushbuttonsthatshow
up.Youwillseesomethinglikefigure
2.1.
IfFibonaccihadnotspecified
amonth
forthenew
bornpair
tomature,he
would
nothaveasequence
named
after
him
.Thenumber
ofpairswould
simply
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
17
18Chapter2.
Fibon
acci
Numbers
Figure
2.1.Fibonacci’srabbits.
double
each
month.After
nmonthstherewould
be2n
pairsofrabbits.
That’sa
lotof
rabbits,
butnotdistinctivemathem
atics.
Let
f ndenote
thenumber
ofpairsofrabbitsafter
nmonths.
Thekey
fact
isthatthenumber
ofrabbitsattheendofamonth
isthenumber
atthebeginning
ofthemonth
plusthenumber
ofbirthsproducedbythemature
pairs:
f n=
f n−1+f n
−2.
Theinitialconditionsarethatin
thefirstmonth
thereison
epairofrabbitsandin
thesecondtherearetw
opairs:
f 1=
1,
f 2=
2.
Thefollow
ingMatlabfunction,storedin
afilefibonacci.mwitha.msuffix,
producesavectorcontainingthefirstnFibonaccinumbers.
functionf
=fibonacci(n)
%FIBONACCI
Fibonacci
sequence
%f
=FIBONACCI(n)
generates
the
first
nFibonacci
numbers.
19
f=
zeros(n,1);
f(1)=
1;
f(2)=
2;
fork
=3:n
f(k)=
f(k-1)
+f(k-2);
end
Withtheseinitialconditions,thean
swer
toFibonacci’soriginalquestionaboutthe
size
oftherabbitpopulationafteroneyearis
given
by
fibonacci(12)
This
produces
1 2 3 5 8
13
21
34
55
89
144
233
Theansw
eris
233pairsofrabbits.
(Itwou
ldbe4096pairsifthenumber
doubled
everymonth
for12
months.)
Let’s
look
carefullyatfibonacci.m.It’s
agoodexample
ofhow
tocreate
aMatlabfunction.Thefirstlineis
functionf
=fibonacci(n)
Thefirstword
onthefirstlinesaysfibonacci.mis
afunction,notascript.
The
remainder
ofthefirstlinesaysthis
particularfunctionproducesoneoutputresult,
f,an
dtakes
oneinputargument,n.Thenameofthefunctionspecified
onthefirst
lineis
notactuallyused,because
Matlablooksforthenameofthefile
witha.m
suffixthatcontainsthefunction,butitiscommonpracticeto
havethetw
omatch.
Thenexttw
olines
are
comments
thatprovidethetextdisplayed
when
youask
for
help.
helpfibonacci
produces
FIBONACCI
Fibonaccisequence
f=
FIBONACCI(n)
generates
the
first
nFibonacci
numbers.
20Chapter2.
Fibon
acci
Numbers
Thenameof
thefunctionis
inuppercase
because
historicallyMatlab
was
case
insensitiveandranonterm
inals
withon
lyasingle
font.
Theuse
ofcapitalletters
may
beconfusingto
somefirst-timeMatlabusers,buttheconventionpersists.
Itisim
portantto
repeattheinputandou
tputarguments
inthesecomments
because
thefirstlineis
notdisplayedwhen
youask
forhelponthefunction.
Thenextline
f=
zeros(n,1);
createsan
n-by-1
matrix
containingallzerosandassignsit
tof.In
Matlab,a
matrix
withonly
onecolumnisacolumnvectorandamatrix
withonly
onerow
isarow
vector.
Thenexttw
olines,
f(1)=
1;
f(2)=
2;
providetheinitialconditions.
Thelast
threelines
are
theforstatementthatdoes
allthework.
fork
=3:n
f(k)=
f(k-1)
+f(k-2);
end
Weliketo
use
threespacesto
indentthebodyofforandifstatements,butother
people
prefertw
oor
fourspaces,oratab.Youcanalso
puttheentire
construction
ononelineifyouprovideacommaafter
thefirstclause.
This
particularfunctionlooksalotlikefunctionsin
other
programminglan-
guages.
Itproducesavector,
butit
does
notuse
anyof
theMatlab
vectoror
matrix
operations.
Wewillseesomeoftheseoperationssoon.
Hereis
another
Fibonaccifunction,fibnum.m.Itsoutputis
simply
thenth
Fibonaccinumber.
functionf
=fibnum(n)
%FIBNUM
Fibonacci
number.
%FIBNUM(n)
generates
the
nth
Fibonacci
number.
if
n<=
1
f=1;
else f=
fibnum(n-1)
+fibnum(n-2);
end
Thestatement
fibnum(12)
produces
ans=
233
21
Thefibnumfunctionis
recursive.
Infact,theterm
recursiveis
usedin
both
amathem
aticalandacomputerscience
sense.
Inmathem
atics,therelationship
f n=
f n−1+
f n−2is
arecursionrelationIn
computerscience,afunctionthatcalls
itselfis
arecursivefunction.
Arecursiveprogram
iselegant,
butexpensive.
You
canmeasure
execution
timewithticandtoc.Try
tic,fibnum(24),toc
Donottry
tic,fibnum(50),toc
FibonacciMee
tsGolden
Ratio
TheGolden
Ratioφcanbeexpressed
asaninfinitecontinued
fraction.
φ=
1+
1
1+
11+
11+
···
.
Toverifythis
claim
,suppose
wedid
notknow
thevalueof
this
fraction.Let
x=
1+
1
1+
11+
11+
···
.
Wecanseethefirstdenominatoris
just
another
copyofx.In
other
words.
x=
1+
1 x
This
immed
iately
leadsto
x2−
x−
1=
0
whichis
thedefiningquadraticequationforφ,
OurexmfunctiongoldfractgeneratesaMatlabstringthatrepresents
the
firstnterm
softheGolden
Ratiocontinued
fraction.Hereis
thefirstsectionof
codein
goldfract.
p=
’1’;
fork
=2:n
p=
[’1
+1/(’
p’)’];
end
display(p)
Westart
withasingle
’1’,whichcorrespondsto
n=
1.Wethen
repeatedly
make
thecurrentstringthedenominatorin
alonger
string.
Hereis
theoutputfrom
goldfract(n)when
n=
7.
1+
1/(1
+1/(1
+1/(1
+1/(1
+1/(1
+1/(1))))))
22Chapter2.
Fibon
acci
Numbers
You
canseethattherearen-1plussignsandn-1pairsof
matchingparentheses.
Let
φndenote
thecontinued
fractiontruncatedafter
nterm
s.φnisarational
approxim
ationto
φ.Let’s
express
φnasaconventionalfracton,theratioof
two
integers φn=
pn q n
p=
1;
q=
0;
fork
=2:n
t=
p;
p=
p+
q;
q=
t;
end
Now
compare
theresultsproducedbygoldfract(7)andfibonacci(7).The
firstcontainsthefraction21/1
3whilethesecondendswith13and21.This
isnot
just
acoincidence.
Thecontinued
fraction
fortheGolden
Ratiois
collapsed
by
repeatingthestatement
p=
p+
q;
whiletheFibonaccinumbersaregeneratedby
f(k)=
f(k-1)
+f(k-2);
Infact,ifweletφndenote
thegolden
ratiocontinued
fractiontruncatedatnterm
s,then
φn=
f nf n
−1
Intheinfinitelimit,theratioof
successiveFibonaccinumbersap
proaches
thegolden
ratio:
lim
n→
∞f n
f n−1=
φ.
Toseethis,compute
40Fibonaccinumbers.
n=
40;
f=
fibonacci(n);
Then
compute
theirratios.
r=
f(2:n)./f(1:n-1)
This
takes
thevectorcontainingf(2)throughf(n)and
divides
it,elem
entby
elem
ent,bythevectorcontainingf(1)throughf(n-1).Theoutputbeginswith
23
2.00000000000000
1.50000000000000
1.66666666666667
1.60000000000000
1.62500000000000
1.61538461538462
1.61904761904762
1.61764705882353
1.61818181818182
andendswith
1.61803398874990
1.61803398874989
1.61803398874990
1.61803398874989
1.61803398874989
Doyouseewhywechose
n=
40?Compute
phi=
(1+sqrt(5))/2
r-
phi
Whatis
thevalueofthelast
elem
ent?
Thefirstfew
oftheseratioscanalsobeusedto
illustrate
therationaloutput
form
at. formatrat
r(1:10)
ans=
2 3/2
5/3
8/5
13/8
21/13
34/21
55/34
89/55
Thepopulationof
Fibonacci’srabbit
pen
doesn’t
double
everymonth;it
ismultiplied
bythegolden
ratioeverymonth.
AnAnalyticExp
ression
Itis
possible
tofind
aclosed-form
solution
totheFibonaccinumber
recurrence
relation.Thekey
isto
look
forsolutionsof
theform
f n=
cρn
24Chapter2.
Fibon
acci
Numbers
forsomeconstants
candρ.Therecurrence
relation
f n=
f n−1+f n
−2
becomes cρn=
cρn−1+
cρn−2
Dividingboth
sides
bycρ
n−2gives
ρ2=
ρ+1.
We’veseen
thisequationin
thechapterontheGolden
Ratio.Thereare
twopossible
values
ofρ,namelyφand1−φ.Thegeneralsolutionto
therecurrence
is
f n=
c 1φn+c 2(1
−φ)n.
Theconstants
c 1andc 2
are
determined
byinitialconditions,
whicharenow
convenientlywritten
f 0=
c 1+c 2
=1,
f 1=
c 1φ+c 2(1
−φ)=
1.
Oneoftheexercisesasksyouto
use
theMatlabbackslash
operatorto
solvethis
2-by-2
system
ofsimultaneouslinearequations,butitismay
beeasier
tosolvethe
system
byhand:
c 1=
φ
2φ−1,
c 2=
−(1
−φ)
2φ−1.
Insertingthesein
thegeneralsolutiongives
f n=
1
2φ−1(φ
n+1−
(1−φ)n
+1).
This
isanamazingequation.Theright-handsideinvo
lves
pow
ersan
dquo-
tients
ofirrationalnumbers,buttheresultisasequence
ofintegers.
You
cancheck
this
withMatlab.
n=
(1:40)’;
f=
(phi.^(n+1)
-(1-phi).^(n+1))/(2*phi-1)
f=
round(f)
The.^operatoris
anelem
ent-by-elementpow
erop
erator.
Itis
notnecessary
touse
./forthefinaldivisionbecause
(2*phi-1)isascalarquantity.Roundofferror
prevents
theresultsfrom
beingexact
integers,
sotheroundfunction
isused
toconvert
floatingpointquantities
tonearest
integers.
Theresultingfbeginswith
f=
1
25
2 3 5 8
13
21
34
andendswith
5702887
9227465
14930352
24157817
39088169
63245986
102334155
165580141
Rec
ap
%%Fibonacci
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Fibonacci
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %fibonacci_recap
%edit
fibonacci_recap
%publish
fibonacci_recap
%%Related
EXM
Programs
% %fibonacci.m
%fibnum.m
%rabbits.m
%%Functions
%Save
in
file
sqrt1px.m
% %function
y=
sqrt1px(x)
%%
SQRT1PX
Sample
function.
%%
Usage:
y=
sqrt1px(x)
% %y
=sqrt(1+x);
%%Create
vector
n=
8;
26Chapter2.
Fibon
acci
Numbers
f=
zeros(1,n)
t=
1:n
s=
[1
23
58
13
21
34]
%%
Subscripts
f(1)=
1;
f(2)=
2;
fork
=3:n
f(k)=
f(k-1)
+f(k-2);
end
f
%%
Recursion
%function
f=
fibnum(n)
%if
n<=
1
%f=
1;
%else
%f=
fibnum(n-1)
+fibnum(n-2);
%end
%%
Tic
and
Toc
formatshort
tic
fibnum(24);
toc
%%
Element-by-element
array
operations
f=
fibonacci(5)’
fpf=
f+f
ftf=
f.*f
ff
=f.^2
ffdf=
ff./f
cosfpi=
cos(f*pi)
even=
(mod(f,2)
==
0)
formatrat
r=
f(2:5)./f(1:4)
%%
Strings
hello_world
27
Exe
rcises
2.1
Rabbits.
Explain
whatourrabbitssimulation
dem
onstrates.
Whatdothe
differentfiguresandcolors
onthepushbuttonssignify?
2.2
Waltz.
WhichFibonaccinumbersareeven?W
hy?
2.3
Primes.Use
theMatlabfunctionisprimeto
discover
whichofthefirst40
Fibonaccinumbersareprime.
You
donotneedto
use
aforloop.Instead,check
out
helpisprime
helplogical
2.4
Backslash.Use
theMatlabbackslash
operatorto
solvethe2-by-2
system
of
simultaneouslinearequations
c 1+c 2
=1,
c 1φ+c 2(1
−φ)=
1
forc 1
andc 2.Youcanfindoutabou
tthebackslash
operatorbytakingapeekat
theLinearEquationschapter,
orwiththecommands
help\
helpslash
2.5
Logarithmic
plot.
Thestatement
semilogy(fibonacci(18),’-o’)
makes
alogarithmic
plotofFibonaccinumbersversustheirindex.Thegraphis
close
toastraightline.
Whatis
theslopeofthis
line?
2.6
Execution
time.
How
does
theexecution
timeoffibnum(n)dep
end
on
the
executiontimeforfibnum(n-1)andfibnum(n-2)?Use
thisrelationship
toobtain
anapproxim
ate
form
ula
fortheexecutiontimeoffibnum(n)asafunctionofn.
Estim
ate
how
longitwould
takeyourcomputerto
compute
fibnum(50).Warning:
You
probably
donotwan
tto
actuallyrunfibnum(50).
2.7
Overfl
ow.W
hatis
theindex
ofthelargestFibonaccinumber
thatcanberep-
resentedexactly
asaMatlab
double-precisionquantity
withoutroundoff
error?
Whatistheindex
ofthelargestFibonaccinumber
thatcanberepresentedapprox-
imately
asaMatlabdouble-precisionquantity
withoutoverflow
ing?
2.8
Slower
maturity.W
hatifrabbitstooktw
omonthsto
mature
insteadofone?
28Chapter2.
Fibon
acci
Numbers
Thesequence
would
bedefined
by
g 1=
1,
g 2=
1,
g 3=
2
and,forn>
3,
g n=
g n−1+
g n−3
(a)Modifyfibonacci.mandfibnum.mto
compute
this
sequence.
(b)How
manypairsof
rabbitsare
thereafter
12months?
(c)g n
≈γn.W
hat
isγ?
(d)Estim
ate
how
longit
wou
ldtakeyourcomputerto
compute
fibnum(50)with
this
modified
fibnum.
2.9
Mortality.W
hatifrabbitstookonemonth
tomature,butthen
diedafter
six
months.
Thesequence
would
bedefined
by
dn
=0,
n<=
0
d1=
1,
d2=
1
and,forn>
2,
dn=
dn−1+dn−2−
dn−7
(a)Modifyfibonacci.mandfibnum.mto
compute
this
sequence.
(b)How
manypairsof
rabbitsare
thereafter
12months?
(c)dn≈
δn.W
hat
isδ?
(d)Estim
ate
how
longit
wou
ldtakeyourcomputerto
compute
fibnum(50)with
this
modified
fibnum.
2.10
Hello
World.
Programming
languages
aretraditionally
introduced
by
the
phrase
”hello
world”.Anscriptin
exmthatillustratessomefeaturesin
Matlabis
available
with
hello_world
Explain
whateach
ofthefunctionsan
dcommandsin
hello_worlddo.
2.11Fibonaccipower
series.TheFibonaccinumbers,f n
,canbeusedascoeffi
cients
inapow
erseries
definingafunctionofx.
F(x)
=∞ ∑ n=1
f nxn
=x+2x2+3x3+5x4+8x5+13x6+...
29
Ourfunctionfibfun1isafirstattemptat
aprogram
tocompute
thisseries.It
sim-
ply
invo
lves
addinganaccumulatingsum
tofibonacci.m.Theheader
offibfun1.m
includes
thehelpentries.
function[y,k]=
fibfun1(x)
%FIBFUN1
Power
series
with
Fibonacci
coefficients.
%y
=fibfun1(x)
=sum(f(k)*x.^k).
%[y,k]
=fibfun1(x)
also
gives
the
number
of
terms
required.
Thefirstsectionofcodeinitializesthevariablesto
beused.Thevalueofnis
the
index
wheretheFibonaccinumbersoverflow
.
\excise
\emph{Fibonaccipowerseries}.
TheFibonaccinumbers,$f_n$,canbe
used
as
coefficients
in
a
powerseriesdefininga
function
of
$x$.
\begin{eqnarray*}
F(x)&
=&
\sum_{n
=1}^\infty
f_n
x^n
\\
&=
&x
+2
x^2
+3
x^3
+5
x^4
+8
x^5
+13
x^6
+...
\end{eqnarray*}
Ourfunction#fibfun1#is
afirst
attempt
at
aprogram
to
computethisseries.
It
simply
involves
adding
an
accumulating
sumto
#fibonacci.m#.
Theheaderof
#fibfun1.m#
includes
the
help
entries.
\begin{verbatim}
function[y,k]=
fibfun1(x)
%FIBFUN1
Power
series
with
Fibonacci
coefficients.
%y
=fibfun1(x)
=sum(f(k)*x.^k).
%[y,k]
=fibfun1(x)
also
gives
the
number
of
terms
required.
Thefirstsectionofcodeinitializesthevariablesto
beused.Thevalueof
nis
the
index
wheretheFibonaccinumbersoverflow
.
n=
1476;
f=
zeros(n,1);
f(1)=
1;
f(2)=
2;
y=
f(1)*x
+f(2)*x.^2;
t=
0;
Themain
bodyof
fibfun1im
plements
theFibonaccirecu
rren
cean
dincludes
atest
forearlyterm
inationoftheloop.
fork
=3:n
f(k)=
f(k-1)
+f(k-2);
y=
y+
f(k)*x.^k;
if
y==
t
return
end
30Chapter2.
Fibon
acci
Numbers
t=y;
end
Thereareseveralobjectionsto
fibfun1.Thecoeffi
cientarrayof
size
1476
isnotactuallynecessary.Therepeatedcomputationof
pow
ers,
x^k,is
ineffi
cient
because
once
somepow
erofxhasbeencomputed,thenextpow
ercanbeobtained
withonemultiplication.W
hen
theseries
converges
thecoeffi
cients
f(k)increase
insize,butthepow
ersx^kdecrease
insize
more
rapidly.
Theterm
sf(k)*x^k
approach
zero,buthugef(k)preventtheircomputation.
Amore
efficientandaccurate
approach
involves
combiningthecomputation
oftheFibonaccirecurrence
andthepow
ersof
x.Let
pk=
f kxk
Then,since
f k+1xk+1=
f kxk+f k
−1xk−1
theterm
spksatisfy
pk+1
=pkx+pk−1x2
=x(p
k+xpk−1)
Thisisthebasisforourfunctionfibfun2.Theheader
isessentiallythesame
asfibfun1
function[yk,k]=
fibfun2(x)
%FIBFUN2
Power
series
with
Fibonacci
coefficients.
%y
=fibfun2(x)
=sum(f(k)*x.^k).
%[y,k]
=fibfun2(x)
also
gives
the
number
of
terms
required.
Theinitialization.
pkm1=
x;
pk
=2*x.^2;
ykm1=
x;
yk
=2*x.^2
+x;
k=
0;
Andthecore.
whileany(abs(yk-ykm1)>
2*eps(yk))
pkp1=
x.*(pk
+x.*pkm1);
pkm1=
pk;
pk=
pkp1;
ykm1=
yk;
yk=
yk
+pk;
k=
k+1;
end
31
Thereis
noarrayofcoeffi
cients.Only
threeofthepkterm
sarerequired
foreach
step.Thepow
erfunction^is
notnecessary.Computationof
thepow
ersis
incor-
poratedin
therecurrence.Consequently,
fibfun2is
both
more
efficientan
dmore
accurate
thanfibfun1.
Butthereisaneven
betterway
toevaluate
thisparticularseries.It
ispossible
tofindaan
alyticexpressionfortheinfinitesum.
F(x)
=
∞ ∑ n=1
f nxn
=x+2x2+3x3+5x4+8x5+...
=x+(1
+1)x2+(2
+1)x
3+(3
+2)x
4+(5
+3)x
5+...
=x+x2+x(x
+2x
2+3x3+5x4+
...)+
x2(x
+2x2+3x3+...)
=x+x2+xF(x)+
x2F(x)
So
(1−x−x2)F
(x)=
x+x2
Finally F(x)=
x+x2
1−
x−
x2
Itis
noteven
necessary
tohavea.mfile.A
one-liner
does
thejob.
fibfun3=
@(x)
(x
+x.^2)./(1
-x
-x.^2)
Compare
thesethreefibfun’s.
32Chapter2.
Fibon
acci
Numbers
Chapter3
CalendarsandClock
s
Computationsinvolvingtime,
dates,
biorhythmsandEaster.
Calendars
are
interestingmathem
aticalob
jects.
TheGregoriancalendarwas
firstproposed
in1582.
Ithasbeen
graduallyadopted
byvariouscountriesan
dchurches
over
thefourcenturies
since
then
.TheBritish
Empire,
includingthe
coloniesin
NorthAmerica,adopteditin
1752.Turkey
did
notadoptituntil1923.
TheGregoriancalendarisnow
themost
widelyusedcalendarin
theworld,butby
nomeanstheonly
one.
IntheGregoriancalendar,ayearyisaleapyearifandon
lyifyisdivisibleby4
andnotdivisible
by100,orisdivisible
by400.In
Matlabthefollow
ingexpression
must
betrue.Thedouble
ampersands,
’&&’,mean“and”andthedouble
vertical
bars,’||’,mean“or”.
mod(y,4)==
0&&
mod(y,100)
~=
0||
mod(y,400)
==
0
For
example,2000wasaleapyear,
but2100willnotbealeapyear.
This
rule
implies
thattheGregoriancalendarrepeats
itselfevery400years.In
that400-year
period,thereare
97leapyears,4800months,
20871weeks,
and146097day
s.The
averagenumber
ofday
sin
aGregoriancalendaryearis
365+
97
400=
365
.2425.
TheMatlabfunctionclockreturnsasix-elementvectorcwithelem
ents
c(1)=
year
c(2)=
month
c(3)=
day
c(4)=
hour
c(5)=
minute
c(6)=
seconds
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
33
34Chapter3.
CalendarsandClocks
Thefirstfiveelem
ents
are
integers,whilethesixth
elem
enthasafractionalpart
that
isaccurate
tomilliseconds.
Thebestway
toprintaclockvectoristo
use
fprintf
orsprintfwithaspecified
form
atstringthathasboth
integer
andfloatingpoint
fields. f=
’%6d
%6d
%6d
%6d
%6d
%9.3f\n’
Iam
revisingthis
chapteronAugust
2,2011,atafew
minutesafter2:00pm,so
c=
clock;
fprintf(f,c);
produces
2011
82
14
219.470
Inother
words,
year=
2011
month=
8
day=
2
hour=
14
minute=
2
seconds=
19.470
TheMatlabfunctionsdatenum,datevec,datestr,andweekdayuse
clock
andfactsabouttheGregoriancalendarto
facilitate
computationsinvolvingcalendar
dates.
Datesarerepresentedbytheirserialdate
number,
whichis
thenumber
of
day
ssince
thetheoreticaltimean
dday
over
20centuries
agowhen
clockwould
havebeensixzeros.
Wecan’t
pin
thatdow
nto
anactualdate
because
different
calendars
wou
ldhavebeenin
use
atthattime.
Thefunctiondatenum
returnsthedate
number
foran
yclock
vector.
For
example,usingthevectorcthatIjust
found,thecurrentdate
number
is
datenum(c)
is
734717.585
This
indicatesthatthecu
rrenttimeis
alittle
over
halfway
throughday
number
734717.Iget
thesameresultfrom
datenum(now)
orjust now
Thedatenumfunctionalsoworkswithagiven
year,month
andday,oradate
specified
asastring.Forexample
35
datenum(2011,8,2)
and
datenum(’Aug.2,
2011’)
bothreturn
734717
Thesameresult
isobtained
from
fix(now)
Intw
oandahalfday
sthedate
number
willbe
datenum(fix(now+2.5))
734720
Computingthedifference
betweentw
odate
numbersgives
anelapsedtime
measuredin
day
s.How
manyday
sare
left
betweentoday
andthefirstday
ofnext
year?
datenum(’Jan1,
2012’)
-datenum(fix(now))
ans=
152
Theweekdayfunctioncomputestheday
oftheweek,as
both
aninteger
be-
tween1and7andastring.Forexample
both
[d,w]=
weekday(datenum(2011,8,2))
and
[d,w]=
weekday(now)
bothreturn
d=
3
w= Tue
Sotoday
isthethirdday
oftheweek,aTuesday.
Friday
the13th
Friday
the13th
isunlucky,
butisitunlikely?W
hatistheprobabilitythatthe13th
day
ofan
ymonth
fallsonaFriday
?Thequickansw
eris
1/7,butthatis
notquite
right.
Thefollow
ingcodecounts
thenumber
oftimes
thatFriday
occurs
onthe
variousweekdaysin
a400yearcalendarcycleandproducesfigure
3.1.(Y
oucan
alsorunfriday13yourself.)
36Chapter3.
CalendarsandClocks
SuM
TuW
ThF
Sa68
0
681
682
683
684
685
686
687
688
689
690
Figure
3.1.The13th
ismore
likely
tobe
onFridaythananyother
day.
c=
zeros(1,7);
fory
=1601:2000
form
=1:12
d=
datenum([y,m,13]);
w=
weekday(d);
c(w)=
c(w)
+1;
end
end
c bar(c)
axis([08
680
690])
avg=
4800/7;
line([08],[avgavg],’linewidth’,4,’color’,’black’)
set(gca,’xticklabel’,{’Su’,’M’,’Tu’,’W’,’Th’,’F’,’Sa’})
c=
687
685
685
687
684
688
684
Sothe13th
day
ofamonth
ismore
likelyto
beon
aFriday
thananyother
day
oftheweek.Theprobabilityis688/4800=
.143333.This
probabilityis
close
to,
butslightlylarger
than,1/7=
.142857.
Biorhythms
Biorhythmswereinventedover
100years
agoandenteredourpopularculture
inthe
1960s.
YoucanstillfindmanyWeb
sitestoday
thatoff
erto
prepare
personalized
biorhythms,
orthatsellsoftware
tocompute
them
.Biorhythmsare
basedonthe
notion
thatthreesinusoidalcycles
influence
ourlives.
Thephysicalcyclehasa
37
periodof23day
s,theem
otionalcyclehasaperiodof28day
s,an
dtheintellectual
cyclehasaperiodof
33day
s.Foran
yindividual,thecycles
areinitializedatbirth.
Figure
3.2
ismybiorhythm,whichbeginsonAugust
17,1939,plotted
foran
eight-weekperiod
centeredaround
thedate
this
isbeingrevised,July
27,2011.
Itshow
sthatImust
bein
prettygoodshape.
Today,Iam
nearthepeakofmy
intellectualcycle,
andmyphysicalandem
otionalcycles
peaked
onthesameday
less
thanaweekago.
06/2
907
/06
07/1
307
/20
07/2
708
/03
08/1
008
/17
08/2
4−1
00−50050100
07/2
7/11
birt
hday
: 08/
17/3
9
Phys
ical
Emot
iona
lIn
telle
ctua
l
Figure
3.2.Mybiorhythm.
Asearch
oftheUnited
StatesGovernmentPatentand
Tradem
ark
Office
database
ofUSpatents
issued
between1976and2010finds147patents
thatare
baseduponormentionbiorhythms.
(Therewerejust
113in
2007.)
TheWeb
site
is
http://patft.uspto.gov
Thedate
andgraphicsfunctionsin
Matlabmake
thecomputationanddis-
playofbiorhythmsparticularlyconvenient.
Thefollow
ingcodesegmentis
part
of
ourprogram
biorhythm.mthatplots
abiorhythm
foraneight-weekperiodcentered
onthecurrentdate.
t0
=datenum(’Aug.
17,
1939’)
t1
=fix(now);
t=
(t1-28):1:(t1+28);
y=
100*[sin(2*pi*(t-t0)/23)
sin(2*pi*(t-t0)/28)
sin(2*pi*(t-t0)/33)];
plot(t,y)
You
seethatthetimevariable
tis
measuredin
day
sandthatthetrig
functions
takearguments
measuredin
radians.
38Chapter3.
CalendarsandClocks
When
isEaster?
Easter
Day
ison
eofthemost
importantevents
intheChristiancalendar.
Itisalso
oneof
themost
mathem
aticallyelusive.
Infact,regularizationof
theobservan
ceofEaster
was
oneoftheprimary
motivationsforcalendarreform
.Theinform
al
rule
isthatEaster
Day
isthefirstSunday
after
thefirstfullmoonafter
thevernal
equinox
.Buttheecclesiasticalfullmoonandequinox
invo
lved
inthis
rule
are
not
alwaysthesameas
thecorrespondingastronomicalevents,which,afterall,dep
end
upon
thelocation
oftheobserver
ontheearth.
Computingthedate
ofEaster
isfeatured
inDon
Knuth’s
classic
TheArt
ofComputerProgramming
and
has
consequentlybecomeafrequentexercise
inprogrammingcourses.
OurMatlab
versionof
Knuth’s
program,easter.m,is
thesubject
ofseveralexercisesin
this
chapter.
Rec
ap
%%
Calendar
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Calendar
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %calendar_recap
%edit
calendar_recap
%publish
calendar_recap
% %Related
EXM
programs
% %biorhythm.m
%easter.m
%clockex.m
%friday13.m
%%
Clock
and
fprintf
formatbank
c=
clock
f=
’%6d
%6d
%6d
%6d
%6d
%9.3f\n’
fprintf(f,c);
%%
Modular
arithmetic
y=
c(1)
is_leapyear=
(mod(y,4)
==
0&&
mod(y,100)
~=
0||
mod(y,400)
==
0)
%%
Date
functions
c=
clock;
dnum=
datenum(c)
dnow=
fix(now)
39
xmas=
datenum(c(1),12,25)
days_till_xmas=
xmas
-dnow
[~,wday]=
weekday(now)
%%Count
Friday
the
13th’s
c=
zeros(1,7);
fory
=1:400
form
=1:12
d=
datenum([y,m,13]);
w=
weekday(d);
c(w)=
c(w)
+1;
end
end
formatshort
c
%%Biorhythms
bday=
datenum(’8/17/1939’)
t=
(fix(now)-bday)
+(-28:28);
y=
100*[sin(2*pi*t/23)
sin(2*pi*t/28)
sin(2*pi*t/33)];
plot(t,y)
axistight
Exe
rcises
3.1
Microcentury.Theoptimum
length
ofaclassroom
lecture
ison
emicrocentury.
How
longis
that?
3.2
π·1
07.A
goodestimate
ofthenumber
ofsecondsin
ayearis
π·1
07.How
accurate
isthis
estimate?
3.3
datestr.W
hatdoes
thefollow
ingprogram
do?
fork
=1:31
disp(datestr(now,k))
end
3.4
calendar(a)W
hatdoes
thecalendarfunctiondo?(IfyouhavetheFinancial
Toolbox
,helpcalendarwillalsotellyouaboutit.)
Try
this:
y=
your
birth
year
%An
integer,
1900
<=
y<=
2100.
m=
your
birth
month
%An
integer,
1<=
m<=
12.
40Chapter3.
CalendarsandClocks
calendar(y,m)
(b)W
hatdoes
thefollow
ingcodedo?
c=
sum(1:30);
s=
zeros(12,1);
form
=1:12
s(m)=
sum(sum(calendar(2011,m)));
end
bar(s/c)
axis([013
0.80
1.15])
3.5
Another
clock.
Whatdoes
thefollow
ingprogram
do?
clf
set(gcf,’color’,’white’)
axisoff
t=
text(0.0,0.5,’
’,’fontsize’,16,’fontweight’,’bold’);
while1
s=
dec2bin(fix(86400*now));
set(t,’string’,s)
pause(1)
end
Try
helpdec2bin
ifyouneedhelp.
3.6
How
long?
You
should
nottryto
runthefollow
ingprogram.Butifyou
were
torunit,how
longwould
ittake?
(Ifyouinsist
onrunningit,changeboth
3’s
to5’s.)
c=
clock
d=
c(3)
whiled
==
c(3)
c=
clock;
end
3.7
First
datenum.Thefirstcountriesto
adopttheGregoriancalendarwereSpain,
Portugalandmuch
ofItaly.They
did
soonOctober
15,1582,ofthenew
calendar.
Thepreviousday
wasOctober
4,1582,usingtheold,Julian,calendar.
SoOctober
5throughOctober
14,1582,did
notexistin
thesecountries.
WhatistheMatlab
serialdate
number
forOctober
15,1582?
3.8
Future
datenum’s.Use
datestrto
determinewhen
datenumwillreach
750,000.
When
willit
reach
1,000,000?
41
3.9
Yourbirthday.
Onwhichday
oftheweekwereyou
born?In
a400-yearGrego-
riancalendarcycle,whatistheprobabilitythatyourbirthday
occurs
onaSaturday
?W
hichweekday
isthemost
likelyforyourbirthday
?
3.10Opspercentury.W
hichdoes
more
operations,
ahumancomputerdoingon
eoperationper
secondforacentury,oranelectronic
computerdoingon
eoperation
per
microsecondforaminute?
3.11Julianday.
TheJulianDay
Number
(JDN)is
commonly
usedto
date
astro-
nomicalobservations.
FindthedefinitionofJulianDay
ontheWeb
andexplain
why
JDN=
datenum
+1721058.5
Inparticular,
whydoes
theconversionincludean0.5
fractionalpart?
3.12Unix
time.
TheUnix
operatingsystem
andPOSIX
operatingsystem
standard
measure
timein
secondssince
00:00:00UniversaltimeonJanuary
1,1970.There
are
86,400secondsin
oneday.Consequently,
Unix
time,time_t,canbecomputed
inMatlabwith
time_t=
86400*(datenum(y,m,d)
-datenum(1970,1,1))
SomeUnix
system
sstore
thetimein
an32-bit
signed
integer
register.
When
will
time_texceed
231andoverflow
onsuch
system
s.
3.13Easter.
(a)Thecomments
ineaster.m
use
theterm
s“golden
number”,“epact”,an
d“metonic
cycle”.Findthedefinitionsof
theseterm
sontheWeb.
(b)Plotabargraphof
thedatesofEaster
duringthe21-stcentury.
(c)How
manytimes
duringthe21-stcentury
does
Easter
occurin
Marchandhow
manyin
April?
(d)Onhow
manydifferentdatescanEaster
occur?
Whatis
theearliest?W
hatis
thelatest?
(e)Is
thedate
ofEaster
aperiodic
functionoftheyearnumber?
3.14Biorhythms.
(a)Use
biorhythmto
plotyou
row
nbiorhythm,basedonyou
rbirthday
andcentered
aroundthecurrentdate.
(b)Allthreebiorhythm
cycles
start
atzero
when
youwereborn.W
hen
dothey
return
tothis
initialcondition?
Compute
them,theleast
commonmultiple
of23,
28,and33.
m=
lcm(lcm(23,28),33)
Now
try
biorhythm(fix(now)-m)
42Chapter3.
CalendarsandClocks
Whatis
specialaboutthis
biorhythm?
How
old
wereyou,or
willyou
be,
mdays
after
youwereborn?
(c)Is
itpossible
forallthreebiorhythm
cycles
toreach
theirmaxim
um
atexactly
thesametime?
Whyorwhynot?
Try
t=
17003
biorhythm(fix(now)-t)
Whatis
specialaboutthis
biorhythm?How
manyyears
istday
s?Atfirstglan
ce,
itappears
thatallthreecycles
arereachingmaxim
aat
thesametime,t.Butifyou
lookmore
closely
at
sin(2*pi*t./[2328
33])
you
willseethatthevalues
arenotallexactly
1.0.Atwhattimes
neart=
17003
dothethreecycles
actuallyreach
theirmaxim
a?
Thethreetimes
arehow
many
hours
apart?
Are
therean
yother
values
oftbetween0and
theleast
common
multiple,m,from
thepreviousexercise
wherethecycles
comeso
close
toob
taining
asimultaneousmaxim
um?This
codewillhelpyouansw
erthesequestions.
m=
lcm(lcm(23,28),33);
t=
(1:m)’;
find(mod(t,23)==6&
mod(t,28)==7
&mod(t,33)==8)
(d)Is
itpossible
forallthreebiorhythm
cycles
toreach
theirminim
um
atexactly
thesametime?
Whyor
whynot?
When
dothey
nearlyreach
asimultaneous
minim
um?
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
16−S
ep−2
010
Figure
3.3.clockex
3.15clockex.
This
exercise
isabouttheclockexprogram
inourexmtoolbox
and
show
nin
figure
3.3.
(a)W
hydoes
clockexuse
trig
functions?
43
(b)Makeclockexruncounter-clockwise.
(c)W
hydothehourandminute
handsin
clockexmovenearlycontinuouslywhile
thesecondhandmoves
indiscretesteps.
(d)Thesecondhandsometim
eskipstw
omarks.
Why?How
often?
(e)Modifyclockexto
haveadigitaldisplayofyourow
ndesign.
44Chapter3.
CalendarsandClocks
Chapter4
Matrices
Matlabbeganasamatrix
calculator.
TheCartesiancoordinate
system
wasdeveloped
inthe17th
century
bythe
French
mathem
aticianandphilosopher
Ren
eDescartes.A
pair
ofnumberscorre-
spondsto
apointin
theplane.
Wewilldisplaythecoordinatesin
avectoroflength
two.In
order
towork
properly
withmatrix
multiplication,wewan
tto
thinkofthe
vectorasacolumnvector,
So
x=
( x1
x2
)den
otesthepointxwhose
firstcoordinate
isx1andsecondcoordinate
isx2.W
hen
itisinconvenientto
write
avectorin
thisverticalform
,wecananticipate
Matlab
notationanduse
asemicolonto
separate
thetw
ocompon
ents,
x=
(x1;x2)
Forexample,thepointlabeled
xin
figure
4.1hasCartesiancoordinates
x=
(2;
4)
Arithmetic
operationsonthevectors
are
defined
innaturalways.
Additionis
defined
by
x+
y=
( x1
x2
) +
( y 1 y 2
) =
( x1+y 1
x2+y 2
)Multiplicationbyasingle
number,orscalar,
isdefined
by
sx=
( sx1
sx2
)
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
45
46Chapter4.
Matrices
A2-by-2
matrix
isanarray
offournumbersarranged
intw
orowsan
dtw
ocolumns.
A=
( a1,1
a1,2
a2,1
a2,2
)or
A=
(a1,1
a1,2;a2,1
a2,2
)
Forexample
A=
(4
−3−2
1
)
−10
−8−6
−4−2
02
46
810
−10−8−6−4−20246810
x
Ax
Figure
4.1.Matrix
multiplicationtransform
slines
through
xto
lines
through
Ax.
Matrix-vectormultiplicationbya2-by-2
matrix
Atransform
savectorxto
avectorAx,accordingto
thedefinition
Ax=
( a1,1x1+
a1,2x2
a2,1x1+
a2,2x2
)Forexample
(4
−3−2
1
)(2 4
) =
(4·2
−3·4
−2·2
+1·4
) =
( −4 0
)Thepointlabeled
xin
figure
4.1is
transform
edto
thepointlabeled
Ax.Matrix-
vectormultiplicationsproduce
lineartransform
ations.
This
meansthatforscalars
sandtandvectors
xandy,
A(sx+
ty)=
sAx+
tAy
47
Thisim
plies
thatpoints
nearxare
transform
edto
points
nearAxandthatstraight
lines
intheplanethroughxare
transform
edto
straightlines
throughAx.
Ourdefinitionofmatrix-vectormultiplicationis
theusualon
einvolvingthe
dotproduct
oftherowsofA,denotedai,:,withthevectorx.
Ax=
( a1,:·x
a2,:·x
)
Analternate,andsometim
esmore
revealing,definitionuseslinearcombinationsof
thecolumnsofA,denotedbya:,j.
Ax=
x1a:,1+x2a:,2
For
example
(4
−3−2
1
)(2 4
) =2
(4 −2) +
4
( −3 1
) =
( −4 0
)
Thetranspose
ofacolumnvectoris
arow
vector,
denotedbyxT.Thetran
s-pose
ofamatrix
interchanges
itsrowsandcolumns.
Forexample,
xT=
(2
4)
AT=
(4
−2−3
1
)
Vector-matrix
multiplicationcanbedefined
by
xTA
=A
Tx
Thatisprettycryptic,
soifyouhavenever
seen
itbefore,youmighthaveto
ponder
itabit.
Matrix-m
atrix
multiplication,AB,canbethoughtofas
matrix-vectormulti-
plicationinvo
lvingthematrixAandthecolumnsvectors
from
B,oras
vector-matrix
multiplicationinvo
lvingtherow
vectors
from
Aandthematrix
B.It
isim
portan
tto
realize
thatAB
isnotthesamematrix
asBA.
Matlabstarted
itslife
as“Matrix
Laboratory”,so
itsveryfirstcapabilities
involved
matrices
andmatrix
multiplication.Thesyntaxfollow
sthemathem
atical
notationclosely.Weuse
square
bracketsinsteadofroundparentheses,anasterisk
toden
ote
multiplication,andx’forthetranspose
ofx.
Theforegoingexample
becomes
x=
[2;
4]
A=
[4
-3;
-2
1]
A*x
This
produces
x=
2 4
48Chapter4.
Matrices
A=
4-3
-2
1
ans= -4 0
Thematrices
A’*AandA*A’are
notthesame.
A’*A=
20
-14
-14
10
while A*A’=
25
-11
-11
5
Thematrix
I=
( 10
01
)is
the2-by-2
iden
tity
matrix.It
hastheim
portantproperty
thatforany2-by-2
matrix
A,
IA
=AI=
A
Originally,
Matlabvariable
names
werenotcase
sensitive,
soiandIwere
thesamevariable.
Since
iis
frequentlyusedasasubscript,
aniterationindex,
andsqrt(-1),wecould
not
use
Ifortheidentity
matrix.Instead,wechose
touse
thesound-alikeword
eye.Today,Matlabis
case
sensitiveandhasmanyusers
whose
nativelanguageis
notEnglish,butwecontinueto
use
eye(n,n)to
den
ote
then-by-n
identity.(T
heMetro
inWashington,DC,usesthesamepun–“Istreet”
is“eyestreet”ontheirmaps.)
2-by-2Matrix
Transform
ations
Theexmtoolbox
includes
afunctionhouse.Thestatement
X=
house
producesa2-by-11matrix,
X= -6
-6
-7
07
66
-3
-3
00
-7
21
81
2-7
-7
-2
-2
-7
ThecolumnsofXare
theCartesiancoordinatesof
the11points
show
nin
figure
4.2.
Doyou
remem
ber
the“dotto
dot”
game?
Try
itwiththesepoints.Finishoff
by
connectingthelast
pointback
tothefirst.
Thehouse
infigure
4.2
isconstructed
from
Xby
49
−10
−50
510
−10−8−6−4−20246810
123
4
56 7
8910 11
−10
−50
510
−10−8−6−4−20246810
Figure
4.2.Connectthedots.
dot2dot(X)
Wewantto
investigate
how
matrix
multiplicationtransform
sthis
house.In
fact,ifyouhaveyou
rcomputerhandy,
trythis
now
.
wiggle(X)
Ourgoalis
toseehow
wiggleworks.
Hereare
fourmatrices.
A1
= 1/2
0
01
A2
= 10
01/2
A3
= 01
1/2
0
A4
= 1/2
0
0-1
Figure
4.3usesmatrix
multiplicationA*Xanddot2dot(A*X)to
show
theeff
ectof
theresultinglineartransform
ationsonthehouse.Allfourmatrices
are
diagonal
orantidiagonal,so
they
just
scale
andpossibly
interchangethecoordinates.
The
coordinatesare
notcombined
inanyway.Thefloor
andsides
ofthehouse
remain
at
50Chapter4.
Matrices
−10
010
−10−50510
A1
−10
010
−10−50510
A2
−10
010
−10−50510
A3
−10
010
−10−50510
A4
Figure
4.3.Theeff
ectofmultiplicationby
scalingmatrices.
rightan
glesto
each
other
andparallel
totheaxes.Thematrix
A1shrinksthefirst
coordinate
toreduce
thewidth
ofthehouse
whiletheheightremainsunchanged.
ThematrixA2shrinksthesecondcoordinate
toreduce
theheight,butnotthewidth.
Thematrix
A3interchanges
thetw
ocoordinateswhileshrinkingoneofthem
.The
matrix
A4shrinksthefirstcoordinate
andchanges
thesignof
thesecond.
Thedeterminantofa2-by-2
matrix
A=
( a1,1
a1,2
a2,1
a2,2
)is
thequantity
a1,1a2,2−a1,2a2,1
Ingeneral,determinants
are
notveryusefulin
practicalcomputationbecause
they
haveatrociousscalingproperties.But2-by-2
determinants
canbeusefulin
under-
standingsimple
matrix
properties.If
thedeterminantofamatrix
ispositive,
then
multiplicationbythatmatrix
preserves
left-orright-handedness.
Thefirsttw
oof
ourfourmatrices
havepositivedeterminants,so
thedoorremainson
theleft
side
ofthehouse.Theother
twomatrices
havenegativedeterminants,so
thedoor
istransform
edto
theother
sideof
thehouse.
TheMatlabfunctionrand(m,n)generatesan
m-by-n
matrix
withrandom
entriesbetween0an
d1.Sothestatement
51
R=
2*rand(2,2)
-1
generatesa2-by-2
matrix
withrandom
entriesbetween-1
and1.
Herearefourof
them
. R1
=
0.0323
-0.6327
-0.5495
-0.5674
R2
=
0.7277
-0.5997
0.8124
0.7188
R3
=
0.1021
0.1777
-0.3633
-0.5178
R4
= -0.8682
0.9330
0.7992
-0.4821
−10
010
−10−50510
R1
−10
010
−10−50510
R2
−10
010
−10−50510
R3
−10
010
−10−50510
R4
Figure
4.4.Theeff
ectofmultiplicationby
random
matrices.
52Chapter4.
Matrices
Figure
4.4show
stheeff
ectofmultiplicationbythesefourmatrices
onthehouse.
Matrices
R1andR4havelargeoff
-diagonalentriesan
dnegativedeterminants,so
they
distort
thehouse
quiteabitandflip
thedoorto
therightside.
Thelines
are
still
straight,
butthewallsare
notperpendicularto
thefloor.
Lineartransform
ations
preservestraightlines,butthey
donotnecessarily
preservethean
glesbetweenthose
lines.Matrix
R2is
close
toarotation,whichwewilldiscu
ssshortly.Matrix
R3is
nearlysingular;itsdeterminantisequalto
0.0117.Ifthedeterminantwereexactly
zero,thehouse
wou
ldbeflattened
toaon
e-dim
ensionalstraightline.
Thefollow
ingmatrix
isaplanerotation.
G(θ)=
( cosθ
−sinθ
sinθ
cosθ
)
Weuse
theletter
Gbecause
Wallace
Givenspioneeredtheuse
ofplanerotations
inmatrix
computationin
the1950s.
MultiplicationbyG(θ)rotatespoints
inthe
planethroughanangle
θ.Figure
4.5show
stheeff
ectofmultiplicationbytheplane
rotationswithθ=
15◦ ,
45◦ ,
90◦ ,
and215◦ .
−10
010
−10−50510
G15
−10
010
−10−50510
G45
−10
010
−10−50510
G90
−10
010
−10−50510
G21
5
Figure
4.5.
Theaffectofmultiplication
byplanerotationsthough
15◦ ,
45◦ ,
90◦ ,
and215◦.
G15= 0.9659
-0.2588
53
0.2588
0.9659
G45= 0.7071
-0.7071
0.7071
0.7071
G90=
0-1
10
G215=
-0.8192
0.5736
-0.5736
-0.8192
You
canseethatG45is
fairly
close
totherandom
matrix
R2seen
earlierandthat
itseff
ectonthehouse
issimilar.
Matlabgeneratesaplanerotationforanglesmeasuredin
radianswith
G=
[cos(theta)
-sin(theta);
sin(theta)
cos(theta)]
andforanglesmeasuredin
degrees
with
G=
[cosd(theta)
-sind(theta);
sind(theta)
cosd(theta)]
Ourexmtoolbox
functionwiggleusesdot2dotandplanerotationsto
pro-
duce
ananim
ationofmatrix
multiplication.Hereiswiggle.m,withouttheHandle
Graphicscommands.
functionwiggle(X)
thetamax=
0.1;
delta=
.025;
t=
0;
whiletrue
theta=
(4*abs(t-round(t))-1)
*thetamax;
G=
[cos(theta)
-sin(theta);
sin(theta)
cos(theta)]
Y=
G*X;
dot2dot(Y);
t=
t+
delta;
end
Since
thisversiondoes
nothaveastopbutton,itwou
ldrunforever.Thevariablet
advancessteadilybyincrem
entof
delta.Astincreases,
thequantity
t-round(t)
varies
between−1
/2and1/2
,so
thean
gle
θcomputedby
theta=
(4*abs(t-round(t))-1)
*thetamax;
varies
inasawtooth
fashionbetween-thetamaxandthetamax.Thegraphofθas
afunctionoftis
show
nin
figure
4.6.Each
valueofθproducesacorresponding
planerotationG(θ).
Then
54Chapter4.
Matrices
00.
51
1.5
22.
53
3.5
4
−0.10
0.1
Figure
4.6.Wiggleangleθ
Y=
G*X;
dot2dot(Y)
applies
therotationto
theinputmatrix
Xandplots
thewigglingresult.
Vectors
andMatrices
Hereis
aquicklookatafew
ofthemanyMatlab
operationsinvolvingvectors
andmatrices.Try
topredictwhatoutputwillbeproducedbyeach
ofthefollow
ing
statements.
Youcanseeif
you
arerightbyusingcutan
dpasteto
execute
the
statement,orbyrunning
matrices_recap
Vectors
arecreatedwithsquare
brackets.
v=
[0
1/4
1/2
3/4
1]
Row
sofamatrix
areseparatedbysemicolonsor
new
lines.
A=
[8
16;
35
7;
49
2]
Thereareseveralfunctionsthatcreate
matrices.
Z=
zeros(3,4)
E=
ones(4,3)
I=
eye(4,4)
M=
magic(3)
R=
rand(2,4)
[K,J]=
ndgrid(1:4)
Acoloncreatesuniform
lyspacedvectors.
v=
0:0.25:1
n=
10
y=
1:n
Asemicolonattheendofalinesuppresses
output.
n=
1000;
y=
1:n;
55
Matrix
arithmetic
Matrix
additionandsubtractionare
denotedby+
and-.Theoperations
A+
B
and
A-
B
requireAandBto
bethesamesize,orto
bescalars,whichare
1-by-1
matrices.
Matrix
multiplication,denotedby*,
follow
stherulesoflinearalgebra.The
operation
A*
B
requires
thenumber
ofcolumnsofAto
equalthenumber
ofrow
B,that
is
size(A,2)==
size(B,1)
Rem
ember
thatA*Bis
usuallynotequalto
B*A
Ifpis
aninteger
scalar,
theexpression
A^p
denotesrepeatedmultiplicationofAbyitselfptimes.
Theuse
ofthematrix
divisionoperationsin
Matlab,
A\
B
and
A/
B
isdiscussed
inour“LinearEquations”
chapter
Array
arithmetic
.Weusuallytryto
distinguishbetweenmatrices,whichbehaveaccordingto
therulesoflinearalgebra,an
darrays,which
are
just
rectangularcollectionsof
numbers.
Element-by-elementoperationsarray
operationsaredenotedby+
,-,.*
,./
,.and.^.Forarray
multiplicationA.*Bis
equalto
B.*A
K.*J
v.^2
Anapostrophedenotesthetranspose
ofarealarray
andthecomplexconjugate
transpose
ofacomplexarray.
v=
v’
inner_prod=
v’*v
outer_prod=
v*v’
Z=
[1
2;
3+4i
5]’
Z=
[1
2;
3+4i
5].’
56Chapter4.
Matrices
Figure
4.7.ThecoverofGilbert
Strang’stextbookshowsaquiltby
ChrisCurtis.
Further
Rea
ding
Ofthedozensof
good
book
son
matrices
and
linearalgebra,wewould
liketo
recommendonein
particular.
GilbertStrang,Introductionto
LinearAlgebra,Wellesley-C
ambridge
Press,Wellesley,MA,2003.
http://www.wellesleycambridge.com
Besides
itsexcellenttechnicalcontentandexposition,it
hasaterrificcover.
The
house
thatwehaveusedthroughoutthis
chaptermadeitsdebutin
Strang’s
book
in1993.Thecoverofthefirsteditionlooked
somethinglikeou
rfigure
4.4.Chris
Curtis
saw
thatcoverand
created
agorgeousquilt.
Apicture
ofthequilthas
appearedonthecoverofallsubsequenteditionsofthebook
.
Rec
ap
%%
Matrices
Chapter
Recap
57
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Matrices
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %matrices_recap
%edit
matrices_recap
%publish
matrices_recap
% %Related
EXM
Programs
% %wiggle
%dot2dot
%house
%hand
%%Vectors
and
matrices
x=
[2;
4]
A=
[4
-3;
-2
1]
A*x
A’*A
A*A’
%%Random
matrices
R=
2*rand(2,2)-1
%%Build
ahouse
X=
house
dot2dot(X)
%%Rotations
theta=
pi/6
%radians
G=
[cos(theta)
-sin(theta);
sin(theta)
cos(theta)]
theta=
30
%degrees
G=
[cosd(theta)
-sind(theta);
sind(theta)
cosd(theta)]
subplot(1,2,1)
dot2dot(G*X)
subplot(1,2,2)
dot2dot(G’*X)
%%More
on
Vectors
and
Matrices
%Vectors
are
created
with
square
brackets.
v=
[0
1/4
1/2
3/4
1]
%Rows
of
amatrix
are
separated
by
semicolons
or
new
lines.
58Chapter4.
Matrices
A=
[8
16;
35
7;
49
2]
A=
[8
16
357
492]
%%
Creating
matrices
Z=
zeros(3,4)
E=
ones(4,3)
I=
eye(4,4)
M=
magic(3)
R=
rand(2,4)
[K,J]=
ndgrid(1:4)
%%
Colons
and
semicolons
%A
colon
creates
uniformally
spaced
vectors.
v=
0:0.25:1
n=
10
y=
1:n
%A
semicolon
at
the
end
of
aline
suppresses
output.
n=
1000;
y=
1:n;
%%
Matrix
arithmetic.
%Addition
and
subtraction,
+and
-,
are
element-by-element.
%Multiplication,
*,
follows
the
rules
of
linear
algebra.
%Power,
^,
is
repeated
matrix
multiplication.
KJ
=K*J
JK
=J*K
%%
Array
arithmetic
%Element-by-element
operations
are
denoted
by
%+
,-
,.*
,./
,.\
and
.^
.
K.*J
v.^2
%%
Transpose
%An
apostrophe
denotes
the
transpose
of
areal
array
59
%and
the
complex
conjugate
transpose
of
acomplex
array.
v=
v’
inner_prod=
v’*v
outer_prod=
v*v’
Z=
[1
2;
3+4i
5]’
Z=
[1
2;
3+4i
5].’
Exe
rcises
4.1
Multiplication.
(a)W
hich2-by-2
matrices
haveA
2=
I?
(b)W
hich2-by-2
matrices
haveA
TA
=I?
(c)W
hich2-by-2
matrices
haveA
TA
=AA
T?
4.2
Inverse.
Let
A=
( 34
23
)
Findamatrix
Xso
thatAX
=I.
4.3
Powers.
Let
A=
(0.99
0.01
−0.01
1.01
)
Whatis
An?
4.4
Powers.
Let
A=
( 11
10
)
Whatis
An?
4.5
Parametrizedproduct.Let
A=
( 12
x3
)(4
56
7
)
Whichelem
ents
ofA
dep
enduponx?Is
itpossible
tochoose
xso
thatA
=A
T?
4.6
Product
oftwosymmetricmatrices.It
turnsou
tthatanymatrix
istheproduct
oftw
osymmetricmatrices.Let
A=
( 34
810
)
60Chapter4.
Matrices
Express
Aastheproduct
oftw
osymmetricmatrices.
4.7
Given
srotations.
(a)W
hatis
thedeterminantof
G(θ)?
(b)Explain
whyG(θ)2
=G(2θ).
(c)Explain
whyG(θ)n
=G(nθ).
4.8
X8.Findareal2-by-2
matrix
Xso
thatX
8=
−I.
4.9
GT.W
hatis
theeff
ectonpoints
intheplaneofmultiplicationbyG(θ)T
?
4.10G.(a)W
hatis
theeff
ectonpoints
intheplaneofmultiplicationby
G(θ)=
( cosθ
sinθ
sinθ
−cosθ
)
(b)W
hatis
thedeterminantof
G(θ)?
(c)W
hat
hap
pensifyou
modifywiggle.mto
use
Ginsteadof
G?
4.11Goldie.W
hatdoes
thefunctiongoldiein
theexmtoolbox
do?
4.12Transform
ahand.Rep
eattheexperim
ents
inthis
chapterwith
X=
hand
insteadof
X=
house
Figure
4.8
show
s
dot2dot(hand)
4.13Mirrorim
age.Finda2-by-2
matrix
Rso
that
dot2dot(house)
and
dot2dot(R*house)
aswellas
dot2dot(hand)
and
dot2dot(R*hand)
61
−10
−50
510
−10−8−6−4−20246810
Figure
4.8.A
hand.
are
mirrorim
ages
ofeach
other.
4.14Transform
yourownhand.Rep
eattheexperim
ents
inthischapterusingaplot
ofyou
row
nhand.Start
with
figure(’position’,get(0,’screensize’))
axes(’position’,[00
11])
axis(10*[-11
-1
1])
[x,y]=
ginput;
Place
yourhandonthecomputerscreen.Use
themouse
toselect
afewdozenpoints
outliningyourhand.Terminate
theginputwithacarriagereturn.Youmightfind
iteasier
totrace
yourhandonapiece
ofpaper
andthen
putthepaper
onthe
computerscreen.Youshould
beable
toseetheginputcursorthroughthepaper.
Thedata
you
havecollected
form
stw
ocolumn
vectors
with
entriesin
the
rangefrom
-10to
10.You
canarrangethedata
astw
orowsin
asingle
matrix
with
H=
[x
y]’;
Then
youcanuse
dot2dot(H)
dot2dot(A*H)
wiggle(H)
andso
on.
Youcansave
you
rdata
inthefile
myhand.matwith
savemyhandH
andretrieveit
inalaterMatlabsessionwith
loadmyhand
62Chapter4.
Matrices
4.15Wiggler.
Make
wiggler.m,yourow
nversionof
wiggle.m,withtw
osliders
thatcontrolthespeedandam
plitude.
Intheinitialization,replace
thestatements
thetamax=
0.1;
delta=
.025;
with
thetamax=
uicontrol(’style’,’slider’,’max’,1.0,
...
’units’,’normalized’,’position’,[.25.01.25.04]);
delta=
uicontrol(’style’,’slider’,’max’,.05,
...
’units’,’normalized’,’position’,[.60.01.25.04]);
Thequantities
thetamaxanddeltaare
now
thehandlesto
thetw
osliders.
Inthe
bodyoftheloop
,replace
thetamaxby
get(thetamax,’value’);
andreplace
deltaby
get(delta,’value’);
Dem
onstrate
yourwiggleronthehouseandthehand.
Chapter5
Linea
rEquations
Themost
importanttask
intechnicalcomputing.
Iam
thinkingoftw
onumbers.
Theiraverageis
3.W
hatare
thenumbers?
Please
remem
ber
thefirstthingthatpop
sinto
yourhead.Iwillget
back
tothis
problem
inafew
pages.
Solvingsystem
sofsimultaneouslinearequationsis
themost
importanttask
intechnicalcomputing.It
isnotonly
importantin
itsow
nright,
itis
alsoafun-
damental,an
doften
hidden,componentof
other
more
complicatedcomputational
tasks.Theverysimplest
linearequationsinvolveonly
oneunknow
n.Solve
7x=
21
Thean
swer,ofcourse,
is
x=
21 7=
3
Now
solve
ax=
b
Thean
swer,ofcourse,
is
x=
b a
Butwhat
ifa=
0?Then
wehaveto
look
atb.
Ifb�=
0then
thereisnovalue
ofxthatsatisfies
0x=
b
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
63
64Chapter5.
LinearEquations
Thesolutiondoes
notexist.
Ontheother
hand,ifb=
0then
anyvalueofxsatisfies
0x=
0
Thesolutionis
notunique.
Mathem
aticianshavebeenthinkingab
outexistence
anduniquenessforcenturies.Wewillseethattheseconcepts
are
also
importantin
moderntechnicalcomputing.
Hereis
atoystory
problem.
Alice
buysthreeapples,adozenbananas,andon
ecantaloupefor$2.36.
Bobbuysadozenapplesandtw
ocantaloupes
for$5.26.Carolbuystw
obananasan
dthreecantaloupes
for$2.77.How
much
dosingle
piecesof
each
fruitcost?
Let
x1,x
2,andx3denote
theunknow
nprice
ofeach
fruit.Wehavethreeequations
inthreeunknow
ns.
3x1+12
x2+
x3=
2.36
12x1
+2x3=
5.26
2x2+3x3=
2.77
Because
matrix-vectormultiplicationhasbeendefined
theway
ithas,
theseequa-
tionscanbewritten
⎛ ⎝3
12
112
02
02
3
⎞ ⎠⎛ ⎝x1
x2
x3
⎞ ⎠ =⎛ ⎝2.
365.27
2.77
⎞ ⎠Or,
simply
Ax=
b
whereA
isagiven
3-by-3
matrix,bis
agiven
3-by-1
columnvector,
andx
isa
3-by-1
columnvectorwithunknow
nelem
ents.
Wewantto
solvethis
equation.If
youknow
anythingaboutmatrices,you
know
thattheequationcanbesolved
usingA
−1,theinverseof
A,
x=
A−1b
Thisisafineconcepttheoretically,butnotso
goodcomputationallyWedon’treally
needA
−1,wejust
wantto
findx.
Ifyou
donotknow
anythingaboutmatrices,you
mightbetemptedto
divide
bothsides
oftheequationbyA.
x=
b A
This
isaterrible
idea
theoretically–youcan’t
dividebymatrices
–butit
isthe
beginningofago
odidea
computationally.
Tofindthesolutionto
alinearsystem
ofequationswithMatlab,startby
enteringthematrix
ofcoeffi
cients.
65
A=
[3
12
1;
12
02;
02
3]
Since
alltheelem
ents
ofAare
integers,thematrix
isprintedwithaninteger
form
at.
A=
312
1
12
02
02
3
Next,entertherighthandsideasacolumnvector.
b=
[2.36
5.26
2.77]’
Theelem
ents
ofbare
notintegers,so
thedefaultform
atshow
sfourdigitsafter
the
decim
alpoint.
b=
2.3600
5.2600
2.7700
Matlab
hasan
outputform
atintended
forfinancialcalculations,
likethis
fruit
price
calculation.Thecommand
formatbank
changes
theoutputto
show
only
twodigitsafter
thedecim
alpoint.
b=
2.36
5.26
2.77
InMatlabthesolutionto
thelinearsystem
ofequations
Ax=
b
isfoundusingthebackslash
operator.
x=
A\b
Thinkof
this
as“dividing”bothsides
oftheequationbyA.Theresult
is
x=
0.29
0.05
0.89
Thisgives
usthesolutionto
ourstory
problem
–applescost
29cents
each,bananas
are
anickel
each,an
dcantaloupes
are
89cents
each.
Veryrarely,system
soflinearequationscomein
theform
xA
=b
wherebandxare
row
vectors.In
thiscase,thesolutionisfoundusingtheforw
ard
slash
operator.
66Chapter5.
LinearEquations
x=
b/A
Thetw
ooperationsA\bandb/Aare
sometim
escalled
leftandrightmatrix
division.
Inbothcases,
thecoeffi
cientmatrix
isin
the“d
enominator”.Forscalars,left
and
rightdivisionarethesamething.Thequantities
7\21and21/7are
both
equalto
3. Singularmatrix
Let’s
changeou
rstory
problem
abit.Assumenow
thatCarolbuyssixapplesand
onecantaloupefor$2.77.Thecoeffi
cientmatrix
andrighthandsidebecome
A=
312
1
12
02
60
1
and
b=
2.36
5.26
2.77
Atfirstglance,this
does
notlook
likemuch
ofachange.
How
ever,
x=
A\b
produces
Warning:Matrixis
singular
to
working
precision.
x= NaN
-Inf
Inf
Infand-Infstandforplusandminusinfinityandresultfrom
divisionof
nonzero
numbersbyzero.NaNstandsfor“Not-a-N
umber”andresultsfrom
doingarithmetic
involvinginfinities.
Thesourceof
thediffi
cultyisthatthenew
inform
ationab
outCarol’spurchase
isinconsistentwiththeearlierinform
ationab
outAlice’s
andBob’s
purchases.
We
havesaid
thatCarolboughtexactly
halfasmuch
fruit
asBob.Butshepaid
2.77
when
halfofBob’spay
mentwou
ldhavebeenonly
2.63.Thethirdrow
ofAisequal
toone-halfofthesecondrow,butb(3)is
notequalto
one-halfofb(2).
Forthis
particularmatrix
Aandvectorb,
thesolutionto
thelinearsystem
ofequations
Ax=
bdoes
notexist.
WhatifwemakeCarol’spurchase
price
consistentwithBob’s?
WeleaveA
unchanged
andrevisebwith
b(3)=
2.63
67
so
b=
2.36
5.26
2.63
Now
wedonothaveenoughinform
ation.Ourlast
twoequationsare
scalarmultiples
ofeach
other.
12x1+2x3=
5.26
6x1+
x3=
2.63
Onepossible
solutionis
thesolutionto
theoriginalproblem.
x=
0.29
0.05
0.89
A*x=
2.36
5.26
2.63
Butwecanpickanarbitrary
valueforanycomponentof
thesolutionandthen
use
thefirstequationandoneofthelast
twoequationsto
compute
theother
compo-
nents.Theresultisasolutionto
allthreeequations.
Forexample,hereisasolution
withitsthirdcomponentequalto
zero.
y=
A(1:2,1:2)\b(1:2);
y(3)=
0
y=
0.44
0.09
0.00
A*y=
2.36
5.26
2.63
Thereareinfinitelymanymore.
Forthis
particularmatrix
Aand
vectorb,
thesolution
toAx
=bis
not
unique.
Thefamilyofpossible
solutionsis
generatedbythenullvectorofA.This
isanonzero
vectorzforwhich
Az=
0
Thegeneralform
ofthesolutionis
oneparticularsolution,sayourvectorx,plus
anyarbitrary
parameter
times
thenullvector.
Foranyvalueoftthevector
y=
x+tz
68Chapter5.
LinearEquations
isasolutionto
Ay=
b
InMatlab
z=
null(A)
A*z
t=
rand
y=
x+
t*z
A*y
You
canseethatA*zis
zero
andA*yis
equalto
b.
“Theirave
rageis
three.”
Let’s
return
tothequestionthatIasked
youto
consider
atthebeginningofthis
chapter.
I’m
thinkingof
twonumbers.
Theiraverageis
three.
Whatare
the
numbers?
Whatpop
ped
into
you
rhead?Youprobably
realizedthatIhadn’tgiven
you
enoughinform
ation.Butyoumust
havethoughtof
somesolution.In
myexperi-
ence,themost
frequentresponse
is“2and4”.Let’s
seewhatMatlabresponds.
Myproblem
ison
elinearequationin
twounknow
nsThematrix
andright
handsideare
A=
[1/2
1/2]
b=
3
Wewan
tto
solveAx=
b.Thisisnow
anunderdetermined
system
.Thereare
fewer
equationsthanunknow
ns,
sothereareinfinitelymanysolutions.
Backslash
offersonepossible
solution
x=
A\b
x=
6 0
Ibet
youdidn’tthinkofthis
solution.
Ifwetryinv(A)*bweget
anerrormessagebecause
rectangularmatrices
do
nothaveinverses.Butthereis
somethingcalled
thepseudoinverse.
Wecantry
that. x
=pinv(A)*b
x=
3 3
69
Did
thatsolutionoccurto
you?
Thesetw
ox’s
are
just
twomem
bersoftheinfinitefamilyofsolutions.
Ifwe
wan
tedMatlab
tofindthe[24]’solution,wewould
haveto
pose
aproblem
wherethesolutionis
constrained
tobeapair
ofintegers,
close
toeach
other,but
notequal.
Itis
possible
tosolvesuch
problems,
butthatwould
takeustoofar
afield.
MyRules.
This
chapterillustratestw
ofundamentalfactsabou
ttechnicalcomputing.
•Thehardestquantities
tocompute
are
ones
thatdonotexist.
•Thenexthardestare
ones
thatarenotunique.
Rec
ap
%%Linear
Equations
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Linear
Equations
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %linear_recap
%edit
linear_recap
%publish
linear_recap
%%Backslash
formatbank
A=
[3
12
1;
12
02;
02
3]
b=
[2.36
5.26
2.77]’
x=
A\b
%%Forward
slash
x=
b’/A’
%%Inconsistent
singular
system
A(3,:)=
[6
01]
A\b
%%Consistent
singular
system
b(3)=
2.63
%%One
particular
solution
x=
A(1:2,1:2)\b(1:2);
x(3)=
0
A*x
70Chapter5.
LinearEquations
%%
Null
vector
z=
null(A)
A*z
%%
General
solution
t=
rand
%Arbitrary
parameter
y=
x+
t*z
A*y
Exe
rcises
5.1
Two-by-two.Use
backslash
totryto
solveeach
ofthesesystem
sof
equations.
Indicate
ifthesolutionexists,andifitunique.
(a)
( 10
02
) x=
( −1 2
)(b)
( 12
34
) x=
( 1 1
)(c)
( 12
36
) x=
( 1 2
)(d)
( 12
36
) x=
( 1 3
)
5.2
Three-by-three.
Use
backslash
totryto
solveeach
ofthesesystem
sof
equations.
Indicate
ifthesolutionexists,andifitunique.
(a)
⎛ ⎝10
00
20
00
3
⎞ ⎠ x=
⎛ ⎝1 1 1
⎞ ⎠(b)
⎛ ⎝9
−36
30−3
6192
−180
30−1
80180
⎞ ⎠ x=
⎛ ⎝1 0 0
⎞ ⎠
71
(c)
⎛ ⎝12
34
12
67
812
⎞ ⎠ x=
⎛ ⎝3 12
15
⎞ ⎠(d)
⎛ ⎝81
63
57
49
2
⎞ ⎠ x=
⎛ ⎝1 1 1
⎞ ⎠(e)
⎛ ⎝12
34
56
78
9
⎞ ⎠ x=
⎛ ⎝1 1 1
⎞ ⎠(f)
⎛ ⎝12
34
56
78
9
⎞ ⎠ x=
⎛ ⎝1 0 0
⎞ ⎠
5.3
Nullvector.
Findanonzero
solutionzto
⎛ ⎝12
34
56
78
9
⎞ ⎠ z=
⎛ ⎝0 0 0
⎞ ⎠
5.4
Matrix
equations.
Backslash
canbeusedto
solvematrix
equationsoftheform
AX
=B
whereB
hasseveralcolums.
Doyourecognizethesolutionto
thefollow
ingequa-
tion?⎛ ⎝
53
−52
23−2
28
38
−768
−37
⎞ ⎠ X=
⎛ ⎝360
00
0360
00
0360
⎞ ⎠
5.5
More
Fruit.Alice
buysfourap
ples,
twodozenbananas,
andtw
ocantaloupes
for$4.14.Bobbuysadozenapplesandtw
ocantaloupes
for$5.26.Carolbuysa
halfdozenbananasandthreecantaloupes
for$2.97.How
much
dosingle
piecesof
each
fruitcost?(Y
oumightwantto
setformatbank.)
5.6
Truss.Figure
5.1
dep
icts
aplanetruss
hav
ing13mem
bers(thenumbered
lines)
connecting8joints
(thenumbered
circles).Theindicatedloads,in
tons,are
applied
72Chapter5.
LinearEquations
12
56
8
34
7
13
57
911
12
26
1013
48
1015
20
Figure
5.1.A
planetruss.
atjoints
2,5,and6,
andwewan
tto
determinetheresultingforceoneach
mem
ber
ofthetruss.
Forthetruss
tobein
staticequilibrium,theremust
benonet
force,
hor-
izontallyorvertically,
atan
yjoint.
Thus,
wecan
determinethemem
ber
forces
byequatingthehorizontalforces
totheleft
andrightat
each
joint,
andsimilarly
equatingtheverticalforces
upward
and
dow
nward
ateach
joint.
Fortheeight
joints,this
would
give16equations,
whichis
more
thanthe13unknow
nfactors
tobedetermined.Forthetruss
tobestaticallydeterminate,thatis,forthereto
beauniquesolution,weassumethatjoint1is
rigidly
fixed
both
horizontallyand
verticallyandthatjoint8isfixed
vertically.
Resolvingthemem
ber
forces
into
hor-
izontalan
dverticalcomponents
anddefiningα
=1/√ 2
,weobtain
thefollow
ing
system
ofequationsforthemem
ber
forces
f i:
Joint2:
f 2=
f 6,
f 3=
10;
Joint3:
αf 1
=f 4
+αf 5,
αf 1
+f 3
+αf 5
=0;
Joint4:
f 4=
f 8,
f 7=
0;
Joint5:
αf 5
+f 6
=αf 9
+f 1
0,
αf 5
+f 7
+αf 9
=15;
Joint6:
f 10=
f 13,
f 11=
20;
Joint7:
f 8+
αf 9
=αf 1
2,
αf 9
+f 1
1+
αf 1
2=
0;
Joint8:
f 13+αf 1
2=
0.
Solvethis
system
ofequationsto
findthevectorfofmem
ber
forces.
73
5.7
Circuit.Figure
5.2
isthecircuit
diagram
forasm
allnetworkofresistors.
12
3
4
5
r 23r 34 r 45
r 25
r 13
r 12r 14
r 35
v s
i 1
i 2i 3
i 4
Figure
5.2.A
resistornetwork.
Thereare
fivenodes,eightresistors,andon
econstantvoltagesource.
Wewan
tto
compute
thevoltagedropsbetweenthenodes
andthecurrents
aroundeach
of
theloops.
Severaldifferentlinearsystem
sofequationscanbeform
edto
describethis
circuit.
Let
v k,k
=1,...,4,denote
thevoltagedifference
between
each
ofthe
firstfournodes
andnodenumber
5andleti k,k
=1,...,4,denote
theclockwise
currentaroundeach
oftheloop
sin
thediagram.Ohm’s
law
saysthatthevoltage
dropacross
aresistoris
theresistance
times
thecurrent.
Forexample,thebranch
betweennodes
1and2gives
v 1−v 2
=r 1
2(i
2−i 1).
Usingtheconductance,whichisthereciprocaloftheresistance,g k
j=
1/r
kj,Ohm’s
law
becomes
i 2−i 1
=g 1
2(v
1−v 2).
Thevoltagesourceis
included
intheequation
v 3−v s
=r 3
5i 4.
Kirchhoff’s
voltage
law
saysthatthesum
ofthevoltagedifferencesaround
each
loopis
zero.For
example,aroundloop1,
(v1−
v 4)+(v
4−v 5)+(v
5−v 2)+(v
2−v 1)=
0.
Combiningthevoltagelaw
with
Ohm’s
law
leadsto
theloopequationsforthe
currents:
Ri=
b.
Hereiis
thecurrentvector,
i=
⎛ ⎜ ⎝i 1 i 2 i 3 i 4
⎞ ⎟ ⎠,
74Chapter5.
LinearEquations
bis
thesourcevoltagevector,
b=
⎛ ⎜ ⎝0 0 0 v s
⎞ ⎟ ⎠,andR
istheresistance
matrix,
⎛ ⎜ ⎝r 25+r 1
2+r 1
4+r 4
5−r
12
−r14
−r45
−r12
r 23+r 1
2+r 1
3−r
13
0−r
14
−r13
r 14+r 1
3+
r 34
−r34
−r45
0−r
34
r 35+
r 45+
r 34
⎞ ⎟ ⎠.Kirchhoff’s
curren
tlawsaysthatthesum
ofthecurrents
ateach
nodeiszero.
For
example,at
node1,
(i1−i 2)+(i
2−
i 3)+(i
3−i 1)=
0.
Combiningthecurrentlaw
withtheconductance
versionofOhm’slaw
leadsto
the
nodalequationsforthevoltages:
Gv=
c.
Herevis
thevoltagevector,
v=
⎛ ⎜ ⎝v 1 v 2 v 3 v 4
⎞ ⎟ ⎠,cis
thesourcecurrentvector,
c=
⎛ ⎜ ⎝0 0
g 35v s 0
⎞ ⎟ ⎠,andG
istheconductance
matrix,
⎛ ⎜ ⎝g 12+
g 13+g 1
4−g
12
−g13
−g14
−g12
g 12+g 2
3+g 2
5−g
23
0−g
13
−g23
g 13+g 2
3+g 3
4+g 3
5−g
34
−g14
0−g
34
g 14+g 3
4+g 4
5
⎞ ⎟ ⎠.You
cansolvethelinearsystem
obtained
from
theloopequationsto
compute
thecurrents
andthen
use
Ohm’s
law
torecover
thevoltages.
Oryou
cansolve
thelinearsystem
obtained
from
thenodeequationsto
compute
thevoltages
and
then
use
Ohm’slaw
torecover
thecurrents.You
rassignmentisto
verifythatthese
twoap
proaches
produce
thesameresultsforthiscircuit.You
canchooseyou
row
nnumericalvalues
fortheresistancesandthevoltagesource.
Chapter6
FractalFern
Thefractalfern
involves
2-by-2matrices.
Theprogramsfernandfinitefernin
theexmtoolbox
produce
theFractal
Ferndescribed
byMichael
Barnsley
inFractalsEverywhere[?].They
generate
and
plotapotentiallyinfinitesequence
ofrandom,butcarefullychoreographed,points
intheplane.
Thecommand
fern
runsforever,producinganincreasingly
dense
plot.
Thecommand
finitefern(n)
generatesnpoints
andaplotlikeFigure
6.1.Thecommand
finitefern(n,’s’)
show
sthegenerationofthepoints
oneatatime.
Thecommand
F=
finitefern(n);
generates,
butdoes
notplot,
npoints
andreturnsanarray
ofzerosandones
for
use
withsparsematrix
andim
age-processingfunctions.
Theexmtoolbox
also
includes
fern.jpg,a768-by-1024colorim
agewithhalf
amillionpoints
thatyoucanview
withabrowserorapaintprogram.Youcanalso
view
thefile
with
F=
imread(’fern.png’);
image(F)
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
75
76Chapter6.
Fractal
Fern
Figure
6.1.Fractalfern.
Ifyou
liketheim
age,
youmighteven
choose
tomakeit
yourcomputerdesktop
background.How
ever,you
should
reallyrunfernonyou
row
ncomputerto
seethe
dynamicsoftheem
ergingfern
inhighresolution.
Thefern
isgeneratedbyrepeatedtransform
ationsof
apointin
theplane.
Let
xbeavectorwithtw
ocompon
ents,x1andx2,representingthepoint.
Thereare
fourdifferenttransform
ations,
allof
them
oftheform
x→
Ax+b,
withdifferentmatrices
Aandvectors
b.Theseare
know
nasaffinetransform
ations.
Themost
frequentlyusedtransform
ationhas
A=
( 0.85
0.04
−0.04
0.85
) ,b=
( 0 1.6
) .
This
transform
ationshortensandrotatesxalittle
bit,then
adds1.6
toitssecond
component.
Rep
eatedapplicationof
thistransform
ationmoves
thepointupandto
theright,
headingtoward
theupper
tipofthefern.Everyonce
inawhile,
oneof
theother
threetransform
ationsis
picked
atrandom.Thesetransform
ationsmove
77
thepointinto
thelower
subfern
ontheright,
thelower
subfern
ontheleft,orthe
stem
. Hereis
thecomplete
fractalfern
program.
functionfern
%FERN
MATLABimplementationof
the
Fractal
Fern
%MichaelBarnsley,FractalsEverywhere,AcademicPress,1993
%Thisversionrunsforever,or
until
stop
is
toggled.
%Seealso:FINITEFERN.
shg
clfreset
set(gcf,’color’,’white’,’menubar’,’none’,...
’numbertitle’,’off’,’name’,’FractalFern’)
x=
[.5;
.5];
h=
plot(x(1),x(2),’.’);
darkgreen=
[0
2/3
0];
set(h,’markersize’,1,’color’,darkgreen,’erasemode’,’none’);
axis([-33
010])
axisoff
stop=
uicontrol(’style’,’toggle’,’string’,’stop’,
...
’background’,’white’);
drawnow
p=
[.85
.92
.99
1.00];
A1
=[
.85
.04;
-.04
.85];
b1
=[0;
1.6];
A2
=[
.20
-.26;
.23
.22];
b2
=[0;
1.6];
A3
=[-.15
.28;
.26
.24];
b3
=[0;
.44];
A4
=[
00
;0
.16];
cnt=
1;
tic
while~get(stop,’value’)
r=
rand;
if
r<
p(1)
x=
A1*x
+b1;
elseifr
<p(2)
x=
A2*x
+b2;
elseifr
<p(3)
x=
A3*x
+b3;
else x=
A4*x;
end
set(h,’xdata’,x(1),’ydata’,x(2));
cnt=
cnt
+1;
drawnow
78Chapter6.
Fractal
Fern
end
t=
toc;
s=
sprintf(’%8.0f
points
in
%6.3f
seconds’,cnt,t);
text(-1.5,-0.5,s,’fontweight’,’bold’);
set(stop,’style’,’pushbutton’,’string’,’close’,...
’callback’,’close(gcf)’)
Let’s
examinethis
program
afew
statements
atatime.
shg
standsfor“show
graphwindow
.”It
bringsan
existinggraphicswindow
forw
ard,
orcreatesanew
oneifnecessary.
clfreset
resets
most
ofthefigure
properties
totheirdefault
values.
set(gcf,’color’,’white’,’menubar’,’none’,...
’numbertitle’,’off’,’name’,’FractalFern’)
changes
thebackgroundcolorof
thefigure
window
from
thedefault
gray
towhite
andprovides
acustomized
titleforthewindow
.
x=
[.5;
.5];
provides
theinitialcoordinatesofthepoint.
h=
plot(x(1),x(2),’.’);
plots
asingle
dotin
theplanean
dsavesahandle,h,so
thatwecanlatermodify
theproperties
oftheplot.
darkgreen=
[0
2/3
0];
defines
acolorwheretheredandbluecompon
ents
arezero
andthegreencomponent
istw
o-thirdsofitsfullintensity.
set(h,’markersize’,1,’color’,darkgreen,’erasemode’,’none’);
makes
thedotreferencedbyhsm
aller,changes
itscolor,andspecifies
thattheim
age
ofthedotonthescreen
should
notbeerasedwhen
itscoordinatesare
changed.A
record
oftheseold
points
iskeptbythecomputer’sgraphicshardware(untilthe
figure
isreset),butMatlabitselfdoes
notremem
ber
them
.
axis([-33
010])
axisoff
specifies
thattheplotshould
cover
theregion
−3≤
x1≤
3,0≤
x2≤
10,
butthattheax
esshould
notbedrawn.
79
stop=
uicontrol(’style’,’toggle’,’string’,’stop’,
...
’background’,’white’);
createsatoggle
userinterface
control,
labeled
’stop’andcoloredwhite,
inthe
defaultpositionnearthelower
left
corner
ofthefigure.Thehandle
forthecontrol
issavedin
thevariable
stop.
drawnow
causestheinitialfigure,includingtheinitialpoint,
toactuallybeplotted
onthe
computerscreen.
Thestatement
p=
[.85
.92
.99
1.00];
sets
upavectorofprobabilities.
Thestatements
A1
=[
.85
.04;
-.04
.85];
b1
=[0;
1.6];
A2
=[
.20
-.26;
.23
.22];
b2
=[0;
1.6];
A3
=[-.15
.28;
.26
.24];
b3
=[0;
.44];
A4
=[
00
;0
.16];
definethefouraffinetransform
ations.
Thestatement
cnt=
1;
initializesacounterthatkeepstrack
ofthenumber
ofpoints
plotted.Thestatement
tic
initializesastopwatchtimer.Thestatement
while~get(stop,’value’)
beginsawhileloopthatrunsaslongasthe’value’property
ofthestoptoggle
isequalto
0.Clickingthestoptoggle
changes
thevaluefrom
0to
1andterm
inates
theloop
. r=
rand;
generatesapseudorandom
valuebetween0and1.Thecompoundifstatement
if
r<
p(1)
x=
A1*x
+b1;
elseifr
<p(2)
x=
A2*x
+b2;
elseifr
<p(3)
x=
A3*x
+b3;
else x=
A4*x;
end
80Chapter6.
Fractal
Fern
pickson
eofthefouraffi
netransform
ations.
Because
p(1)is0.85,thefirsttrans-
form
ationis
chosen85%
ofthetime.
Theother
threetransform
ationsare
chosen
relativelyinfrequently.
set(h,’xdata’,x(1),’ydata’,x(2));
changes
thecoordinatesof
thepointhto
thenew
(x1,x
2)andplots
thisnew
point.
Butget(h,’erasemode’)is’none’,so
theold
pointalsoremainson
thescreen.
cnt=
cnt
+1;
counts
onemore
point.
drawnow
tellsMatlabto
takethetimeto
redraw
thefigure,show
ingthenew
pointalong
withalltheold
ones.W
ithoutthiscommand,nothingwould
beplotted
untilstop
istoggled.
end
matches
thewhileatthebeginningoftheloop
.Finally,
t=
toc;
readsthetimer.
s=
sprintf(’%8.0f
points
in
%6.3f
seconds’,cnt,t);
text(-1.5,-0.5,s,’fontweight’,’bold’);
displaystheelapsedtimesince
ticwas
called
andthefinalcountofthenumber
ofpoints
plotted
.Finally,
set(stop,’style’,’pushbutton’,’string’,’close’,...
’callback’,’close(gcf)’)
changes
thecontrolto
apush
buttonthatclosesthewindow
.
Rec
ap
%%
Fern
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Fern
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %fern_recap
%edit
fern_recap
%publish
fern_recap
% %Related
EXM
programs
%
81
%fern
%finitefern
%%fern.jpg
F=
imread(’fern.png’);
image(F)
%%A
few
graphics
commands
shg
clfreset
set(gcf,’color’,’white’)
x=
[.5;
.5];
h=
plot(x(1),x(2),’.’);
darkgreen=
[0
2/3
0];
set(h,’markersize’,1,’color’,darkgreen,’erasemode’,’none’);
set(h,’xdata’,x(1),’ydata’,x(2));
axis([-33
010])
axisoff
stop=
uicontrol(’style’,’toggle’,’string’,’stop’,’background’,’white’);
drawnow
cnt=
12345;
t=
5.432;
s=
sprintf(’%8.0f
points
in
%6.3f
seconds’,cnt,t);
text(-1.5,-0.5,s,’fontweight’,’bold’);
set(stop,’style’,’pushbutton’,’string’,’close’,’callback’,’close(gcf)’)
Exe
rcises
6.1
Ferncolor.
Changethefern
colorschem
eto
use
pinkonablack
background.
Don’tforget
thestopbutton.
6.2
Flipthefern.Flipthefern
byinterchangingitsx-an
dy-coordinates.
6.3
Erase
mode.
(a)W
hathappensifyouresize
thefigure
window
whilethefern
isbeinggenerated?
Why?
(b)Theexmprogram
finiteferncanbeusedto
produce
printedoutputofthe
fern.Explain
whyprintingis
possible
withfinitefern.mbutnotwithfern.m.
6.4
Fernstem
.(a)W
hathappensto
thefern
ifyouchangetheon
lynonzero
elem
entin
thematrix
A4?
(b)W
hatarethecoordinatesofthelower
endofthefern’s
stem
?
82Chapter6.
Fractal
Fern
6.5
Fern
tip.
Thecoordinatesof
thepointat
theupper
tipendofthefern
can
becomputedbysolvingacertain
2-by-2
system
ofsimultaneouslinearequations.
Whatis
thatsystem
andwhatarethecoordinatesofthetip?
6.6
Trajectories.Thefern
algorithm
involves
repeatedrandom
choices
from
four
differentform
ulasforad
vancingthepoint.
Ifthekth
form
ula
isusedrepeatedly
byitself,withoutrandom
choices,itdefi
nes
adeterministictrajectory
inthe(x,y)
plane.
Modifyfinitefern.mso
thatplots
ofeach
ofthesefourtrajectories
are
superim
posedontheplotof
thefern.Start
each
trajectory
atthepoint(−
1,5).
Ploto’s
connectedwithstraightlines
forthestepsalongeach
trajectory.Takeas
manystepsas
areneeded
toshow
each
trajectory’slimitpoint.
You
cansuperim
pose
severalplots
with
plot(...)
holdon
plot(...)
plot(...)
holdoff
6.7
Sierpinski’striangle.
Modifyfern.mor
finitefern.mso
thatitproducesSier-
pinski’striangle.
Start
at
x=
( 0 0
) .
Ateach
iterativestep
,thecurrentpointxisreplacedwithAx+b,wherethematrix
Ais
always
A=
( 1/2
00
1/2
)
andthevectorbis
chosenatrandom
withequalprobabilityfrom
amongthethree
vectors b=
( 0 0
) ,b=
( 1/2 0
) ,and
b=
( 1/4
√ 3/4
) .
Chapter7
PageR
ank
Theworld’s
largestmatrix
computation.(T
his
chapteris
outofdate
andneedsa
majoroverhaul.)
OneofthereasonswhyGoogle
TM
issuch
an
effectivesearch
engineis
the
PageR
ank
TM
algorithm
developed
by
Google’s
founders,
Larry
Pageand
Sergey
Brin,when
they
weregraduate
students
atStanford
University.PageR
ankis
de-
term
ined
entirely
bythelinkstructure
oftheWorldW
ideWeb
.It
isrecomputed
aboutonce
amonth
anddoes
notinvo
lvetheactualcontentofanyWeb
pages
orindividualqueries.
Then,foranyparticularquery,
findsthepages
onthe
Web
thatmatchthatqueryandliststhose
pages
intheorder
oftheirPageR
ank.
ImaginesurfingtheWeb,goingfrom
pageto
pagebyrandomly
choosingan
outgoinglinkfrom
onepageto
get
tothenext.
This
can
lead
todead
endsat
pages
with
nooutgoinglinks,
orcycles
around
cliques
ofinterconnected
pages.
So,acertain
fraction
ofthetime,
simply
choose
arandom
pagefrom
theWeb.
This
theoreticalrandom
walk
isknow
nasaMarkovchain
orMarkovprocess.The
limitingprobabilitythatan
infinitelyded
icatedrandom
surfer
visitsan
yparticular
pageis
itsPageR
ank.A
pagehashighrankifother
pages
withhighranklinkto
it.
Let
WbethesetofWeb
pages
thatcanbereached
byfollow
ingachain
of
hyperlinksstartingatsomerootpage,
andletnbethenumber
ofpages
inW
.For
Google,thesetW
actuallyvaries
withtime,
butbyJune2004,nwasover
4billion.
Let
Gbethen-by-n
connectivity
matrix
ofaportionoftheWeb,thatis,g i
j=
1ifthereis
ahyperlinkto
pageifrom
pagejandg i
j=
0otherwise.
Thematrix
Gcanbehuge,
butit
isverysparse.
Itsjthcolumnshow
sthelinksonthejthpage.
Thenumber
ofnonzerosin
Gis
thetotalnumber
ofhyperlinksin
W.
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
83
84Chapter7.
Goog
lePageR
ank
Let
r iandc j
betherow
andcolumnsumsof
G:
r i=
∑ j
g ij,
c j=
∑ i
g ij.
Thequantities
r jandc j
are
thein-degreeandout-degreeofthejthpage.
Let
pbe
theprobabilitythattherandom
walk
follow
salink.A
typicalvalueis
p=
0.85.
Then
1−pistheprobabilitythatsomearbitrary
pageischosenandδ=
(1−p)/n
istheprobabilitythataparticularrandom
pageis
chosen.Let
Abethen-by-n
matrix
whose
elem
ents
are
aij=
{ pg i
j/c
j+
δ:
c j�=
01/n
:c j
=0.
Notice
thatA
comes
from
scalingtheconnectivitymatrix
byitscolumnsums.
The
jthcolumnis
theprobabilityofjumpingfrom
thejthpageto
theother
pages
ontheWeb
.If
thejthpageis
adeadend,thatis
hasnoou
t-links,
then
weassigna
uniform
probabilityof1/nto
alltheelem
ents
initscolumn.Most
oftheelem
ents
ofA
are
equalto
δ,theprobabilityof
jumpingfrom
onepageto
another
without
follow
ingalink.Ifn=
4·109andp=
0.85,then
δ=
3.75·10−11.
Thematrix
Ais
thetransitionprobabilitymatrix
oftheMarkov
chain.Its
elem
ents
are
allstrictly
betweenzero
andoneanditscolumnsumsare
allequalto
one.
Anim
portantresultin
matrix
theory
know
nas
thePerron–Frobeniustheorem
applies
tosuch
matrices.It
concludes
thatanonzero
solutionof
theequation
x=
Ax
exists
andis
uniqueto
within
ascalingfactor.
Ifthis
scalingfactoris
chosenso
that
∑ i
xi=
1,
then
xis
thestate
vectoroftheMarkov
chain
and
isGoogle’s
PageR
ank.
The
elem
ents
ofxare
allpositiveandless
thanone.
Thevectorxis
thesolutionto
thesingular,
homogeneouslinearsystem
(I−A)x
=0.
For
modestn,an
easy
way
tocompute
xin
Matlabisto
start
withsomeapprox-
imate
solution,such
asthePageR
anksfrom
thepreviousmonth,or
x=
ones(n,1)/n
Then
simply
repeattheassignmentstatement
x=
A*x
untilsuccessivevectors
agreeto
within
aspecified
tolerance.This
isknow
nas
the
power
method
andis
abou
ttheon
lypossible
approach
forverylargen.
Inpractice,
thematrices
GandA
are
never
actuallyform
ed.Onestep
ofthe
pow
ermethodwould
bedonebyonepass
over
adatabase
ofWeb
pages,updating
weightedreference
counts
generatedbythehyperlinksbetweenpages.
85
Thebestway
tocompute
PageR
ankin
Matlabis
totakeadvantageofthe
particularstructure
oftheMarkov
matrix.Hereis
anapproach
thatpreserves
the
sparsityofG.Thetransitionmatrix
canbewritten
A=
pGD
+ez
T
whereD
isthediagonalmatrix
form
edfrom
thereciprocals
oftheoutdegrees,
djj=
{ 1/c j
:c j
�=0
0:
c j=
0,
eis
then-vectorofallon
es,andzis
thevectorwithcomponents
z j=
{δ
:c j
�=0
1/n
:c j
=0.
Therank-onematrix
ezT
accounts
fortherandom
choices
ofWeb
pages
thatdo
notfollow
links.
Theequation
x=
Ax
canbewritten
(I−pGD)x
=γe
where γ
=zTx.
Wedonotknow
thevalueof
γbecause
itdep
endsupontheunknow
nvectorx,but
wecantemporarily
takeγ=
1.Aslongaspisstrictly
less
thanone,
thecoeffi
cient
matrix
I−pGD
isnonsingularan
dtheequation
(I−pGD)x
=e
canbesolved
forx.Then
theresultingxcanberescaledso
that
∑ i
xi=
1.
Notice
thatthevectorzis
notactuallyinvolved
inthis
calculation.
Thefollow
ingMatlabstatements
implementthis
approach
c=
sum(G,1);
k=
find(c~=0);
D=
sparse(k,k,1./c(k),n,n);
e=
ones(n,1);
I=
speye(n,n);
x=
(I
-p*G*D)\e;
x=
x/sum(x);
Thepow
ermethodcanalso
beim
plementedin
away
thatdoes
notactually
form
theMarkov
matrix
andso
preserves
sparsity.
Compute
86Chapter7.
Goog
lePageR
ank
G=
p*G*D;
z=
((1-p)*(c~=0)
+(c==0))/n;
Start
with
x=
e/n
Then
repeatthestatement
x=
G*x
+e*(z*x)
untilxsettlesdow
nto
severaldecim
alplaces.
Itis
alsopossible
touse
analgorithm
know
nas
inverseiteration.
A=
p*G*D
+delta
x=
(I
-A)\e
x=
x/sum(x)
Atfirstglance,this
appears
tobeaverydangerousidea.Because
I−
Ais
the-
oreticallysingular,
with
exact
computation
somediagonalelem
entof
theupper
triangularfactorofI−
Ashould
bezero
andthis
computationshould
fail.But
withroundofferror,
thecomputedmatrix
I-
Ais
probably
notexactly
singular.
Even
ifit
issingular,
roundoff
duringGaussianelim
inationwillmost
likelypre-
ventanyexact
zero
diagonalelem
ents.Weknow
thatGaussianelim
inationwith
partialpivotingalwaysproducesasolutionwithasm
allresidual,
relativeto
the
computedsolution,even
ifthematrix
isbadly
conditioned.Thevectorobtained
withthebackslash
operation,(I
-A)\e
,usuallyhasverylargecomponents.If
itis
rescaledbyitssum,theresidualis
scaledbythesamefactorandbecomes
very
small.Consequently,
thetw
ovectors
xandA*xequaleach
other
towithin
roundoff
error.
Inthis
setting,solvingthesingularsystem
withGaussianelim
inationblows
up,butit
blowsupin
exactly
therightdirection.
Figure
7.1
isthegraphforatinyexample,withn=
6insteadof
n=
4·1
09.
Pages
ontheWeb
areidentified
bystringsknow
nas
uniform
resourcelocators,
orURLs.
Most
URLsbegin
withhttpbecause
they
use
thehypertext
transfer
protocol.In
Matlab,wecanstore
theURLsasanarray
ofstringsin
acellarray.
This
example
invo
lves
a6-by-1
cellarray.
U=
{’http://www.alpha.com’
’http://www.beta.com’
’http://www.gamma.com’
’http://www.delta.com’
’http://www.rho.com’
’http://www.sigma.com’}
Twodifferentkindsofindexinginto
cellarraysare
possible.Parentheses
denote
subarray
s,includingindividualcells,
andcu
rlybracesden
ote
thecontents
ofthe
cells.
Ifkisascalar,then
U(k)isa1-by-1
cellarray
consistingofthekth
cellin
U,
whileU{k}isthestringin
thatcell.ThusU(1)isasinglecellandU{1}isthestring
’http://www.alpha.com’.
Thinkofmailbox
eswithad
dresses
onacity
street.
B(502)is
thebox
atnumber
502,whileB{502}is
themailin
thatbox
.
87
alph
a
beta
gam
ma
delta
sigm
arh
o
Figure
7.1.A
tinyWeb.
Wecan
generate
theconnectivitymatrix
byspecifyingthepairsofindices
(i,j)ofthenonzero
elem
ents.Because
thereisalinkto
beta.comfrom
alpha.com,
the(2,1)elem
entofG
isnonzero.Thenineconnectionsare
described
by
i=
[2
63
44
56
11]
j=
[1
12
23
33
46]
Asparsematrix
isstoredin
adata
structure
thatrequires
mem
ory
only
forthe
nonzero
elem
ents
andtheirindices.This
ishardly
necessary
fora6-by-6
matrix
withonly
27zero
entries,
butit
becomes
cruciallyim
portantforlarger
problems.
Thestatements
n=
6
G=
sparse(i,j,1,n,n);
full(G)
generate
thesparserepresentationofann-by-n
matrix
withones
inthepositions
specified
bythevectors
iandjanddisplayitsfullrepresentation.
00
01
01
10
00
00
01
00
00
01
10
00
00
10
00
10
10
00
Thestatement
c=
full(sum(G))
88Chapter7.
Goog
lePageR
ank
computesthecolumnsums
c=
22
31
01
Notice
thatc(5)=
0because
the5th
page,
labeled
rho,hasnoout-links.
Thestatements
x=
(I
-p*G*D)\e
x=
x/sum(x)
solvethesparselinearsystem
toproduce
x=
0.3210
0.1705
0.1066
0.1368
0.0643
0.2007
12
34
56
0
0.050.1
0.150.2
0.250.3
0.35
Page
Ran
k
Figure
7.2.Page
RankforthetinyWeb
Thebargraphof
xisshow
nin
figure
7.2.IftheURLsaresorted
inPageR
ank
order
andlisted
alongwiththeirin-andout-degrees,theresultis
page-rank
in
out
url
10.3210
22
http://www.alpha.com
60.2007
21
http://www.sigma.com
20.1705
12
http://www.beta.com
40.1368
21
http://www.delta.com
30.1066
13
http://www.gamma.com
50.0643
10
http://www.rho.com
89
Weseethatalphahasahigher
PageR
ankthandeltaorsigma,even
thoughthey
allhavethesamenumber
ofin-links.
Arandom
surfer
willvisitalphaover
32%
of
thetimeandrhoonly
about6%
ofthetime.
Forthis
tinyexample
withp=
.85,thesm
allestelem
entof
theMarkov
tran-
sitionmatrix
isδ=
.15/6=
.0250.
A=
0.0250
0.0250
0.0250
0.8750
0.1667
0.8750
0.4500
0.0250
0.0250
0.0250
0.1667
0.0250
0.0250
0.4500
0.0250
0.0250
0.1667
0.0250
0.0250
0.4500
0.3083
0.0250
0.1667
0.0250
0.0250
0.0250
0.3083
0.0250
0.1667
0.0250
0.4500
0.0250
0.3083
0.0250
0.1667
0.0250
Notice
thatthecolumnsumsofAare
allequalto
one.
Theexmtoolbox
includes
theprogram
surfer.A
statementlike
[U,G]=
surfer(’http://www.xxx.zzz’,n)
startsataspecified
URLandtriesto
surf
theWeb
untilit
hasvisited
npages.If
successful,itreturnsann-by-1
cellarrayof
URLsan
dann-by-n
sparseconnectivity
matrix.Thefunctionusesurlread,whichwasintroducedin
Matlab6.5,along
withunderlyingJavautilities
toaccess
theWeb
.SurfingtheWeb
automatically
isadangerousundertakingandthis
functionmust
beusedwithcare.SomeURLs
contain
typographicalerrors
and
illegalcharacters.
Thereis
alist
ofURLsto
avoid
thatincludes
.gif
filesand
Web
sitesknow
nto
cause
diffi
culties.
Most
importantly,
surfercanget
completely
bogged
dow
ntryingto
readapagefrom
asite
thatappears
toberesponding,butthatnever
deliversthecomplete
page.
When
this
happens,
itmay
benecessary
tohavethecomputer’soperatingsystem
ruthlessly
term
inate
Matlab.W
iththeseprecautionsin
mind,youcanuse
surfer
togenerate
yourow
nPageR
ankexamples.
Thestatement
[U,G]=
surfer(’http://www.harvard.edu’,500)
accessesthehomepageofHarvard
University
andgeneratesa500-by-500test
case.
Thegraphgeneratedin
August
2003isavailablein
theexmtoolbox
.Thestatements
loadharvard500
spy(G)
produce
aspyplot(F
igure
7.3)thatshow
sthenonzero
structure
oftheconnectivity
matrix.Thestatement
pagerank(U,G)
computespageranks,
producesabargraph(F
igure
7.4)of
theranks,
andprints
themost
highly
ranked
URLsin
PageR
ankorder.
For
theharvard500data,thedozenmost
highly
ranked
pages
are
90Chapter7.
Goog
lePageR
ank
010
020
030
040
050
0
0 50 100
150
200
250
300
350
400
450
500
nz =
263
6
Figure
7.3.Spyplotoftheharvard500graph.
page-rank
in
out
url
10.0843
195
26
http://www.harvard.edu
10
0.0167
21
18
http://www.hbs.edu
42
0.0166
42
0http://search.harvard.edu:8765/
custom/query.html
130
0.0163
24
12
http://www.med.harvard.edu
18
0.0139
45
46
http://www.gse.harvard.edu
15
0.0131
16
49
http://www.hms.harvard.edu
90.0114
21
27
http://www.ksg.harvard.edu
17
0.0111
13
6http://www.hsph.harvard.edu
46
0.0100
18
21
http://www.gocrimson.com
13
0.0086
91
http://www.hsdm.med.harvard.edu
260
0.0086
26
1http://search.harvard.edu:8765/
query.html
19
0.0084
23
21
http://www.radcliffe.edu
TheURL
wherethesearchbegan,www.harvard.edu,dominates.
Likemost
uni-
versities,Harvardis
organized
into
variouscolleges
HarvardMedicalSchool,the
HarvardBusinessSchool,an
dtheRadcliffeInstitute.Youcanseethatthehome
pages
oftheseschools
havehighPageR
ank.W
ithadifferentsample,such
asthe
onegeneratedbyGoogle
itself,therankswou
ldbedifferent.
91
050
100
150
200
250
300
350
400
450
500
0
0.00
2
0.00
4
0.00
6
0.00
8
0.01
0.01
2
0.01
4
0.01
6
0.01
8
0.02
Page
Ran
k
Figure
7.4.PageRankoftheharvard500graph.
Further
Rea
ding
Further
readingonmatrix
computationincludes
book
sbyDem
mel
[?],Goluban
dVan
Loan
[?],Stewart
[?,?],andTrefethen
andBau[?].Thedefinitivereferenceson
Fortranmatrix
computationsoftware
are
theLAPACK
Users’GuideandWeb
site
[?].TheMatlabsparsematrix
data
structure
andoperationsare
described
in[?].
Inform
ationavailable
onWeb
sitesab
outPageR
ankincludes
abrief
explanationat
Google[?],atechnicalreport
byPage,Brin,andcolleagues
[?],andacomprehensive
survey
byLangville
andMeyer
[?].
Rec
ap
%%
PageRank
Chapter
Recap
%Thisis
an
executable
program
that
illustrates
the
statements
%introducedin
the
Page
Rank
Chapter
of
"Experiments
in
MATLAB".
%Youcan
access
it
with
% %pagerank_recap
%edit
pagerank_recap
%publish
pagerank_recap
% %Related
EXM
programs
% %surfer
%pagerank
%%
Sparsematrices
92Chapter7.
Goog
lePageR
ank
n=
6
i=
[2
63
44
56
11]
j=
[1
12
23
33
46]
G=
sparse(i,j,1,n,n)
spy(G)
%%
Page
Rank
p=
0.85;
delta=
(1-p)/n;
c=
sum(G,1);
k=
find(c~=0);
D=
sparse(k,k,1./c(k),n,n);
e=
ones(n,1);j
I=
speye(n,n);
x=
(I
-p*G*D)\e;
x=
x/sum(x)
%%
Conventional
power
method
z=
((1-p)*(c~=0)
+(c==0))/n;
A=
p*G*D
+e*z;
x=
e/n;
oldx=
zeros(n,1);
whilenorm(x-
oldx)
>.01
oldx=
x;
x=
A*x;
end
x=
x/sum(x)
%%
Sparse
power
method
G=
p*G*D;
x=
e/n;
oldx=
zeros(n,1);
whilenorm(x-
oldx)
>.01
oldx=
x;
x=
G*x
+e*(z*x);
end
x=
x/sum(x)
%%
Inverse
iteration
x=
(I
-A)\e;
x=
x/sum(x)
%%
Bar
graph
bar(x)
title(’PageRank’)
93
Exe
rcises
7.1
Use
surferandpagerankto
compute
PageR
anksforsomesubsetoftheWeb
thatyou
choose.Doyouseeanyinterestingstructure
intheresults?
7.2
Suppose
thatU
and
Gare
theURL
cell
array
and
theconnectivity
matrix
producedbysurferandthatkis
aninteger.Explain
what
U{k},U(k),G(k,:),G(:,k),U(G(k,:)),U(G(:,k))
are.
7.3
Theconnectivitymatrix
fortheharvard500data
sethasfoursm
all,alm
ost
entirely
nonzero,submatrices
thatproduce
dense
patches
nearthediagonalofthe
spyplot.
Youcanuse
thezoom
buttonto
findtheirindices.Thefirstsubmatrix
hasindices
around
170and
theother
threehaveindices
inthe200sand
300s.
Mathem
atically,
agraphwitheverynodeconnectedto
everyother
nodeis
know
nas
aclique.
Identify
theorganizationswithin
theHarvard
community
thatare
responsible
forthesenearcliques.
7.4
AWeb
connectivitymatrix
Ghasg i
j=
1ifit
ispossible
toget
topageifrom
pagejwithoneclick.Ifyoumultiply
thematrix
byitself,theentriesofthematrix
G2countthenumber
ofdifferentpathsof
length
twoto
pageifrom
pagej.
The
matrix
pow
erG
pshow
sthenumber
ofpathsof
length
p.
(a)Fortheharvard500data
set,findthepow
erpwherethenumber
ofnonze-
rosstopsincreasing.In
other
words,foranyqgreaterthanp,nnz(G^q)isequalto
nnz(G^p).
(b)W
hatfraction
oftheentriesin
Gpare
nonzero?
(c)Use
subplotandspyto
show
thenonzerosin
thesuccessivepow
ers.
(d)Is
thereasetofinterconnectedpages
thatdonotlinkto
theother
pages?
7.5
Thefunctionsurferusesasubfunction,hashfun,to
speedupthesearchfora
possibly
new
URLin
thelist
ofURLsthathavealreadybeenprocessed.Findtw
odifferentURLsonTheMathWorkshomepagehttp://www.mathworks.comthat
havethesamehashfunvalue.
7.6
Figure
7.5
isthegraphofanother
six-nodesubsetoftheWeb.In
thisexample,
thereare
twodisjointsubgraphs.
(a)W
hatis
theconnectivitymatrix
G?
(b)W
hatarethePageR
anksifthehyperlinktransitionprobabilitypis
the
defaultvalue0.85?
(c)Describewhathappenswiththisexample
toboth
thedefinitionofPageR
-ankandthecomputationdonebypagerankin
thelimit
p→
1.
7.7
Thefunctionpagerank(U,G)computesPageR
anksbysolvingasparselinear
94Chapter7.
Goog
lePageR
ank
alph
a
beta
gam
ma
delta
sigm
arh
o
Figure
7.5.Another
tinyWeb.
system
.It
then
plots
abargraphan
dprints
thedominantURLs.
(a)Create
pagerank1(G)bymodifyingpagerankso
thatitjust
computesthe
PageR
anks,
butdoes
notdoan
yplottingorprinting.
(b)Create
pagerank2(G)bymodifyingpagerank1to
use
inverse
iteration
insteadof
solvingthesparselinearsystem
.Thekey
statements
are
x=
(I
-A)\e
x=
x/sum(x)
Whatshould
bedonein
theunlikelyeventthatthebackslash
operationinvo
lves
adivisionbyzero?
(c)Create
pagerank3(G)bymodifyingpagerank1to
use
thepow
ermethod
insteadof
solvingthesparselinearsystem
.Thekey
statements
are
G=
p*G*D
z=
((1-p)*(c~=0)
+(c==0))/n;
whiletermination_test
x=
G*x
+e*(z*x)
end
Whatis
anappropriate
test
forterm
inatingthepow
eriteration?
(d)Use
you
rfunctionsto
compute
thePageR
anksof
thesix-nodeexample
discussed
inthetext.
Makesure
you
getthecorrectresultfrom
each
ofyou
rthree
functions.
7.8
Hereis
yet
another
functionforcomputingPageR
ank.This
versionusesthe
pow
ermethod,butdoes
notdoan
ymatrix
operations.
Only
thelinkstructure
of
theconnectivitymatrix
isinvo
lved.
function[x,cnt]=
pagerankpow(G)
95
%PAGERANKPOW
PageRank
by
power
method.
%x
=pagerankpow(G)
is
the
PageRank
of
the
graph
G.
%[x,cnt]
=pagerankpow(G)
%counts
the
number
of
iterations.
%Link
structure
[n,n]=
size(G);
forj
=1:n
L{j}=
find(G(:,j));
c(j)=
length(L{j});
end
%Power
method
p=
.85;
delta=
(1-p)/n;
x=
ones(n,1)/n;
z=
zeros(n,1);
cnt=
0;
whilemax(abs(x-z))>
.0001
z=
x;
x=
zeros(n,1);
forj
=1:n
if
c(j)
==
0
x=
x+
z(j)/n;
else x(L{j})=
x(L{j})
+z(j)/c(j);
end
end
x=
p*x
+delta;
cnt=
cnt+1;
end (a)How
dothestoragerequirem
ents
andexecutiontimeofthisfunctioncom-
pare
withthethreepagerankfunctionsfrom
thepreviousexercise?
(b)Use
thisfunctionasatemplate
towrite
afunctionthatcomputesPageR
-ankin
someother
programminglanguage.
96Chapter7.
Goog
lePageR
ank
Chapter8
Exp
onen
tialFunction
Thefunctionez.
Ihopeyou
havealiveMatlab
and
theexm
functionshandy.
Enterthe
statement
expgui
Click
onthebluelinewithyourmouse.Moveituntilthegreen
lineisontopofthe
blueline.
Whatis
theresultingvalueofa?
Theexponentialfunctionisdenotedmathem
aticallybyet
andin
Matlabby
exp(t).This
functionis
thesolutionto
theworld’s
simplest,an
dperhapsmost
important,differentialequation,
y=
ky
Thisequationisthebasisforanymathem
aticalmodeldescribingthetimeevolution
ofaquantity
witharate
ofproductionthatis
proportionalto
thequantity
itself.
Such
modelsincludepopulations,
investm
ents,feedback,an
dradioactivity.
We
are
usingtfortheindep
endentvariable,yforthedep
endentvariable,kforthe
proportionality
constant,and
y=
dy dt
fortherate
ofgrowth,or
derivative,
withrespectto
t.Weare
lookingforafunction
thatis
proportionalto
itsow
nderivative.
Let’s
start
byexaminingthefunction
y=
2t
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
97
98Chapter8.
Exp
onential
Function
00.
20.
40.
60.
81
1.2
1.4
1.6
1.8
20
0.51
1.52
2.53
3.54
Figure
8.1.Thebluecurveis
thegraphofy=
2t.Thegreencurveis
the
graphoftherate
ofgrowth,y=
dy/dt.
Weknow
what2t
meansiftis
aninteger,2t
isthet-th
pow
erof2.
2−1=
1/2,
20=
1,
21=
1,22
=4,...
Wealsoknow
what2t
meansif
t=
p/qis
arationalnumber,theratiooftw
ointegers,
2p/qis
theq-th
rootofthep-thpow
erof2.
21/2=
√ 2=
1.4142...,
25/3=
3√ 25=
3.1748...,
2355/113=
113√ 2
355=
8.8250
...
Inprincipal,forfloatingpointarithmetic,this
isallweneedto
know
.Allfloating
pointnumbersare
ratiosof
twointegers.
Wedonothaveto
beconcerned
yet
about
thedefinitionof2t
forirrationalt.
IfMatlabcancompute
pow
ersan
droots,we
canplotthegraphof2t,thebluecu
rvein
figure
8.1
Whatisthederivativeof2t?May
beyou
havenever
considered
thisquestion,
ordon’t
remem
ber
theansw
er.
(Becareful,
itis
nott2
t−1.)
Wecan
plotthe
graphoftheapproxim
ate
derivative,usingastep
size
ofsomethinglike0.0001.The
follow
ingcodeproducesfigure
8.1,thegraphsofbothy=
2tanditsapproxim
ate
derivative,y.
t=
0:.01:2;
h=
.00001;
y=
2.^t;
ydot=
(2.^(t+h)
-2.^t)/h;
plot(t,[y;ydot])
Thegraphofthederivative
hasthesameshapeasthegraphoftheoriginal
function.Let’s
lookattheirratio,y(t)/y(t).
99
00.
20.
40.
60.
81
1.2
1.4
1.6
1.8
20.
5
0.550.6
0.650.7
0.750.8
0.850.9
Figure
8.2.Theratio,y/y
.
plot(t,ydot./y)
axis([02
.5
.9])
Weseethattheratioofthederivativeto
thefunction,show
nin
figure
8.2,hasa
constantvalue,
y/y=
0.6931...,thatdoes
notdep
endupont.
Now
,if
you
are
follow
ingalongwith
aliveMatlab,repeatthepreceding
calculationswithy=
3tinsteadof
y=
2t.Youshould
findthattheratiois
again
indep
endentoft.
This
timey/y=
1.0986....Betteryet,experim
entwithexpgui.
Ifwetakeanyvalueaandlook
aty=
at,wefindthat,
numericallyatleast,
theratioy/yis
constant.
Inother
words,
yis
proportionalto
y.
Ifa=
2,the
proportionality
constantis
less
thanone.
Ifa=
3,theproportionality
constantis
greaterthanone.
Canwefindanaso
thaty/yis
actuallyequalto
one?
Ifso,we
havefoundafunctionthatis
equalto
itsow
nderivative.
Theapproxim
ate
derivative
ofthefunctiony(t)=
atis
y(t)=
at+
h−at
h
This
canbefactoredandwritten
y(t)=
ah−1
hat
Sotheratioofthederivative
tothefunctionis
y(t)
y(t)=
ah−1
h
Theratiodep
endsuponh,butnotupon
t.If
wewanttheratioto
beequalto
1,
weneedto
findaso
that
ah−
1
h=
1
100
Chapter8.
Exp
onential
Function
Solvingthis
equationfora,wefind
a=
(1+
h)1
/h
Theap
proxim
ate
derivative
becomes
more
accurate
ash
goes
tozero,so
weare
interested
inthevalueof
(1+h)1
/h
ash
approaches
zero.
This
involvestakingnumbersveryclose
to1and
raising
them
toverylargepow
ers.
Thesurprisingfact
isthatthis
limitingprocess
defi
nes
anumber
thatturnsoutto
beon
eofthemost
importantquantities
inmathem
atics
e=
lim
h→
0(1
+h)1
/h
Hereis
thebeginningandendofatable
ofvalues
generatedbyrepeatedly
cutting
hin
half.
formatlong
formatcompact
h=
1;
whileh
>2*eps
h=
h/2;
e=
(1
+h)^(1/h);
disp([he])
end
0.500000000000000
2.250000000000000
0.250000000000000
2.441406250000000
0.125000000000000
2.565784513950348
0.062500000000000
2.637928497366600
0.031250000000000
2.676990129378183
0.015625000000000
2.697344952565099
...
...
0.000000000000014
2.718281828459026
0.000000000000007
2.718281828459036
0.000000000000004
2.718281828459040
0.000000000000002
2.718281828459043
0.000000000000001
2.718281828459044
0.000000000000000
2.718281828459045
Thelast
lineof
outputinvolves
avalueofhthatisnotzero,butisso
smallthatit
prints
asastringofzeros.
Weare
actuallycomputing
(1+2−
51)2
51
whichis (1+
1
2251799813685248
)2251799813685248
101
Theresultgives
usthenumericalvalueofecorrectto
16significantdecim
aldigits.
It’s
easy
toremem
ber
therepeatingpatternofthefirst10
significantdigits.
e=
2.718281828...
Let’s
deriveamore
usefulrepresentationoftheexponentialfunction.Start
byputtingtback
inthepicture.
et=
(lim
h→
0(1
+h)1
/h)t
=lim
h→
0(1
+h)t/h
Hereis
theBinomialTheorem.
(a+b)
n=
an+
nan−1b+
n(n
−1)
2!an−2b2
+n(n
−1)(n−2)
3!
an−3b3
+...
Ifnisaninteger,thisterm
inatesafter
n+1term
swithbn.Butifnisnotaninteger,
theexpansionisaninfiniteseries.Apply
thebinonim
ialtheorem
witha=
1,b=
handn=
t/h.
(1+
h)t/h
=1+(t/h)h
+(t/h)(t/h−1)
2!
h2+
(t/h)(t/h−1)(t/h−2)
3!h3+
...
=1+
t+
t(t−h)
2!
+t(t−h)(t−2h
)
3!
+...
Now
lethgoto
zero.Weget
thepow
erseries
fortheexponentialfunction.
et=
1+t+
t2 2!+
t3 3!+
...+
tn n!+...
This
series
isarigorousmathem
aticaldefinitionthatapplies
toan
yt,
positiveor
negative,
rationalorirrational,realor
complex.Then+
1-stterm
istn/n!.
As
nincreases,
thetn
inthenumeratoris
eventuallyoverwhelmed
bythen!in
the
denominator,so
theterm
sgoto
zero
fast
enoughthattheinfiniteseries
converges.
Itisalm
ost
possibleto
use
thispow
erseries
foractualcomputationof
et.Here
isanexperim
entalMatlabprogram.
functions
=expex(t)
%EXPEX
Experimental
version
of
EXP(T)
s=
1;
term=
1;
n=
0;
r=
0;
whiler
~=
s
r=
s;
n=
n+
1;
term=
(t/n)*term;
s=
s+
term;
end
102
Chapter8.
Exp
onential
Function
Notice
thatthereare
nopow
ersor
factorials.
Each
term
isobtained
from
the
previousoneusingthefact
that
tn n!=
t n
tn−1
(n−1)!
Thepotentiallyinfiniteloopis
term
inatedwhen
r==s,thatis
when
thefloating
pointvalues
oftw
osuccessivepartialsumsareequal.
Thereare“only”tw
othingswrongwith
this
program
–itsspeed
and
its
accuracy.Theterm
sin
theseries
increase
aslongas
|t/n|≥
1,then
decrease
after
nreaches
thepointwhere|t/
n|<
1.Soif|t|
isnottoolarge,
say|t|
<2,everything
isOK;only
afew
term
sare
required
andthesum
iscomputedaccurately.
But
larger
values
oftrequiremore
term
sandtheprogram
requires
more
time.
This
isnotaveryseriousdefectiftis
realandpositive.
Theseries
converges
sorapidly
thattheextratimeis
hardly
noticeable.
How
ever,iftisrealandnegativethecomputedresultmay
beinaccurate.The
term
salternate
insignandcanceleach
other
inthesum
toproduce
asm
allvalue
foret.
Take,forexample,t=
−20.
Thetruevalueofe−
20is
roughly
2·1
0−9.
Unfortunately,thelargestterm
sin
theseries
are(−
20)1
9/19!and
(−20)2
0/20!,
whichareoppositein
signandboth
ofsize
4·107.Thereis16ordersofmagnitude
difference
betweenthesize
ofthelargestterm
san
dthesize
ofthefinalsum.W
ith
only
16digitsofaccuracy,welose
everything.Thecomputedvalueob
tained
from
expex(-20)is
completely
wrong.
For
real,
negativetit
ispossible
togetanaccurate
result
from
thepow
erseries
byusingthefact
that
et=
1 e−t
For
complext,
thereis
nosuch
easy
fixfortheaccuracy
diffi
cultiesofthepow
erseries. In
contrast
toitsmore
famouscousin,π,theactualnumericalvalueofeis
notveryim
portant.
It’s
theexpon
entialfunction
et
that’sim
portant.
Infact,Matlabdoesn’thavethevalueofebuiltin.Nevertheless,
wecanuse
e=
expex(1)
tocompute
anap
proxim
ate
valuefore.
Only
seventeen
term
sare
required
toget
floatingpointaccuracy.
e=
2.718281828459045
After
computinge,youcould
then
use
e^t,butexp(t)is
preferable.
Logarithms
Thelogarithm
istheinversefunctionoftheexponential.If
y=
et
103
then
log e(y)=
t
Thefunctionlog e(y)isknow
nasthenaturallogarithm
andisoftendenotedbylny.
More
generally,
if
y=
at
then
log a(y)=
t
Thefunctionlog 1
0(y)is
know
nasthecommon
logarithm.Matlabuseslog(y),
log10(y),an
dlog2(y)forloge(y),log 1
0(y),andlog 2(y).
Exp
onen
tialGrowth
Theterm
exponen
tialgrowth
isoften
usedinform
allyto
describean
ykindof
rapid
growth.Mathem
atically,
theterm
refers
toanytimeevolution,y(t),wheretherate
ofgrowth
isproportionalto
thequantity
itself.
y=
ky
Thesolutionto
this
equationis
determined
foralltbyspecifyingthevalueof
yat
oneparticulart,usuallyt=
0.
y(0)=
y 0
Then
y(t)=
y 0ek
t
Suppose,attimet=
0,wehaveamillionE.colibacteria
inatest
tubeunder
ideallaboratory
conditions.
Twenty
minuteslatereach
bacterium
hasfissioned
toproduce
another
one.
Soatt=
20,thepopulationistw
omillion.Every20minutes
thepopulationdoubles.
Att=
40,it’s
fourmillion.Att=
60,it’s
eightmillion.
Andso
on.Thepopulation,measuredin
millionsofcells,
y(t),is
y(t)=
2t/
20
Let
k=
ln2/20=
.0347.Then,withtmeasuredin
minutesan
dthepopulationy(t)
measuredin
millions,
wehave
y=
ky,
y(0)=
1
Consequently
y(t)=
ekt
Thisisexpon
entialgrow
th,butitcannotgo
onforever.Eventually,
thegrowth
rate
isaffectedbythesize
ofthecontainer.Initiallyatleast,thesize
ofthepop
ulation
ismodelledbytheexpon
entialfunction.
104
Chapter8.
Exp
onential
Function
Suppose,attimet=
0,youinvest
$1000in
asavingsaccountthatpay
s5%
interest,compou
nded
yearly.
Ayearlater,
att=
1,thebankadds5%
of$1000
toyou
raccount,
givingyouy(1)=
1050.Another
yearlateryou
get5%
of1050,
whichis
52.50,givingy(2)=
1102.50.If
y(0)=
1000is
you
rinitialinvestment,
r=
0.05istheyearlyinterest
rate,tismeasuredin
years,andhisthestep
size
for
thecompou
ndinterest
calculation,wehave
y(t+h)=
y(t)+
rhy(t)
Whatiftheinterest
iscompounded
monthly
insteadof
yearly?
Attheendofthe
each
month,you
get.05/12times
yourcurrentbalance
added
toyouraccount.
The
sameequation
applies,butnow
with
h=
1/1
2instead
ofh
=1.
Rew
rite
the
equationas
y(t+
h)−
y(t)
h=
ry(t)
andlethtendto
zero.Weget
y(t)=
ry(t)
This
defines
interest
compounded
continuously.
Theevolutionof
yourinvestment
isdescribed
by
y(t)=
y(0)e
rt
HereisaMatlabprogram
thattabulatesthegrow
thof
$1000invested
at5%
over
a20
yearperiod,withinterest
compounded
yearly,
monthly,andcontinuously.
formatbank
r=
0.05;
y0
=1000;
fort
=0:20
y1
=(1+r)^t*y0;
y2
=(1+r/12)^(12*t)*y0;
y3
=exp(r*t)*y0;
disp([ty1
y2
y3])
end
Thefirstfew
andlast
few
lines
ofoutputare
tyearly
monthly
continuous
01000.00
1000.00
1000.00
11050.00
1051.16
1051.27
21102.50
1104.94
1105.17
31157.63
1161.47
1161.83
41215.51
1220.90
1221.40
51276.28
1283.36
1284.03
..
.......
.......
.......
16
2182.87
2221.85
2225.54
105
17
2292.02
2335.52
2339.65
18
2406.62
2455.01
2459.60
19
2526.95
2580.61
2585.71
20
2653.30
2712.64
2718.28
Compoundinterest
actuallyqualifies
asexponentialgrowth,althoughwithmodest
interest
rates,
most
people
would
notuse
thatterm
.Let’s
borrow
money
tobuyacar.
We’lltakeouta$20,000carloanat10%
per
yearinterest,makemonthly
pay
ments,an
dplanto
pay
offtheloanin
3years.
Whatis
ourmonthly
pay
ment,
p?
Each
monthly
transactionaddsinterest
toour
currentbalance
andsubtractsthemonthly
pay
ment.
y(t+
h)=
y(t)+
rhy(t)−p
=(1
+rh
)y(t)−
p
Apply
this
repeatedly
fortw
o,three,
then
nmonths.
y(t+2h)=
(1+
rh)y(t+h)−p
=(1
+rh
)2y(t)−
((1+
rh)+1)p
y(t+3h)=
(1+
rh)3y(t)−
((1+
rh)2
+(1
+rh
)+1)p
y(t+
nh)=
(1+
rh)ny(0)−((1+
rh)n
−1+...+(1
+rh
)+1)p
=(1
+rh
)ny(0)−((1+
rh)n
−1)/(1+rh
−1)p
Solveforp
p=
(1+
rh)n/((1+
rh)n
−1)rh
y 0
Use
Matlabto
evaluate
this
forourcarloan.
y0
=20000
r=
.10
h=
1/12
n=
36
p=
(1+r*h)^n/((1+r*h)^n-1)*r*h*y0
Wefindthemonthly
pay
mentwould
be
p=
645.34
Ifwedidn’t
haveto
pay
interest
ontheloanandjust
made36monthly
pay
ments,
they
would
be
y0/n
=555.56
It’shard
tothinkaboutcontinuouscompoundingforaloanbecause
wewould
have
tofigure
outhow
tomakeinfinitelymanyinfinitelysm
allpay
ments.
106
Chapter8.
Exp
onential
Function
Complexex
ponen
tial
Whatdowemeanbyez
ifziscomplex?Thebehav
iorisverydifferentfrom
etfor
realt,butequallyinterestingandim
portant.
Let’sstart
withapurely
imaginary
zandsetz=
iθwhereθisreal.Wethen
makethedefi
nition
eiθ=
cosθ+
isinθ
This
form
ula
isremarkab
le.It
defines
theexpon
entialfunctionforan
imaginary
argumentin
term
sof
trig
functionsof
arealargument.
Thereare
severalreasons
whythisisareasonabledefintion.First
ofall,itbehaves
likean
exponentialshould.
Weexpect
eiθ+iψ
=ei
θei
ψ
This
behav
ioris
aconsequence
ofthedouble
angle
form
ulasfortrig
functions.
cos(θ+ψ)
=cosθcosψ−sinθsinψ
sin(θ
+ψ)
=cosθsinψ+sinθcosψ
Secondly,derivativesshould
behaveasexpected.
d dθei
θ=
ieiθ
d2
dθ2
eiθ
=i2ei
θ=
−eiθ
Inwords,
thesecondderivativeshould
bethenegativeof
thefunctionitself.This
worksbecause
thesameistrueofthetrig
functions.
Infact,thiscould
bethebasis
forthedefintionbecause
theinitialconditionsarecorrect.
e0=
1=
cos0+
isin0
Thepow
erseries
isan
other
consideration.Replace
tbyiθ
inthepow
erseries
for
et.Rearrangingterm
sgives
thepow
erseries
forcosθandsinθ.
For
Matlabespecially,
thereisanim
portantconnectionbetweenmultiplica-
tionbyacomplexexpon
entialan
dtherotationmatrices
thatweconsidered
inthe
chapteron
matrices.Let
w=
x+
iybeanyother
complexnumber.W
hatis
eiθw
?Let
uandvbetheresultofthe2-by-2
matrix
multiplication
( u v
) =
( cosθ
−sinθ
sinθ
cosθ
)(x y
)
Then
eiθw
=u+iv
Thissaysthatmultiplicationof
acomplexnumber
byei
θcorrespondsto
arotation
ofthatnumber
inthecomplexplanebyan
angle
θ.
107
Figure
8.3.Twoplots
ofei
θ.
When
theMatlabplotfunctionsees
acomplexvectorasitsfirstargument,
itunderstandsthecompon
ents
tobepoints
inthecomplexplane.
Sotheoctagon
intheleft
halfoffigure
8.3
canbedefined
andplotted
usingei
θwith
theta=
(1:2:17)’*pi/8
z=
exp(i*theta)
p=
plot(z);
Thequantity
pis
thehandle
totheplot.
This
allowsusto
complete
thegraphic
with set(p,’linewidth’,4,’color’,’red’)
axissquare
axisoff
Anexercise
asksyouto
modifythis
codeto
produce
thefive-pointedstarin
the
righthalfofthefigure.
Once
wehavedefined
eiθforrealθ,
itis
clearhow
todefineez
forageneral
complexz=
x+
iy,
ez=
ex+iy
=exei
y
=ex(cosy+
isiny)
Finally,
settingz=
iπ,wegetafamousrelationship
invo
lvingthreeofthe
most
importantquantities
inmathem
atics,e,
i,an
dπ
eiπ=
−1Let’s
checkthatMatlabandtheSymbolicToolbox
get
this
right.
>>
exp(i*pi)
ans=
-1.0000+
0.0000i
108
Chapter8.
Exp
onential
Function
>>
exp(i*sym(pi))
ans=
-1
Rec
ap
%%
Exponential
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Exponential
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %exponential_recap
%edit
exponential_recap
%publish
exponential_recap
% %Related
EXM
programs
% %expgui
%wiggle
%%
Plot
a^t
and
its
approximate
derivative
a=
2;
t=
0:.01:2;
h=
.00001;
y=
2.^t;
ydot=
(2.^(t+h)
-2.^t)/h;
plot(t,[y;ydot])
%%
Compute
e
formatlong
formatcompact
h=
1;
whileh
>2*eps
h=
h/2;
e=
(1
+h)^(1/h);
disp([he])
end
%%
Experimental
version
of
exp(t)
t=
rand
s=
1;
term=
1;
n=
0;
r=
0;
whiler
~=
s
109
r=
s;
n=
n+
1;
term=
(t/n)*term;
s=
s+
term;
end
exp_of_t=
s
%%Value
of
e
e=
expex(1)
%%Compound
interest
fprintf(’
tyearly
monthly
continuous\n’)
formatbank
r=
0.05;
y0
=1000;
fort
=0:20
y1
=(1+r)^t*y0;
y2
=(1+r/12)^(12*t)*y0;
y3
=exp(r*t)*y0;
disp([ty1
y2
y3])
end
%%Payments
for
acar
loan
y0
=20000
r=
.10
h=
1/12
n=
36
p=
(1+r*h)^n/((1+r*h)^n-1)*r*h*y0
%%Complex
exponential
theta=
(1:2:17)’*pi/8
z=
exp(i*theta)
p=
plot(z);
set(p,’linewidth’,4,’color’,’red’)
axissquareoff
%%Famous
relation
between
e,
iand
pi
exp(i*pi)
%%Use
the
Symbolic
Toolbox
exp(i*sym(pi))
110
Chapter8.
Exp
onential
Function
Exe
rcises
8.1
ecubed.Thevalueof
e3is
close
to20.How
close?
Whatis
thepercentage
error?
8.2
expgui.
(a)W
ithexpgui,thegraphof
y=
at,theblueline,
alwaysintercep
tsthey-axisat
y=
1.W
heredoes
thegraphofdy/dx,thegreen
line,
intercep
tthey-axis?
(b)W
hat
hap
pensifyou
replace
plotbysemilogyin
expgui?
8.3
Computinge.
(a)Ifwetryto
compute
(1+h)1
/hforsm
allvalues
ofhthatare
inversepow
ersof10,
itdoesn’tworkverywell.Since
inversepow
ersof10cannotberepresentedexactly
asbinary
floatingpointnumbers,
theportionofhthateff
ectivelygets
added
to1
isdifferentthanthevalueinvolved
inthecomputationof1/h.That’swhyweused
inversepow
ersof2in
thecomputationshow
nin
thetext.
Try
this:
formatlong
formatcompact
h=
1;
whileh
>1.e-15
h=
h/10;
e=
(1
+h)^(1/h);
disp([he])
end
How
close
doyou
getto
computingthecorrectvalueof
e?(b)Now
trythis
instead:
formatlong
formatcompact
h=
1;
whileh
>1.e-15
h=
h/10;
e=
(1
+h)^(1/(1+h-1));
disp([he])
end
How
welldoes
this
work?W
hy?
8.4
expex.
Modifyexpexbyinserting
disp([terms])
asthelast
statementinsidethewhileloop.
Changetheoutputyou
seeatthe
commandline.
111
formatcompact
formatlong
Explain
whatyouseewhen
youtryexpex(t)forvariousrealvalues
oft.
expex(.001)
expex(-.001)
expex(.1)
expex(-.1)
expex(1)
expex(-1)
Try
someim
aginary
values
oft.
expex(.1i)
expex(i)
expex(i*pi/3)
expex(i*pi)
expex(2*i*pi)
Increase
thewidth
oftheoutputwindow
,changetheoutputform
atandtrylarger
values
oft.
formatlonge
expex(10)
expex(-10)
expex(10*pi*i)
8.5
Instrumen
texpex.
Investigate
both
thecost
andtheaccuracy
ofexpex.Modify
expexso
thatitreturnsboth
thesum
sandthenumber
ofterm
srequired
n.Assess
therelativeerrorbycomparingtheresult
from
expex(t)withtheresult
from
the
built-in
functionexp(t).
relerr=
abs((exp(t)
-expex(t))/exp(t))
Makeatable
show
ingthatthenumber
ofterm
srequired
increasesandtherelative
errordeterioratesforlarget,particularlynegativet.
8.6
Complexwiggle.
Revisewiggleanddot2dotto
create
wigglezanddot2dotz
thatuse
multiplicationbyei
θinsteadof
multiplicationbytw
o-by-twomatrices.The
cruxofwiggleis
G=
[cos(theta)
sin(theta);
-sin(theta)
cos(theta)];
Y=
G*X;
dot2dot(Y);
Inwigglezthis
willbecome
w=
exp(i*theta)*z;
dot2dotz(w)
112
Chapter8.
Exp
onential
Function
Youcanuse
wigglezwithascaledoctogon.
theta=
(1:2:17)’*pi/8
z=
exp(i*theta)
wigglez(8*z)
Or,
withourhouseexpressed
asacomplexvector.
H=
house;
z=
H(1,:)
+i*H(2,:);
wigglez(z)
8.7
Make
thestar.
Recreate
thefive-pointedstarin
therighthalfoffigure
8.3.The
points
ofthestarcanbetraversedin
thedesired
order
with
theta=
(0:3:15)’*(2*pi/5)
+pi/2
Chapter9
TPuzzle
Aclassic
puzzle
dem
onstratescomplexarithmetic.
Figure
9.1.Thewooden
Tpuzzle.Photo
courtesyofShopNew
Zeland,
http://www.shopnewzealand.co.nz.
Ifirstsaw
thewooden
Tpuzzle
show
nin
figure
9.1atPuzzlingWorld
in
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
113
114
Chapter9.
TPuzzle
Wan
aka,New
Zealand.They
told
methatitwas
theirmost
pop
ularpuzzle.Ihave
since
learned
thatit
was
awell-know
ntoyin
the1800sand
anad
vertisingtool
intheearly1900s.
Theunderlyingmathem
atics
involves
geometry,trigonometry,
andarithmetic
withcomplexnumbers.
Thet_puzzleprogram
intheexmtoolbox
dem
onstratessomeusefulprogrammingtechniques.
Figure
9.2.Thefourpieces.
Figure
9.2show
stheelectronic
versionofthefourpieces.
They
allhavethe
samewidth,butdifferentheights.Oneof
them
hasanunshapelychunkcutou
tof
it,resultingin
anirregularpentagon.
Figure
9.3.TheT.
Itturnsou
tto
bepossible
toarrangethefourpiecesto
form
thecapital“T”
shapeshow
nin
figure
9.3,as
wellas
thearrow
andtherhombusshapes
show
nin
figure
4.4.W
hathappened
tothose
allof
those
45◦
anglesandwhathappened
tothatchunk?
Ifyou
doaGoogle
searchon
“T-puzzle”you
canquickly
seehow
tosolvethe
puzzle
andform
theT,butplease
tryt_puzzleforawhilebefore
yougosurfing
forthesolution.Ifyouclicknearthecenterof
oneofthefourpieces,you
canmove
115
Figure
9.4.Thearrow
andtherhombus.
ithorizontallyandvertically.
Ifyouclicknearoneofthevertices,
you
canrotate
apiece
aboutitscenter.
Ifyou
clickwiththerightmouse
button,or,
onaone-
buttonmouse,hold
dow
nthecontrolkey
whileyouclick,you
canflip
apiece
over
horizontally,
changingitsright/left-handed
orientation.If
youclickin
thesubplot
onthelower
left
ofthewindow
,youcancyclethroughim
ages
ofthethreeshapes.
Thekey
toeff
ectivecomputationinvo
lvingtheT-puzzle
piecesis
theuse
of
complexarithmetic.A
complexnumber
isform
edfrom
apair
ofrealnumbersand
theim
aginary
uniti=
√ −1.For
example,
z=
3+4i
Therealpart
is3andtheim
aginary
part
is4.ThisistheCartesianrepresentation
ofacomplexnumber.
Additionofcomplexnumberssimply
involves
additionoftherealpartsand
oftheim
aginary
parts.
(3+4i)+(5
−i)
=(3
+5)
+(4i−i)
=8+
3i
Multiplicationofcomplexnumbersis
more
interestingbecause
itmakes
use
ofthefact
thati2
=−1
.
(3+4i)·(5−
i)
=(3
·5+(4i)·(−i
))+(3
·(−i
)+(4i)·5)
=(15+
4)+(−
3+20)i
=19+
17i
Afundamentalfact
involvingcomplexnumbersis
Euler’sform
ula.
eiφ=
cosφ+isinφ
Ifyouare
notfamiliarwitheorEuler’sform
ula,seeourchapteronthe“Expon
ential
Function”an
dtheW
ikiped
iaentryon“Euler’sIdentity”.Or,
youcanjust
accept
theform
ula
asconvenientnotation.
116
Chapter9.
TPuzzle
Settingφ=
πleadsto
eiπ
=cosπ+isinπ
=−1
Mov
ingthe−1
totheleft
handsideproducesabeautifulequationinvolvingfive
basicmathem
aticalquantities,e,
i,π,1,
and0.
eiπ+1=
0
Thepolarform
ofacomplexnumber
is
z=
x+iy
=re
iφ
where r
=|z|
=√ x
2+
y2
x=
rcosφ
y=
rsinφ
OurT
puzzle
program
usesthefact
thatmultiplicationbyei
θrotatesacom-
plexnumber
throughan
angle
θ.Toseethis,let
w=
eiθ
z=
reiφ
then
w·z
=ei
θ·rei
φ=
rei(θ+φ)
InMatlabtheletter
iis
canbeusedin
anyof
threedifferentroles.
Itcan
beaniterationcounter,
fori
=1:10
orasubscript,
A(i,k)
ortheim
aginary
unit.
z=
3+
4i
Thepolarform
ofacomplexnumber
zisobtained
withtheMatlabfunctions
abs(z)andangle(z).Thequan
tity
eiφis
written
exp(i*phi).Forexample
z=
3+
4i
r=
abs(z)
phi=
angle(z)
w=
r*exp(i*phi)
produces
117
z=
3.0000+
4.0000i
r=
5
phi= 0.9273
w=
3.0000+
4.0000i
−10
12
0123
Figure
9.5.
Thecomplexcoordinatesofthevertices
are
0+
0i,1+
0i,
1+2i,
and0+3i.
ThelargestofthefourpiecesoftheT
puzzle
canberepresentedin
Matlab
bythestatement
z=
[0
11+2i
3i
0]
Thevectorzhasfivecomplexentires.
Thefirsttw
oelem
ents
are
0and1;
their
imaginary
partsare
zero.
Thethirdelem
entis
1+
2i;itsrealpart
is1andits
imaginary
part
is2.After
thatcomes
3i;itsrealpart
iszero
anditsim
aginary
part
is3.Thelast
elem
entofzis
arepeatof
thefirstso
thatthelinedrawnbythe
statement
line(real(z),imag(z))
returnsto
theorigin.Theresult
isfigure
9.5.
Withthisrepresentation,thetranslationsandrotationsrequired
bythegraph-
icsin
t_puzzlecanbeprogrammed
easily
andeffi
ciently.
Translationisveryeasy.
Thestatement
z=
z-
(3-i)/2
repositionsthepiece
infigure
9.5
atoneofthecorners.
Doyouseewhichcorner?
Rotationsarealsoeasily
done.
Thestatements
118
Chapter9.
TPuzzle
−10
12
0123
Figure
9.6.Rotatingthrough
multiplesofninedegrees.
mu
=mean(z(1:end-1))
theta=
pi/20
omega=
exp(i*theta)
z=
omega*(z
-mu)
+mu
rotate
thepiece
aboutitscenterthrough9◦
inthecounterclockwisedirection.Fig-
ure
9.6show
stheresult
ofrepeatingthesestatements
severaltimes.Let’s
look
ateach
step
more
carefully.
Thestatement
mu
=mean(z(1:end-1))
dropsthelast
elem
entof
zbecause
itisarepeatofthefirstan
dthen
computesthe
complexaverageofthecoordinatesofthevertices.
This
gives
thecoordinatesof
whatwecancallthecenterof
thepolygon.Theangle
θ=
π/20
is9◦in
radians.
Thestatements
omega=
exp(i*theta)
z=
omega*(z
-mu)
+mu
translate
thepiece
sothatitscenteris
attheorigin,doacomplexscalartimes
vectormultiplicationto
produce
therotation,andthen
translate
theresultback
toitsoriginalcenter.
Rec
ap
%%
TPuzzle
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
TPuzzle
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
119
% %puzzle_recap
%edit
puzzle_recap
%publish
puzzle_recap
% %Related
EXM
programs
% %t_puzzle
%%The
TPuzzle
closeall
figure
t_puzzle
%%Polar
Form
of
aComplex
Number.
z=
3+
4i
r=
abs(z)
phi=
angle(z)
z_again=
r*exp(i*phi)
%%A
Puzzle
Piece.
figure
z=
[0
11+2i
3i
0]
line(real(z),imag(z),’color’,’red’)
axis([-2.51.5-.54.0])
%%Translation.
z=
z-
(3-i)/2
line(real(z),imag(z),’color’,’blue’)
%%Rotation.
mu
=mean(z(1:end-1));
theta=
pi/10;
omega=
exp(i*theta);
z=
omega*(z
-mu)
+mu
line(real(z),imag(z),’color’,’green’)
Exe
rcises
9.1
Complexnumbers.Express
each
ofthesenumbersin
polarform
.
i −4 3−4i
120
Chapter9.
TPuzzle
8+15i
Express
each
ofthesenumbersin
Cartesianform
.
e−iπ
eiπ/3
ei e3
9.2
Stopsign
.Try
this
forn=8andforother
smallinteger
values
ofn.Describe
andexplain
theresults.
n=
8
z=
exp(2*pi*i*(0:n)’/n)
plot(z,’-o’)
axissquare
s=
sum(z)
9.3
Strip.How
manydifferentwaysare
thereto
form
theshapein
figure
9.7
with
theT-puzzle
pieces?
Figure
9.7.A
strip.
9.4
Area.If
thewidth
ofeach
oftheT-puzzle
piecesis
oneunit,whatare
their
areas?
9.5
Sym
metry.W
hichoneof
ourthreeshapes
–T,arrow
andrhombus–does
not
haveanaxis
ofsymmetry?
9.6
Jumpyrotation.Click
nearavertexofoneof
theT-puzzle
piecesandrotate
thepiece
slow
ly.Youshould
seethattherotationis
notsm
ooth,butproceedsin
discretejumps.
Why?
How
largearethejumps?
How
does
t_puzzleaccomplish
this?
9.7
Snappytranslation.Dragon
eoftheT-puzzle
piecesuntilitisclose
to,butnot
exactly
touching,another.W
hen
you
release
themouse
buttonyousometim
essee
thepiece
snapinto
place.Under
whatcircumstancesdoes
this
happen?How
does
t_puzzleaccomplish
it?
121
9.8
Rotation.Reproduce
figure
9.6.
9.9
Differen
tpuzzles.
DoaGoogle
searchon“T-puzzle”.Includethequotesan
dhyphen
inthesearchstringso
thatyouget
anexact
match.SomeoftheWeb
pages
havepieceswithdifferentsizesthantheon
eswehavedescribed
here.
(a)How
manydifferentversionsoftheT-puzzle
arethereontheWeb?
(b)Canyou
makeallthreeof
ourshapes
–theT,arrow
,andrhombus–withthe
piecesshow
non
theseWeb
sites.
(c)Modifyou
rt_puzzleto
use
thesetofpiecesfrom
oneoftheWeb
sites.
122
Chapter9.
TPuzzle
Chapter10
Magic
Squares
Withoriginsin
centuries
old
recreationalmathem
atics,magicsquaresdem
onstrate
Matlabarrayoperations.
Figure
10.1.LoShu.(T
hanks
toByerlyWiser
Cline.)
Magic
squarespredate
recorded
history.AnancientChineselegendtellsofa
turtle
emergingfrom
theLoriver
duringaflood.Theturtle’s
shellshow
edavery
unusualpattern–athree-by-threegridcontainingvariousnumbersofspots.Of
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
123
124
Chapter10
.Magic
Squares
course,
wedonothavean
yeye-witnessaccounts,so
wecanonly
imaginethatthe
turtle
looked
likefigure
10.1.Each
ofthethreerows,
thethreecolumns,
andthe
twodiagonals
contain
atotalof15spots.Referencesto
LoShuan
dtheLoShu
numericalpatternoccurthroughoutChinesehistory.Today,itisthemathem
atical
basisforFen
gShui,
thephilosophyof
balance
andharm
onyin
oursurroundings
andlives.
An
n-by-n
magic
square
isan
arraycontainingtheintegersfrom
1to
n2,
arranged
sothateach
oftherows,
each
ofthecolumns,
and
thetw
oprincipal
diagonals
havethesamesum.
Foreach
n>
2,therearemanydifferentmagic
squaresof
order
n,buttheMatlabfunctionmagic(n)generatesaparticularone.
Matlabcangenerate
LoShuwith
A=
magic(3)
whichproduces
A=
81
6
35
7
49
2
Thecommand
sum(A)
sumstheelem
ents
ineach
columnto
produce
15
15
15
Thecommand
sum(A’)’
transposesthematrix,sumsthecolumnsofthetranspose,
andthen
transposes
the
resultsto
produce
therow
sums
15
15
15
Thecommand
sum(diag(A))
sumsthemain
diagonalofA,whichrunsfrom
upper
left
tolower
right,to
produce
15
Theop
posite
diagonal,whichrunsfrom
upper
rightto
lower
left,is
less
important
inlinearalgeb
ra,so
findingitssum
isalittle
trickier.
Oneway
todoit
makes
use
ofthefunctionthat“flips”
amatrix
“upside-dow
n.”
125
sum(diag(flipud(A)))
produces 15
This
verifies
thatAhasequalrow,column,an
ddiagonalsums.
Whyis
themagic
sum
equalto
15?Thecommand
sum(1:9)
tellsusthatthesum
oftheintegersfrom
1to
9is45.Iftheseintegersare
allocated
to3columnswithequalsums,
thatsum
must
be
sum(1:9)/3
whichis15.
Thereareeightpossible
waysto
place
atransparency
onan
overheadprojec-
tor.
Sim
ilarly,
thereare
eightmagic
squaresoforder
threethatare
rotationsand
reflectionsofA.Thestatements
fork
=0:3
rot90(A,k)
rot90(A’,k)
end
displayalleightof
them
.
81
68
34
35
71
59
49
26
72
67
24
92
15
93
57
83
48
16
29
42
76
75
39
51
61
84
38
43
86
18
95
17
53
27
62
94
Theseare
allthemagic
squaresoforder
three.
The5is
alwaysin
thecenter,
the
other
oddnumbersare
alwaysin
thecenters
oftheedges,andtheeven
numbers
are
alwaysin
thecorners.
MelancholiaIis
afamousRenaissance
engravingbytheGermanartistand
amateurmathem
aticianAlbrechtDurer.
Itshow
smanymathem
aticalobjects,
in-
cludingasphere,
atruncatedrhombohedron,an
d,in
theupper
righthandcorner,
amagic
square
oforder
4.Youcanseetheengravingin
ourfigure
10.2.Betteryet,
issuetheseMatlabcommands
126
Chapter10
.Magic
Squares
..
Figure
10.2.Albrect
Durer’sMelancolia,1514.
loaddurer
whos
Youwillsee
X648x509
2638656
doublearray
caption
2x28
112
chararray
map
128x3
3072
doublearray
Theelem
ents
ofthearray
Xare
indices
into
thegray
-scale
colormapnamed
map.
Theim
ageis
displayedwith
image(X)
colormap(map)
axisimage
Click
themagnifyingglass
witha“+
”in
thetoolbaranduse
themouse
tozoom
inonthemagic
square
intheupper
right-handcorner.Thescanningresolution
becomes
evidentas
youzoom
in.Thecommands
loaddetail
127
..
Figure
10.3.Detailfrom
Melancolia.
image(X)
colormap(map)
axisimage
displaythehigher
resolutionscanof
theareaaroundthemagic
square
thatwehave
infigure
10.3.
Thecommand
A=
magic(4)
producesa4-by-4
magic
square.
A=
16
23
13
511
10
8
97
612
414
15
1
Thecommands
sum(A),sum(A’),sum(diag(A)),sum(diag(flipud(A)))
yield
enough34’s
toverifythatAis
indeedamagic
square.
The4-by-4
magic
square
generatedbyMatlabis
notthesameasDurer’s
magic
square.Weneedto
interchangethesecondandthirdcolumns.
A=
A(:,[1
32
4])
128
Chapter10
.Magic
Squares
changes
Ato
A=
16
32
13
510
11
8
96
712
415
14
1
Interchangingcolumnsdoes
notchangethecolumnsumsortherowsums.
Itusually
changes
thediagonalsums,butin
thiscase
bothdiagonalsumsare
still34.Sonow
ourmagic
square
matches
theon
ein
Durer’setching.Durerprobably
chose
this
particular4-by-4
square
because
thedate
hedid
thework,1514,occurs
inthe
middle
ofthebottom
row.
Theprogram
durerperminterchanges
rowsandcolumnsin
theim
ageproduced
from
detailbyinterchanginggroupsofrowsandcolumnsin
thearrayX.This
isnotespeciallyim
portantor
useful,butitprovides
aninterestingexercise.
Wehaveseen
twodifferent4-by-4
magic
squares.
Itturnsou
tthatthereare
880differentmagic
squaresoforder
4and
275305224differentmagic
squaresof
order
5.Determiningthenumber
ofdifferentmagic
squaresoforder
6orlarger
isanunsolved
mathem
aticalproblem.
For
amagic
square
oforder
n,themagic
sum
is
μ(n)=
1 n
n2 ∑ k=1
k
whichturnsoutto
be
μ(n)=
n3+
n
2.
Hereis
thebeginningofatable
ofvalues
ofthemagic
sum.
nμ(n)
315
434
565
6111
7175
8260
You
cancompute
μ(n)in
Matlabwitheither
sum(diag(magic(n)))
or
(n^3+
n)/2
Thealgorithmsforgeneratingmatrix
square
fallinto
threedistinct
cases:
129
odd,nis
odd.
singly-even
,nis
divisible
by2,butnotby4.
doubly-even
,nis
divisible
by4.
Thebestknow
nalgorithm
forgeneratingmagic
squaresofoddorder
isdeLa
Loubere’smethod.Sim
ondeLaLouberewas
theFrench
ambassadorto
Siam
inthe
late
17th
century.Isometim
esreferto
his
methodasthe”nor’easter
algorithm”,
after
thewinterstorm
sthatmovenortheasterly
upthecoast
ofNew
England.You
canseewhyifyoufollow
theintegerssequentiallythroughmagic(9).
47
58
69
80
112
23
34
45
57
68
79
911
22
33
44
46
67
78
810
21
32
43
54
56
77
718
20
31
42
53
55
66
617
19
30
41
52
63
65
76
16
27
29
40
51
62
64
75
5
26
28
39
50
61
72
74
415
36
38
49
60
71
73
314
25
37
48
59
70
81
213
24
35
Theintegersfrom
1to
n2are
inserted
alongdiagonals,startingin
themiddle
offirstrowan
dheadingin
anortheasterly
direction.W
hen
you
gooff
anedgeofthe
array,whichyoudoattheveryfirststep,continuefrom
theop
posite
edge.
When
youbumpinto
acellthatis
alreadyoccupied,dropdow
nonerow
andcontinue.
Weusedthis
algorithm
inMatlabformanyyears.Hereis
thecode.
A=
zeros(n,n);
i=
1;
j=
(n+1)/2;
fork
=1:n^2
is
=i;
js
=j;
A(i,j)=
k;
i=
n-
rem(n+1-i,n);
j=
rem(j,n)
+1;
if
A(i,j)
~=
0
i=
rem(is,n)
+1;
j=
js;
end
end
Abig
diffi
cultywiththis
algorithm
andresultingprogram
isthatit
inserts
theelem
ents
oneatatime–itcannotbevectorized.
Afew
years
agowediscoveredanalgorithm
forgeneratingthesamemagic
squaresofoddorder
asdeLaLoubere’smethod,butwithjust
fourMatlabmatrix
operations.
[I,J]=
ndgrid(1:n);
130
Chapter10
.Magic
Squares
A=
mod(I+J+(n-3)/2,n);
B=
mod(I+2*J-2,n);
M=
n*A
+B
+1;
Let’s
seehow
this
workswithn
=5.Thestatement
[I,J]=
ndgrid(1:n)
producesapair
ofmatrices
whose
elem
ents
arejust
therow
andcolumnindices,i
andj. I
=
11
11
1
22
22
2
33
33
3
44
44
4
55
55
5
J=
12
34
5
12
34
5
12
34
5
12
34
5
12
34
5
Usingtheseindices,wegenerate
twomore
matrices.Thestatements
A=
mod(I+J+1,n)
B=
mod(I+2*J-2,n)
produce
A=
34
01
2
40
12
3
01
23
4
12
34
0
23
40
1
B=
13
02
4
24
13
0
30
24
1
41
30
2
02
41
3
Both
AandBare
fledglingmagicsquares.
They
haveequalrow,columnan
ddiagonal
sums.
Buttheirelem
ents
arenottheintegersfrom
1to
n2.Each
hasduplicated
elem
ents
between0an
dn−
1.Thefinalstatement
131
M=
n*A+B+1
producesamatrix
whose
elem
ents
are
integersbetween1an
dn2andwhichhas
equalrow,columnan
ddiagonalsums.
Whatis
notobvious,
butis
true,
isthat
therearenoduplicates.
SoMmust
contain
alloftheintegersbetween1andn2and
consequentlyis
amagic
square.
M=
17
24
18
15
23
57
14
16
46
13
20
22
10
12
19
21
3
11
18
25
29
Thedoubly-even
algorithm
isalsoshort
andsw
eet,andtricky.
M=
reshape(1:n^2,n,n)’;
[I,J]=
ndgrid(1:n);
K=
fix(mod(I,4)/2)
==
fix(mod(J,4)/2);
M(K)=
n^2+1
-M(K);
Let’s
look
atou
rfriendmagic(4).Thematrix
Mis
initiallyjust
theintegers
from
1to
16storedsequentiallyin
4rows.
M=
12
34
56
78
910
11
12
13
14
15
16
Thelogicalarray
Kis
trueforhalfoftheindices
andfalsefortheother
halfin
apatternlikethis.
K=
10
01
01
10
01
10
10
01
Theelem
ents
whereKis
false,
thatis
0,are
left
alone.
.2
3.
5.
.8
9.
.12
.14
15
.
Theelem
ents
whereKis
true,
thatis
1,are
reversed.
16
..
13
.11
10
.
.7
6.
4.
.1
132
Chapter10
.Magic
Squares
Thefinalresult
merges
thesetw
omatrices
toproduce
themagic
square.
Thealgorithm
forsingly
even
order
isthemost
complicatedandso
wewillgive
just
aglimpse
ofhow
itworks.
Ifnis
singly
even
,then
n/2is
oddandmagic(n)
canbeconstructed
from
fourcopiesofmagic(n/2).
Forexample,magic(10)is
obtained
from
A=
magic(5)byform
ingablock
matrix.
[A
A+50
A+75
A+25]
Thecolumnsumsareallequalb
ecause
sum(A)+
sum(A+75)equalssum(A+50)+
sum(A+25).
Buttherowssumsare
notquiteright.
Thealgorithm
must
finishbydoingafew
swapsof
piecesofrowsto
cleanuptherow
sums.
For
thedetails,
issuethecom-
mand. typemagic 9
10
1112
Figure
10.4.Surf
plots
ofmagicsquaresoforder
9,10,11,12.
Let’s
concludethis
chapterwithsomegraphics.
Figure
10.4
show
s
surf(magic(9))
surf(magic(10))
surf(magic(11))
surf(magic(12))
You
canseethethreedifferentcases–on
theleft,theupper
right,
andthelower
right.
Ifyou
increase
ofeach
oftheordersby4,you
getmore
cells,
buttheglobal
shapes
remain
thesame.
Theoddncase
ontheleft
remindsmeofOrigami.
133
Further
Rea
ding
ThereasonswhyMatlabhasmagic
squarescanbetracedback
tomyjuniorhigh
schoolday
swhen
Idiscoveredaclassic
book
byW
.W
.Rouse
Ball,Mathem
atical
RecreationsandEssays.Balllived
from
1850until1925.Hewas
afellow
ofTrinity
College,
Cambridge.
Thefirstedition
ofhis
book
on
mathem
aticalrecreations
was
published
in1892andthetenth
editionin
1937.Latereditionswererevised
andupdatedbyanother
famousmathem
atician,H.S.M
Cox
eter.Thethirteenth
edition,published
byDover
in1987,is
available
from
manybooksellers,
including
Pow
ells
andAmazon.
http://www.powells.com/cgi-bin/biblio?inkey=17-0486253570-0
http://www.amazon.com/Mathematical-Recreations-Essays-Dover-Books/
dp/0486253570
TherearedozensofinterestingWeb
pages
abou
tmagic
squares.
Hereare
afew
authors
andlinksto
theirpages.
MutsumiSuzu
ki
http://mathforum.org/te/exchange/hosted/suzuki/MagicSquare.html
EricWeisstein
http://mathworld.wolfram.com/MagicSquare.html
KwonYou
ngShin
http://user.chollian.net/~brainstm/MagicSquare.htm
WalterTrump
http://www.trump.de/magic-squares
Rec
ap
%%Magic
Squares
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Magic
Squares
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %magic_recap
%edit
magic_recap
%publish
magic_recap
% %Related
EXM
programs
% %magic
%ismagical
%%A
Few
Elementary
Array
Operations.
formatshort
A=
magic(3)
sum(A)
134
Chapter10
.Magic
Squares
sum(A’)’
sum(diag(A))
sum(diag(flipud(A)))
sum(1:9)/3
fork
=0:3
rot90(A,k)
rot90(A’,k)
end
%%
Durer’s
Melancolia
clearall
closeall
figure
loaddurer
whos
image(X)
colormap(map)
axisimage
%%
Durer’s
Magic
Square
figure
loaddetail
image(X)
colormap(map)
axisimage
A=
magic(4)
A=
A(:,[1
32
4])
%%
Magic
Sum
n=
(3:10)’;
(n.^3+
n)/2
%%
Odd
Order
n=
5
[I,J]=
ndgrid(1:n);
A=
mod(I+J+(n-3)/2,n);
B=
mod(I+2*J-2,n);
M=
n*A
+B
+1
%%
Doubly
Even
Order
n=
4
M=
reshape(1:n^2,n,n)’;
[I,J]=
ndgrid(1:n);
K=
fix(mod(I,4)/2)
==
fix(mod(J,4)/2);
M(K)=
n^2+1
-M(K)
135
%%Rank
figure
forn
=3:20
r(n)=
rank(magic(n));
end
bar(r)
axis([221
020])
%%Ismagical
helpismagical
forn
=3:10
ismagical(magic(n))
end
%%Surf
Plots
figure
forn
=9:12
subplot(2,2,n-8)
surf(rot90(magic(n)))
axistightoff
text(0,0,20,num2str(n))
end
set(gcf,’color’,’white’)
Exe
rcises
10.1
ismagic.
Write
aMatlab
functionismagic(A)thatchecksifAis
amagic
square.
10.2
Magicsum.Show
that
1 n
n2 ∑ k=1
k=
n3+
n
2.
10.3
durerperm.Investigate
thedurerpermprogram.Click
ontw
odifferentelem
ents
tointerchangerowsan
dcolumns.
Dotheinterchanges
preserverow
andcolumn
sums?
Dotheinterchanges
preservethediagonalsums?
10.4
Colorm
aps.
Try
this.
clear
loaddetail
whos
136
Chapter10
.Magic
Squares
You
willseethreematrices
inyou
rworkspace.
You
can
lookatall
ofmapand
caption.
map
caption
Thematrix
Xis
359-by-371.That’s133189elem
ents.Look
atjust
apiece
ofit.
X(101:130,101:118)
Theelem
ents
ofXare
integersin
therange
min(min(X))
max(max(X))
Thecommands
image(X)
axisimage
produce
aprettycolorfuldisplay.
That’sbecause
theelem
ents
ofXare
beingused
asindices
into
thedefault
colorm
ap,jet(64).Youcanuse
theintended
colorm
ap
instead.
colormap(map)
Thearray
mapis
a64-by-3
array.Each
row,map(k,:),specifies
intensities
ofred,
green
andblue.
Thecolorusedatpoint(i,j)ismap(X(i,j),:).In
thiscase,the
colorm
apthatcomes
withdetailhasallthreecolumnsequalto
each
other
andso
isthesameas
colormap(gray(64))
Now
experim
entwithother
colorm
aps
colormap(hot)
colormap(cool)
colormap(copper)
colormap(pink)
colormap(bone)
colormap(flag)
colormap(hsv)
You
caneven
cyclethrough101colorm
aps.
forp
=0:.001:1
colormap(p*hot+(1-p)*pink)
drawnow
end
You
canplotthethreecolorcompon
ents
ofacolorm
aplikehotwith
137
rgbplot(hot)
This
iswhatTV
mov
iechannelsdowhen
they
colorize
old
black
andwhitefilm
s.
10.5
Knight’stour.
Doyou
know
how
aknightisallow
edto
moveon
achessboard?
Theexmfunctionknightstourgen
eratesthis
matrix,K.
K=
50
11
24
63
14
37
26
35
23
62
51
12
25
34
15
38
10
49
64
21
40
13
36
27
61
22
952
33
28
39
16
48
760
120
41
54
29
59
445
853
32
17
42
647
257
44
19
30
55
358
546
31
56
43
18
Ifyou
follow
theelem
ents
innumericalorder,youwillbetaken
onakn
ight’stour
ofK.Even
thestep
from
64back
to1is
aknight’smove.
IsKamagic
square?W
hyorwhynot?
Try
this.
image(K)
colormap(pink)
axissquare
Selectthedata
cursoricononthefigure
tourbar.
Now
use
themouse
totakethe
knight’stourfrom
dark
tolightontheim
age.
10.6
ismagical.Theexmfunctionismagicalchecksforfourdifferentmagicalprop-
erties
ofsquare
array
s.Sem
imagic:allofthecolumnsandallofrowshavethesamesum.
Magic:allofthecolumns,
allofrowsandboth
principaldiagonals
havethesame
sum.
Panmagic:allofthecolumns,
allofrowsandallofthediagonals,includingthe
broken
diagonals
inboth
directions,
havethesamesum.
Associative:allpairsof
elem
ents
onopposite
sides
ofthecenterhavethesamesum.
For
example,this
matrix
thathasallfourproperties.
M=
10
18
114
22
11
24
720
3
17
513
21
9
23
619
215
412
25
816
Hereis
oneof
thebroken
diagonals.Itssum
isμ(5)=
65.
138
Chapter10
.Magic
Squares
..
.14
.
..
..
3
17
..
..
.6
..
.
..
25
..
Allof
thebroken
diagonalsin
both
directionshavethesamesum,so
Mispanmagic.
Onepairof
elem
ents
onop
posite
sides
ofthecenteris24
and2.Theirsum
istw
ice
thecentervalue.
Allpairsof
elem
ents
onop
posite
sides
ofthecenterhavethissum,
soMis
associative.
(a)Use
ismagicalto
verifythatMhasallfourproperties.
(b)Use
ismagicalto
investigate
themagicalproperties
ofthematrices
generated
bytheMatlabmagicfunction.
(c)Use
ismagicalto
investigate
themagicalproperties
ofthematrices
generated
bythis
algorithm
forvariousoddnandvariousvalues
ofa0andb0,
a0
=...
b0
=...
[I,J]=
ndgrid(1:n);
A=
mod(I+J-a0,n);
B=
mod(I+2*J-b0,n);
M=
n*A
+B
+1;
(d)Use
ismagicalto
investigate
themagicalproperties
ofthematrices
generated
bythis
algorithm
forvariousoddnandvariousvalues
ofa0andb0,
a0
=...
b0
=...
[I,J]=
ndgrid(1:n);
A=
mod(I+2*J-a0,n);
B=
mod(I+3*J-b0,n);
M=
n*A
+B
+1;
10.7
Inverse.
Ifyou
havestudiedmatrix
theory,youhaveheard
ofmatrix
inverses.
Whatis
thematrix
inverseofamagic
square
oforder
n?
Itturnsoutto
dep
end
uponwhether
nisoddoreven
.Foroddn,thematrices
magic(n)are
nonsingular.
Thematrices
X=
inv(magic(n))
donothavepositive,
integer
entries,butthey
dohaveequalrow
andcolumnsums.
But,
foreven
n,thedeterminant,det(magic(n)),is
0,andtheinverse
does
notexist.
IfA=
magic(4)tryingto
compute
inv(A)producesanerrormessage.
10.8
Rank.
Ifyou
havestudiedmatrix
theory,youknow
thattherankof
amatrix
isthenumber
oflinearlyindep
endentrowsandcolumns.
Ann-by-n
matrix
issingular
ifitsrank,r,
isnotequalto
itsorder.This
codecomputestherankof
themagic
squaresupto
order
20,generatedwith
139
forn
=3:20
r(n)=
rank(magic(n));
end
Theresultsare
n=
34
56
78
910
11
12
13
14
15
16
17
18
19
20
r=
33
55
73
97
11
313
915
317
11
19
3
Doyouseethepattern?May
bethebargraphin
figure
10.5
willhelp.Youcansee
24
68
1012
1416
1820
02468101214161820 Figure
10.5.Rankofmagicsquares.
thatthethreedifferentalgorithmsusedto
generate
magicsquaresproduce
matrices
withdifferentrank.
nrank
odd
n
even,notdivisibleby
4n/2+2
divisibleby
43
140
Chapter10
.Magic
Squares
Chapter11
TicTacT
oeMagic
Threesimple
games
are
relatedin
asurprisingway.
And,theprogrammingofthe
gameplayis
instructive.
Thefirstofthethreegames
isPick15.Harold
Stark,whowas
then
atthe
University
ofMichigan,told
meab
outthegamein
thelate
1960s.
Isuspectthat
this
isthefirsttimeyouhaveheard
ofit.
Thegameinvo
lves
twoplayers.Youstart
bylistingthesingle
digit
numbers
from
1to
9.Youthen
taketurnsselectingnumbersfrom
thelist,attem
ptingto
acquirethreenumbersthataddupto
15.Each
number
canbechosenonly
once.
Youmay
eventuallyacquiremore
thanthreenumbers,
butyoumust
use
exactly
threeof
them
tototal15.If
neither
playercanachieve
thedesired
total,thegame
isadraw.
Forexample,suppose
thatGreen
andBlueare
playing.They
start
withthe
list.
List
:1
23
45
67
89
Green
:
Blue
:
Suppose
Green
hasthefirstmoveandchooses8.Then
Bluechooses
4an
dGreen
chooses2.Now
Blueshould
respondbychoosing5to
preventGreen
from
getting
2+5+8=
15.Hereis
thesituationafter
thefirsttw
orounds.
List
:1
2/3
4/5/
67
8/9
Green
:2
8
Blue
:4
5
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
141
142
Chapter11
.TicTacToeMagic
Now
Green
chooses
6to
block
4+
5+
6=
15from
Blue.
This
isactually
an
advantageousmoveforGreen
because
itgives
her
twodifferentway
sto
win,1+6+8
and2+
6+7.
Bluecannotblock
both.If
Bluechooses7,then
Green
chooses
1to
win.If
Bluechooses
1,then
Green
chooses
7to
win.Thefinalpositionmightbe
List
:1/
2/3
4/5/
6/7/
8/9
Green
:2
67
8
Blue
:1
45
Note
thatGreen
alsohas7+8=
15,butthisdoes
notcountbecause
thereare
only
twonumbersin
thesum.
Figure
11.1
show
sthestartingpositionforthePick15op
tionin
ourMatlab
tictactoeprogram.W
hen
youplayagainst
thecomputer,
yourmoves
are
show
nin
green
andtheresponsesfrom
theprogram
are
show
nin
blue.
Figure
11.1.StartingpositionforPick15.
Figure
11.2
show
sthepositionforourexample
gameaftertw
omoves
from
each
player.
Green
now
hasto
choose6to
block
Blue.
Figure
11.2.Positionafter
twomovesfrom
each
player.
Figure
11.3
show
sthefinalpositionforourexample
game.
Green
haswon
with2+6+7=
15.
Figure
11.3.FinalpositionforPick15.
Please
taketimeoutfrom
readingthis
chapterto
trythis
gameafew
times
you
rself,playingagainst
afriendonpaper
oragainst
ourMatlabprogram.Ithink
you
willfindthatPick15is
more
challengingthanit
soundsatfirst.
Thesecondofourthreegames
isfamiliarworldwide.
Itiscalled
“TicTacT
oe”
intheUnited
States,“Noughts
andCrosses”in
GreatBritain,andhasmanyother
names
inmanyother
countriesan
dlanguages.Ourprogram
tictactoeusesgreen
andblueinsteadofX’s
andO’s.Theobjective,
ofcourse,
isto
get
yourcoloron
allthreesquaresin
arow,column,or
diagonal.Figure
11.6
show
stypicalou
tput.
143
Figure
11.4.Typicaloutputfrom
agameofTicTacT
oe.Green
haswon
withthethirdcolumn.
Ourprogram
tictactoeusesanaivethree-step
strategy.
•Ifpossible,makeawinningmove.
•Ifnecessary,block
apossible
winningmovebytheopponent.
•Otherwise,
pickarandom
empty
square.
This
strategywillleadto
awin
only
when
theop
ponentmakes
aerror.
And,even
thoughit
ispossible
fortheplayerwiththesecondmoveto
alwaysforceadraw,
this
strategywillnotaccomplish
that.
Ourthirdgame,
Magic15,introducestheLo-Shu3-by-3
magic
square.Now
weseethatPick15is
actuallyTicTacT
oeplayed
onamagic
square.
Therows,
columnsandmain
diagonalsofthemagic
square
provideallpossible
waysofhav
ing
threedistinct
numbersthatsum
to15.W
inningmoves
inPick15correspondto
winningmoves
inTicTacToe.
Allthreegames
are
actuallythesamegamewith
differentdisplays.
Figure
11.5.InitialconfigurationforagameofMagic3.
144
Chapter11
.TicTacToeMagic
Figure
11.6.Green
haswonwith6+7+2=
15in
thethirdcolumn.
GamePlay
Thetictactoeprogram
operatesthreedifferentviewsofthesamegame.
You
are
thegreenplayerandthecomputeris
theblueplayer.
Thestate
ofthegameis
carriedin
a3-by-3
matrix
Xwhose
entriesare
+1forcellsoccupiedbygreen,-1
forcellsoccupiedbyblue,
and0foras
yet
unoccupiedcells.
Obviously,
thegame
beginswithX=
zeros(3,3)
Hereis
theportionofmain
program
thatis
executedwhen
you,as
thegreen
player,
clickon
oneofthebuttonsorcells.
Itretrieves
iandj,thematrix
indices
correspondingto
theactivebuttonandthen
checksifthecurrentgamealreadyhas
awinner.If
not,
itmakes
yourmovebysettingX(i,j)=
1andrecu
rsivelycalls
tictactoeto
letbluetakeaturn.
switch
case’green’
[i,j]=
find(gcbo
==
B);
ifwinner(X)
return
end
X(i,j)=
1;
tictactoe(’blue’)
Hereis
thecorrespondingblueportionof
themain
program.If
thecurrent
gamedoes
notalreadyhaveawinner,it
callsthestrategyfunctionto
get
the
indices
foramoveandmakes
thatmove.
case’blue’
ifwinner(X)
return
end
[i,j]=
strategy(X,-1);
X(i,j)=
-1;
...
end
145
Briefl
y,then
,thegreen
moves
are
determined
byuserbuttonclicksandthe
bluemoves
are
determined
bythestrategyfunction.
Hereis
thefunctionthatchecksto
seeifthereis
awinner.It
looksforany
column,row,ordiagonalwhose
elem
ents
sum
to3*pwherep=1forgreen
and
p=-1forblue.
Theon
lyway
asum
canbe3*pis
ifallthreeelem
ents
are
equal
top. functionp
=winner(X)
%p
=winner(X)
returns
%p
=0,
no
winner
yet,
%p
=-1,
blue
has
won,
%p
=1,
green
has
won,
%p
=2,
game
is
adraw.
forp
=[-1
1]
s=
3*p;
win=
any(sum(X)
==
s)
||
any(sum(X’)
==
s)
||
...
sum(diag(X))==
s||
sum(diag(fliplr(X)))
==
s;
if
win
return
end
end
p=
2*all(X(:)
~=
0);
Hereis
thenaive,
butfairly
effectivestrategythatthecomputerusesagainst
you
.It
firstchecksto
seeifit
canmakeawinningmove.
Ifnot,
itthen
checksto
seeifit
needsto
block
you
rpendingwinningmove.
Itneither
ofthesehappen
s,it
simply
choosesarandom
empty
cell.Thiscrudestrategymakes
theplayinteresting.
Itis
possible
foryouto
beatthecomputerifitsrandom
choices
are
notoptimal.
Theexercisessuggestsomeim
provements
inthis
strategy.
function[i,j]=
strategy(X,p);
%[i,j]
=strategy(X,p)
is
abetter,
but
not
perfect,
move
for
player
p.
%Appear
to
think.
pause(0.5)
%If
possible,
make
awinning
move.
[i,j]=
winningmove(X,p);
%Block
any
winning
move
by
opponent.
if
isempty(i)
[i,j]=
winningmove(X,-p);
end
%Otherwise,
make
arandom
move.
if
isempty(i)
146
Chapter11
.TicTacToeMagic
[i,j]=
find(X
==
0);
m=
ceil(rand*length(i));
i=
i(m);
j=
j(m);
end Finally,
hereis
thefunctionthatseeksawinningmovebyeither
player.
Itlooksforrows,
columns,
ordiagonals
withsumsequalto
2*pforeither
valueofp.
function[i,j]=
winningmove(X,p);
%[i,j]
=winningmove(X,p)
finds
any
winning
move
for
player
p.
s=
2*p;
if
any(sum(X)
==
s)
j=
find(sum(X)
==
s);
i=
find(X(:,j)
==
0);
elseifany(sum(X’)==
s)
i=
find(sum(X’)
==
s);
j=
find(X(i,:)
==
0);
elseifsum(diag(X))==
s
i=
find(diag(X)
==
0);
j=
i;
elseifsum(diag(fliplr(X)))==
s
i=
find(diag(fliplr(X))
==
0);
j=
4-
i;
else i=
[];
j=
[];
end Theremainder
ofthetictactoefunctionis
responsible
forthegu
i.Let’sseehow
allthisworksonanexample.Webegin
withthepositionshow
nin
figure
11.2,after
twomoves
byeach
player.
Thestate
ofthegameat
this
point
is
X=
10
0
0-1
0
-1
01
Itis
green’s
move.
Green
needsto
clickthe6buttonto
preventbluefrom
hav
ing
4+5+6=
15.Or,
itis
probably
easier
tolookatM
=magic(3)
M=
81
6
35
7
49
2
147
You
seethat4and5are
ontheantidiagonal,so
greenneedsthe6to
preventawin
intheTicTacToeversionofthisgame.
Either
way,thiscorrespondsto
(i,j)=
(1,3)
andsettingX(i,j)=
1.
X=
10
1
0-1
0
-1
01
Now
itisblue’sturn.Both
thefirstrow
andthethirdcolumnofXhavesums
equalto
2,correspondingto
thefact
thatgreenhastw
opendingwinningmoves.
Withp=-1,thestrategyfunctionlooksforablock
bycallingwinningmove(X,-p).
Theresultis(i,j)=
(2,3),althoughitjust
aswellcould
havebeen(i,j)=
(1,2)
ifwinningmovemadeitschecksin
another
order.This
leadsto
X=
10
1
0-1
-1
-1
01
Onthenextturn,green
findsthatthesum
alongthefirstrow
is2,an
dso
sets
X(1,2)=
1to
makethesum
3.This
gives
X=
11
1
0-1
-1
-1
01
andgreenproclaim
sawin.
Itwould
bepossible
toim
plementrecursivebacktrackingstrategieslikewe
describein
theSudokuchapter.
Butthereare
only
afew
hundredpossible
games,
sothebacktrackingwould
beexhaustive.
Andboring–allthegames
would
be
draws.
Rec
ap
%%TicTacToe
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
TicTacToe
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %tictactoe_recap
%edit
tictactoe_recap
%publish
tictactoe_recap
% %Related
EXM
programs
% %tictactoe
148
Chapter11
.TicTacToeMagic
%%
tictactoe/winner
%function
p=
winner(X)
%%
p=
winner(X)
returns
%%
p=
0,
no
winner
yet,
%%
p=
-1,
blue
has
won,
%%
p=
1,
green
has
won,
%%
p=
2,
game
is
adraw.
% %for
p=
[-1
1]
%s
=3*p;
%win
=any(sum(X)
==
s)
||
any(sum(X’)
==
s)
||
...
%sum(diag(X))==
s||
sum(diag(fliplr(X)))
==
s;
%if
win
%return
%end
%end
%p
=2*all(X(:)
~=
0);
%%
tictactoe/strategy
%function
[i,j]
=strategy(X,p);
%%
[i,j]
=strategy(X,p)
is
amove
for
player
p.
% %%
Appear
to
think.
%pause(0.5)
% %%
If
possible,
make
awinning
move.
%[i,j]
=winningmove(X,p);
% %%
Block
any
winning
move
by
opponent.
%if
isempty(i)
%[i,j]
=winningmove(X,-p);
%end
% %%
Otherwise,
make
arandom
move.
%if
isempty(i)
%[i,j]
=find(X
==
0);
%m
=ceil(rand*length(i));
%i
=i(m);
%j
=j(m);
%end
%%
tictactoe/winningmove
149
%function
[i,j]
=winningmove(X,p);
%%
[i,j]
=winningmove(X,p)
finds
any
winning
move
for
player
p.
% %s
=2*p;
%if
any(sum(X)
==
s)
%j
=find(sum(X)
==
s);
%i
=find(X(:,j)
==
0);
%elseif
any(sum(X’)
==
s)
%i
=find(sum(X’)
==
s);
%j
=find(X(i,:)
==
0);
%elseif
sum(diag(X))
==
s
%i
=find(diag(X)
==
0);
%j
=i;
%elseif
sum(diag(fliplr(X)))
==
s
%i
=find(diag(fliplr(X))
==
0);
%j
=4
-i;
%else
%i
=[];
%j
=[];
%end
Exe
rcises
11.1
Traditional.
Modifytictactoe.mso
thatitusestraditionalX’s
andO’s.
11.2
Win.Is
itpossible
towin
against
tictactoewithitsnaivestrategy?
11.3
First
move.Modifytictactoeso
thatthecomputertakes
thefirstmove.
11.4
Cen
tersquare.Modifythestrategyusedbytictactoe.mso
that,before
taking
arandom
move,
ittakes
thecentersquare
ifit
isavailable.Does
this
improve
the
program’s
chancesofwinningorforcingadraw?
11.5
xkcd.Im
plementthecomplete
tictactocstrategyavailable
from
theWeb
comic
stripxkcdbyRandalMunroeat
http://xkcd.com/832
11.6
Computerversuscomputer.
Modifytictactoe.mso
thatthecomputerplays
against
itself.Countthenumber
ofdrawsan
dwinsforboth
sides.Runalarge
number
ofgames,withandwithoutthead
ditionofthecentersquare
strategy.
150
Chapter11
.TicTacToeMagic
Chapter12
GameofLife
Conway’sGameofLifemakesuse
ofsparsematrices.
The“GameofLife”
wasinvented
byJohn
Horton
Conway,aBritish-born
mathem
aticianwhoisnow
aprofessorat
Princeton.Thegamemadeitspublicdebut
intheOctober
1970issueofScien
tificAmerican,in
the“Mathem
aticalGames”
columnwritten
byMartin
Gardner.Atthetime,
Gardner
wrote
Thismonth
weconsider
Conway’slatest
brainchild,afantastic
solitaire
pastim
ehecalls“life”.Because
ofitsanalogieswiththerise,falland
alternationsofasocietyoflivingorganisms,itbelongsto
agrowingclass
ofwhatare
called
“simulationgames”–games
thatresemble
real-life
processes.Toplaylife
youmust
haveafairly
largecheckerboard
anda
plentifulsupply
offlatcounters
oftw
ocolors.
Ofcourse,
today
wecanrunthesimulationson
ourcomputers.
Theuniverse
isaninfinite,
two-dim
ensionalrectangulargrid.Thepopulation
isacollection
ofgrid
cellsthatare
marked
asalive.
Thepop
ulationevolvesat
discretetimestep
sknow
nasgenerations.
Ateach
step,thefate
ofeach
cell
isdetermined
bythevitality
ofitseightnearest
neighbors
andthis
rule:
•A
livecellwithtw
oliveneighbors,
oranycellwiththreeliveneighbors,
isalive
atthenextstep
.
ThefascinationofConway’sGameofLifeisthatthisdeceptivelysimple
rule
leads
toanincredible
varietyofpatterns,
puzzles,
andunsolved
mathem
aticalproblems
–just
likereallife.
Iftheinitialpopulationconsistsofon
lyoneortw
olivecells,itexpires
inone
step.Iftheinitialpopulationconsistsofthreelivecellsthen,because
ofrotational
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
151
152
Chapter12
.Gam
eof
Life
Figure
12.1.A
pre-block
andablock.
andreflexivesymmetries,thereare
only
twodifferentpossibilities–thepopulation
iseither
L-shaped
orI-shaped.Theleft
halfoffigure
12.1
show
sthreelivecellsin
anL-shape.
Allthreecellshavetw
oliveneighbors,
sothey
survive.
Thedeadcell
thatthey
alltouch
hasthreeliveneighbors,
soit
springsto
life.Noneof
theother
deadcellshaveenoughliveneighborsto
cometo
life.Sotheresult,afteronestep,
isthepopulationshow
nin
therighthalfoffigure
12.1.This
four-cellpopulation,
know
nastheblock,
isstationary.Each
ofthelivecellshasthreeliveneighborsand
solives
on.Noneof
theother
cellscancometo
life.
Figure
12.2.A
blinkerblinking.
Theother
three-cellinitialpop
ulationis
I-shaped.Thetw
opossible
orienta-
tionsareshow
nin
firsttw
ostepsof
figure
12.2.Ateach
step,tw
oendcellsdie,the
middle
cellstay
salive,
andtw
onew
cellsare
born
togivetheorientationshow
nin
thenextstep
.If
nothingdisturbsit,this
blinkerkeepsblinkingforever.It
repeats
itselfin
twostep
s;this
isknow
nasitsperiod.
Onepossible
four-cellinitialpopulationis
theblock.Discoveringthefate
of
theother
four-cellinitialpop
ulationsis
left
toanexercise.
Thebeginningoftheevolutionofthemost
importantfive-cellinitialpop
ula-
tion,know
nastheglider,isshow
nin
figure
12.3.Ateach
step
twocellsdie
andtw
onew
ones
areborn.After
fourstepstheoriginalpop
ulationreappears,butit
has
moved
diagonallydow
nandacross
thegrid.It
continues
tomovein
this
direction
forever,eventuallydisappearingoutof
ourfieldofview,butcontinuingto
existin
theinfiniteuniverse.
ThefascinationoftheGameof
Lifecannotbecapturedin
thesestaticfig-
ures.
Computergraphicslets
you
watchthedynamic
development.
Wewillshow
just
more
onestaticsnapshotoftheevolutionofanim
portantlarger
population.
Figure
12.4
istheglider
gundeveloped
byBillGosper
atMIT
in1970.Theportion
ofthepopulationbetweenthetw
ostaticblocksoscillatesback
andforth.Every30
steps,aglider
emerges.Theresultisaninfinitestream
ofglidersthatflyou
tofthe
fieldofview.
153
Figure
12.3.A
glider
gliding.
Figure
12.4.Gosper’s
glider
gun.
Matlabisaconvenientenvironmentforim
plementingtheGameofLife.
The
universe
isamatrix.Thepopulationis
thesetofnonzero
elem
ents
inthematrix.
Theuniverse
isinfinite,
butthepopulationis
finiteandusuallyfairly
small.So
wecanstore
thepopulationin
afinitematrix,most
ofwhose
elem
ents
arezero,
andincrease
thesize
ofthematrix
ifnecessary
when
thepopulationexpands.
This
istheidealsetupforasparsematrix.Conventionalstorageofann-by-n
matrix
requires
n2mem
ory.Butsparsestorageofamatrix
Xrequires
just
threevectors,
oneinteger
andonefloatingpointvectoroflength
nnz(X)–thenumber
ofnonzero
154
Chapter12
.Gam
eof
Life
elem
ents
inX–an
doneinteger
vectoroflength
n,not
n2,to
representthestart
ofeach
column.Forexample,thesnapshotof
theGosper
glider
gunin
figure
12.4
isrepresentedbyan
85-by-85matrix
with68
nonzero
entries.
Conventionalfull
matrix
storagewou
ldrequire852
=7225elem
ents.Sparsematrix
storagerequires
only
2·65+85
=221elem
ents.Thisad
vantageofsparseover
fullstorageincreases
asmore
glidersarecreated,thepopulationexpands,
andnincreases.
Theexm
toolbox
includes
aprogram
called
lifex.
(Matlab
itself
hasa
simplerdem
oprogram
called
life.)
Theinitialpop
ulation
isrepresented
bya
matrix
of0’s
and1’s.For
example,
G=
[1
11;
10
0;
01
0]
producesasingle
glider
G=
11
1
10
0
01
0
Theuniverse
isrepresentedbyasparsen-by-n
matrix
Xthatis
initiallyallzero.
Wemightstart
withn=
23so
therewillbea10
cellwideborder
arou
nda3-by-3
center.
Thestatements
n=
23;
X=
sparse(n,n)
produce
X=
Allzerosparse:23-by-23
Theinitialpopulationis
injected
inthecenteroftheuniverse.withthestatement
X(11:13,11:13)=
G
This
producesalist
ofthenonzero
elem
ents
X= (11,11)
1
(12,11)
1
(11,12)
1
(13,12)
1
(11,13)
1
Weare
now
readyto
takethefirststep
inthesimulation.W
hether
cellsstay
alive,
die,orgenerate
new
cellsdep
endsuponhow
manyof
theireightneighborsare
alive.
Thestatements
n=
size(X,1);
p=
[1
1:n-1];
q=
[2:n
n];
155
generate
index
vectors
thatincrease
ordecrease
thecenteredindex
byon
e,thereby
accessingneighbors
totheleft,right,up,dow
n,an
dso
on.Thestatement
Y=
X(:,p)
+X(:,q)
+X(p,:)
+X(q,:)
+...
X(p,p)+
X(q,q)
+X(p,q)
+X(q,p)
producesasparsematrix
withinteger
elem
ents
between0an
d8thatcounts
how
manyoftheeightneighbors
ofeach
interiorcellare
alive.
Inourexample
withthe
firststep
oftheglider,thecellswithnonzero
counts
are
Y=
(9,9)
1(10,9)
2
(11,9)
2(12,9)
1
(9,10)
2(10,10)
2
(11,10)
3(12,10)
2
(13,10)
1(9,11)
3
(10,11)
3(11,11)
5
(12,11)
1(13,11)
1
(9,12)
2(10,12)
1
(11,12)
3(12,12)
1
(13,12)
1(9,13)
1
(10,13)
1(11,13)
1
Thebasicrule
ofLifeis
Alivecellwithtw
oliveneighbors,or
anycellwiththreeliveneighbors,
isaliveatthenextstep
.
This
isim
plementedwiththesingle
Matlabstatement
X=
(X
&(Y
==
2))
|(Y
==
3)
Thetw
ocharacters’==’mean“isequalto”.The’&’character
means“and”.The
’|’means“or”.Theseoperationsare
doneforallthecellsin
theinteriorofthe
universe.In
thisexample,therearefourcellswhereYisequalto
3,so
they
survive
orcomealive.
Thereis
onecellwhereXis
equalto
1andYis
equal
to2,
soit
survives.Theresult
is
X= (11,11)
1
(12,11)
1
(10,12)
1
(11,12)
1
(12,13)
1
Ourglider
hastaken
itsfirststep.
Oneway
touse
lifexis
toprovideyourow
ninitialpopulation,aseither
afullor
asparsematrix.Forexample,youcreate
yourow
nfleetofglidersfleetwith
G=
[1
11;
10
0;
01
0]
156
Chapter12
.Gam
eof
Life
S=
sparse(15,15);
fori
=0:6:12
forj
=0:6:12
S(i+(1:3),j+(1:3))=
G;
end
end
lifex(S)
TheWeb
page
http://www.argentum.freeserve.co.uk/lex_home.htm
isthehomeof
the“LifeLexicon”,maintained
byStephen
Silver.Amongthousands
ofthefactsofLife,
this
160-pagedocumentlistsnearly450differentinitialpopu-
lations,
together
withtheirhistory
andim
portantproperties.Wehaveincluded
atextcopyoftheLexiconwiththeexmtoolbox
inthefile
exm/lexicon.txt
Lifexcanreadinitialpop
ulationsfrom
theLexicon.Callinglifexwithno
arguments,
lifex
picksarandom
initialpop
ulationfrom
theLexicon.Either
lifex(’xyz’)
or
lifexxyz
willlookforapop
ulationwhose
namebeginswithxyz.Forexample,thestatements
lifexpre-block
lifexblock
lifexblinker
lifexglider
start
with
thesimple
populationsthatwehaveused
inthis
introduction.
The
statement
lifexGosper
provides
Gosper’s
glider
gun.
Bydefault,theinitialpop
ulationis
surrounded
byastripof
20deadborder
cellsto
provideaviewingwindow
.Youcanchangethis
touse
bborder
cellswith
lifex(’xyz’,b)
Ifthepopulationexpandsduringthesimulationandcellstravelbeyondthisviewing
window
,they
continueto
livean
dparticipate
even
thoughthey
cannotbeseen.
157
Further
Rea
ding
TheW
ikiped
iaarticle
isagoodintroduction.
http://en.wikipedia.org/wiki/Conway’s_Game_of_Life
Another
goodintroductionisavailablefrom
Math.com,althoughthereare
annoy
ing
pop
upsandad
s.
http://www.math.com/students/wonders/life/life.html
Ifyou
findyourselfatallinterested
intheGameofLife,
takeagoodlookatthe
Lexicon,either
byreadingourtextversionorbyvisitingtheWeb
page.
http://www.argentum.freeserve.co.uk/lex_home.htm
Rec
ap
%%Life
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Life
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %life_recap
%It
does
not
work
so
well
with
edit
and
publish.
% %Related
EXM
programs
% %lifex
%Generate
arandom
initial
population
X=
sparse(50,50);
X(21:30,21:30)=
(rand(10,10)
>.75);
p0
=nnz(X);
%Loop
over
100
generations.
fort
=1:100
spy(X)
title(num2str(t))
drawnow
%Whether
cells
stay
alive,
die,
or
generate
new
cells
depends
%upon
how
many
of
their
eight
possible
neighbors
are
alive.
%Index
vectors
increase
or
decrease
the
centered
index
by
one.
n=
size(X,1);
p=
[1
1:n-1];
158
Chapter12
.Gam
eof
Life
q=
[2:n
n];
%Count
how
many
of
the
eight
neighbors
are
alive.
Y=
X(:,p)
+X(:,q)
+X(p,:)
+X(q,:)
+...
X(p,p)+
X(q,q)
+X(p,q)
+X(q,p);
%A
live
cell
with
two
live
neighbors,
or
any
cell
with
%three
live
neigbhors,
is
alive
at
the
next
step.
X=
(X
&(Y
==
2))
|(Y
==
3);
end
p100=
nnz(X);
fprintf(’%5d%5d%8.3f\n’,p0,p100,p100/p0)
Exe
rcises
12.1
life
vs.
lifex
TheMatlab
dem
ostoolbox
hasan
oldprogram
called
life,
withoutan
x.In
whatwaysisitthesameas,an
din
whatwaysdoes
itdiffer
from,
ourexmguilifex?
12.2
Four-cellinitialpopulations.
Whatare
allofthepossible
four-cellinitialpop
u-
lations,andwhataretheirfates?
You
cangenerate
oneof
thefour-cellpopulations
with
L=
[1
11;
10
0];
lifex(L,4)
12.3
Lexicon.Describethebehav
iorof
each
ofthesepop
ulationsfrom
theLexicon.
Ifanyis
periodic,whatis
itsperiod?
ants
B-52
blinkerpuffer
diehard
Canadagoose
glidersby
the
dozen
Kok’sgalaxy
rabbits
R2D2
spacefiller
wasp
washerwoman
159
12.4
Glider
collisions.
Whathappenswhen:
Aglider
collides
head-onwithablock?
Aglider
side-sw
ipes
ablock?
Twogliderscollidehead-on?
Twoglidersclip
each
otherswings?
Fourgliderssimultaneouslyleavethecornersofasquare
andheadto-
wardsitscenter?
See
also:lifex(’4-8-12’).
12.5
Factory.How
manystepsdoes
ittakethefactoryto
makeoneglider?
12.6
R-pen
tomino.Ofallthepossible
five-cellinitialpopulations,theonly
onethat
requires
acomputerto
determineitsultim
ate
behav
ioris
theonethatConway
dubbed
theR-pen
tomino.It
isshow
nin
figure
12.5
andcanbegeneratedby
R=
[0
11;
11
0;
01
0]
Figure
12.5.TheR-pen
tomino.
Asthesimulationproceeds,
thepopulationthrowsoff
afew
gliders,
butotherwise
remainsbounded.Ifyou
makeblargeenough,thestatement
lifex(R,b)
show
sallofthebounded
behav
ior.
How
largedoes
this
bhaveto
be?
Whatis
themaxim
um
pop
ulationduringtheevolution?
How
manyglidersare
produced?
How
manystepsdoes
ittakeforthepop
ulationto
stabilize?
How
manyblinkers
are
presentin
thestabilized
pop
ulation?W
hatissize
ofthestabilized
pop
ulation?
12.7
Executiontime.
Displayactualcomputerexecutiontimebyinsertingticand
tocin
lifex.m.Place
thesingle
statement
tic
before
thestart
oftheinner
loop.Changethecallofcaptionto
caption(t,nnz(X),toc)
Maketheap
propriate
modificationsin
thecaption
subfunction
attheend
of
lifex.m.Dem
onstrate
yourmodified
program
onafew
interestingexamples.
12.8
TheArk.Run
160
Chapter12
.Gam
eof
Life
lifex(’ark’,128)
forafew
minutes.
Abouthow
much
timedoes
ittakeonyou
rcomputerto
doone
step?Accordingto
theLexicon,thearkrequires
736692stepsto
stabilize.
About
how
much
timewillittakeonyou
rcomputerforthearkto
stabilize?
12.9
Houses.
Checkou
t
lifex(houses)
12.10Checkerboards.
This
codegeneratesan
n-by-n
checkerboard
of0’sand1’s.
[I,J]=
meshgrid(1:n);
C=
(mod(I+J,2)==0);
Whatare
thestabilization
times
and
finalpopulationsforn-by-n
checkerboard
initialpopulationswith3<=
n<=
30?
12.11Earlyexit.W
hereshould
thesecodesegments
Xsave=
X;
and
if
isequal(X,Xsave)
break
end
beinserted
inlife_recap?
12.12Sym
metry.W
hatis
theeff
ectofinsertingthestatement
X=
(X
+X’
>0)
after
thegenerationofX
atthebeginningoflife_recap.
12.13Liveordie?
Insert
thesestatements
before
thefirstexecutable
statementof
life_recap.
P=
zeros(10000,2);
fors
=1:10000
Insert
thesestatements
after
thelast
executable
statement.
P(s,:)=
[p0
p100];
end
Deactivateorremovethespy,title,pauseandfprintfstatements.
Runthe
resultingprogram
andthen
generate
histogramsof
theresults.
161
R=
P(:,2)./P(:,1);
hist(P(:,1),30)
hist(P(:,2),30)
hist(R,30)
(a)Is
thetypicalpopulationgrow
ingordecliningafter100generations?
(b)Doyourecognizetheshapeofthehistogram
oftheinitialpopulation,P(:,1)?
Whatismean(P(:,1))?W
hy?
(c)Idon’t
know
enoughab
outprobabilityto
describethedistributionofthefinal
population,P(:,2).Cananybodyhelpmeou
t?
12.14
Gosper’s
glider
gun.
Replace
therandom
initialpop
ulation
generatorin
life_recapbythis
code:
%Gosper
glider
gun
gun=
[’
+’
’+
+’
’++
++
++’
’+
+++
++’
’++
++
++
’
’++
++
++
++
’
’+
++
’
’+
+’
’++
’];
X=
sparse(76,76);
X(34:42,20:57)=
(gun==’+’);
spy(X)
(a)How
manystepsdoes
ittakeforthegunto
emitaglider?
(b)W
hathappenswhen
aglider
meets
thebou
ndary?
How
isthis
differentfrom
lifex(’gosper’)?
162
Chapter12
.Gam
eof
Life
Chapter13
MandelbrotSet
Fractals,topology,complexarithmetic
andfascinatingcomputergraphics.
Benoit
MandelbrotwasaPolish/French/American
mathem
atician
whohas
spentmost
ofhiscareer
attheIB
MWatsonResearchCenterin
YorktownHeights,
N.Y
.Hecoined
theterm
fractalandpublished
averyinfluentialbook
,TheFractal
Geometry
ofNature,in
1982.Anim
ageofthenow
famousMandelbrotsetappeared
onthecoverofScien
tificAmericanin
1985.Thiswas
abou
tthetimethatcomputer
graphicaldisplayswerebecomingwidelyavailable.Since
then,theMandelbrotset
hasstim
ulateddeepresearchtopicsin
mathem
atics
andhasalsobeenthebasisfor
anuncountable
number
ofgraphicsprojects,
hardwaredem
os,
andWeb
pages.
Toget
inthemoodfortheMandelbrotset,consider
theregionin
thecomplex
planeof
trajectories
generatedbyrepeatedsquaring,
z k+1=
z2 k,
k=
0,1,...
For
whichinitialvalues
z 0does
this
sequence
remain
bounded
ask→
∞?
Itis
easy
toseethatthis
setis
simply
theunit
disc,
|z 0|<
=1,
show
nin
figure
13.1.
If|z 0
|<=
1,thesequence
z kremainsbounded.Butif|z 0
|>1,thesequence
isunbou
nded.Theboundary
oftheunit
discis
theunit
circle,|z 0
|=1.Thereis
nothingverydiffi
cult
orexcitinghere.
Thedefinition
istheMandelbrotsetis
only
slightlymore
complicated.
Itinvolves
repeatedly
addingin
theinitialpoint.
TheMandelbrotsetistheregionin
thecomplexplaneconsistingofthevalues
z 0forwhichthetrajectories
defi
ned
by
z k+1=
z2 k+z 0,
k=
0,1,...
remain
bou
nded
atk→
∞.That’sit.That’stheentire
defi
nition.It’s
amazing
thatsuch
asimple
definitioncanproduce
such
fascinatingcomplexity.
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
163
164
Chapter13
.Mandelbrot
Set
−1.5
−1−0.5
00.5
11.5
−1.5−1
−0.50
0.51
1.5
Figure
13.1.Theunit
discis
shownin
red.Theboundary
issimply
the
unitcircle.Thereis
nointricate
fringe.
−2−1
.5−1
−0.5
00.
51
−1.5−1
−0.50
0.51
1.5
Figure
13.2.TheMandelbrotsetis
shownin
red.Thefringe
just
outside
theset,shownin
black,is
aregionofrich
structure.
165
Figure
13.3.Twotrajectories.z0
=.25-.54igeneratesacycleoflength
four,
whilenearbyz0
=.22-.54igeneratesanunbounded
trajectory.
Figure
13.2
show
stheoverallgeometry
oftheMandelbrotset.
How
ever,this
view
does
nothavetheresolutionto
show
therichly
detailed
structure
ofthefringe
just
outsidetheboundary
oftheset.
Infact,thesethastinyfilaments
reachinginto
thefringeregion,even
thoughthefringeappears
tobesolidblack
inthefigure.It
hasrecentlybeenprovedthattheMandelbrotsetismathem
aticallyconnected,but
theconnectedregionis
sometim
esso
thin
thatwecannotresolveit
onagraphics
screen
oreven
compute
itin
areasonable
amountoftime.
Toseehow
thedefinitionworks,
enter
z0
=.25-.54i
z=
0
into
Matlab.Then
use
theup-arrow
key
torepeatedly
execute
thestatement
z=
z^2
+z0
Thefirstfew
lines
ofoutputare
0.2500-
0.5400i
0.0209-
0.8100i
-0.4057-
0.5739i
0.0852-
0.0744i
0.2517-
0.5527i
...
Thevalues
eventuallysettle
into
acycle
0.2627-
0.5508i
0.0156-
0.8294i
-0.4377-
0.5659i
0.1213-
0.0446i
166
Chapter13
.Mandelbrot
Set
0.2627-
0.5508i
...
This
cyclerepeats
forever.Thetrajectory
remainsbounded
.This
tellsusthatthe
startingvaluevalue,
z0=
.25-.54i,is
intheMandelbrotset.
Thesamecycleis
show
nin
theleft
halfoffigure
13.3.
Ontheother
hand,start
with
z0
=.22-.54i
z=
0
andrepeatedly
execute
thestatement
z=
z^2
+z0
Youwillsee
0.2200-
0.5400i
-0.0232-
0.7776i
-0.3841-
0.5039i
0.1136-
0.1529i
0.2095-
0.5747i
...
Then,after24iterations,
...
1.5708-
1.1300i
1.4107-
4.0899i
-14.5174-12.0794i
6.5064e+001+3.5018e+002i
-1.1840e+005+4.5568e+004i
Thetrajectory
isblowinguprapidly.After
afew
more
iterations,thefloatingpoint
numbersoverflow
.Sothisz0is
notin
theMandelbrotset.
Thesameunbou
nded
trajectory
isshow
nin
therighthalf
offigure
13.3.
Weseethatthefirstvalue,
z0
=.25-.54i,is
intheMandelbrotset,
whilethesecondvalue,
z0
=.22-.54i,
whichis
nearby,
isnot.
Thealgorithm
doesn’t
haveto
wait
untilzreachsfloatingpointoverflow
.As
soonaszsatisfies
abs(z)>=
2
subsequentiterationswillessentialsquare
thevalueof|z|
andit
willbeh
avelike
22k
.Try
ityourself.
Putthesestatements
ononeline.
z0
=...
z=
0;
whileabs(z)<
2
z=
z^2+z0;
disp(z),
end
167
Use
theuparrow
andbackspace
keysto
retrievethestatementandchangez0to
differentvalues
near.25-.54i.
Ifyouhaveto
hit
<ctrl>-cto
breakou
tofan
infiniteloop,then
z0isin
theMandelbrotset.
Ifthewhileconditioniseventually
falseandtheloopterm
inateswithoutyourhelp,then
z0is
notin
theset.
Thenumber
ofiterationsrequired
forzto
escapethediscofradius2provides
thebasisforshow
ingthedetailin
thefringe.
Let’s
addaniterationcounterto
the
loop.A
quantity
wecalldepthspecifies
themaxim
um
interationcountan
dthereby
determines
boththelevelofdetailandtheoverallcomputationtime.
Typicalvalues
ofdepthare
severalhundredorafew
thousand.
z0
=...
z=
0;
k=
0;
whileabs(z)<
2&&
k<
depth
z=
z^2+z0;
k=
k+
1;
end
Themaxim
um
valueof
kis
depth.If
thevalueof
kis
less
thandepth,then
z0
isoutsidetheset.
Largevalues
ofkindicate
thatz0is
inthefringe,
close
tothe
bou
ndary.Ifkreaches
depththen
z0is
declaredto
beinsidetheMandelbrotset.
Hereisasm
alltable
ofiterationcounts
sz0ranges
over
complexvalues
near
0.22-0.54i.Wehavesetdepth=
512.
0.2050.2100.2150.2200.2250.2300.2350.2400.245
-0.520
512
512
512
512
512
512
44
512
512
-0.525
512
512
512
512
512
36
51
512
512
-0.530
512
512
512
512
35
31
74
512
512
-0.535
512
512
512
512
26
28
57
512
512
-0.540
512
139
113
26
24
73
56
512
512
-0.545
512
199
211
21
22
25
120
512
512
-0.550
33
25
21
20
20
25
63
85
512
-0.555
34
20
18
18
19
21
33
512
512
-0.560
62
19
17
17
18
33
162
40
344
Weseethatabouthalfof
thevalues
are
less
thandepth;they
correspondto
points
outsideof
theMandelbrotset,
inthefringeneartheboundary.Theother
halfof
thevalues
are
equalto
depth,correspondingto
points
thatare
regarded
asin
the
set.
Ifwewereto
redothecomputationwithalarger
valueof
depth,theentries
thatare
less
than512in
this
table
would
notchange,
butsomeoftheentriesthat
are
now
capped
at512mightincrease.
Theiterationcounts
canbeusedasindices
into
anRGB
colormapof
size
depth-by-3.Thefirstrow
ofthis
mapspecifies
thecolorassigned
toanypoints
onthez0gridthatlieoutsidethediscofradius2.Thenextfew
rowsprovidecolors
forthepoints
onthez0gridthatgenerate
trajectories
thatescapequickly.The
last
row
ofthemapis
thecolorof
thepoints
thatsurvivedepthiterationsandso
are
intheset.
168
Chapter13
.Mandelbrot
Set
Themapusedin
figure
13.2
emphasizesthesetitselfan
ditsboundary.The
maphas12rowsofwhiteat
thebeginning,on
erow
ofdark
redattheend,and
black
inbetween.Im
ages
thatem
phasize
thestructure
inthefringeareachieved
when
thecolormapvaries
cyclicly
over
afew
dozencolors.Oneoftheexercises
asksyou
toexperim
entwithcolormaps.
Figure
13.4.Im
provingresolution.
Array
operations.
Ourscriptmandelbrot_recapshow
show
Matlabarray
arithmeticoperatesagrid
ofcomplexnumberssimultaneouslyandaccumulatesan
arrayof
iterationcounters,
producingim
ages
likethose
infigure
13.4
Thecodebeginsbydefiningtheregion
inthecomplexplaneto
besampled.A
step
size
of0.05gives
thecoarseresolution
show
non
therightin
thefigure.
x=
0:
0.05:
0.80;
y=
x’;
Thenextsectionof
codeusesanelegant,buttricky,
bitofMatlabindexing
know
nasTony’s
Trick.Thequantities
xandyare
one-dim
ensionalrealarray
sof
length
n,on
eacolumnvectorandtheother
arow
vector.
Wewantto
create
atw
o-
dim
ensionaln-by-n
array
withelem
ents
form
edfrom
allpossible
sumsof
elem
ents
from
xandy.
z k,j=
xk+
y ji,
i=
√ −1,k,j
=1,...,n
Thiscanbedonebygeneratingavectoreoflength
nwithallelem
ents
equalto
one.
Then
thequantity
x(e,:)isatw
o-dim
ensionalarray
form
edbyusingx,whichisthe
sameasx(1,:),ntimes.Sim
ilarly,
y(:,e)is
atw
o-dim
ensionalarrraycontaining
ncopiesofthecolumnvectory.
169
n=
length(x);
e=
ones(n,1);
z0
=x(e,:)
+i*y(:,e);
Ifyoufindithard
toremem
ber
Ton
y’sindexingtrick,thefunctionmeshgrid
does
thesamethingin
twostep
s.
[X,Y]=
meshgrid(x,y);
z0
=X
+i*Y;
Now
initialize
twomore
array
s,on
eforthecomplexiteratesan
doneforthe
counts.
z=
zeros(n,n);
c=
zeros(n,n);
HereistheMandelbrotiterationrepeateddepthtimes.W
itheach
iterationwealso
keeptrack
oftheiteratesthatare
stillwithin
thecircle
ofradius2.
depth=
32;
fork
=1:depth
z=
z.^2
+z0;
c(abs(z)<
2)
=k;
end
Weare
now
finished
withz.Theactualvalues
ofzare
notim
portant,
only
the
counts
are
needed
tocreate
theim
age.
Ourgridissm
allenoughthatwecanactually
printoutthecounts
c.
c
Theresultsare
c= 32
32
32
32
32
32
11
76
54
33
32
22
32
32
32
32
32
32
32
96
54
33
32
22
32
32
32
32
32
32
32
32
75
43
33
22
2
32
32
32
32
32
32
32
32
27
54
33
32
22
32
32
32
32
32
32
32
32
30
64
33
32
22
32
32
32
32
32
32
32
32
13
74
33
32
22
32
32
32
32
32
32
32
32
14
75
33
22
22
32
32
32
32
32
32
32
32
32
17
43
32
22
2
32
32
32
32
32
32
32
16
818
43
32
22
2
32
32
32
32
32
32
32
11
65
43
32
22
2
32
32
32
32
32
32
32
19
65
43
22
22
2
32
32
32
32
32
32
32
23
84
43
22
22
2
32
32
32
19
11
13
14
15
14
43
22
22
22
22
32
12
32
77
714
64
32
22
22
2
12
98
75
55
17
43
32
22
22
1
32
76
55
44
43
32
22
22
11
17
75
44
44
33
22
22
22
11
170
Chapter13
.Mandelbrot
Set
Weseethatpoints
intheupper
left
ofthegrid,withfairly
smallinitialz0values,
havesurvived
32iterationswithoutgoingoutsidethecircle
ofradiustw
o,while
points
inthelower
right,withfairly
largeinitialvalues,havelasted
only
oneortw
oiterations.
Theinterestinggridpoints
are
inbetween,they
are
onthefringe.
Now
comes
thefinalstep,makingtheplot.
Theimagecommanddoes
the
job,even
thoughthisisnotan
imagein
theusualsense.Thecountvalues
incare
usedasindices
into
a32-by-3
array
ofRGB
colorvalues.In
this
example,thejet
colorm
apisreversedto
givedark
redas
itsfirstvalue,
pass
throughshades
ofgreen
andyellow,andfinishwithdark
blueasits32-ndandfinalvalue.
image(c)
axisimage
colormap(flipud(jet(depth)))
Exercisesask
youto
increase
theresolutionbydecreasingthestep
size,thereby
producingtheother
halfof
figure
13.4,to
investigate
theeff
ectofchangingdepth,
andto
invesigate
other
colormaps.
MandelbrotGUI
Theexmtoolbox
functionmandelbrotis
yourstartingpointforexplorationofthe
Mandelbrotset.
Withnoarguments,thestatement
mandelbrot
provides
thumbnailiconsofthetw
elve
regionsfeaturedin
this
chapter.
Thestate-
ment
mandelbrot(r)
withrbetween1and12startswithther-thregion.Thestatement
mandelbrot(center,width,grid,depth,cmapindx)
explorestheMandelbrotsetin
asquare
regionof
thecomplexplanewiththespeci-
fied
centerandwidth,usingagrid-by-gridgrid,an
iterationlimitofdepth,and
thecolormapnumber
cmapindx.Thedefaultvalues
oftheparametersare
center=
-0.5+0i
width=
3
grid=
512
depth=
256
cmapindx=
1
Inother
words,
mandelbrot(-0.5+0i,3,
512,
256,
1)
generatesfigure
13.2,butwiththejetscolormap.Changingthelast
argument
from
1to
6generatestheactualfigure
13.2
withthefringecolormap.Onmy
laptop,thesecomputationseach
takeabouthalfasecond.
Asimple
estimate
oftheexecutiontimeis
proportionalto
171
grid^2*
depth
Sothestatement
mandelbrot(-0.5+0i,3,
2048,
1024,
1)
could
take
(2048/512)2
·(1024/256)=
64
times
aslongasthedefault.How
ever,this
isanoverestimate
andtheactualexe-
cutiontimeis
about11
seconds.
Most
ofthecomputationaltimerequired
tocompute
theMandelbrotsetis
spentupdatingtw
oarray
szandkzbyrepeatedly
executingthestep
z=
z.^2
+z0;
j=
(abs(z)
<2);
kz(j)=
d;
Thiscomputationcanbecarriedoutfaster
bywritingafunctionmandelbrot_step
inC
andcreatingasaMatlabexecutableor
c-mex
file.Differentmachines
and
operatingsystem
srequiredifferentversionsof
amex
file,so
youshould
seefiles
withnames
likemandelbrot_step.mexw32andmandelbrot_step.glnx64in
the
exmtoolbox
.ThemandelbrotguiturnsontheMatlabzoom
feature.Themouse
pointer
becomes
asm
allmagnifyingglass.You
canclickandrelease
onapointto
zoom
by
afactoroftw
o,oryoucanclickanddragto
delineate
anew
region.
Themandelbrotguiprovides
severaluicontrols.Try
theseasyoureadalong.
Thelistboxatthebottom
oftheguiallow
syouto
select
anyof
thepredefined
region
sshow
nin
thefigu
resin
this
chapter.
depth.Increase
thedepth
byafactorof3/2or
4/3.
grid.
Refinethegrid
byafactorof3/2or4/3.
Thedep
than
dgrid
size
are
alwaysapow
eroftw
oor
threetimes
apow
eroftw
o.Twoclicksonthedepthor
gridbuttondoublestheparameter.
color.
Cyclethroughseveralcolormaps.
jetsandhotsare
cyclic
repetitions
ofshort
copiesof
thebasicMatlabjetandhotcolormaps.
cmykcycles
through
eightbasiccolors,blue,
green,red,cyan
,magenta,yellow,gray,
andblack.fringe
isanoncyclic
mapusedforim
ages
likefigure
13.2.
exit.Close
thegu
i.
TheMandelbrotsetis
selfsimilar.
Smallregionsin
thefringerevealfeatures
thatare
similarto
theoriginalset.
Figure
13.6,whichwehavedubbed
“Mandelbrot
172
Chapter13
.Mandelbrot
Set
Figure
13.5.Thefiguresin
this
chapter,
andthepredefi
ned
regionsin
our
mandelbrotprogram,show
theseregionsin
thefringe
just
outsidetheMandelbrot
set.
Junior”,is
oneexample.Figure
13.7,whichwecallthe“‘Plaza”,usesourflag
colorm
apto
revealfinedetailin
red,whitean
dblue.
Theportionoftheboundary
oftheMandelbrotsetbetweenthetw
olarge,
nearlycircularcentralregionsis
know
nas“TheValleyof
theSeahorses”.
Fig-
ure
13.8
show
stheresult
ofzoomingin
onthepeninsula
betweenthetw
onearly
circularregionsof
theset.
Thefigure
canbegenerateddirectlywiththecommand
mandelbrot(-.7700-.1300i,0.1,1024,512)
Wedecided
tonametheim
agein
figure
13.9
the“WestW
ing”because
itresembles
theX-w
ingfighterthatLukeSkywalker
fliesin
StarWars
andbecause
itislocated
neartheleftmost,orfarwestern,portionof
theset.
Themagnificationfactoris
arelativelymodest104,so
depthdoes
notneedto
beverylarge.
Thecommandto
gen
erate
theWestW
ingis
mandelbrot(-1.6735-0.0003318i,1.5e-4,1024,160,1)
Oneof
thebestknow
nexamplesof
selfsimilarity,the“Buzzsaw”,isshow
nin
figure
13.11.It
canbegeneratedwith
mandelbrot(0.001643721971153+0.822467633298876i,...
173
4.0e-11,1024,2048,2)
Takingwidth=
4.0e-11correspondsto
amagnificationfactorofalm
ost
101
1.To
appreciate
thesize
ofthis
factor,
iftheoriginalMandelbrotsetfillsthescreen
onyou
rcomputer,
theBuzzsaw
issm
aller
thantheindividualtransistors
inyou
rmachine’smicroprocessor.
Wecallfigure
13.12“Nebula”because
itremindsusofinterstellardust.It
isgeneratedby
mandelbrot(0.73752777-0.12849548i,4.88e-5,1024,2048,3)
Thenextthreeim
ages
are
obtained
bycarefullyzoomingonon
elocation.We
callthem
the“Vortex”,the“M
icrobug”,andthe“Nucleus”.
mandelbrot(-1.74975914513036646-0.00000000368513796i,...
6.0e-12,1024,2048,2)
mandelbrot(-1.74975914513271613-0.00000000368338015i,...
3.75e-13,1024,2048,2)
mandelbrot(-1.74975914513272790-0.00000000368338638i,...
9.375e-14,1024,2048,2)
Themost
intricate
andcolorfulim
ageamongourexamplesis
figure
13.16,
the“Geode”.It
invo
lves
afinegridandalargevalueofdepthandconsequently
requires
afew
minutesto
compute.
mandelbrot(0.28692299709-0.01218247138i,6.0e-10,2048,4096,1)
Theseexamplesare
just
atinysamplingof
thestructure
oftheMandelbrot
set.
Further
Rea
ding
Wehighly
recommend
arealtimefractalzoomer
called
“XaoS”,developed
by
ThomasMarsh,JanHubicka
andZoltanKovacs,assistedbyaninternationalgroup
ofvolunteers.
See
http://wmi.math.u-szeged.hu/xaos/doku.php
IfyouareexpertatusingyourWeb
browserandpossibly
dow
nloadingan
obscure
video
codec,takealook
attheW
ikiped
iavideo
http://commons.wikimedia.org/wiki/...
Image:Fractal-zoom-1-03-Mandelbrot_Buzzsaw.ogg
It’s
terrificto
watch,butitmay
bealotoftrouble
togetworking.
Rec
ap
%%Mandelbrot
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
174
Chapter13
.Mandelbrot
Set
%introduced
in
the
Mandelbrot
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %mandelbrot_recap
%edit
mandelbrot_recap
%publish
mandelbrot_recap
% %Related
EXM
programs
% %mandelbrot
%%
Define
the
region.
x=
0:
0.05:
0.8;
y=
x’;
%%
Create
the
two-dimensional
complex
grid
using
Tony’s
indexing
trick.
n=
length(x);
e=
ones(n,1);
z0
=x(e,:)
+i*y(:,e);
%%
Or,
do
the
same
thing
with
meshgrid.
[X,Y]=
meshgrid(x,y);
z0
=X
+i*Y;
%%
Initialize
the
iterates
and
counts
arrays.
z=
zeros(n,n);
c=
zeros(n,n);
%%
Here
is
the
Mandelbrot
iteration.
depth=
32;
fork
=1:depth
z=
z.^3
+z0;
c(abs(z)<
2)
=k;
end
%%
Create
an
image
from
the
counts.
c image(c)
axisimage
%%
Colors
colormap(flipud(jet(depth)))
175
Exe
rcises
13.1
Explore.Use
themandelbrotguito
findsomeinterestingregionsthat,
asfar
asyou
know
,havenever
beenseen
before.Givethem
yourow
nnames.
13.2
depth.Modifymandelbrot_recapto
reproduce
ourtable
ofiterationcounts
forx
=.205:.005:.245andy=
-.520:-.005:-.560.First,use
depth=
512.
Then
use
larger
values
ofdepthandseewhichtable
entrieschange.
13.3
Resolution.Reproduce
theim
agein
therighthalfoffigure
13.4.
13.4
Big
picture.Modifymandelbrot_recapto
displaytheentire
Mandelbrotset.
13.5
Colormaps.
Investigate
colormaps.
Use
mandelbrot_recapwithasm
aller
step
size
andalargevalueofdepthto
produce
anim
age.
Findhow
mandelbrot
computesthecyclic
colormapscalled
jets,hotsandsepia.Then
use
those
maps
onyourim
age.
13.6
p-thpower.In
either
mandelbrot_recaporthemandelbrotgui,changethe
pow
erin
theMandelbrotiterationto
z k+1=
zp k+
z 0
forsomefixed
p�=
2.If
youwan
tto
tryprogrammingHandle
Graphics,
adda
buttonto
mandelbrotthatlets
you
setp.
13.7
Too
much
magn
ification.W
hen
thewidth
oftheregiongetsto
besm
aller
than
about10−15,ourmandelbrotguidoes
notwork
verywell.W
hy?
13.8
Spin
thecolormap.Thismightnotwork
verywellon
yourcomputerbecause
itdep
endsonwhatkindofgraphicshardwareyouhave.
When
youhaveaninteresting
regionplotted
inthefigure
window
,bringupthecommandwindow
,resize
itso
that
itdoes
notcoverthefigure
window
,andenterthecommand
spinmap(10)
Iwon’ttryto
describewhathappens–youhaveto
seeitforyourself.
Theeff
ectis
most
dramaticwiththe“seahorses2”region.Enter
helpspinmap
formore
details.
176
Chapter13
.Mandelbrot
Set
Figure
13.6.
Region
#2,“MandelbrotJunior”.
Thefringe
aroundthe
Mandelbrotsetin
self
similar.
Small
versionsofthesetappearatall
levels
of
magn
ification.
Figure
13.7.Region#3,“Plaza”,withtheflagcolormap.
177
Figure
13.8.Region#4.“Valley
oftheSeahorses”.
Figure
13.9.Region#5.Our“WestWing”
islocatedjust
offtherealaxis
inthethin
farwestern
portionoftheset,nearreal(z)
=-1.67.
178
Chapter13
.Mandelbrot
Set
Figure
13.10.Region#6.“DuelingDragons”.
Figure
13.11.Region#7.The“Buzzsaw”requires
amagn
ificationfactor
of10
11to
revealatinycopyoftheMandelbrotset.
179
Figure
13.12.Region#8.“Nebula”.Interstellardust.
Figure
13.13.Region#9.A
vortex,notfarfrom
theWestWing.
Zoom
inononeofthecircular“microbugs”neartheleftedge.
180
Chapter13
.Mandelbrot
Set
Figure
13.14.
Region#10.A
1013magn
ificationfactorreveals
a“mi-
crobug”
within
thevortex.
Figure
13.15.Region#11.Thenucleusofthemicrobug.
181
Figure
13.16.Region#12.“Geode”.This
colorfulim
age
requires
a2048-
by-2048grid
anddepth
=8192.
182
Chapter13
.Mandelbrot
Set
Chapter14
Sudoku
Humanslookforpatterns,
butmachines
use
backtracking.
Thefascinationthatpeople
haveforsolvingSudoku
puzzleswithoutacom-
puterderives
from
thediscoveryandmasteryofamyriadof
subtlecombinations
andpatternsthatprovidetipstoward
thesolution.TheWeb
hashundredsof
sites
describingthesepatterns,
whichhavenames
like“h
idden
quads”,“X-w
ing”an
d“squirmbag”.
Itisnotourintentionto
describeaMatlabprogram
thatduplicatesthesehu-
manpatternrecognitioncapabilities.
Instead,ourprogram,likemost
other
Sudoku
computercodes,takes
averydifferentapproach,onethatrelies
onthemachine’s
alm
ost
limitless
capacity
tocarry
outbrute
forcetrialand
error.
Weuse
only
onepattern,singletons,
together
withafundamentalcomputerscience
technique,
recursivebacktracking.
IntheSudokuworld,backtrackingisregardingas“guessing”an
disconsidered
badform
.Butas
acomputationaltechnique,
itis
easy
tounderstand,straightfor-
wardto
implementan
dguaranteed
tosucceed.
Inthis
chapter,wewilldescribefiveMatlabprogramsforexperim
ents
with
Sudoku.
•sudoku.Aninteractiveprogram
thatlets
you
follow
thecomputersolution
process.
•sudoku_basic.Thesamesolutionprocess
assudoku,butwithoutthegraph-
icsinterface.
Much
easier
toread,an
dmuch
faster
ifyou
just
wantthe
solution.
•sudoku_all.
Enumerate
all
solutionsto
aSudoku
puzzle.
Avalid
puzzle
should
haveonly
onesolution.
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
183
184
Chapter14
.Sudoku
•sudoku_assist.
Aninteractiveprogram
thatlets
you
controlthesolution
process,asyou
would
byhandonpaper.
•sudoku_puzzle.A
collectionof
interestingpuzzles.
Sudoku
isactuallyanAmericaninvention.It
firstappeared,withthename
Number
Place,in
theDellPuzzle
Magazinein
1979.
Thecreatorwas
probably
How
ardGarns,
anarchitectfrom
Indianapolis.
AJapanesepublisher,Nikoli,took
thepuzzle
toJapanin
1984andeventuallygaveit
thenameSudoku,whichis
akindof
kanjiacronym
for“numbersshould
besingle,unmarried.”
TheTim
esof
Londonbeganpublishingthepuzzle
intheUK
in2004andit
wasnotlongbefore
itspreadback
totheUSan
daroundtheworld.
8 3 4 1
1 5 9 2
6 7 2 3
1 8 3 4
1 5 9
6 7 2 3
1 8 3 4
8 1 5 9
3 6 7 2
Figure
14.1.A
Sudoku
puzzle
featuringLo-Shu,themagicsquare
oforder
3.
You
probably
alreadyknow
therulesofSudoku.Figure
20isan
exampleofan
initial9-by-9
grid,withafew
specified
digitsknow
nastheclues.This
particular
puzzle
reflects
ourinterest
inmagic
squares.
Itis
thesum
ofthetw
omatrices
show
nin
figure
20.
Thematrix
ontheleft
isgenerated
bythefunctionskron,
which
computessomethingknow
nastheKroneckerproduct,an
dmagic,which
generatesmagic
squares.
X=
kron(eye(3),magic(3))
Inthis
case,aKroneckerproduct
involvingtheidentity
matrix
producesablock
matrix
withthreecopiesofmagic(3)onthediagonal.
Byitself,this
array
isnot
avalidSudokupuzzle,because
thereis
more
thanoneway
tocomplete
it.The
solutionis
notunique.
Inorder
togetapuzzle
withauniquesolution,wehave
added
thematrix
ontherightin
figure
20.
Figure
14.3
isthefinalcompleted
grid.
Each
row,each
column,and
each
major3-by-3
block,must
contain
exactly
thedigits1through9.
Incontrast
to
185
8 3 4
1 5 9
6 7 2
8 3 4
1 5 9
6 7 2
8 3 4
1 5 9
6 7 21
2
3
1
3
1
8
3
Figure
14.2.OurLo-Shubasedpuzzle
isthesum
ofthesetwomatrices.
8 3 4 2 9 7 5 6 1
1 5 9 4 8 6 7 2 3
6 7 2 3 1 5 9 4 8
5 6 1 8 3 4 2 9 7
7 2 3 1 5 9 4 8 6
9 4 8 6 7 2 3 1 5
2 9 7 5 6 1 8 3 4
4 8 6 7 2 3 1 5 9
3 1 5 9 4 8 6 7 2
Figure
14.3.Thecompletedpuzzle.Thedigitshave
been
inserted
sothat
each
row,each
column,andeach
major3-by-3blockcontains1through
9.
The
originalclues
are
shownin
blue.
magic
squaresandother
numeric
puzzles,
noarithmetic
isinvolved
.Theelem
ents
inaSudoku
gridcould
just
aswellbeninelettersofthealphabet,oranyother
distinct
symbols.
186
Chapter14
.Sudoku
1
2
3
4
Figure
14.4.Shidoku
1 2
4
3 4
3
3 4
4 2 1 3
2
4 3 1 1 2
2
1 2
1 3 4
Figure
14.5.Candidates
1 2
4
4 3
3 4
4 2 1
2
4 3 1 1 2
2
1 2
1 3 4
Figure
14.6.Insert
singleton
1 2 4 3
3 4 2 1
4 3 1 2
2 1 3 4
Figure
14.7.Solution
Shidoku
Toseehow
oursudokuprogram
works,
wecan
use
Shidoku
instead
ofSudoku.
“Shi”
isJapanesefor“four”.Thepuzzlesuse
a4-by-4
gridandarealm
ost
trivialto
solvebyhand.Figure
20isou
rfirstShidoku
puzzle
andthenextthreefiguresshow
stepsin
itssolution.In
figure
20,thepossible
entries,
orcandidates,
are
show
nby
smalldigits.
For
example,row
twocontainsa3an
dcolumnonecontainsa1so
the
candidatesin
position(2,1)are
2and4.
Fourof
thecellshaveonly
onecandidate
each.Theseare
thesingletons,
show
nin
red.In
figure
20,wehaveinserted
the
singleton3in
the(4,1)cellan
drecomputedthecandidates.
Infigure
14.7,wehave
inserted
theremainingsingletonsas
they
aregeneratedto
complete
thesolution.
187
1
2
3
4
Figure
14.8.diag(1:4)
1 3 4
2
4
2
3
3 4 2 1 4
1 3
2
4
1 4 3 1 2
2
3 1 3 1 2 4
Figure
14.9.Nosingletons.
1 32
3
4
Figure
14.10.Backtrack
step.
1 3 4 2
4 2 1 3
2 4 3 1
3 1 2 4
Figure
14.11.Solutionis
notunique.
Theinputarray
forfigure
14.8
isgeneratedbytheMatlabstatement
X=
diag(1:4)
Asfigure
14.9
show
s,thereare
nosingletons.
So,weem
ployabasiccomputerscience
technique,
recursivebacktracking.Weselect
oneof
theem
pty
cellsandtentatively
insert
oneofitscandidates.
Allof
theem
pty
cellshavetw
ocandidates,
sowepick
thefirston
eandtentatively
insert
a3in
cell(2,1).
This
createsanew
puzzle,
show
nin
figure
14.10.Ourprogram
isthen
called
recursively.
Inthisexample,the
new
puzzle
iseasily
solved
andtheresult
isshow
nin
figure
14.11.How
ever,the
solutiondep
endsuponthechoices
thatwemadebefore
therecursivecall.Other
choices
canleadto
differentsolutions.
Forthis
simple
diagonalinitialcondition,
188
Chapter14
.Sudoku
thesolutionis
notunique.
Therearetw
opossible
solutions,
whichhappen
tobe
matrix
transposesofeach
other.
Y=
shidoku(diag(1:4))
Y=
14
23
32
41
41
32
23
14
Z=
shidoku(diag(1:4)’)’
Z=
13
42
42
13
24
31
31
24
Existen
ceanduniquen
ess
Mathem
aticiansare
alwaysconcerned
abou
texistence
anduniquenessin
thevarious
problemsthatthey
encounter.
For
Sudoku,neither
existence
noruniquenesscan
bedetermined
easily
from
theinitialclues.It
wou
ldbeveryfrustratingifapuzzle
withnosolution,or
withmore
thanon
esolution,wereto
show
upin
yourdaily
new
spaper.
Uniquenessis
anelusiveproperty.
Infact,most
descriptionsof
Sudoku
do
notspecifythattherehasto
beexactly
onesolution.Theonly
way
thatIknow
tocheckuniquen
essis
toexhaustivelyenumerate
allpossibilities.
For
ourmagicsquare
puzzlein
figure
20,ifwewereto
replace
the1in
the(9,1)
cellbya5,6or
7,therow,columnandblock
conditionswould
stillbesatisfied.
Itturnsoutthata5would
produce
another
validpuzzle
withauniquesolution,
a6would
produce
apuzzle
withtw
opossible
solutions,
anda7would
produce
apuzzle
withnosolution.Anexercise
asksyou
toverifytheseclaim
s.It
takes
sudoku_allalittle
over
halfan
houron
mylaptopto
determinethat
themagic
square
puzzle
generatedby
X=
kron(eye(3),magic(3))
withouttheaddionalentrieson
rightin
figure
20has283576solutions.
Thepuzzle
generatedby
X=
sudoku_puzzle(13)
hasan
interestingspiralpattern.Butthereareover
300solutions.
Anexercise
asks
you
tofindoutexactly
how
many.
189
Anumber
ofoperationsonaSudokugridcanchangeitsvisualap
pearance
withoutchangingitsessentialcharacteristics.Allofthevariationsare
basicallythe
samepuzzle.Theseequivalence
operationscanbeexpressed
asarray
operationsin
Matlab.Forexample
p=
randperm(9)
z=
find(X
>0)
X(z)=
p(X(z))
permutesthedigitsrepresentingtheelem
ents.Other
operationsinclude
X’
rot90(X,k)
flipud(X)
fliplr(X)
X([4:91:3],:)
X(:,[randperm(3)4:9])
Thealgorithm
Ifwedonotcountthecomments
andGUI,sudoku.minvolves
less
than40lines
of
code.
Theoutlineofthemain
program
is:
•Fillin
allsingletons.
•Exitifacellhasnocandidates.
•Fillin
atentativevalueforanem
pty
cell.
•Calltheprogram
recursively.
sudokuandsudokubasic
Hereis
theheader
forsudoku_basic.
function[X,steps]=
sudoku_basic(X,steps)
%SUDOKU_BASIC
Solve
the
Sudoku
puzzle
using
recursive
backtracking.
%sudoku_basic(X),
for
a9-by-9
array
X,
solves
the
Sudoku
puzzle
for
X
%without
providing
the
graphics
user
interface
from
sudoku.m
%[X,steps]
=sudoku_basic(X)
also
returns
the
number
of
steps.
%See
also
sudoku,
sudoku_all,
sudoku_assist,
sudoku_puzzle.
Thisfirstbitofcodechecksnargin,thenumber
ofinputarguments,andinitializes
thestep
counteronthefirstentry.
Anysubsequentrecursivecallswillhaveasecond
argument.
if
nargin
<2
steps=
0;
end
190
Chapter14
.Sudoku
Thesecomments
describetheprimary
variablesin
theprogram.
%C
is
the
array
of
candidate
values
for
each
cell.
%N
is
the
vector
of
the
number
of
candidates
for
each
cell.
%s
is
the
index
of
the
first
cell
with
the
fewest
candidates.
Theprogram
fillsin
anysingletons,
oneat
atime.
Thecandidatesarerecomputed
after
each
step.Thissectionof
code,
byitself,cansolvepuzzlesthatdonotrequire
backtracking.Such
puzzlescanbeclassified
as“easy”.
[C,N]=
candidates(X);
whileall(N>0)&
any(N==1)
s=
find(N==1,1);
X(s)=
C{s};
steps=
steps
+1;
[C,N]=
candidates(X);
end
Iffillingin
thesingletonsdoes
notsolvethepuzzle,wereach
thefollow
ingsection
ofcodethatim
plements
backtracking.Thebacktrackinggeneratesmanyim
possi-
ble
configurations.
Therecursionis
term
inatedbyencounteringapuzzle
withno
solution.
if
all(N>0)
Y=
X;
s=
find(N==min(N),1);
fort
=[C{s}]
%Iterate
over
the
candidates.
X=Y;
X(s)=
t;
%Insert
atentative
value.
steps=
steps
+1;
[X,steps]=
sudoku_basic(X,steps);
%Recursive
call.
ifall(X(:)
>0)
%Found
asolution.
break
end
end
end
Allof
thebookkeepingrequired
bybacktrackingis
handledbytherecursivecall
mechanism
inMatlabandtheunderlingoperatingsystem
.Thekey
internalfunctionforsudokuiscandidates.Hereis
thepream
ble.
function[C,N]=
candidates(X)
%C
=candidates(X)
is
a9-by-9
cell
array
of
vectors.
%C{i,j}
is
the
vector
of
allowable
values
for
X(i,j).
%N
is
arow
vector
of
the
number
of
candidates
for
each
cell.
%N(k)
=Inf
for
cells
that
already
have
values.
191
Thefollow
ingstatementcreatesaninternalfunctionnamed
trithatcomputes
theindices
forblocks.
For
example,tri(1),tri(2)andtri(3)are
allequalto
1:3because
theblock
thatincludes
acellwithrow
orcolumnindex
equalto
1,2
or3hasrow
orcolumnindices
1:3.
tri=
@(k)
3*ceil(k/3-1)
+(1:3);
Hereis
thecore
ofcandidates.
C=
cell(9,9);
forj
=1:9
fori
=1:9
if
X(i,j)==0
z=
1:9;
z(nonzeros(X(i,:)))=
0;
z(nonzeros(X(:,j)))=
0;
z(nonzeros(X(tri(i),tri(j))))=
0;
C{i,j}=
nonzeros(z)’;
end
end
end
Foreach
empty
cell,thisfunctionstartswithz
=1:9andusesthenumeric
values
intheassociatedrow,columnandblock
tozero
elem
ents
inz.Thenonzerosthat
remain
are
thecandidates.
Forexample,consider
the(1,1)cellin
figure
20.We
start
with
z=
12
34
56
78
9
Thevalues
inthefirstrow
changezto
z=
10
00
56
78
9
Then
thefirstcolumnchanges
zto
z=
10
00
50
70
9
The(1,1)block
does
notmakeanyfurther
changes,so
thecandidatesforthis
cell
are
C{1,1}=
[1
57
9].
Thecandidatesfunctionconcludes
withthefollow
ingstatements.Thenum-
ber
ofcandidates,
N(i,j)forcell(i,j)is
thelength
ofC{i,j}.
Ifcell(i,j)
alreadyhasavalue,
then
X(i,j)is
nonzero,C{i,j}is
empty
andso
N(i,j)is
zero.ButwemaketheseN(i,j)infiniteto
distinguishthem
from
thecellsthat
signalanim
possible
puzzle.
N=
cellfun(@length,C);
N(X>0)=
Inf;
N=
N(:)’;
192
Chapter14
.Sudoku
Anex
ample,puzzle
number
one
Figure
14.12show
sthekey
stepsin
thesolutionof
ourLo-Shupuzzle.Theinitial
candidatesincludejust
onesingleton,thered5in
position(3,9).
Fivestep
swith
fivesingletonsallow
usto
fillin
thecellswithgreen
5’sand8’s.Wenow
consider
the
firstcellwithtw
ocandidates,
inposition(6,1).
Thefirststep
inthebacktracking
puts
atentative
6in
position(6,1).
Bluevalues
are
theinitialclues,cyanvalues
are
generatedbythebacktrackingan
dgreen
values
are
implied
bytheothers.
After
five
stepswereach
anim
possible
situationbecause
thereare
nocandidatesforposition
(6,2).
Try
again,witha7in
position(6,1).
Thispuzzleiseasily
completedbyfilling
insingletons.
Thetitleonthefinalsolutionsshow
sthatittook50step
s,including
theabortiveones
inthefirstbacktrack,to
reach
thesolution.
Referen
ces
[1]W
ikiped
ia’s
Mathem
atics
ofSudoku.
http://en.wikipedia.org/wiki/Mathematics_of_Sudoku
[2]GordonRoy
le,Minim
um
Sudoku.
http://mapleta.maths.uwa.edu.au/~gordon/sudokumin.php
[3]Nikoli.
http://www.nikoli.co.jp/en
[4]SudokuSquaresan
dChromaticPolynomials.
http://www.ams.org/notices/200706/tx070600708p.pdf
[5]Sudokuweb
site
modeled
after
Wikiped
ia.
http://www.sudopedia.org/wiki
[6]Ruud’s
Guide,
SudoCue.
http://www.sudocue.net/guide.php
Rec
ap
%%
Sudoku
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Sudoku
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %sudoku_recap
%edit
sudoku_recap
%publish
sudoku_recap
% %Related
EXM
programs
% %sudoku
%sudoku_all
%sudoku_assist
%sudoku_basic
%sudoku_puzzle
193
%%Disclaimer
%Our
Sudoku
chapter
and
Sudoku
program
depend
heavily
%upon
recursion,
which
cannot
be
done
by
this
script.
%%Sudoku
puzzle
incorporating
the
Lo-Shu
magic
square.
X=
kron(eye(3),magic(3))
C=
full(sparse([9
84
37
62
1],[1:4
6:9],
[1
23
13
18
3]))
X=
X+
C
%Also
available
as
X=
sudoku_puzzle(1);
%%Transforms
of
aSudoku
puzzle
T=
X;
p=
randperm(9);
z=
find(X
>0);
T(z)=
p(X(z))
X’
rot90(X,-1)
flipud(X)
fliplr(X)
X([4:91:3],:)
X(:,[randperm(3)4:9])
%%Candidates
%C
=candidates(X)
is
acell
array
of
vectors.
%C{i,j}
is
the
set
of
allowable
values
for
X(i,j).
C=
cell(9,9);
tri=
@(k)
3*ceil(k/3-1)
+(1:3);
forj
=1:9
fori
=1:9
if
X(i,j)==0
z=
1:9;
z(nonzeros(X(i,:)))=
0;
z(nonzeros(X(:,j)))=
0;
z(nonzeros(X(tri(i),tri(j))))=
0;
C{i,j}=
nonzeros(z)’;
end
end
end
C
%%First
singleton
and
first
empty.
194
Chapter14
.Sudoku
%N
=number
of
candidates
in
each
cell.
%s
=first
cell
with
only
one
candidate.
%e
=first
cell
with
no
candidates.
N=
cellfun(@length,C)
s=
find(X==0
&N==1,1)
e=
find(X==0
&N==0,1)
%%
Sudoku
puzzles
helpsudoku_puzzle
forp
=1:16
sudoku_puzzle(p)
end
Exe
rcises
14.1
xkcd.Solvethebinary
Sudokupuzzle
byRandalMunroein
theWeb
comic
stripxkcdat
http://xkcd.com/74
14.2
Solve.
Use
sudokuto
solveapuzzle
from
anew
spaper,magazine,
orpuzzle
book
.Use
sudoku_assistto
solvethesamepuzzle
byhand.
14.3
sudoku
puzzle.Theexmprogram
sudoku_puzzlegenerates16
differentpuzzles.
Thecomments
intheprogram
describetheoriginsof
thepuzzles.
How
manysteps
are
required
bysudoku_basicto
solveeach
ofthepuzzles?
14.4
Byhand.Use
sudoku_assistto
solveou
rLo-Shubasedpuzzle
byhand.
X=
sudoku_puzzle(1);
sudoku_assist(X)
14.5
Modifypuzzle
#1.Thefollow
ingprogram
modifies
thelower
left
handcluein
ourLo-Shubasedpuzzle.W
hathappenswitheach
oftheresultingpuzzles?
X=
sudoku_puzzle(1);
fort
=[0
15
67]
X(9,1)=
t;
sudoku_all(X)
end
195
14.6
Try
this.Try
touse
sudoku_assistto
solvethis
puzzle
byhand.
X=
sudoku_puzzle(1);
X(9,1)=
7;
sudoku_assist(X)
14.7
Puzzle
#3.How
manysolutionsdoes
sudoku_puzzle(3)have?
How
dothey
differ
from
each
other?
14.8
Puzzle
#13.How
manysolutionsdoes
sudoku_puzzle(13)have?
Inaddition
totheinitialclues,whatvalues
are
thesamein
allthesolutions?
14.9Allsolutions.
Whatarethedifferencesbetweensudoku_basicandsudoku_all?
14.10Combine.
Combinesudoku_allandsudoku_basicto
create
aprogram
that
returnsallpossible
solutionsandthenumber
ofstepsrequired
tofindeach
one.
Try
yourprogram
onsudoku_puzzle(13).W
hatis
theaveragenumber
ofsteps
required
?Hints:ifSisacellarray
whereeach
cellcontainsonenumeric
value,
then
s=
[S{:}]createsthecorrespondingvector.
And,helpmean.
14.11Searchstrategy
Insudoku_basic,thestatement
s=
find(N==min(N),1);
determines
thesearchstrategy.
Whathappenswhen
youchangethis
to
s=
find(N==min(N),’last’);
or
s=
find(X==0,’first’);
14.12Patterns.
Addsomehumanpuzzle
solvingtechniques
tosudoku.m.Thiswill
complicate
theprogram
andrequiremore
timeforeach
step,butshould
result
infewer
totalsteps.
14.13sudoku
alpha.In
sudoku.m,changeint2str(d)to
char(’A’+d-1)so
that
thedisplayusestheletters’A
’through’I’insteadofthedigits1through9.See
figure
14.13.Does
this
makeit
easier
orharder
tosolvepuzzlesbyhand.
14.14sudoku16.Modifysudoku.mto
solve16-by-16puzzleswith4-by-4
blocks.
196
Chapter14
.Sudoku
0
8 3 4 2
5
7 9
2
6
9
5 6
7 5
7 9
6
9
1
1 5 9
4 7 4 6
8
6
7 8
4 7
2 3
6
7 8
6 7 2 31 4 8 9
5
8
4 5
9
4 8 9
5
8
2
5
7 9
2
6
9
1 8 3 4 2
5
7 9
6
9
5 6
7
2
4 7 2
4 6
3
6
7 8 1 5 9
2
4 7 4 6
8
6
7 8
4 5
9
4 9
5
8 6 7 2 3
1 4 8 9
5
8
2
5
7 9
2
6
9
5 6
7 2
5
7 9
2
6
9
1 8 3 4
2
4 7
8
6
7 2
4 7 2
4 6
3
6
7
1 5 9
31 4 9
5
4 5
9
4 8 9
5
8 6 7 2
6
8 3 4 2
7 9
2
6
9
6
7 5
7 9
6
9
1
1 5 9
4 7 4 6
8
6
7 8
4 7
2 3
6
7
6 7 2 31 4 9
5
4 9
4 9
8
2
5
7 9
2
6
9
1 8 3 4 2
7 9
6
9
6
7
2
4 7 2
4 6
3
6
7
1 5 9 2
4 7 4 6
8
6
7
4 9
4 9
8 6 7 2 31 4 9
5
2
7 9
2
6
9
6
7 2
5
7 9
2
6
9
1 8 3 4
2
4 7
8
6
7 2
4 7 2
4 6
3
6
7
1 5 9
31 4 9
5
4 9
4 8 9
8 6 7 2
7
8 3 4 6 1
1 5 9 2
6 7 2 3 5 8
1 8 3 4
1 5 9
8 6 7 2 3 5
1 8 3 4
8 1 5 9
3 5 8 6 7 2
10
8 3 4 7 2 6
5 9 1
1 5 9
4 4 8
4 7
2 3
6
7
6 7 2 31 4 9
5
4 4
8
2
5
7 9
2
6
9
1 8 3 4 2
7 9
6
6
7
2
4 7 2
4 6
3
6
7
1 5 9 2
4 7 4 6
8
6
7
4 9
4 9
8 6 7 2 31 4
5
2
7 9
2
6
9
6
7 2
5
9
6
9
1 8 3 4
2
4 7
8
6
7 2
4 4 6
3
7
1 5 9
31 4 9
5
4 9
4 9
8 6 7 2
11
8 3 4 7 1
1 5 9 2
6 7 2 3 5 8
1 8 3 4
1 5 9
8 6 7 2 3 5
1 8 3 4
8 1 5 9
3 5 8 6 7 2
50
8 3 4 2 9 7 5 6 1
1 5 9 4 8 6 7 2 3
6 7 2 3 1 5 9 4 8
5 6 1 8 3 4 2 9 7
7 2 3 1 5 9 4 8 6
9 4 8 6 7 2 3 1 5
2 9 7 5 6 1 8 3 4
4 8 6 7 2 3 1 5 9
3 1 5 9 4 8 6 7 2
Figure
14.12.Thekeystepsin
thesolutionofourLo-Shupuzzle.
197
H C D A
A E I B
F G B C
A H C D
A E I
F G B C
A H C D
H A E I
C F G B
Figure
14.13.Use
theletters’A
’through
’I’insteadofthedigits1through
9.
198
Chapter14
.Sudoku
Chapter15
OrdinaryDifferen
tial
Equations
Mathem
aticalmodelsin
manydifferen
tfields.
System
sofdifferentialequationsform
thebasisofmathem
aticalmodelsin
awiderangeof
fields–from
engineeringan
dphysicalsciencesto
finance
andbiological
sciences.
Differentialequationsare
relationsbetweenunknow
nfunctionsan
dtheir
derivatives.Computingnumericalsolutionsto
differentialequationsis
oneofthe
most
importanttasksin
technicalcomputing,andon
eofthestrengthsofMatlab.
Ifyou
havestudiedcalculus,youhavelearned
akindofmechanicalprocess
for
differentiatingfunctionsrepresentedbyform
ulasinvo
lvingpow
ers,
trig
functions,
andthelike.
Youknow
thatthederivativeofx3is3x2andyoumay
remem
ber
that
thederivative
oftanxis
1+
tan2x.Thatkindof
differentiationis
importantand
useful,butnotou
rprimary
focushere.
Weareinterested
insituationswherethe
functionsarenotknow
nandcannotberepresentedbysimple
form
ulas.
Wewill
compute
numericalapproxim
ationsto
thevalues
ofafunctionatenoughpoints
toprintatable
orplotagraph.
Imagineyou
are
travelingonamountain
road.Youraltitudevaries
asyou
travel.Thealtitudecanberegarded
asafunctionoftime,
orasafunctionoflongi-
tudeandlatitude,
orasafunctionofthedistance
youhavetraveled.Let’sconsider
thelatter.Let
xden
ote
thedistance
traveled
andy=
y(x)den
ote
thealtitude.
Ifyouhappen
tobecarryinganaltimeter
withyou,oryouhaveadeluxeGPSsystem
,youcancollectenoughvalues
toplotagraphofaltitudeversusdistance,likethe
firstplotin
figure
15.1.
Suppose
youseeasignsayingthatyouare
ona6%
uphillgrade.
Forsome
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
199
200
Chapter15
.OrdinaryDifferential
Equations
6000
6010
6020
6030
6040
6050
6060
6070
950
1000
1050
1100
1150
altitude
6000
6010
6020
6030
6040
6050
6060
6070
−20
−1001020
slope
dist
ance
Figure
15.1.Altitudealongamountain
road,andderivative
ofthatalti-
tude.
Thederivative
iszero
atthelocalmaximaandminim
aofthealtitude.
valueofxnearthesign,andforh=
100,you
willhave
y(x
+h)−y(x)
h=
.06
Thequotientontheleft
istheslope
oftheroadbetweenxandx+h.
Now
imaginethatyou
hadsignseveryfew
meterstellingyouthegradeat
those
points.Thesesignswould
provideap
proxim
ate
values
oftherate
ofchange
ofaltitudewith
respectto
distance
traveled
,This
isthederivative
dy/dx.
You
could
plotagraphof
dy/dx,likethesecondplotin
figure
15.1,even
thoughyou
do
nothaveclosed-form
form
ulasforeither
thealtitudeoritsderivative.This
ishow
Matlabsolves
differentialequations.
Note
thatthederivativeispositivewherethe
altitudeisincreasing,negativewhereitisdecreasing,zero
atthelocalmaxim
aan
dminim
a,an
dnearzero
ontheflatstretches.
Hereis
asimple
example
illustratingthenumericalsolutionofasystem
ofdifferentialequations.
Figure
15.2
isascreen
shotfrom
Spacewar,
theworld’s
firstvideo
game.
Spacewarwas
written
bySteve“Slug”Russellan
dsomeofhis
buddiesatMIT
in1962.It
ranon
thePDP-1,DigitalEquipmentCorporation’s
firstcomputer.
Twospace
ships,controlled
byplayers
usingsw
itches
onthePDP-1
console,shoot
space
torpedoes
ateach
other.
Thespace
shipsan
dthetorpedoes
orbit
around
acentralstar.
Russell’s
program
needed
tocompute
circularan
dellipticalorbits,likethepath
ofthetorpedo
inthescreen
shot.
Atthetime,
therewasnoMatlab.Programswerewritten
interm
sof
individualmachineinstructions.
Floating-pointarithmetic
wasso
slow
thatit
was
desirable
toavoidevaluationoftrig
functionsin
theorbit
calculations.
Theorbit-generatingprogram
looked
somethinglikethis.
x=
0
y=
32768
201
Figure
15.2.
Spacewar,
theworld’s
firstvideo
game.
Thegravitational
pullofthecentralstarcausesthetorpedoto
move
inanellipticalorbit.
L:
plot
xy
loady
shiftright2
addx
storein
x
changesign
shiftright2
addy
storein
y
go
to
L
Whatdoes
this
program
do?
Thereare
notrig
functions,
nosquare
roots,no
multiplicationsordivisions.
Everythingis
donewith
shifts
and
additions.
The
initialvalueofy,whichis
215,serves
asanoverallscale
factor.
Allthearithmetic
involves
asingle
integer
register.
The“shiftright2”commandtakes
thecontents
ofthis
register,
divides
itby22
=4,
anddiscardsan
yremainder.
IfSpacewarorbit
generatorwerewritten
today
inMatlab,it
would
look
somethingthefollow
ing.Weare
nolonger
limited
tointeger
values,so
wehave
changed
thescale
factorfrom
215to
1.
x=
0;
y=
1;
h=
1/4;
n=
2*pi/h;
plot(x,y,’.’)
fork
=1:n
202
Chapter15
.OrdinaryDifferential
Equations
−1−0
.50
0.5
1
−1
−0.8
−0.6
−0.4
−0.20
0.2
0.4
0.6
0.81
Figure
15.3.The25bluepoints
are
generatedby
theSpacewarorbitgen-
eratorwithastep
size
of1/4.The201greenpoints
are
generatedwithastep
size
of1/32. x
=x
+h*y;
y=
y-
h*x;
plot(x,y,’.’)
end
Theou
tputproducedbythis
program
withh
=1/4andn
=25is
show
nbythe
bluedots
infigure
15.3.Theblueorbit
isactuallyanellipse
thatdeviatesfrom
anexact
circle
byab
out7%.Theoutputproducedwithh=
1/32andn=
201is
show
nbythegreen
dots.Thegreen
orbit
isan
other
ellipse
thatdeviatesfrom
anexact
circle
byless
than1%.
Thinkofxandyasfunctionsof
time,
t.Weare
computingx(t)andy(t)at
discretevalues
oft,increm
entedbythestep
size
h.Thevalues
ofxandyattime
t+
hare
computedfrom
thevalues
attimetby
x(t+h)=
x(t)+
hy(t)
y(t+h)=
y(t)−hx(t+h)
This
canberewritten
as
x(t+
h)−x(t)
h=
y(t)
y(t+h)−y(t)
h=
−x(t+
h)
You
haveprobably
noticedthattherighthandsideofthispairofequationsinvo
lves
203
twodifferentvalues
ofthetimevariable,tandt+
h.Thatfact
turnsoutto
be
important,butlet’signore
itfornow
.Lookattheleft
handsides
ofthelast
pair
ofequations.
Thequotients
are
approxim
ationsto
thederivatives
ofx(t)andy(t).
Weare
look
ingfortw
ofunctions
withtheproperty
thatthederivativeofthefirstfunctionisequalto
thesecondand
thederivativeof
thesecondfunctionis
equalto
thenegativeof
thefirst.
Ineff
ect,
theSpacewarorbit
generatoris
usingasimple
numericalmethod
involvingastep
size
hto
compute
anapproxim
ate
solutionto
thesystem
ofdiffer-
entialequations
x=
y
y=
−xThedotover
xandydenotesdifferentiationwithrespectto
t.
x=
dx dt
Theinitialvalues
ofxandyprovidetheinitialconditions
x(0)=
0
y(0)=
1
Theexact
solutionto
thesystem
is
x(t)=
sint
y(t)=
cost
Toseewhy,
recallthetrig
identities
sin(t+h)=
sintcosh+costsinh
cos(t+h)=
costcosh−sintsinh
For
smallvalues
ofh,
sinh≈
h,
cosh≈
1
Consequently
sin(t+h)−sint
h≈
cost,
cos(t+h)−cost
h≈
−sint,
Ifyouplotx(t)an
dy(t)as
functionsof
t,youget
thefamiliarplots
ofsine
andcosine.
Butifyoumakeaphase
planeplot,
thatis
y(t)versusx(t),
yougeta
circle
ofradius1.
ItturnsoutthatthesolutioncomputedbytheSpacewarorbitgeneratorwith
afixed
step
size
his
anellipse,notan
exact
circle.Decreasinghandtakingmore
204
Chapter15
.OrdinaryDifferential
Equations
stepsgeneratesabetterap
proxim
ationto
acircle.Actually,
thefact
thatx(t+
h)
isusedinsteadof
x(t)in
thesecondhalfofthestep
meansthatthemethodis
not
quiteas
simple
asitmightseem
.Thissubtlechangeisresponsible
forthefact
that
themethodgeneratesellipsesinsteadof
spirals.Oneoftheexercisesasksyou
toverifythis
fact
experim
entally.
Mathem
aticalmodelsinvolvingsystem
sofordinary
differentialequationshave
oneindepen
den
tvariable
andon
eormore
depen
den
tvariables.
Theindep
endent
variable
isusuallytimeandis
den
otedbyt.
Inthis
book,wewillassem
ble
allthe
dep
endentvariablesinto
asinglevectory.Thisissometim
esreferred
toas
thestate
ofthesystem
.Thestate
canincludequantities
likeposition,velocity,temperature,
concentration,andprice.
InMatlab
asystem
ofodes
takes
theform
y=
F(t,y)
Thefunction
Falwaystakes
twoarguments,thescalarindep
endentvariable,t,
andthevectorofdep
endentvariables,
y.A
program
thatevaluatesF(t,y)hou
ldcompute
thederivatives
ofallthestate
variablesandreturn
them
inanother
vector.
Inourcircle
generatingexample,thestate
issimply
thecoordinatesof
the
point.
This
requires
achangeofnotation.Wehavebeenusingx(t)andy(t)to
denote
position,now
weare
goingto
use
y 1(t)andy 2(t).
ThefunctionF
defines
thevelocity.
y(t)=
( y 1(t)
y 2(t)
)
=
(y 2(t)
−y1(t)
)
Matlabhasseveralfunctionsthatcompute
numericalapproxim
ationsto
solu-
tionsof
system
sofordinary
differentialequations.
Thesuiteof
odesolversincludes
ode23,ode45,ode113,ode23s,ode15s,ode23t,an
dode23tb.Thedigitsin
the
names
referto
theorder
oftheunderlyingalgorithms.
Theorder
isrelatedto
the
complexityan
daccuracy
ofthemethod.Allof
thefunctionsau
tomaticallydeter-
minethestep
size
required
toob
tain
aprescribed
accuracy.Higher
order
methods
requiremore
work
per
step,butcantakelarger
steps.
Forexampleode23compares
asecondorder
methodwithathirdorder
methodto
estimate
thestep
size,while
ode45comparesafourthorder
methodwithafifthorder
method.
Theletter
“s”in
thenameofsomeoftheodefunctionsindicatesastiff
solver.
Thesemethodssolveamatrix
equationat
each
step,so
they
domore
workper
step
thanthenonstiff
methods.
Butthey
cantakemuch
larger
stepsforproblemswhere
numericalstabilitylimitsthestep
size,so
they
canbemore
efficientoverall.
You
can
use
ode23
formost
oftheexercisesin
this
book,butif
you
are
interested
intheseeinghow
theother
methodsbehave,
please
experim
ent.
Allof
thefunctionsin
theodesuitetakeatleast
threeinputarguments.
•F,thefunctiondefiningthedifferentialequations,
•tspan,thevectorspecifyingtheintegrationinterval,
205
01
23
45
6−1
−0.8
−0.6
−0.4
−0.20
0.2
0.4
0.6
0.81
Figure
15.4.Graphsofsineandcosinegeneratedby
ode23.
•y0,thevectorofinitialconditions.
Thereareseveralwaysto
write
thefunctiondescribingthedifferentialequa-
tion.Anticipatingmore
complicatedfunctions,
wecancreate
aMatlabprogram
forourcircle
generatorthatextractsthetw
odep
endentvariablesfrom
thestate
vector.
Savethis
inafile
named
mycircle.m.
functionydot=
mycircle(t,y)
ydot=
[y(2);
-y(1)];
Notice
thatthisfunctionhastw
oinputarguments,tandy,even
thoughtheoutput
inthis
example
does
notdep
endupont.
Withthis
functiondefi
nitionstoredin
mycircle.m,thefollow
ingcodecalls
ode23to
compute
thesolutionover
theinterval0≤
t≤
2π,startingwithx(0)=
0andy(0)=
1.
tspan=
[0
2*pi];
y0
=[0;
1];
ode23(@mycircle,tspan,y0)
Withnooutputarguments,theodesolversautomaticallyplotthesolutions.
Figure
15.4
istheresult
forourexample.
Thesm
all
circlesin
theplotare
not
equallyspaced.They
show
thepoints
chosenbythestep
size
algorithm.
Toproduce
thephase
plotplotshow
nin
figure
15.5,capture
theoutputand
plotityou
rself.
tspan=
[0
2*pi];
y0
=[0;
1];
[t,y]=
ode23(@mycircle,tspan,y0)
plot(y(:,1),y(:,2)’-o’)
206
Chapter15
.OrdinaryDifferential
Equations
−1−0
.50
0.5
1
−1
−0.8
−0.6
−0.4
−0.20
0.2
0.4
0.6
0.81
Figure
15.5.Graphofacircle
generatedby
ode23.
axis([-1.11.1-1.11.1])
axissquare
Thecircle
generatorexample
isso
simple
thatwecanbypass
thecreationof
thefunctionfile
mycircle.mandwrite
thefunctionin
oneline.
acircle=
@(t,y)
[y(2);
-y(1)]
Theexpressioncreatedon
therightbythe“@
”symbol
isknow
nasananonym
ous
functionbecause
itdoes
nothaveanameuntilitisassigned
toacircle.Since
the
“@”signis
included
inthedefinitionofacircle,youdon’t
needit
when
you
call
anodesolver.
Once
acirclehasbeendefined,thestatement
ode23(acircle,tspan,y0)
automaticallyproducesfigure
15.4.And,thestatement
[t,y]=
ode23(acircle,tspan,y0)
capturestheou
tputso
you
canprocess
ityou
rself.
Manyadditionaloptionsfortheodesolverscanbesetvia
thefunctionodeset.
Forexample
opts=
odeset(’outputfcn’,@odephas2)
ode23(acircle,tspan,y0,opts)
axissquare
axis([-1.11.1-1.11.1])
207
willalsoproduce
figure
15.5.
Use
thecommand
docode23
toseemore
detailsabouttheMatlab
suiteofodesolvers.
Consult
theODE
chapterin
ourcompanionbook,NumericalComputingwithMATLAB,formore
of
themathem
aticalbackgroundoftheodealgorithms,
andforode23tx,atextbook
version
ofode23.
Hereisaverysimpleexamplethatillustrateshow
thefunctionsin
theodesuite
work.Wecallit
“ode1”because
itusesonly
oneelem
entary
firstorder
algorithm,
know
nas
Euler’smethod.Thefunctiondoes
notem
ploytw
odifferentalgorithms
toestimate
theerroranddeterminethestep
size.Thestep
size
his
obtained
by
dividingtheintegrationinterval
into
200equalsizedpieces.
This
wou
ldap
pearto
beappropriate
ifwejust
wantto
plotthesolutiononacomputerscreen
witha
typicalresolution,butwehavenoidea
oftheactualaccuracy
oftheresult.
function[t,y]=
ode1(F,tspan,y0)
%ODE1
World’s
simplest
ODE
solver.
%ODE1(F,[t0,tfinal],y0)
uses
Euler’s
method
to
solve
%dy/dt=
F(t,y)
%with
y(t0)
=y0
on
the
interval
t0
<=
t<=
tfinal.
t0
=tspan(1);
tfinal=
tspan(end);
h=
(tfinal
-t0)/200;
y=
y0;
fort
=t0:h:tfinal
ydot=
F(t,y);
y=
y+
h*ydot;
end How
ever,even
with200stepsthiselem
entary
firstorder
methoddoes
nothave
satisfactory
accuracy.Theoutputfrom
[t,y]]=
ode1(acircle,tspan,y0)
is
t=
6.283185307179587
y=
0.032392920185564
1.103746317465277
Wecanseethatthefinalvalueoftis
2*pi,butthefinalvalueofyhasmissed
returningto
itsstartingvaluebymore
than10
percent.
Manymore
smaller
step
swou
ldberequired
toget
graphicalaccuracy.
208
Chapter15
.OrdinaryDifferential
Equations
Rec
ap
%%
Ordinary
Differential
Equations
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Ordinary
Differential
Equations
Chapter
%of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %odes_recap
%edit
odes_recap
%publish
odes_recap
% %Related
EXM
programs
% %ode1
%%
Spacewar
Orbit
Generator.
x=
0;
y=
1;
h=
1/4;
n=
2*pi/h;
plot(x,y,’.’)
holdon
fork
=1:n
x=
x+
h*y;
y=
y-
h*x;
plot(x,y,’.’)
end
holdoff
axissquare
axis([-1.11.1-1.11.1])
%%
An
Anonymous
Function.
acircle=
@(t,y)
[y(2);
-y(1)];
%%
ODE23
Automatic
Plotting.
figure
tspan=
[0
2*pi];
y0
=[0;
1];
ode23(acircle,tspan,y0)
%%
Phase
Plot.
figure
tspan=
[0
2*pi];
y0
=[0;
1];
[t,y]=
ode23(acircle,tspan,y0)
209
plot(y(:,1),y(:,2),’-o’)
axissquare
axis([-1.11.1-1.11.1])
%%ODE23
Automatic
Phase
Plot.
opts=
odeset(’outputfcn’,@odephas2)
ode23(acircle,tspan,y0,opts)
axissquare
axis([-1.11.1-1.11.1])
%%ODE1
implements
Euler’s
method.
%ODE1
illustrates
the
structure
of
the
MATLAB
ODE
solvers,
%but
it
is
low
order
and
employs
acoarse
step
size.
%So,
even
though
the
exact
solution
is
periodic,
the
final
value
%returned
by
ODE1
misses
the
initial
value
by
asubstantial
amount.
typeode1
[t,y]=
ode1(acircle,tspan,y0)
err=
y-
y0
Exe
rcises
15.1
Walkingto
class.You
leavehome(oryourdorm
room)attheusualtimein
the
morningan
dwalk
towardyou
rfirstclass.Abouthalfway
toclass,yourealize
that
youhaveforgotten
yourhomew
ork.You
runback
home,
getyou
rhomew
ork,run
toclass,andarriveatyou
rusualtime.
Sketch
aroughgraphbyhandshow
ingyou
rdistance
from
homeasafunctionoftime.
Makeasecondsketch
ofyou
rvelocity
asafunctionoftime.
Youdonothaveto
assumethatyou
rwalkingan
drunning
velocities
are
constant,orthatyourreversals
ofdirectionare
instantaneous.
15.2
Divided
differen
ces.
Create
yourow
ngraphic
likeou
rfigure
15.1.Makeup
yourow
ndata,xandy,fordistance
andaltitude.
Youcanuse
subplot(2,1,1)
and
subplot(2,1,2)
toplace
twoplots
inon
efigure
window
.Thestatement
d=
diff(y)./diff(x)
computesthedivided
differen
ceapproxim
ationto
thederivativeforuse
inthesecond
subplot.
Thelength
ofthevectordis
oneless
thanthelength
ofxandy,so
you
canad
donemore
valueattheendwith
210
Chapter15
.OrdinaryDifferential
Equations
d(end+1)=
d(end)
For
more
inform
ationab
outdiffandsubplot,use
helpdiff
helpsubplot
15.3
Orbitgenerator.
Hereisacomplete
Matlabprogram
fortheorbitgenerator,
includingappropriate
settingofthegraphicsparameters.
Investigate
thebehav
ior
ofthis
program
forvariousvalues
ofthestep
size
h.
axis(1.2*[-11
-1
1])
axissquare
boxon
holdon
x=
0;
y=
1;
h=
...
n=
2*pi/h;
plot(x,y,’.’)
fork
=1:n
x=
x+
h*y;
y=
y-
h*x;
plot(x,y,’.’)
end
15.4
Modified
orbit
generator.
Hereis
aMatlab
program
thatmakes
asimpler
approxim
ationfortheorbit
generator.
Whatdoes
itdo?Investigate
thebehav
ior
forvariousvalues
ofthestep
size
h.
axis(1.5*[-11
-1
1])
axissquare
boxon
holdon
x=
0;
y=
1;
h=
1/32;
n=
6*pi/h;
plot(x,y,’.’)
fork
=1:n
savex=
x;
x=
x+
h*y
y=
y-
h*savex;
plot(x,y,’.’)
end
211
15.5
Linearsystem
Write
thesystem
ofdifferentialequations
y 1=
y 2
y 2=
−y1
inmatrix-vectorform
,
y=
Ay
whereyis
avector-valued
functionoftime,
y(t)=
( y 1(t)
y 2(t)
)
andA
isaconstant2-by-2
matrix.Use
ourode1aswellasode23to
experim
ent
withthenumericalsolutionofthesystem
inthis
form
.
15.6
Example
from
ode23.Thefirstexample
inthedocumentationforode23is
y 1=
y 2y 3
y 2=
−y1y 3
y 3=
−0.51y 1
y 2
withinitialconditions
y 1(0)=
0
y 2(0)=
1
y 3(0)=
1
Compute
thesolutionto
this
system
ontheinterval0≤
t≤
12.Reproduce
the
graphincluded
inthedocumentationprovided
bythecommand
docode23
15.7
Acubicsystem
.Makeaphase
planeplotofthesolutionto
theodesystem
y 1=
y3 2
y 2=
−y3 1
withinitialconditions
y 1(0)=
0
y 2(0)=
1
ontheinterval
0≤
t≤
7.4163
Whatis
specialaboutthefinalvalue,
t=
7.4163?
212
Chapter15
.OrdinaryDifferential
Equations
15.8
Aquinticsystem
.Makeaphase
planeplotof
thesolutionto
theodesystem
y 1=
y5 2
y 2=
−y5 1
withinitialconditions
y 1(0)=
0
y 2(0)=
1
onaninterval
0≤
t≤
T
whereT
isthevaluebetween7and8determined
bytheperiodicitycondition
y 1(T
)=
0
y 2(T
)=
1
15.9
Aquadraticsystem
.W
hathappensto
solutionsof
y 1=
y2 2
y 2=
−y2 1
Whydosolutionsof
y 1=
yp 2
y 2=
−yp 1
havesuch
differentbehav
iorifpis
oddoreven
?
Chapter16
Predator-PreyModel
Modelsofpopulationgrowth.
Thesimplest
model
forthegrow
th,ordecay,ofapopulationsaysthatthe
growth
rate,or
thedecay
rate,is
proportionalto
thesize
ofthepop
ulationitself.
Increasingordecreasingthesize
ofthepopulationresultsin
aproportionalincrease
ordecrease
inthenumber
ofbirthsanddeaths.
Mathem
atically,
this
isdescribed
bythedifferentialequation
y=
ky
Theproportionality
constantkrelatesthesize
ofthepopulation,y(t),
toitsrate
ofgrowth,y(t).
Ifk
ispositive,
thepop
ulation
increases;
ifk
isnegative,
the
populationdecreases.
Asweknow
,thesolutionto
thisequationisafunctiony(t)thatisproportional
totheexponentialfunction
y(t)=
ηek
t
whereη=
y(0).
Thissimplemodelisap
propriate
intheinitialstages
ofgrowth
when
thereare
norestrictionsorconstraints
onthepopulation.Asm
allsampleofbacteria
inalarge
Petridish,forexample.Butin
more
realistic
situationsthereare
limitsto
grow
th,
such
asfinitespace
orfoodsupply.A
more
realistic
model
saysthatthepopulation
competes
withitself.Asthepopulationincreases,itsgrow
thrate
decreaseslinearly.
Thedifferentialequationis
sometim
escalled
thelogistic
equation.
y=
k(1
−y μ)y
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
213
214
Chapter16
.Predator-PreyModel
01
23
45
67
80246810121416182022
Figure
16.1.Exponen
tialgrowth
andlogistic
growth.
Thenew
parameter
μis
thecarryingcapacity.Asy(t)ap
proaches
μthegrowth
rate
approaches
zero
andthegrow
thultim
ately
stops.
Itturnsoutthatthesolution
is
y(t)=
μηek
t
ηek
t+
μ−
η
You
caneasily
verifyforyourselfthatas
tapproaches
zero,y(t)ap
proaches
ηand
thatastapproaches
infinity,
y(t)ap
proaches
μ.If
youknow
calculus,
then
with
quiteabitmore
effort,youcanverifythaty(t)actuallysatisfies
thelogisticequation.
Figure
16.1
show
sthetw
osolutionswhen
both
ηandkare
equalto
one.
The
expon
entialfunction
y(t)=
et
gives
therapidly
grow
inggreen
curve.
Withcarryingcapacity
μ=
20,thelogistic
function
y(t)=
20et
et+19
gives
themore
slow
lygrowingbluecurve.
Both
curves
havethesameinitialvalue
andinitialslope.
Theexponentialfunctiongrowsexponentially,
whilethelogistic
functionapproaches,butnever
exceeds,
itscarryingcapacity.
Figure
16.1
wasgeneratedwiththefollow
ingcode.
k=
1
eta=
1
mu
=20
t=
0:1/32:8;
215
y=
mu*eta*exp(k*t)./(eta*exp(k*t)
+mu
-eta);
plot(t,[y;exp(t)])
axis([08
022])
Ifyoudon’t
havetheform
ula
forthesolutionto
thelogisticequationhandy,
you
cancompute
anumericalsolutionwithode45,oneoftheMatlab
ordinary
differentialequationsolvers.
Try
runningthefollow
ingcode.
Itwillautomatically
produce
aplotsomethinglikethebluecurvein
figure
16.1.
k=
1
eta=
1
mu
=20
ydot=
@(t,y)
k*(1-y/mu)*y
ode45(ydot,[08],eta)
The@signand@(t,y)specifythatyou
aredefiningafunctionoftandy.Thet
isnecessary
even
thoughit
doesn’t
explicitlyappearin
this
particulardifferential
equation.
Thelogisticequation
and
itssolution
occurin
many
differentfields.
The
logisticfunctionis
alsoknow
nasthesigm
oid
functionanditsgraphis
know
nas
theS-curve.
Populationsdonotlivein
isolation.Everybodyhasafew
enem
ieshereand
there.
TheLotka-Volterra
predator-preymodel
isthesimplest
descriptionofcom-
petitionbetweentw
ospecies.
Thinkofrabbitsandfoxes,orzebrasandlions,
orlittle
fish
andbig
fish.
Theidea
isthat,ifleft
tothem
selves
withaninfinitefoodsupply,therabbits
orzebraswould
livehappilyandexperience
exponentialpop
ulationgrow
th.Onthe
other
hand,ifthefoxes
orlionswereleft
withnopreyto
eat,they
wou
lddie
faster
thanthey
could
reproduce,andwould
experience
expon
entialpopulationdecline.
Thepredator-preymodel
isapairofdifferentialequationsinvolvingapairof
competingpopulations,
y 1(t)andy 2(t).
Thegrow
thrate
fory 1
isalinearfunction
ofy 2
andviceversa.
y 1=
(1−
y 2 μ2)y
1
y 2=
−(1−
y 1 μ1)y
2
Weare
usingnotationy 1(t)an
dy 2(t)insteadof,say,
r(t)
forrabbitsandf(t)for
foxes,because
ourMatlabprogram
usesatw
o-componentvectory.
Theextraminussignin
thesecondequationdistinguishes
thepredators
from
theprey.
Notethatify 1
ever
becomes
zero,then
y 2=
−y2
andthepredators
are
introuble.Butify 2
ever
becomes
zero,then
y 1=
y 1
andthepreypopulationgrow
sexponentially.
216
Chapter16
.Predator-PreyModel
Wehaveaform
ula
forthesolutionofthesinglespecieslogisticmodel.How
ever
itis
notpossible
toexpress
thesolutionto
this
predator-preymodel
interm
sof
expon
ential,trigonmetric,
oran
yother
elem
entary
functions.
Itis
necessary,but
easy,to
compute
numericalsolutions.
05
1015
200
100
200
300
400
500
600
Figure
16.2.A
typicalsolutionofthepredator-preyequations.
Therearefourparameters,
thetw
oconstants
μ1andμ2,andthetw
oinitial
conditions,
η 1=
y 1(0)
η 2=
y 2(0)
Ifwehappen
tostart
withη 1
=μ1andη 2
=μ2,then
both
y 1andy 2
are
zero
and
thepopulationsremain
constantattheirinitialvalues.In
other
words,
thepoint
(μ1,μ
2)isanequilibrium
point.
Theorigin,(0,0)isanother
equilibrium
point,but
notaveryinterestingone.
Thefollow
ingcodeusesode45to
automaticallyplotthetypicalsolutionshow
nin
figure
16.2.
mu
=[300
200]’
eta=
[400
100]’
signs=
[1
-1]’
pred_prey_ode=
@(t,y)
signs.*(1-flipud(y./mu)).*y
period=
6.5357
ode45(pred_prey_ode,[03*period],eta)
217
Therearetw
otrickypartsofthis
code.
Matlab
vectoroperationsare
usedto
defi
nepred_prey_ode,thedifferentialequationsin
oneline.
And,thecalculation
thatgeneratesfigure
16.3
provides
thevalueassigned
toperiod.Thisvaluespecifies
avalueoftwhen
thepopulationsreturn
totheirinitialvalues
given
byeta.The
codeintegratesover
threeofthesetimeintervals,andso
attheendweget
back
towherewestarted.
Thecirclessuperim
posedontheplots
infigure
16.2
show
thepoints
where
ode45computesthesolution.
Theplots
looksomethingliketrig
functions,
but
they’renot.
Notice
thatthecurves
neartheminim
aarebroader,andrequiremore
stepsto
compute,then
thecurves
nearthemaxim
a.Theplotof
sintwou
ldlook
thesameat
thetopas
thebottom.
010
020
030
040
050
060
070
080
00
200
400
prey
predator
Dra
g ei
ther
dot
06.
5357
13.0
713
0
200
400
600
800
time
prey
pred
ator
perio
d Figure
16.3.Thepredpreyexperimen
t.
OurMatlabprogram
exm/predpreyshow
sareddotattheequilibrium
point,
(μ1,μ
2),
andablue-green
dotattheinitialpoint,
(η1,η
2).
When
you
drageither
dotwiththemouse,thesolutionisrecomputingbyode45andplotted.Figure
16.3
show
sthattw
oplots
areproduced—
aphase
planeplotofy 2(t)versusy 1(t)an
datimeseries
plotofy 1(t)an
dy 2(t)versust.
Figures16.2
and16.3
havethesame
parameters,an
dconsequentlyshow
thesamesolution,butwithdifferentscalingof
218
Chapter16
.Predator-PreyModel
theaxes.
Theremarkab
leproperty
oftheLotka-Volterra
model
isthatthesolutionsare
alwaysperiodic.Thepopulationsalway
sreturn
totheirinitialvalues
andrepeat
thecycle.
Thisproperty
isnotobviousandnoteasy
toprove.It
israre
fornonlinear
modelsto
haveperiodic
solutions.
Thediffi
cultaspectofcomputingthesolutionto
thepredator-preyequations
isdeterminingthelength
oftheperiod.Ourpredpreyprogram
usesafeature
ofthe
MatlabODE
solverscalled
“eventhandling”to
compute
thelength
ofaperiod.
Iftheinitialvalues
(η1,η
2)are
close
totheequilibrium
point(μ
1,μ
2),
then
thelength
oftheperiodisclose
toafamilarvalue.
Anexercise
asksyouto
discover
thatvalueexperim
entally.
Rec
ap
%%
Predator-Prey
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Preditor
Prey
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %predprey_recap
%edit
predprey_recap
%publish
predprey_recap
% %Related
EXM
programs
% %predprey
%%
Exponential
and
Logistic
Growth.
closeall
figure
k=
1
eta=
1
mu
=20
t=
0:1/32:8;
y=
mu*eta*exp(k*t)./(eta*exp(k*t)
+mu
-eta);
plot(t,[y;exp(t)])
axis([08
022])
title(’Exponentialandlogisticgrowth’)
xlabel(’t’)
ylabel(’y’)
%%
ODE45
for
the
Logistic
Model.
figure
k=
1
219
eta=
1
mu
=20
ydot=
@(t,y)
k*(1-y/mu)*y
ode45(ydot,[08],eta)
%%ODE45
for
the
Predator-Prey
Model.
figure
mu
=[300
200]’
eta=
[400
100]’
signs=
[1
-1]’
pred_prey_ode=
@(t,y)
signs.*(1-flipud(y./mu)).*y
period=
6.5357
ode45(pred_prey_ode,[03*period],eta)
%%Our
predprey
gui.
figure
predprey
Exe
rcises
16.1
Plot.
Makeamore
fewplots
likefigures16.1
and16.2,butwithdifferentvalues
oftheparametersk,η,andμ.
16.2
Decay.
Compare
exponentialandlogisticdecay.Makeaplotlikefigure
16.1
withnegativek.
16.3
Differen
tiate.
Verifythatourform
ula
fory(t)actuallysatisfies
thelogistic
differentialequations.
16.4
Easy
aspie.
Inpredprey,if
theredan
dblue-green
dots
are
close
toeach
other,then
thelength
oftheperiodisclose
toafamilarvalue.
Whatisthatvalue?
Does
thatvaluedep
endupon
theactuallocationofthedots,orjust
theirrelative
closeness?
16.5
Period.In
predprey,iftheredandblue-green
dots
are
farapart,does
the
length
oftheperiodgetlonger
orshorter?Is
itpossible
tomaketheperiodshorter
thanthevalueithasnearequilibrium?
16.6
Phase.
Iftheinitialvalueis
neartheequilibrium
point,
thegraphsofthe
predatorandpreypopulationsare
nearlysinusoidal,withaphase
shift.
Inother
words,
after
thepreypopulationreaches
amaxim
um
orminim
um,thepredator
pop
ulationreaches
amaxim
um
orminim
um
somefractionof
theperiodlater.
What
isthatfraction?
220
Chapter16
.Predator-PreyModel
16.7
Pitstop.
Thepredpreysubfunctionpitstopis
involved
inthe“eventhan-
dling”thatode45usesto
compute
theperiod.pitstop,in
turn,usesatan2to
compute
anglestheta0andtheta1.W
hatisthedifference
betweenthetw
oMat-
labfunctionsatan2,whichtakes
twoarguments,andatan,whichtakes
only
one?
Whathappensifatan2(v,u)is
replacedbyatan(v/u)in
predprey?
Draw
asketch
show
ingtheanglestheta0andtheta1.
16.8
tfinal.Thecallto
ode45in
predpreyspecifies
atimeintervalof[0
100].W
hat
isthesignificance
ofthevalue100?W
hathappensifyouchangeit?
16.9
Lim
itgrowth.Modifypredpreyto
includeagrowth
limitingterm
fortheprey,
similarto
onein
thelogisticequation.
Avoidanother
parameter
bymakingthe
carryingcapacity
twicetheinitialvalue.
Theequationsbecome
y 1=
(1−
y 1 2η 1
)(1−
y 2 μ2)y
1
y 2=
−(1−
y 1 μ1)y
2
Whathappensto
theshapeof
thesolutioncurves?Are
thesolutionsstillperiodic?
Whathappensto
thelength
oftheperiod?
Chapter17
Orbits
Dyn
amicsofmany-bodysystem
s.
Manymathem
aticalmodelsinvolve
thedynamicsofobjectsunder
theinflu-
ence
ofboth
theirmutualinteractionandthesurroundingenvironment.
Theobjects
mightbeplanets,molecules,vehicles,orpeople.Theultim
ate
goalofthischapteris
toinvestigate
then-bodyproblem
incelestialmechanics,whichmodelsthedynamics
ofasystem
ofplanets,
such
asou
rsolarsystem
.Butfirst,
welookattw
osimpler
modelsandprograms,
abouncingballandBrownianmotion.
Theexmprogram
bounceris
amodel
ofabouncingball.Theballis
tossed
into
theair
and
reactsto
thepull
oftheearth’s
gravitation
force.
Thereis
acorrespondingpullof
theballontheearth,buttheearthisso
massivethatwecan
neglect
itsmotion.
Mathem
atically,
weletv(t)an
dz(t)denote
thevelocity
andtheheightofthe
ball.Both
are
functionsof
time.
Highschoolphysics
provides
form
ulasforv(t)an
dz(t),
butwechoosenotto
use
them
because
weareanticipatingmore
complicated
problemswheresuch
form
ulasarenotavailable.Instead,wetakesm
allstepsofsize
δin
time,
computingthevelocity
andheightat
each
step.After
theinitialtoss,
gravitycausesthevelocity
todecrease
ataconstantrate,g.Soeach
step
updates
v(t)with
v(t+δ)
=v(t)−δg
Thevelocity
istherate
ofchangeoftheheight.
Soeach
step
updatesz(t)with
z(t+
δ)=
z(t)+
δv(t)
Hereis
thecore
ofbouncer.m.
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
221
222
Chapter17
.Orbits
[z0,h]=
initialize_bouncer;
g=
9.8;
c=
0.75;
delta=
0.005;
v0
=21;
whilev0
>=
1
v=
v0;
z=
z0;
whileall(z>=
0)
set(h,’zdata’,z)
drawnow
v=
v-
delta*g;
z=
z+
delta*v;
end
v0
=c*v0;
end
finalize_bouncer
Thefirststatement
[z0,h]=
initialize_bouncer;
generatestheplotofasphereshow
nin
figure
17.1
andreturnsz0,thez-coordinates
ofthesphere,
andh,theHandle
Graphics“handle”fortheplot.
Oneoftheexer-
ciseshasyouinvestigate
thedetailsofinitialize_bouncer.Thefigure
show
sthe
situationat
both
thestart
andtheendofthesimulation.Theballisat
rest
andso
thepicture
isprettyboring.Toseewhathappensduringthesimulation,youhave
toactuallyrunbouncer.
Thenextfourstatements
inbouncer.mare
g=
9.8;
c=
0.75;
delta=
0.005;
v0
=21;
Thesestatements
setthevalues
oftheaccelerationof
grav
ityg,anelasticitycoef-
ficientc,thesm
alltimestep
delta,andtheinitialvelocity
fortheball,v0.
Allthecomputationin
bounceris
donewithin
adoubly
nestedwhileloop.
Theouterloopinvolves
theinitialvelocity
v0.
whilev0
>=
1
...
v0
=c*v0;
end
Toachieve
thebouncingaffect,
theinitialvelocity
isrepeatedly
multiplied
byc=
0.75untilit
isless
than1.Each
bou
nce
startswithavelocity
equalto
3/4ofthe
previousone.
Within
theouterloop,thestatements
223
Figure
17.1.Initial,andfinal,positionofabouncingball.Toseewhat
happensin
between,runbouncer.
v=
v0;
z=
z0;
initialize
thevelocity
vto
v0andtheheightzto
z0.Then
theinner
loop
whileall(z>=
0)
set(h,’zdata’,z)
drawnow
v=
v-
delta*g;
z=
z+
delta*v;
end
proceedsuntiltheheightgo
esnegative.
Theplotis
repeatedly
updatedto
reflect
thecurrentheight.
Ateach
step
,thevelocity
visdecreasedbyaconstantamount,
delta*g,therebyaffectingthegravitationaldeceleration.Thisvelocity
isthen
used
tocompute
thechangein
theheightz.Aslongasvispositive,
thezincreaseswith
each
step.W
hen
vreaches
zero,theballhasreached
itsmaxim
um
height.
Then
v
becomes
negativeandzdecreasesuntiltheballreturnsto
heightzero,term
inating
theinner
loop.
After
both
loop
sare
complete,thestatement
finalize_bouncer
activates
apushbuttonthatoff
ersyouthepossibilityof
repeatingthesimulation.
Brownianmotionis
notasobviousasgravityin
ourdailylives,butwedo
encounteritfrequently.
AlbertEinstein’sfirstim
portantscientificpaper
wasabout
Brownianmotion.Thinkofparticplesofdust
suspended
intheairan
dilluminated
224
Chapter17
.Orbits
byabeam
ofsunlight.
Or,
diffusionofodors
throughoutaroom.Or,
abeach
ball
beingtossed
aroundastadium
bythespectators.
InBrownianmotionanobject
–adust
particle,amolecule,or
aball–reacts
tosurroundingrandom
forces.
Oursimulation
oftheseforces
usesthebuilt-in
MATLAB
functionrandnto
generate
norm
allydistributedrandom
numbers.
Each
timethestatement
randn
isexecutedanew
,unpredictable,valueis
produced.Thestatement
randn(m,n)
producesanm-by-n
array
ofrandom
values.Each
timethestatement
hist(randn(100000,1),60)
isexecutedahistogram
plotliketheonein
figure
17.2
isproduced.Try
executing
this
statementseveraltimes.You
willseethatdifferenthistogramsareproduced
each
time,
butthey
allhavethesameshape.
You
mightrecognizethe“bell-shaped
curve”
thatis
know
nmore
form
allyastheGaussianornorm
aldistribution.The
histogram
show
sthatpositivean
dnegativerandom
numbersare
equallylikelyan
dthatsm
allvalues
aremore
likelythanlargeones.This
distributionis
themathe-
maticalheart
ofBrownianmotion.
−50
50
1000
2000
3000
4000
5000
6000
7000
Figure
17.2.Histogram
ofthenorm
alrandom
numbergenerator.
Asimple
example
ofBrownianmotionknow
nas
arandom
walk
isshow
nin
figure
17.3.This
isproducedbythefollow
ingcodefragment.
m=
100;
x=
cumsum(randn(m,1));
y=
cumsum(randn(m,1));
225
plot(x,y,’.-’)
s=
2*sqrt(m);
axis([-ss
-s
s]);
−15
−10
−50
510
15−1
5
−10−5051015
Figure
17.3.A
simple
example
ofBrownianmotion.
Thekey
statementis
x=
cumsum(randn(m,1));
This
statementgeneratesthex-coordinatesofthewalk
byform
ingthesuccessive
cumulative
partialsumsof
theelem
ents
ofthevectorr
=randn(m,1).
x1
=r 1
x2
=r 1
+r 2
x3
=r 1
+r 2
+r 3
...
Asimilarstatementgeneratesthey-coordinates.
Cutandpastethecodefragment
into
theMatlabcommandwindow
.Execute
itseveraltimes.Try
differentvalues
ofm.Youwillseedifferentrandom
walksgoingoffin
differentrandom
directions.
Over
manyexecutions,
thevalues
ofxandyare
just
aslikelyto
bepositiveas
negative.
Wewantto
compute
anaxis
scale
factorsso
thatmost,butnotall,of
thewalksstay
within
theplotboundaries.It
turnsou
tthatasm,thelength
ofthe
walk,increases,
theproper
scale
factorincreaseslike√ m
.A
fancier
Brownianmotionprogram,invo
lvingsimultaneousrandom
walksof
manyparticlesin
threedim
ensions,isavailable
inbrownian3.m.A
snapshotofthe
evolvingmotionis
show
nin
figure
17.4.Hereis
thecore
ofbrownian3.m.
226
Chapter17
.Orbits
−50
510
−50
510−50510
Figure
17.4.A
snapshotoftheoutputfrom
brownian3,showingsimulta-
neousrandom
walksofmanyparticulesin
threedim
ensions.
n=
50;
%Default
number
of
particles
P=
zeros(n,3);
H=
initialize_graphics(P);
while~get(H.stop,’value’)
%Obtain
step
size
from
slider.
delta=
get(H.speed,’value’);
%Normally
distributed
random
velocities.
V=
randn(n,3);
%Update
positions.
P=
P+
delta*V;
update_plot(P,H);
end
Thevariable
nis
thenumber
ofparticles.
Itis
usuallyequalto
50,butsomeother
number
ispossible
withbrownian3(n).
Thearray
Pcontainsthepositionsofn
particlesin
threedim
ensions.
Initially,
alltheparticlesare
locatedattheorigin,
(0,0,0).
Thevariable
His
aMatlabstructure
containinghandlesforalltheuser
227
interface
controls.In
particular,H.stoprefers
toatogglethatterm
inatesthewhile
loopandH.speedrefers
toaslider
thatcontrolsthespeedthroughthevalueof
the
timestep
delta.Thearray
Vis
ann-by-3
arrayof
norm
allydistributedrandom
numbersthatserveastheparticle
velocities
intherandom
walks.
Most
ofthe
complexityof
brownian3is
contained
inthesubfunctioninitialize_graphics.
Inadditionto
thespeedslider
andthestopbutton,theGUIhaspushbuttonsor
togglesto
turn
onatrace,zoom
inandou
t,andchangetheview
point.
Weare
now
readyto
tackle
then-bodyproblem
incelestialmechanics.
Thisis
amodel
ofasystem
ofplanetsandtheirinteractiondescribed
byNew
ton’slawsof
motionandgrav
itationalattraction.Over
fivehundredyears
ago,Johannes
Kepler
realizedthatifthereareonly
twoplanetsin
themodel,theorbitsare
ellipseswitha
commonfocusatthecenterof
mass
ofthesystem
.Thisprovides
afairdescription
ofthemoon
’sorbitaroundtheearth,oroftheearth’sorbitaroundthesun.Butif
youare
planningatrip
tothemoonoramissionto
Mars,youneedmore
accuracy.
Youhaveto
realize
thatthesunaff
ects
themoon’sorbitaroundtheearthandthat
Jupiter
affects
theorbitsofboth
theearthandMars.Furthermore,ifyouwishto
model
more
thantw
oplanets,ananalyticsolutionto
theequationsofmotionisnot
possible.It
isnecessary
tocompute
numericalapproxim
ations.
Ournotationusesvectors
andarray
s.Let
nbethenumber
ofbodiesan
d,
fori=
1,...,n
,letpibethevectordenotingthepositionof
thei-th
body.
For
two-dim
ensionalmotionthei-th
positionvectorhascomponents
(xi,y i).
For
three-
dim
ensionalmotion
itscomponents
are
(xi,y i,z
i).
Thesm
all
system
show
nin
figure
17.5
illustratesthis
notation.Therearethreebodiesmov
ingin
twodim
en-
sions.
Thecoordinate
system
andunitsare
chosenso
thatinitiallythefirstbody,
whichis
gold
ifyouhavecolor,
isattheorigin,
p1=
(0,0)
Thesecond
body,
which
isblue,
isoneunit
away
from
thefirstbodyin
thex
direction,so
p2=
(1,0)
Thethirdbody,
whichisred,isoneunitaw
ayfrom
thefirstbodyin
theydirection,
so
p3=
(0,1)
Wewishto
model
how
thepositionvectors
pivary
withtime,
t.Thevelocity
ofabodyis
therate
ofchangeofitspositionan
dtheacceleration
istherate
ofchangeofitsvelocity.Weuse
oneandtw
odots
over
pito
den
ote
thevelocity
and
accelerationvectors,piandpi.
Ifyouare
familiarwithcalculus,
you
realize
that
thedotmeansdifferentiationwithrespectto
t.Forourthreebodyexample,the
firstbodyisinitiallyheadingaw
ayfrom
theother
twobodies,so
itsvelocity
vector
hastw
onegativecomponents,
p1=
(−0.12,−
0.36)
Theinitialvelocity
ofthesecondbodyis
allin
theydirection,
p2=
(0,0.72)
228
Chapter17
.Orbits
−1.5
−1−0
.50
0.5
11.
5−1
.5−1
−0.50
0.51
1.5
p 1p 2
p 3
Figure
17.5.Initialpositionsandvelocities
ofasm
allsystem
withthree
bodiesin
two-dim
ensionalspace.
andtheinitialvelocity
ofthethirdbodyis
sendingittowardsthesecondbody,
p3=
(0.36,−0
.36)
New
ton’s
law
ofmotion,thefamousF
=ma,saysthatthemass
ofabody
times
itsaccelerationisproportionalto
thesum
oftheforces
actingonit.New
ton’s
law
ofgrav
itationalsaysthattheforcebetweenanytw
obodiesis
proportionalto
theproduct
oftheirmasses
andinverselyproportionalto
thesquare
ofthedistance
betweenthem
.So,theequationsofmotionare
mip
i=
γ∑ j�=i
mim
jpj−pi
||pj−
pi||
3,
i=
1,...,n
Hereγis
thegravitationalconstant,m
iis
themass
ofthei-th
body,
pj−piis
the
vectorfrom
bodyito
bodyjand||p
j−
pi||
isthelength
ornorm
ofthatvector,
whichis
thedistance
betweenthetw
obodies.
Thedenominatorofthefraction
involves
thecubeof
thedistance
because
thenumeratorcontainsthedistance
itself
andso
theresultingquotientinvo
lves
theinverseof
thesquare
ofthedistance.
Figure
17.6
show
sourthreebodyexample
again.Thelength
ofthevector
r 23=
p3−p2isthedistance
betweenp2andp3.Thegravitationforces
betweenthe
bodieslocatedatp2andp3are
directedalongr 2
3and−r
23.
Tosummarize,thepositionof
thei-th
bodyisdenotedbythevectorpi.
The
instantaneouschangein
positionof
thisbodyisgiven
byitsvelocity
vector,denoted
bypi.
Theinstantaneouschangein
thevelocity
isgiven
byitsaccelerationvector,
denotedbypi.
Theaccelerationis
determined
from
thepositionandmasses
ofall
thebodiesbyNew
ton’s
lawsof
motionan
dgravitation.
Thefollow
ingnotationsimplifies
thediscussionofnumericalmethods.
Stack
thepositionvectors
ontopofeach
other
toproduce
ann-by-d
array
wherenisthe
229
−1.5
−1−0
.50
0.5
11.
5−1
.5−1
−0.50
0.51
1.5
p 1p 2
p 3
r 23
Figure
17.6.Thedoublearrowdepicts
thevectors
r 23=
p3−p2and−r
32.
Thelengthofthis
arrow
isthedistance
betweenp2andp3.
number
ofbodiesandd=
2or3is
thenumber
ofspatialdim
ensions..
P=
⎛ ⎜ ⎜ ⎜ ⎝p1
p2 . . . pn
⎞ ⎟ ⎟ ⎟ ⎠Let
Vden
ote
asimilararrayofvelocity
vectors.
V=
⎛ ⎜ ⎜ ⎜ ⎝p1
p2 . . . pn
⎞ ⎟ ⎟ ⎟ ⎠And,letG(P
)denote
thearray
ofgrav
itationforces.
G(P
)=
⎛ ⎜ ⎜ ⎜ ⎝g 1 g 2 . . . g n
⎞ ⎟ ⎟ ⎟ ⎠where g i
=γ∑ j�=i
mj
pj−pi
||pj−pi||
3
Withthis
notation,theequationsofmotioncanbewritten
P=
V
V=
G(P
)
230
Chapter17
.Orbits
Forourthreebodyexample,theinitialvalues
ofP
andV
are
P=
⎛ ⎝00
10
01
⎞ ⎠and
V=
⎛ ⎝−0.12
−0.36
00.72
0.36
−0.36
⎞ ⎠Themasses
inourthreebodyexample
are
m1=
1/2,
m2=
1/3,
m3=
1/6
From
thesequantities,wecancompute
theinitialvalueofthegrav
itationforces,
G(P
). Wewillillustrate
ournumericalmethodsbytryingto
gen
erate
acircle.The
differentialequationsare
x=
y
y=
−xW
ithinitialconditionsx(0)=
0,y(0)=
1,theexact
solutionis
x(t)=
sint,
y(t)=
cost
Theorbitis
aperfect
circle
withaperiodequalto
2π.
Themost
elem
entary
numericalmethod,whichwewillnotactuallyuse,is
know
nastheforw
ard
orexplicitEulermethod.Themethodusesafixed
timestep
δandsimultaneouslyad
vancesboththepositionsandvelocities
from
timet k
totimet k
+1=
t k+δ.
Pk+1=
Pk+δVk
Vk+1=
Vk+
δG(P
k)
Theforw
ardEuler’smethodapplied
tothecircle
generatorproblem
becomes
xk+1=
xk+
δy k
y k+1=
y k−
δxk
Theresultforδ=
2π/30isshow
nin
thefirstplotin
figure
17.7.Insteadof
acircle
weget
agrowingspiral.
Themethodis
unstable
andconsequentlyunsatisfactory,
particularlyforlongtimeperiods.
Smaller
timestep
smerelydelay
theinevitable.
Wewou
ldseemore
complicated,butsimilar,
behav
iorwiththen-bodyequations.
Another
elem
entary
numericalmethodis
know
nas
thebackward
orim
plicit
Eulermethod.In
general,it
invo
lves
somehow
solvinganonlinearsystem
ateach
step
.
Pk+1−δVk+1=
Pk
Vk+1−
δG(P
k+1)=
Vk
231
−10
1
−101
−10
1
−101
−10
1
−101
Figure
17.7.
ThreeversionsofEuler’smethod
forgeneratingacircle.
Thefirstplotshowsthattheforw
ard
method
isunstable.
Thesecondplotshowsthat
thebackward
method
hasexcessivedamping.
Thethirdplotshowsthatsymplectic
method,whichis
acompromisebetween
thefirsttwomethods,
producesanearly
perfectcircle.
For
oursimple
circle
example
theim
plicitsystem
islinear,
soxk+1andy k
+1are
easily
computedbysolvingthe2-by-2
system
xk+1−
δy k
+1=
xk
y k+1+
δxk+1=
y k
Theresult
isshow
nin
thesecondplotin
figure
17.7.Insteadof
acircle
weget
adecay
ingspiral.
Themethodis
stable,butthereis
toomuch
damping.Again,we
wou
ldseesimilarbehav
iorwiththen-bodyequations.
Themethodthatweactuallyuse
isacompromisebetweentheexplicitand
implicitEulermethods.
Itis
themost
elem
entary
instance
ofwhatare
know
nas
symplectic
methods.
Themethodinvolves
twohalf-steps.
Inthefirsthalf-step,the
positionsattimet k
are
usedin
thegrav
itationequationsto
update
ofthevelocities.
Vk+1=
Vk+δG(P
k)
Then,in
thesecondhalf-step,these“new
”velocities
are
usedto
update
theposi-
tions.
Pk+1=
Pk+δVk+1
Thenovel
feature
ofthis
symplectic
methodis
thesubscriptk+
1insteadofkon
theV
term
inthesecondhalf-step.
For
thecircle
gen
erator,
thesymplectic
methodis
xk+1=
xk+δy k
y k+1=
y k−δxk+1
Theresultisthethirdplotin
figure
17.7.Ifyoulook
carefully,
youcanseethatthe
orbitin
notquiteacircle.It’sactuallyanearlycircularellipse.Andthefinalvalue
does
notquitereturn
totheinitialvalue,
sotheperiodis
notexactly
2π.Butthe
importantfact
isthattheorbitis
neither
agrow
ingnoradecay
ingspiral.
232
Chapter17
.Orbits
−1.5
−1−0
.50
0.5
11.
5−1
.5−1
−0.50
0.51
1.5
p 1p 2
p 3
Figure
17.8.Thefirstfew
stepsofourexample
system
.
Therearemore
complicatedsymplectic
algorithmsthataremuch
more
accu-
rate
per
step
thanthis
symplectic
Euler.
Butthesymplectic
Euleris
satisfactory
forgeneratingwellbehaved
graphicaldisplays.
Most
well-know
nnumericalmeth-
ods,
includingRunge-Kuttamethodsand
traditionalmultistep
methods,
donot
havethis
symplectic
stabilityproperty
and,asaresult,arenotas
satisfactory
for
computingorbitsover
longtimespans.
Figure
17.8
show
sthefirstfew
stepsforourexample
system
.Aswenoted
earlier,
theinitialpositionandvelocity
are
P=
00
1.0000
0
01.0000
V=
-0.1200
-0.3600
00.7200
0.3600
-0.3600
After
onestep
withδ=
0.20weobtain
thefollow
ingvalues.
P=
-0.0107
-0.0653
0.9776
0.1464
0.0767
0.9033
V=
-0.0533
-0.3267
-0.1118
0.7318
233
0.3836
-0.4836
Thethreemasses,1/2,1/3,and1/
6,are
notequal,butare
comparable,so
allthree
bodieshavesignificantaffects
oneach
other
andallthreemovenoticeabledistances.
Weseethattheinitialvelocity
ofthefirstbodycausesit
tomoveaw
ayfrom
the
other
two.In
onestep,itspositionchanges
from
(0,0)to
smallnegativevalues,
(−0.0107
,−0.0653)Thesecondbodyisinitiallyatposition(1,0)withvelocity
(0,1)
inthepositiveydirection.In
onestep,itspositionchanges
to(0.9776,0.1464).
The
x-coordinate
haschanged
relativelylittle,whilethey-coordinate
haschanged
by
roughly
0.72
δ.Thethirdbodymoves
inthedirectionindicatedbythevelocity
vectorin
figure
17.5.
After
asecondstep
wehavethefollow
ingvalues.Asexpected,allthetrends
notedin
thefirststep
continue.
P=
-0.0079
-0.1209
0.9325
0.2917
0.1589
0.7793
V=
0.0136
-0.2779
-0.2259
0.7268
0.4109
-0.6198
−1.5
−1−0
.50
0.5
11.
5−1
.5−1
−0.50
0.51
1.5
Figure
17.9.Theinitialtrajectories
ofourexample
system
.
Figure
17.9
show
saninitialsectionofthetrajectories.Youshould
runour
Experim
ents
program
orbits(3)to
seethethreebodiesin
motion.
Thesm
all
bodyandthelargebodyorbitin
aclockwisedirectionaroundeach
other
whilethe
medium-sizebodyorbitsin
acounter-clockwisedirectionaroundtheother
two.
234
Chapter17
.Orbits
−20
0
20
−20
−10
010
20−505
Figure
17.10.Thesolarsystem
,withtheinitialpositionsofalltheplanets
andtheorbitsoftheouterplanets,
Jupiter,Saturn,Uranus,
andNeptune.
OurExperim
ents
program
orbitsmodelsninebodiesin
thesolarsystem
,namelythesunandeightplanets.
Figures17.10and17.11show
snapshots
ofthe
outputfrom
orbitswithtw
odifferentzoom
factors
thatare
necessary
tospanthe
scale
ofthesystem
.Theorbitsforalltheplanetsarein
theproper
proportion.
But,obviously,
thesymbolsforthesunandtheplanetsdonothavethesamescale.
Web
sources
forinform
ationab
outthesolarsystem
are
provided
bytheUniversity
CorporationforAtm
ospheric
Research,theJet
PropulsionLaboratory,andtheUS
NationalAir
andSpace
Museum,
http://www.windows.ucar.edu
http://www.jpl.nasa.gov/solar_system
http://airandspace.si.edu:80/etp/ss/index.htm
235
−1
0
1
−1−0
.50
0.5
1
−0.3
−0.2
−0.10
0.1
0.2
0.3
Figure
17.11.Zoomingin
byafactorof16reveals
theorbitsoftheinner
planets,
Mercury,Ven
us,
EarthandMars.
Rec
ap
%%Orbits
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Orbits
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %orbits_recap
%edit
orbits_recap
%publish
orbits_recap
% %Related
EXM
programs
% %bouncer
%orbits
236
Chapter17
.Orbits
%%
Core
of
bouncer,
simple
gravity.
no
gravity
%Initialize
z0
=eps;
g=
9.8;
c=
0.75;
delta=
0.005;
v0
=21;
y=
[];
%Bounce
whilev0
>=
1
v=
v0;
z=
z0;
whilez
>=
0
v=
v-
delta*g;
z=
z+
delta*v;
y=
[y
z];
end
v0
=c*v0;
end
%Simplified
graphics
closeall
figure
plot(y)
%%
Normal
random
number
generator.
figure
hist(randn(100000,1),60)
%%
Snapshot
of
two
dimensional
Brownian
motion.
figure
m=
100;
x=
cumsum(randn(m,1));
y=
cumsum(randn(m,1));
plot(x,y,’.-’)
s=
2*sqrt(m);
axis([-ss
-s
s]);
237
%%Snapshot
of
three
dimensional
Brownian
motion,
brownian3
n=
50;
delta=
0.125;
P=
zeros(n,3);
fort
=0:10000
%Normally
distributed
random
velocities.
V=
randn(n,3);
%Update
positions.
P=
P+
delta*V;
end
figure
plot3(P(:,1),P(:,2),P(:,3),’.’)
boxon
%%Orbits,
the
n-body
problem.
%{
%ORBITS
n-body
gravitational
attraction
for
n=
2,
3or
9.
%ORBITS(2),
two
bodies,
classical
elliptic
orbits.
%ORBITS(3),
three
bodies,
artificial
planar
orbits.
%ORBITS(9),
nine
bodies,
the
solar
system
with
one
sun
and
8planets.
% %ORBITS(n,false)
turns
off
the
uicontrols
and
generates
astatic
plot.
%ORBITS
with
no
arguments
is
the
same
as
ORBITS(9,true).
%n
=number
of
bodies.
%P
=n-by-3
array
of
position
coordinates.
%V
=n-by-3
array
of
velocities
%M
=n-by-1
array
of
masses
%H
=graphics
and
user
interface
handles
if
(nargin
<2)
gui=
true;
end
if
(nargin
<1);
n=
9;
end
[P,V,M]=
initialize_orbits(n);
H=
initialize_graphics(P,gui);
steps=
20;
%Number
of
steps
between
plots
238
Chapter17
.Orbits
t=
0;
%time
whileget(H.stop,’value’)==
0
%Obtain
step
size
from
slider.
delta=
get(H.speed,’value’)/(20*steps);
fork
=1:steps
%Compute
current
gravitational
forces.
G=
zeros(size(P));
fori
=1:n
forj
=[1:i-1
i+1:n];
r=
P(j,:)
-P(i,:);
G(i,:)=
G(i,:)
+M(j)*r/norm(r)^3;
end
end
%Update
velocities
using
current
gravitational
forces.
V=
V+
delta*G;
%Update
positions
using
updated
velocities.
P=
P+
delta*V;
end
t=
t+
steps*delta;
H=
update_plot(P,H,t,gui);
end
finalize_graphics(H,gui)
end
%}
%%
Run
all
three
orbits,
with
2,
3,
and
9bodies,
and
no
gui.
figure
orbits(2,false)
figure
orbits(3,false)
figure
orbits(9,false)
239
Exe
rcises
17.1
Bouncingball.
(a)W
hatis
themaxim
um
heightofthebouncingball?
(b)How
manytimes
does
theballbounce?
(c)W
hatistheeff
ectofchangingeach
ofthefourbouncervalues
g,c,delta,an
dv0.
17.2
Pluto
andCeres.Changeorbitsto
orbits11byaddingtheerstwhileplanet
Pluto
andtherecentlypromoteddwarf
planet
Ceres.See
Wikipedia:
http://en.wikipedia.org/wiki/Planet
http://en.wikipedia.org/wiki/Ceres_(dwarf_planet)
and
http://orbitsimulator.com/gravity/articles/ceres.html
17.3
Comet.
Addacomet
toorbits.
Findinitialconditionsso
thatthecomet
hasastable,buthighly
ellipticalorbit
thatextendswellbeyondtheorbitsofthe
planets.
17.4
Twin
suns.
Turn
thesunin
orbitsinto
atw
instarsystem
,withtw
osuns
orbitingeach
other
outoftheplaneoftheplanets.
Whateventuallyhappensto
the
planetary
orbits?
Forexample,try
sun1.p=
[1
00];
sun1.v=
[0
0.25
0.25];
sun1.m=
0.5;
sun2.p=
[-1
00];
sun2.v=
[0
-0.25
-0.25];
sun2.m=
0.5;
Try
other
values
aswell.
240
Chapter17
.Orbits
Chapter18
Shallo
wW
aterEquations
Theshallow
waterequationsmodel
tsunamis
andwavesin
bathtubs.
Thischapterismore
advancedmathem
aticallythanearlierchapters,butyou
mightstillfinditinterestingeven
ifyoudonotmaster
themathem
aticaldetails.
Theshallow
waterequationsmodel
thepropagationofdisturbancesin
water
andother
incompressible
fluids.
Theunderlyingassumptionis
thatthedepth
of
thefluid
issm
allcomparedto
thewavelength
ofthedisturbance.For
example,we
donotordinary
thinkoftheIndianOceanasbeingshallow
.Thedepth
istw
oor
threekilometers.
Butthedevastatingtsunamiin
theIndianOceanonDecem
ber
26,2004involved
waves
thatweredozensor
hundredofkilometerslong.Sothe
shallow
waterapproxim
ationprovides
areasonable
model
inthis
situation.
Theequationsare
derived
from
theprinciplesofconservation
ofmass
and
conservationofmomentum.Theindep
endentvariablesare
time,
t,andtw
ospace
coordinates,xandy.Thedep
endentvariablesare
thefluid
heightor
dep
th,h,an
dthetw
o-dim
ensionalfluid
velocity
field,uandv.W
iththeproper
choiceofunits,the
conserved
quantities
aremass,whichis
proportionalto
h,an
dmomentum,which
isproportionalto
uhandvh.Theforceactingon
thefluid
isgravity,
represented
bythegravitationalconstant,g.Thepartialdifferentialequationsare:
∂h ∂t+
∂(uh)
∂x
+∂(vh)
∂y
=0
∂(uh)
∂t
+∂(u
2h+
1 2gh2)
∂x
+∂(uvh)
∂y
=0
∂(vh)
∂t
+∂(uvh)
∂x
+∂(v
2h+
1 2gh2)
∂x
=0
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
241
242
Chapter18
.Shallow
Water
Equations
Inorder
towrite
theequationsin
acompact
form
,introduce
threevectors.
U=
⎛ ⎝h uh
vh
⎞ ⎠
F(U
)=
⎛ ⎝uh
u2h+
1 2gh2
uvh
⎞ ⎠
G(U
)=
⎛ ⎝vh
uvh
v2h+
1 2gh2
⎞ ⎠W
iththisnotation,theshallow
waterequationsareaninstance
ofahyperbolic
conservationlaw.
∂U ∂t+
∂F(U
)
∂x
+∂G(U
)
∂y
=0
Onedelicate
aspectofthis
model
isthebou
ndary
conditions,
especiallyifwe
intendto
model
arealworld
geometry
such
astheIndianOcean.For
oursimple
experim
ent,weconfineourselves
toasquare
regionandspecifyreflective
bou
ndary
conditions,
u=
0ontheverticalsides
ofthesquare
andv=
0onthehorizontal
sides.Theseconditionscause
anywaves
thatreach
thebou
ndary
tobereflected
back
into
theregion.
More
realistic
modelsof
oceansandtsunamisincludeterm
sthatdescribethe
topographyoftheoceanfloor,theCoriolisforceresultingtheearth’srotation,and
possibly
other
externalforces.Buttheequationsweareconsideringherearestill
thebasisofsuch
models.
Figure
18.1.Atthebeginningofatimestep,thevariables
representthe
solutionatthecenters
ofthefinitedifferen
cegrid.
Wewilluse
theLax-Wen
droff
methodto
compute
anumericalap
proxim
ation
tothesolution.Introduce
aregularsquare
finitedifferen
cegridwithavector-valued
solutioncenteredin
thegridcells,
asshow
nin
figure
18.1.Thequantity
Un i,j
243
represents
athreecomponentvectorateach
gridcelli,jthatevolves
withtimestep
n.
Each
timestep
involves
twostages,somethinglikeatw
o-stageRungeKutta
methodforordinary
differentialequations.
Thefirststageis
ahalfstep;it
defines
values
ofU
attimestep
n+
1 2andthemidpoints
oftheedges
ofthegrid,as
show
nin
figure
18.2.
Un+
1 2
i+1 2,j=
1 2(U
n i+1,j+
Un i,j)−
Δt
2Δx(F
n i+1,j−F
n i,j)
Un+
1 2
i,j+
1 2
=1 2(U
n i,j+1+
Un i,j)−
Δt
2Δy(G
n i,j+1−G
n i,j)
Figure
18.2.Thefirststage
computesvalues
thatrepresentthesolution
atthemidpoints
oftheedgesin
thefinitedifferen
cegrid.
Thesecondstagecompletesthetimestep
byusingthevalues
computedin
the
firststageto
compute
new
values
atthecenters
ofthecells,returningto
figure
18.1.
Un+1
i,j
=U
n i,j−
Δt
Δx(F
n+
1 2
i+1 2,j−F
n+
1 2
i−1 2,j)−
Δt
Δy(G
n+
1 2
i,j+
1 2
−G
n+
1 2
i,j−
1 2
)
OurMATLABprogram,exm/waterwave,usesLax-W
endroffto
solvetheshal-
lowwaterequationsonasquare
regionwithreflectiveboundary
conditions.
Initially,
h=
1,u
=0,v
=0over
theentire
region,so
thesolutionis
static.
Then,at
re-
peatedintervals,atw
odim
ensionalGaussianshaped
peakisadded
toh,simulating
anim
pulsivedisturbance
likeawaterdrophittingthesurface.Theresultingwaves
propagate
back
andforthover
theregion.A
few
snapshots
ofthedynamic
graphic
are
show
nin
figure
18.3.TheLax-W
endroffschem
eamplifies
artificial,nonphysi-
caloscillations.
Eventuallythenumericalvalues
overflow
,producingfloatingpoint
Infsan
dNaNs,
whichcause
thesurface
plotto
disappear.
CLAW
PACK,whichstandsforConservationLaw
Package,isalargecollection
ofFortransubroutines
developed
byRandyLeV
equeandhiscolleagues
attheUni-
versity
ofWashington.A
versionofthepackagespecializedto
modelingtsunamis
hasbeendeveloped
byDav
idGeorge.
See:
http://www.clawpack.org
http://www.amath.washington.edu/~dgeorge/tsunamimodeling.html
244
Chapter18
.Shallow
Water
Equations
Figure
18.3.A
waterdropinitiatesawave
thatreflects
offtheboundary.
Rec
ap
%%
Shallow
Water
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Shallow
Water
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %water_recap
%edit
water_recap
%publish
water_recap
% %Related
EXM
programs
% %waterwave
%%
Finite
Differences
%A
simple
example
of
the
grid
operations
in
waterwave.
245
%%Create
atwo
dimensional
grid.
m=
21;
[x,y]=
ndgrid(-1:
2/(m-1):
1);
%%The
water
drop
function
from
waterwave.
U=
exp(-5*(x.^2+y.^2));
%%Surf
plot
of
the
function
clf
shg
h=
surf(x,y,U);
axisoff
ax
=axis;
%%Colormap
c=
(37:100)’;
cyan=
[0*c
cc]/100;
colormap(cyan)
pause(1)
%%Indices
in
the
four
compass
directions.
n=
[2:m
m];
e=
n;
s=
[1
1:m-1];
w=
s;
%%A
relation
parameter.
Try
other
values.
%Experiment
with
omega
slightly
greater
than
one.
omega=
1;
%%Relax.
%Repeatedly
replace
grid
values
by
relaxed
average
of
four
neighbors.
tfinal=
500;
fort
=1:tfinal
U=
(1-omega)*U
+omega*(U(n,:)+U(:,e)+U(s,:)+U(:,w))/4;
set(h,’zdata’,U);
axis(ax)
drawnow
end
246
Chapter18
.Shallow
Water
Equations
Exe
rcises
18.1
Relax.
Modifyomegain
water_recap.Try
values
slightlylarger
than1.0
.
18.2
Boundary
conditions.
Modifytheboundary
conditionsusedin
waterwave.
Chapter19
MorseCode
Morsecodedem
onstratesbinary
treesandcellarrays.
HV
FL
PJ
BX
CY
ZQ
SU
RW
DK
GO
IA
NM
ET
Figure
19.1.Thebinary
tree
defi
ningMorsecode.
Abranch
totheleft
sign
ifies
adotin
thecodeandabranch
totherightis
adash.In
additionto
the
rootnode,
thereare
26nodes
containingthecapitallettersoftheEnglishalphabet.
Thischapterbringstogether
threedisparate
topics:
Morsecode,
binary
trees,
andcellarray
s.Morsecodeis
nolonger
importantcommercially,
butit
stillhas
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
247
248
Chapter19
.Morse
Code
someav
idfansamonghobbyists.Binary
treesare
afundamentaldata
structure
used
throughoutmoderncomputing.Andcellarrays,whichare
uniqueto
Matlab,are
array
swhose
elem
ents
areother
arrays.
You
canget
aheadstart
onou
rinvestigationbyrunningtheexmprogram
morse_gui
Experim
entwiththefourbuttonsin
variouscombinations,
andthetextandcode
box
.This
chapterwillexplain
theiroperation.
Morsecodewasinventedover
150years
ago,notbySamuel
F.B.Morse,
but
byhiscolleague,AlfredVail.Ithasbeenin
widespreaduse
ever
since.Thecodecon-
sistsofshort
dots,’.’,an
dlonger
dashes,’-’,separatedbyshort
andlongspaces.
Youare
certainly
familiarwiththeinternationaldistresssignal,’...---...’,
thecodefor“SOS”,abbreviating“SaveOurShips”
orperhaps“SaveOurSouls”.
Butdid
you
notice
thatsomemoderncellphones
signal’...--
...’,thecode
for“S
MS”,indicatingactivityof
the“Short
MessageService”.
Until2003,alicense
tooperate
anam
ateurradio
required
minim
alproficiency
inMorsecode.
(Fulldisclosure:W
hen
Iwasin
juniorhighschool,Ilearned
Morse
codeto
get
myham
license,andI’venever
forgotten
it.)
Accordingto
Wikipedia,in
2004,theInternationalTelecommunicationUnion
form
allyadded
acodefortheubiquitousem
ailcharacter,@,to
theinternational
Morsecodestandard.This
wasthefirstad
ditionsince
WorldWar
I.
TheMorsetree
Wecould
provideatable
show
ingthat’.-’isthecodeforA,’-...’thecodefor
B,andso
on.Butwe’re
notgoingto
dothat,
andourMatlabprogram
does
not
start
withsuch
atable.Instead,wehavefigure
19.1.This
isabinary
tree,an
dfor
ourpurposes,
itis
thedefi
nition
ofMorsecode.
Incontrast
tonature,computer
scientistsputtheroot
ofatree
ontop.Startingatthisroot,or
anyother
node,
and
mov
ingleft
alongalinksignifies
adot,whilemov
ingrightis
adash.Forexample,
startingattherootandmov
ingonestep
totheleft
follow
edbyoneto
theright
getsusto
A.Sothis
fact,rather
thanatable,tellsus’.-’is
thecodeforA.
Thelength
ofaMorsecodesequence
foraparticularcharacter
isdetermined
by
thefrequency
ofthatcharacter
intypicalEnglish
text.
Themost
frequent
character
is“E
”.Consequently,
itsMorsesequence
isasingle
dotan
dit
islinked
directlyto
theroot
ofou
rtree.Theleast
frequentcharacters,such
as“Z
”and“X”,
havethelongestMorsesequencesandarefarfrom
theroot.
(Wewillconsider
the
fourmissingnodes
inthetree
later.)
Binary
treesare
bestim
plementedin
Matlabbycellarrays,whichare
array
swhose
elem
ents
arethem
selves
other
Matlabobjects,includingother
arrays.
Cell
array
shavetw
okindsofindexingop
erations.
Curlybraces,
{and},areusedfor
constructionandforaccessingindividualcells.
Conventionalsm
ooth
parentheses,
(and),areusedforaccessingsubarray
s.Forexample,
C=
{’A’,’rolling’,’stone’,’gathers’,’momentum’,’.’}
249
producesacellarray
thatcontainssixstringsofdifferentlengths.
This
example
isdisplayedas
C=
’A’’rolling’’stone’’gathers’’momentum’’.’
Thethirdelem
ent,denotedwithcurlybracesbyC{3},is
thestring’stone’.The
thirdsubarray,
denotedwithparentheses
byC(3),isanother
cellarraycontaininga
single
elem
ent,thestring’stone’.Now
goback
andreadthose
last
twosentences
afewmore
times.Thesubtledistinctionbetweenthem
isboththekey
tothepow
erofcellarraysandthesourceofpervasiveconfusion.
Thinkof
astringofmailbox
esalongCstreet.
Assumethey
are
numbered
consecutively.
Then
C(3)is
thethirdmailbox
andC{3}is
themailin
thatbox
.Byitself,Cistheentire
array
ofmailbox
es.TheexpressionC(1:3)isthesubarray
containingthefirstthreemailbox
es.
And
hereis
an
unusualconstruction,with
curlybracesC{1:3}is
acommaseparatedlist,
C{1},C{2},C{3}
Byitself,onthecommandline,
thiswilldothreeassignmentstatements,assigning
thecontents
ofeach
ofthefirstthreemailbox
es,oneatatime,
toans.W
ithmore
curlybraces,{C{1:3}}is
thesameasC(1:3).
Inthecomputerhardware
itself,thereis
adistinction
between
amem
ory
locationwithaparticularad
dress
andthecontents
ofthatlocation.
This
same
distinctionis
preserved
byindexingacellarraywithparentheses
andwithbraces.
Did
youseethe“M
enin
Black”mov
ies?
Inoneofthem
theMIB
headquarters
hasabankofstoragelockers.
Itturnsou
tthateach
locker
containsan
entire
civilization,presumably
withitsow
nlockers.
Attheen
dof
themov
ieitisrevealed
thattheEarthitselfis
astoragelocker
inalarger
civilization.It
ispossible
that
weare
alllivingin
oneelem
entofahugecellarray.
Thebinary
tree
definingMorsecodeis
acellarraywhose
contents
are
char-
actersandother
cellarrays.
Each
cellrepresents
anodein
thetree.
Acell,N,
hasthreeelem
ents,Thefirstelem
ent,N{1},is
astringwithasingle
capitalletter,
X,designatingthenode.
Thesecondelem
ent,
N{2},is
another
cellarray,thedot
branch.Thethirdelem
ent,N{3},is
thedash
branch.Thereare
acouple
ofexcep-
tionalcases.
Theroot
nodedoes
nothaveanassociatedletter,so
itsfirstelem
ent
isanem
pty
string.TheUandRnodes,andtheleafnodes,haveoneortw
oem
pty
cellarraysforbranches.
Inprinciple,wecould
create
theentire
Morsebinary
tree
withasinglegigantic,
butunrealistic,assignmentstatement.
M=
{’’
...
{’E’...
{’I’{’S’{’H’{}
{}}
{’V’
{}
{}}}
...
{’U’{’F’{}
{}}
{}}}
...
{’A’{’R’{’L’{}
{}}
{}}
...
{’W’{’P’{}
{}}
{’J’
{}
{}}}}}
...
{’T’...
250
Chapter19
.Morse
Code
{’N’{’D’
{’B’{}
{}}
{’X’
{}
{}}}
...
{’K’{’C’{}
{}}
{’Y’
{}
{}}}}
...
{’M’{’G’
{’Z’{}
{}}
{’Q’
{}
{}}}
...
{’O’{}
{}}}}}
You
canseethecellarrayswithin
cellarraysandthemanyem
pty
cellarraysat
theleaves.Thisstatementactuallyworks,butitisunreasonable
because
itiserror
pronean
dnearlyim
possible
toextend.Instead,ourfunctionmorse_treebegins
withaheader
functionM
=morse_tree
This
isfollow
edby27
assignmentstatements,tw
elve
atlevel
four.
h=
{’H’
{}
{}};
v=
{’V’
{}
{}};
f=
{’F’
{}
{}};
l=
{’L’
{}
{}};
p=
{’P’
{}
{}};
j=
{’J’
{}
{}};
b=
{’B’
{}
{}};
x=
{’X’
{}
{}};
c=
{’C’
{}
{}};
y=
{’Y’
{}
{}};
z=
{’Z’
{}
{}};
q=
{’Q’
{}
{}};
Eightat
level
three.
s=
{’S’
hv};
u=
{’U’
f{}};
r=
{’R’
l{}};
w=
{’W’
pj};
d=
{’D’
bx};
k=
{’K’
cy};
g=
{’G’
zq};
o=
{’O’
{}
{}};
Fou
ratlevel
two.
i=
{’I’
su};
a=
{’A’
rw};
n=
{’N’
dk};
m=
{’M’
go};
Twoat
level
one.
e=
{’E’
ia};
t=
{’T’
nm};
251
Andfinallyoneassignmentstatementat
level
zero
tocreate
therootnode.
M=
{’’
et};
This
functionis
attheheart
ofallourMorsecodesoftware.
You
cantravel
dow
nthis
tree
byfirstenteringthreecommands.
M=
morse_tree
M=
M{2}
M=
M{3}
You
then
canuse
theup-arrow
onyourkeyboard
torepeatedly
select
andreexecute
thesecommands.
Returningto
thefirstcommandgives
youafreshtree.Executing
thesecondcommandisadotoperation,mov
ingdow
nthetree
totheleft.Executing
thethirdcommandis
adash
operation,mov
ingdow
nthetree
totheright.
For
example,thefivecommands
M=
morse_tree
M=
M{3}
M=
M{2}
M=
M{2}
M=
M{3}
correspondto
theMorsesequence
’-..-’.This
bringsyouto
thenode
’X’
{}
{}
Youhavereached
theXleafofthetree.
Sea
rchingthetree
Returningto
the“Men
inBlack”analogy,
theMorsebinary
tree
isasingle
locker.
When
youop
enthatlocker,you
seeanem
pty
string(because
therootdoes
not
haveaname)
andtw
omore
lockers.
When
youopen
thelocker
ontheleft,yousee
an’E’andtw
omore
lockers.
Now
youhavethreechoices.You
canopen
either
of
thetw
olockersin
theElocker,or
youcangoback
totherootlocker
andopen
the
locker
ontherightto
visittheTlocker.
Rep
eatedly
choosingdifferentlockers,ordifferentbranches
inthetree,corre-
spondsto
traversingthetree
indifferentorders.
Amongthesemanypossibleorders,
twohavestandard
names,“depth
firstsearch”and“breadth
firstsearch”.They
are
show
nin
figures19.2
and19.3
andyoucanseeandhearanim
atedversionswiththe
morse_guiprogram.
Depth
firstsearchvisitseach
branch
assoon
asitsees
it.W
hen
ithasvisited
bothbranches
atanode,
itbacksupto
thefirstnodethathasanavailable
branch.
Figure
19.2
show
stheprogress
ofdepth
firstorder
upto
nodeW.
EI
SH
VU
FA
RL
W
Nothingto
therightofWhasyet
beenvisited.
252
Chapter19
.Morse
Code
E
I
S
HV
U
F
A
R
L
W
..
..
Figure
19.2.A
depth
firstsearchin
progress.
Thenodes
are
visitedfrom
leftto
right.
Breadth
firstsearch
takes
onestep
alongeach
branch
atanodebefore
itcontinues.Theresultisatopto
bottom
search,much
likereadingEnglish
language
text.
Figure
19.3
show
sthebreadth
firstorder
upto
nodeW.
ET
IA
NM
SU
RW
Nothingbelow
Whasyet
beenvisited.
Dep
thfirstsearchusesadata
structure
know
nasastack.
Hereis
acode
segmentwithastack
thatsimply
displaysthenodes
ofthetree
indepth
firstorder.
S=
{morse_tree};
while~isempty(S)
N=
S{1};
S=
S(2:end);
if
~isempty(N)
fprintf(’%s’,N{1})
S=
{N{2}
N{3}
S{:}};
end
end
fprintf(’\n’)
Thestack
Sis
acellarray.
Initially,
ithasonecellcontainingthetree.Thewhile
253
ET
IA
NM
SU
RW
..
..
Figure
19.3.A
breadth
firstsearchin
progress.
Thenodes
are
visitedfrom
topto
bottom.
loopcontinues
aslongas
thestack
isnotem
pty.W
ithin
thebodyoftheloopa
nodeisremoved
from
thetopofthestack
andthestack
shortened
byoneelem
ent.
Ifthenodeisnotem
pty,thesingle
character
representingthenodeisdisplayedand
twonew
nodes,thedotanddash
branches,are
inserted
into
thetopofthestack.
Thetraversalvisitsrecentlydiscoverednodes
before
itreturnsto
older
nodes.
Breadth
firstsearchusesadata
structure
know
nas
aqueue.
Hereis
another
codesegment,thistimewithaqueue,
thatdisplaysthenodes
ofthetree
inbreadth
firstorder.
Q=
{morse_tree};
while~isempty(Q)
N=
Q{1};
Q=
Q(2:end);
if
~isempty(N)
fprintf(’%s’,N{1})
Q=
{Q{:}
N{2}
N{3}};
end
end
fprintf(’\n’)
Thiscodeissimilarto
thestack
code.
Thedistinguishingfeature
isthatnew
nodes
254
Chapter19
.Morse
Code
are
inserted
attheend,rather
thatthebeginningofthequeue.
Aqueueisem
ployinga“First
In,First
Out”,or
FIFO,strategy,
whileastack
isem
ployinga“Last
In,First
Out”,or
LIFO,strategy.
Aqueueis
likealineat
agrocery
store
orticket
office.Customersat
thebeginningofthelineareserved
first
andnew
arrivalswaitattheendoftheline.
Withastack,new
arrivalscrow
din
at
thestart
oftheline.
Ourstack
andqueuecodes
arenotrecursive.
They
simply
loopuntilthereis
nomore
workto
bedone.
Hereis
differentap
proach
thatem
ploysrecursionto
do
adepth
firstsearch.Actually,
this
does
involveahidden
stack
because
computer
system
ssuch
asMatlabuse
stacksto
managerecursion.
functiontraverse(M)
if
nargin
==
0
M=
morse_tree;
%Initial
entry.
end
if
~isempty(M)
disp(M{1})
traverse(M{2})
%Recursive
calls.
traverse(M{3})
end
end%
traverse
Decodeanden
code
Decodingis
theprocess
oftranslatingdots
anddashes
into
text.
Encodingis
the
reverse.W
ithou
rbinary
tree,decodingis
easier
thanencodingbecause
thedots
anddashes
directlydeterminewhichlinksto
follow
.Hereisafunctionthatdecodes
onecharacter’s
worth
ofdots
anddashes.Thefunctionreturnsan
asterisk
ifthe
inputdoes
notcorrespondto
oneofthe26
lettersin
thetree.
functionch
=decode(dd)
M=
morse_tree;
fork
=1:length(dd)
ifdd(k)
==
’.’
M=
M{2};
elseifdd(k)==
’-’
M=
M{3};
end
ifisempty(M)
ch=
’*’;
return
end
end
ch
=M{1};
end%
decode
255
Encodingis
alittle
more
work
because
wehaveto
searchthetree
untilwefind
thedesired
letter.Hereisafunctionthatem
ploysdepth
firstsearchto
encodeon
echaracter.
Astack
ofdots
anddashes
isbuiltupduringthesearch.
Again,an
asteriskis
returned
iftheinputcharacter
isnotin
thetree.
functiondd
=encode(ch)
S=
{morse_tree};
D=
{’’};
while~isempty(S)
N=
S{1};
dd
=D{1};
S=
S(2:end);
D=
D(2:end);
if
~isempty(N)
if
N{1}
==
ch;
return
else S=
{N{2}
N{3}
S{:}};
D=
{[dd
’.’]
[dd
’-’]
D{:}};
end
end
end
dd
=’*’;
end%
encode
Thesetw
ofunctionsarethecore
ofthedecodingandencodingaspectofmorse_gui.
Exten
sion
ET
IA
NM
SU
RW
DK
GO
HV
FÜ
LÄ
PJ
BX
CY
ZQ
Ö×
54
3Ð
2+
À1
67
Ñ8
90
?.
@−
;,
:
Figure
19.4.TheMorsetree
extended
twolevels
toaccommodate
digits,
punctuation,andnon-E
nglishcharacters.
256
Chapter19
.Morse
Code
Abinary
tree
withfourlevelshas25−
1=
31nodes,countingtheroot.
Our
morse_treehasarootandon
ly26
other
nodes.So,thereare
fourem
pty
spots.
You
canseethem
onthedash
branches
ofU
andR
andon
both
branches
ofO.So
far,
ourdefinitionofMorsecodedoes
notprovideforthefoursequences
..--
.-.-
---.
----
How
should
thesebedecoded?Wealsowan
tto
addcodes
fordigitsan
dpunctuation
marks.
And,it
wou
ldbeniceto
provideforatleast
someof
thenon-E
nglish
charactersrepresentedwithumlauts
andother
diacriticalmarks.
Morsecodewasinvented100years
before
moderncomputers.Today,alpha-
bets,
keyboards,
character
sets
andfonts
vary
from
countryto
countryan
deven
from
computerto
computer.
Ourfunction
morse_table_extended
extendsthetree
twomore
levelsandad
ds10digits,
8punctuationcharacters,
and
7non-E
nglish
charactersto
the26charactersin
theoriginalmorse_table.Figure
19.4
show
stheresult.Theextendbuttonin
morse_guiaccessesthis
extension.
Someoftheissues
involved
inrepresentingtheseadditionalcharactersare
pursued
intheexercises.
Morseco
detable
Wepromised
thatwewou
ldnotuse
atable
todefineMorsecode.
Everythinghas
beenbasedonthebinary
tree.W
hen
anactualtable
isdesired
wecangenerate
onefrom
thetree.Ourfunctionisnamed
morse_codeanditem
ploystherecu
rsive
algorithm
traversefrom
theprevioussection.
Therecursion
carriesalongthe
emergingtable
Candagrowingstringddofdots
anddashes.
Indexinginto
thetable
isbasedupontheASCII
codeforcharacters.
The
ASCIIstandard
isan7-bitcodethatisthebasisforcomputerrepresentationoftext.
Seven
bitsprovidefor27
=128characters.
Thefirst32oftheseare
nonprinting
charactersthatwereoriginallyused
forcontrolofteletypes
and
are
now
largely
obsolete.Theremaining96
arethe52
upper
andlower
case
lettersoftheEnglish
alphabet,10digits,
and
32punctuation
marks.
InMatlab
thefunctionchar
convertsanumeric
valueto
acharacter.Forexample
char(65)
ans=
’A’
Thefunctiondoubleconvertsacharacter
toafloatingpointnumber
andthefunc-
tionuint8converts
acharacter
toanunsigned
8-bitinteger.Either
ofthesecanbe
usedasanindex.For
example
double(’A’)
ans=
65
257
Acomputerbyte
is8bitsan
dso
theASCII
standard
isreadilyextended
byan
other
128characters.
Theactualgraphic
printedforsomeof
thesecharactersmay
vary
from
countryto
countryandfrom
fontto
font.
Ourfunctionmorse_codeproducesatableofboth
ASCIIandMorsecodefrom
either
morse_treeormorse_tree_extended.Therecu
rsivetraversealgorithm
isusedforthedepth
firstsearch.ASCII
codes
are
usedasindices
into
a256-element
cellarrayofdots
anddashes.Thesearchinsertsonly
26or51elem
ents
into
the
array,so
afinalstep
createsaprintable
table.
functionC
=morse_code(C,M,dd)
%MORSE_CODE
%C
=morse_code
%C
=morse_code(morse_tree)
%C
=morse_code(morse_tree_extended)
if
nargin
<3
%Choose
binary
tree
if
nargin
==
0
M=
morse_tree;
else M=
C;
end
C=
cell(256,1);
%The
temporary
code
table
dd
=’’;
%dots
and
dashes
end
if
~isempty(M)
%Depth
first
search
if
~isempty(M{1})
C{double(M{1})}=
dd;
%Use
ASCII
value
as
an
index
end
C=
morse_code(C,M{2},[dd
’.’]);
%Recursive
call
C=
morse_code(C,M{3},[dd
’-’]);
%Recursive
call
end
if
nargin
<3
%Final
processing,
convert
to
char.
c=
char(C{:});
k=
find(c(:,1)
~=
’’);
%Find
the
nonblank
entries.
b=
blanks(length(k))’;
C=
[char(k)
bb
int2str(k)
bb
char(C{k})];
end
Theou
tputfrom
morse_code
isthefollow
ing26lines.
A65
.-
B66
-...
258
Chapter19
.Morse
Code
C67
-.-.
D68
-..
E69
.
F70
..-.
G71
--.
H72
....
I73
..
J74
.---
K75
-.-
L76
.-..
M77
--
N78
-.
O79
---
P80
.--.
Q81
--.-
R82
.-.
S83
...
T84
-
U85
..-
V86
...-
W87
.--
X88
-..-
Y89
-.--
Z90
--..
You
should
run
morse_code(morse_tree_extended)
toseewhatoutputis
generatedon
yourcomputer.
Referen
ces
[1]W
ikipedia
article
onMorsecode.
See
thecomplete
Morsecodetree
attheend
ofthearticle.
http://en.wikipedia.org/wiki/Morse_code
Rec
ap
%%
Morse
Code
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Morse
Code
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %morse_recap
%edit
morse_recap
259
%publish
morse_recap
% %Related
EXM
programs
% %morse_gui
%morse_tree
%morse_tree_extended
%morse_code
%%Cell
Arrays
C=
{’A’,’rolling’,’stone’,’gathers’,’momemtum’,’.’}
C{3}
C(3)
C(1:3)
C{1:3}
{C{1:3}}
%%Create
aMorse
Tree
%An
absurd
statement.
For
abetter
way,
see
morse_tree.
M=
{’’
...
{’E’...
{’I’{’S’{’H’{}
{}}
{’V’
{}
{}}}
...
{’U’{’F’{}
{}}
{}}}
...
{’A’{’R’{’L’{}
{}}
{}}
...
{’W’{’P’{}
{}}
{’J’
{}
{}}}}}
...
{’T’...
{’N’{’D’{’B’{}
{}}
{’X’
{}
{}}}
...
{’K’{’C’{}
{}}
{’Y’
{}
{}}}}
...
{’M’{’G’{’Z’{}
{}}
{’Q’
{}
{}}}
...
{’O’{}
{}}}}}
%%Follow
’-..-’
M=
morse_tree
M=
M{3}
M=
M{2}
M=
M{2}
M=
M{3}
%%Depth
first,
with
astack.
S=
{morse_tree};
while~isempty(S)
N=
S{1};
S=
S(2:end);
if
~isempty(N)
260
Chapter19
.Morse
Code
fprintf(’%s’,N{1})
S=
{N{2}
N{3}
S{:}};
end
end
fprintf(’\n’)
%%
Breadth
first,
with
aqueue.
Q=
{morse_tree};
while~isempty(Q)
N=
Q{1};
Q=
Q(2:end);
if
~isempty(N)
fprintf(’%s’,N{1})
Q=
{Q{:}
N{2}
N{3}};
end
end
fprintf(’\n’)
%%
Recursive
traversal.
%function
traverse(M)
%if
nargin
==
0
%M=
morse_tree;
%Initial
entry.
%end
%if
~isempty(M)
%disp(M{1})
%traverse(M{2})
%Recursive
calls.
%traverse(M{3})
%end
%end
%traverse
%%
ASCII
character
set
k=
reshape([32:127
160:255],32,[])’;
C=
char(k)
txt=
text(.25,.50,C,’interp’,’none’);
set(txt,’fontname’,’LucidaSansTypewriter’)
Exe
rcises
19.1
Greetings.W
hatdoes
this
say?
.....
.-..
.-..
---
.--
---
.-.
.-..
-..
19.2
Email.Use
theextendbuttonandtranslate
box
inmorse_guito
translate
you
rem
ailad
dress
into
Morsecode.
261
19.3
dash.W
hydidn’t
Iuse
anunderscore,’_’,insteadofaminussign,’-’,to
representadash?
19.4
Note.W
hatmusicalnote
isusedbythesoundfeature
inmorse_gui?
19.5
Reverse
order.Findthis
statementin
functiondepthin
morse_gui.
S=
{N{2}
N{3}
S{:}};
WhathappensifyouinterchangeN{2}andN{3}like
this?
S=
{N{3}
N{2}
S{:}};
19.6
Exten
d.Usingthetree
attheendoftheW
ikipedia
article,addmore
characters
tomorse_code_extended.Ifyou
livein
alocale
thathasnon-E
nglish
charactersin
thealphabet,besure
toincludethem
.
19.7
Fourdashes.W
hydoes
thecharacter
withMorsecode’----’cause
aunique
diffi
cultyin
morse_code_extended?
19.8
YouTube.
Checkout“Morsecode”
onYouTube.
Besure
tolisten
tothe
“Morsecodesong”.W
howinstheMorsecodeversustextingcontest
onJay
Leno’s
Ton
iteshow
?
19.9
Trinary.
Whatdoes
this
function
do?
Whyis
itnamed
trinary.
What
determines
how
longitruns?
Whatcausesitto
term
inate?W
hyistheifstatement
necessary?
Modifytheprogram
tomakeit
use
adepth
firstsearch.Modifythe
program
tomakeitwork
withmorse_tree_extended.
functiontrinary
T=
[0
00
0];
Q=
{morse_tree};
whileany(T(1,:)<
2)
p=
T(1,:);
y=
polyval(p,3);
if
~isempty(Q{1})
fprintf(’%s
%d%d%d%d
%2d\n’,Q{1}{1},p,y)
Q=
{Q{2:end}
Q{1}{2}
Q{1}{3}};
else Q=
{Q{2:end}
{}
{}};
end
T=
[T(2:end,:);
[T(1,2:end)
1];
[T(1,2:end)
2]];
end
end%
trinary
19.10Cellarrays.Let
262
Chapter19
.Morse
Code
C=
{’A’
’rolling’
’stone’
’gathers’
’momentum’
’.’}
Explain
whyeach
ofthefollow
ingdoes
whatitdoes.
C’
char(C)
size(C)
size(char(C))
double(char(C))
upper(C)
C(end)
C{end}
C{3}(3)
fliplr(C)
[C{:}]
C(5:7)=
{’no’
’moss’
’.’}
19.11Fonts.Experim
entwithvariousfonts.See
theW
ikiped
iaarticleontheASCII
character
set.
Youcangenerate
aprintable
character
table
inMatlabwith
k=
reshape([32:127
160:255],32,[])’;
C=
char(k)
Thefirsthalfof
this
table
isstandard,butthesecondhalfdep
endsuponthefonts
you
areusing.
You
can
changefonts
inthecommand
window
byaccessingthe
“File”
menu,then
selecting“P
references”
and“F
onts”.Highlightthedesktopcode
fontanduse
theupan
ddow
narrow
keys.
Inmyopinionthebestfontforthe
Matlabcommandwindow
isLucidaSansTypew
riter
You
canalsodisplaythecharacter
table
inthefigure
window
with
txt=
text(.25,.50,C,’interp’,’none’);
Tochangedisplayfonts
inthefigure
window
,trycommandsliketheseworkonyou
rcomputer.
set(txt,’fontname’,’LucidaSansTypewriter’)
set(txt,’fontname’,’CourierNew’)
set(txt,’fontname’,’ComicSansMS’)
set(txt,’fontname’,’Wingdings’)
set(txt,’fontname’,’GiGi’)
Use
thefontnames
available
under
thecommandwindow
preferences.
Chapter20
Music
Whatdoes
12√ 2
have
todowithmusic?
Inthetheory
ofmusic,
anoctave
isanintervalwithfrequencies
thatrange
over
afactorof
two.
Inmost
Western
music,
anoctaveis
divided
into
twelve
semitones
withequalfrequency
ratios.
Since
twelvesemitones
comprise
afactorof
two,onesemitoneis
afactorof
12√ 2
.Andbecause
this
quantity
occurs
sooften
inthis
chapter,
let
σ=
12√ 2
OurMatlabprogramsuse
sigma=
2^(1/12)
=1.059463094359295
Thinkofσasanim
portantmathem
aticalconstant,likeπandφ.
Key
board
Figure
20.1
show
sourminiature
pianokeyboard
with25
keys.
This
keyboard
has
twooctaves,withwhitekeyslabeled
CD
...
GA
B,plusanother
Ckey.Counting
bothwhitean
dblack,therearetw
elveskeysin
each
octave.
Thefrequency
ofeach
key
isasemitoneaboveandbelow
itsneighbors.Each
black
key
canberegarded
aseither
thesharp
ofthewhitebelow
itor
theflatofthewhiteaboveit.Sothe
black
key
betweenCandDis
bothC�andD�.
Thereis
noE�/F�orB�/C�.
Aconventionalfullpianokeyboard
has88keys.
Seven
complete
octaves
ac-
countfor7×
12=
84keys.
Thereare
threeadditionalkeysatthelower
left
and
onead
ditionalkey
attheupper
end.Iftheoctaves
are
numbered
0through8,then
Copyright
c ©2011CleveMoler
Matlab
R ©is
aregisteredtradem
ark
ofMath
Works,
Inc.
TM
October
4,2011
263
264
Chapter20
.Music
Figure
20.1.Miniature
pianokeyboard.
Figure
20.2.Middle
C.
akey
letter
follow
edbyanoctavenumber
specifies
auniquekey.In
this
notation,
twoim
portantkeysare
C4andA4.TheC4key
isnearthecenterof
thekeyboard
andso
isalsoknow
nasmiddle
C.A
pianois
usuallytuned
sothatthefrequency
oftheA4key
is440Hz.
C4is
ninekeysto
theleft
ofA4so
itsfrequen
cyis
C4=
440σ−9≈
261.6256
Hz
OurEXMprogram
pianoexusesC4asthecenterof
its25keys,so
thenumber
265
rangeis-12:12.Thestatement
pianoex(0)
generatesan
dplaysthesoundfrom
asinewavewithfrequency
C4.Theresulting
visualdisplayis
show
nin
figure
20.2.
Thisforloopplaysatw
ooctavechromaticscalestartingcoveringall25
notes
onourminiature
keyboard.
forn
=-12:12
pianoex(n)
end
DoReMi
Oneofthefirstsongsyoulearned
tosingwas
DoReMiFaSoLaTiDo
Ifyoustart
atC4,youwould
besingingthemajorscale
inthekey
ofC.Thisscale
isplayed
onapianousingon
lythewhitekeys.
Thenotesare
notequallyspaced.
Most
ofthestepsskip
over
black
keysan
dso
are
twosemitones.
Butthesteps
betweenMiandFaandTiandDoare
thestepsfrom
Eto
FandBto
C.Thereare
nointerveningblack
keysandso
thesestepsare
only
onesemitone.
Interm
sof
σ,
theC-m
ajorscale
is
σ0σ2σ4σ5σ7σ9σ11σ12
Youcanplaythis
scale
onourminiature
keyboard
with
forn
=[0
24
57
911
12]
pianoex(n)
end
Thenumber
ofsemitones
betweenthenotesis
given
bythevector
diff([0
24
57
911
12])
=[2
21
22
21]
Thesequence
offrequencies
inourmost
commonscale
issurprising.W
hyare
there8notesin
theC-m
ajorscale?W
hydon’tthenotesin
thescale
haveuniform
frequency
ratios?
For
thatmatter,whyis
theoctavedivided
into
12semitones?
Thenotesin
“DoReMe”are
sofamiliarthatwedon’t
even
ask
ourselves
these
questions.
Are
theremathem
aticalexplanations?
Idon’t
havedefinitiveansw
ers,
butIcanget
somehints
bylookingatharm
ony,
chords,
andtheratiosofsm
all
integers.
Vibrationsandmodes
Musicalinstruments
create
soundthroughtheactionofvibratingstringsorvibrat-
ingcolumnsofair
that,
inturn,produce
vibrationsin
thebodyoftheinstrument
266
Chapter20
.Music
Figure
20.3.Thefirstninemodes
ofavibratingstring,
andtheirweightedsum.
andthesurroundingair.Mathem
atically,
vibrationscanbemodeled
byweighted
sumsofcharacteristicfunctionsknow
nasmodes
oreigenfunctions.
Differentmodes
vibrate
atdifferentcharacteristicfrequen
cies
oreigenvalues.Thesefrequen
cies
are
determined
byphysicalparameterssuch
asthelength,thicknessandtensionin
astring,orthegeometry
oftheaircavity.
Short,thin,tightlystretched
stringshave
highfrequencies,whilelong,thick,looselystretched
stringshavelow
frequencies.
Thesimplest
model
isaone-dim
ensionalvibratingstring,held
fixed
atits
ends.
Theunitsofthevariousphysicalparameterscanbechosenso
thatthelength
ofthestringis
2π.Themodes
are
then
simply
thefunctions
v k(x)=
sinkx,k=
1,2,...
Each
ofthesefunctionssatisfythefixed
endpointconditions
v k(0)=
v k(2π)=
0
Thetime-dep
endentmodalvibrationsare
uk(x,t)=
sinkxsinkt,
k=
1,2,...
andthefrequen
cyissimply
theinteger
k.(T
wo-andthree-dim
ensionalmodelsare
much
more
complicated,butthis
one-dim
ensionalmodel
isallweneedhere.)
OurEXMprogram
vibrating_stringprovides
adynamic
view.Figure
20.3
isasnapshotshow
ingthefirstninemodes
andtheresultingwavetravelingalong
thestring.Anexercise
asksyou
tochangethecoeffi
cients
intheweightedsum
toproduce
differentwaves.
Lissajousfigures
Lissajousfiguresprovidesomeinsightinto
themathem
aticalbehav
iorof
musical
chords.
Twodim
ensionalLissajousfiguresare
plots
oftheparametriccurves
x=
sin(at+α),
y=
sin(bt+β)
267
Figure
20.4.x=
sint,y=
sin3/2t,z=
sin5/4
t.
Figure
20.5.x=
sint,y=
sinσ7t,z=
sinσ4t.
Threedim
ensionalLissajousfiguresare
plots
oftheparametriccurves
x=
sin(at+α),
y=
sin(bt+β),
z=
sin(ct+γ)
Wecansimplify
ourgraphicsinterface
byjust
considering
x=
sint,
y=
sinat
268
Chapter20
.Music
Figure
20.6.x=
sint,y=
sin5/4t.
Figure
20.7.x=
sint,y=
sinσ4t.
and
x=
sint,
y=
sinat,
z=
sinbt
Theexample
witha=
3/2
andb=
5/4
show
nin
figure
20.4.is
producedbythe
defaultsettingsin
ourexmprogram
lissajous.Thisprogram
allow
syou
tochange
269
aandbbyenteringvalues
inedit
boxesonthefigure.Enteringb
=0resultsin
atw
odim
ensionalLissajousfigure
liketheoneshow
nin
figure
20.6.
Thesimplest,“cleanest”
Lissajousfiguresresultwhen
theparametersaandb
are
ratiosof
smallintegers.
a=
p q,b=
r s,
p,q,r,s
=sm
allintegers
This
isbecause
thethreefunctions
x=
sint,
y=
sinp qt,
z=
sinr st
allreturn
tozero
when
t=
2mπ,
m=
lcm(q,s)
wherelcm(q,s)istheleast
commonmultiple
ofqands.
When
aandbare
fractions
withlargenumerators
anddenominators,thecurves
oscillate
more
rapidly
andtake
longer
toreturn
totheirstartingvalues.
Intheextrem
esituationwhen
aandbare
irrational,thecurves
never
return
totheirstartingvalues
and,in
fact,eventuallyfilluptheentire
square
orcu
be.
Wehaveseen
thatdividingtheoctaveinto
12equalsizedsemitones
resultsin
frequencies
thatarepow
ersofσ,anirrationalvalue.
TheEkeyan
dGkeysare
four
andseven
semitones
aboveC,so
theirfrequen
cies
are
sigma^4
=1.259921049894873
sigma^7
=1.498307076876682
Theclosest
fractionswithsm
allnumeratoranddenominatorare
5/4
=1.250000000000000
3/2
=1.500000000000000
This
iswhywechose
5/4an
d3/2forourdefault
parameters.
Iftheirrational
pow
ersof
sigmaare
usedinstead,theresultsare
figures20.5
and20.7.
Infact,
thesefiguresare
merelytheinitialsnapshots.If
wewereto
lettheprogram
keep
running,theresultswould
notbepleasant.
HarmonyandIntonation
Harm
onyisanelusiveattribute.Dictionary
definitionsinvolveterm
slike“pleasing”,
“congruent”,“fittingtogether”.Forthepurposes
ofthischapter,let’ssaythattw
oormore
musicalnotesare
harm
oniousiftheratiosof
theirfrequencies
arerational
270
Chapter20
.Music
numberswithsm
allnumeratoran
ddenominator.
Thehumanearfindssuch
notes
fittogether
inapleasingmanner.
Strictlyspeaking,amusicalchord
isthreeormore
notessounded
simultane-
ously,
buttheterm
canalsoapply
totw
onotes.
Withthesedefinitions,
chords
madefrom
ascale
withequalsemitones
are
notexactly
harm
onious.
Thefrequency
ratiosare
pow
ersof
σ,whichis
irrational.
Tuningamusicalinstrumentinvolves
adjustingitsphysicalparametersso
thatitplaysharm
oniousmusicbyitselfan
din
conjunctionwithother
instruments.
Tuningapianois
adiffi
cult
process
thatis
doneinfrequently.
Tuningaviolinora
guitarisrelativelyeasy
andcanbedoneeven
duringbreaksin
aperform
ance.The
humansingingvoice
isan
instrumentthatcanundergocontinuousretuning.
For
hundredsof
years,musictheory
hasincluded
thedesignofscalesan
dthe
tuningofinstruments
toproduce
harm
oniouschords.
Ofthemanypossibilities,
let’sconsider
only
two–equaltemperamen
tandjust
intonation.
Equaltemperamentistheschem
ewe’ve
beendescribingso
farin
thischapter.
Thefrequency
ratiobetweenthenotesin
achord
canbeexpressed
interm
sof
σ.
Tuninganinstrumentto
haveequaltemperamentisdoneonce
andforall,without
regard
tothemusicthatwillbeplayed.A
single
base
note
ischosen,usuallyA=
440Hz,
andthatdetermines
thefrequency
ofalltheother
notes.
Pianosare
almost
alwaystuned
tohaveequaltemperament.
Just
intonationmodifies
thefrequen
cies
slightlyto
obtain
more
strictly
har-
moniouschords.
Thetuninganticipatesthekey
ofthemusicab
outto
beplayed.
Barbershopquartetsandacapella
choirscanob
tain
just
intonationdynamically
duringaperform
ance.
Hereis
aMatlabcodesegmentthatcomparesequaltemperamentwithjust
intonationfrom
astrictly
numericalpointof
view.Equaltemperamentis
defined
byrepeatedpow
ersofσ.Just
intonationis
defined
byasequence
offractions.
sigma=
2^(1/12);
k=
(0:12)’;
equal=
sigma.^k;
num=
[1
16
96
54
73
85
715
2]’;
den=
[1
15
85
43
52
53
48
1]’;
just=
num./den;
delta=
(equal
-just)./equal;
T=
[k
equal
num
den
just
delta];
fprintf(’%8d%12.6f%7d/%d%10.6f%10.4f\n’,T’)
kequal
just
delta
01.000000
1/1
1.000000
0.0000
11.059463
16/15
1.066667
-0.0068
21.122462
9/8
1.125000
-0.0023
31.189207
6/5
1.200000
-0.0091
41.259921
5/4
1.250000
0.0079
271
51.334840
4/3
1.333333
0.0011
61.414214
7/5
1.400000
0.0101
71.498307
3/2
1.500000
-0.0011
81.587401
8/5
1.600000
-0.0079
91.681793
5/3
1.666667
0.0090
10
1.781797
7/4
1.750000
0.0178
11
1.887749
15/8
1.875000
0.0068
12
2.000000
2/1
2.000000
0.0000
Thelast
columnin
thetable,delta,istherelativedifference
betweenthetw
o.We
seethatdeltaisless
thanonepercent,exceptforonenote.Butthemore
important
considerationis
how
themusicsounds.
Chords
.
Figure
20.8.Dissonace
andbeats
betweentwoadjacentwhole
notes.
Chordsare
twoormore
notesplayedsimultaneously.
Withacomputerkey-
board
andmouse,wecan’t
clickonmore
thanon
ekey
atatime.
Sochordsare
producedwithpianoexbyselectingthetoggle
switches
labeled
1through12.The
switch
labeled
0is
alway
sselected
.Figure
20.8
show
sthevisualoutputgeneratedwhen
pianoexplaysachord
involvingtw
oadjacentwhitekeys,
inthis
case
CandD.Youcansee,
andhear,
the
phenomenonknow
nasbeating.
Thisoccurs
when
tones
withnearlyequalfrequencies
alternate
betweenadditivereinforcem
entan
dsubtractivecancellation.Therelevant
trig
identity
is
sinat+sinbt
=sina+b
2tcosa−
b
2t
272
Chapter20
.Music
Thesum
oftw
onotesisanote
withtheaverageofthetw
ofrequencies,modulated
byacosineterm
invo
lvingthedifference
ofthetw
ofrequencies.Theplayers
inan
orchestratuneupbylisten
ingforbeats
betweentheirinstruments
andareference
instrument.
Themost
importantthree-note
chord,ortriad,istheC
majorfifth.IfCisthe
lowest,
orroot,
note,thechord
isC-E-G.In
just
intonation,thefrequency
ratios
are
1:5 4:3 2
Theseare
theparameter
values
forourdefault
Lissajousfigure,show
nin
figure
20.4.Figures20.9
and20.10show
thevisualou
tputgeneratedwhen
pianoexplays
aC
majorfifthwithjust
intonationan
dwithequaltemperament.
Youcanseethat
thewaveform
sin
theoscilloscopeare
different,
butcanyouhearanydifference
inthesoundgenerated?
Figure
20.9.C
majorfifthwithjust
intonation.
Syn
thesizingMusic
Ourpianoexprogram
isnotapow
erfulmusicsynthesizer.Creatingsuch
aprogram
isahugeundertaking,way
beyondthescopeofthischapteror
thisbook
.Wemerely
wantto
illustrate
afew
oftheinterestingmathem
aticalconcepts
involved
inmusic.
Thecore
ofthepianoexisthecodethatgeneratesavectoryrepresentingthe
amplitudeofsoundasafunctionoftimet.Hereis
theportionof
thecodethat
handlesequaltemperament.
Thequantity
chordiseither
asingle
note
number,or
avectorlike[0
47]withthesettingsof
thechord
toggles.
sigma=
2^(1/12);
273
Figure
20.10.C
majorfifthwithequaltemperamen
t.
C4
=440*sigma^(-9);
fs
=44100;
t=
0:1/fs:T;
y=
zeros(size(t));
forn
=chord
hz
=C4
*sigma^n;
y=
y+
sin(2*pi*hz*t);
end
y=
y/length(chord);
Hereisthecorrespondingportionofcodeforjust
intonation.Thevectorrofratios
isrepeatedafew
times,scaledbypow
ersof2,
tocoverseveraloctaves.
sigma=
2^(1/12);
C4
=440*sigma^(-9);
fs
=44100;
t=
0:1/fs:T;
r=
[1
16/15
9/8
6/5
5/4
4/3
7/5
3/2
8/5
5/3
7/4
15/8];
r=
[r/2
r2*r
4];
y=
zeros(size(t));
forn
=chord
hz
=C4
*r(n+13);
y=
y+
sin(2*pi*hz*t);
end
y=
y/length(chord);
Asm
allexample
ofwhatafullmusicsynthesizer
would
soundlikeis
provided
by
the“piano”toggle
onpianoex.This
synthesizer
usesasingle
sample
ofan
actual
274
Chapter20
.Music
Figure
20.11.C
majorfifthchord
withsimulatedpiano.
pianoplayingmiddle
C.Thissample
isloaded
duringtheinitializationofpianoex,
S=
load(’piano_c.mat’);
middle_c=
double(S.piano_c)/2^15;
set(gcf,’userdata’,middle_c)
Afunctionfrom
theMatlabSignalProcessingToolbox
isthen
usedto
generate
notesat
differentfrequencies.
middle_c=
get(gcf,’userdata’);
fs
=44100;
t=
0:1/fs:T;
y=
zeros(size(t));
forn
=chord
y=
y+
resamplex(middle_c,2^(n/12),length(y));
end
Figure
20.11displaysthepianosimulationoftheC
majorfith
chord.You
can
seethatthewaveform
ismuch
richer
thantheones
obtained
from
superpositionof
sinewaves.
Further
Rea
ding,andViewing
Thiswon
derfulvideo
show
saperform
ance
ofthe“DoReMi”
songfrom
“TheSound
ofMusic”
intheAntw
erpCentralRailway
Station.(I
hopetheURLpersists.)
http://www.youtube.com/watch?v=7EYAUazLI9k
Wikiped
iahashundredsofarticlesonvariousaspects
ofmusictheory.Hereisone:
275
http://en.wikipedia.org/wiki/Music_and_mathematics
Wikipedia
onLissajouscurves:
http://en.wikipedia.org/wiki/Lissajous_curve
Rec
ap
%%Music
Chapter
Recap
%This
is
an
executable
program
that
illustrates
the
statements
%introduced
in
the
Music
Chapter
of
"Experiments
in
MATLAB".
%You
can
access
it
with
% %music_recap
%edit
music_recap
%publish
music_recap
% %Related
EXM
programs
% %pianoex
%%Size
of
asemitone,
one
twelth
of
an
octave.
sigma=
2^(1/12)
%%Twelve
pitch
chromatic
scale.
forn
=0:12
pianoex(n)
end
%%C
major
scale
forn
=[0
24
57
911
12]
pianoex(n)
end
%%Semitones
in
Cmajor
scale
diff([0
24
57
911
12])
%%Equal
temperament
and
just
intonation
[sigma.^(0:12)
116/15
9/8
6/5
5/4
4/3
7/5
3/2
8/5
5/3
7/4
15/8
2]’
%%C
major
fifth
chord,
equal
temperament
and
just
temperament
276
Chapter20
.Music
[sigma.^[04
7]
15/4
3/2]’
Exe
rcises
20.1
Strings.TheW
ikiped
iapage
http://en.wikipedia.org/wiki/Piano_key_frequencies/
hasatableheaded
“VirtualKeyboard”thatshow
sthefrequencies
ofthepianokeys
aswellasfivestringinstruments.Theopen
stringviolinfrequencies
aregiven
by
v=
[-14
-7
07]’
440*sigma.^v
Whatare
thecorrespondingvectors
fortheother
fourstringinstruments?
20.2
Vibratingstring.
Invibrating_string.m,findthestatement
a=
1./(1:9)
Changeitto
a=
1./(1:9).^2
or
a=
1./sqrt(1:9)
Also,changetheloopcontrol
fork
=1:9
to
fork
=1:2:9
or
fork
=1:3:9
Whateff
ectdothesechanges
haveon
theresultingwave?
20.3
Comet.Try
this:
a=
2/3;
b=
1/2;
277
tfinal=
12*pi;
t=
0:pi/512:tfinal;
x=
sin(t);
y=
sin(a*t);
z=
sin(b*t);
comet3(x,y,z)
Whydid
Ichoose
this
particularvalueoftfinal?
How
does
this
tfinaldep
end
uponaandb?
20.4
DissonantLissajous.
Whatis
theLissajousfigure
correspondingto
twoor
threeadjacentkeys,
oradjacentwhitekeys,
onthepianokeyboard?
20.5Irrationalbiorhythms.
Thebiorhythmsdescribed
inour“C
alendars
andClocks”
chapterarebasedonthepremisethatourlives
are
governed
byperiodicfunctionsof
timewithperiodsof
23,28and33day
s.W
hatistheeff
ectofrevisingbiorhythms.m
sothattheperiodsare
irrationalvalues
nearthese?
20.6
Just
intonation.
Withjust
intonation,theratiosof
frequencies
ofadjacent
notesare
nolonger
equalto
σ.W
hataretheseratios?
20.7
Rationalintonation.Matlabalm
ost
hasthecapabilityto
discoverjust
intona-
tion.TheMatlabfunctionratcomputesrationalapproxim
ations.
Forexample,
thefollow
ingstatementcomputesthenumeratornanddenominatordin
arational
approxim
ationof
π.
[n,d]=
rat(pi)
n=
355
d=
113
This
gives
ustheapproxim
ationπ
≈355/1
13,whichis
accurate
to7significant
figures.
For
aless
accurate
approxim
ation,specifyatolerance
of2percent.
[n,d]=
rat(pi,.02)
n=
22
d=
7
This
gives
usthefamiliarπ≈
22/7.
(a)Let’shaverat,withatolerance
of.02,generate
rationalap
proxim
ationsto
the
pow
ersof
σ.Wecancompare
theresultto
therationalapproxim
ationsusedin
just
intonation.
Inou
rcodethatcomparesequaltemperamentwithjust
intonation,
changethestatements
thatdefinejust
intonationfrom
num=
[...]
278
Chapter20
.Music
den=
[...]
to
[num,den]=
rat(sigma.^k,.02);
You
should
seethatthebestrationalapproxim
ationagrees
withtheon
eusedby
just
intonationformost
ofthenotes.
Only
notesneartheendsofthescale
are
different.
•18/1
7vs.
16/15fork=1
•9/5vs.
7/4
fork
=10
•17/9
vs.
15/8
fork
=11
Theap
proxim
ationsfrom
ratare
more
accurate,butthedenominators
are
primes
orpow
ersofprimes
andso
areless
likelyto
becompatible
withother
de-
nominators.
(b)Changethetolerance
invo
lved
inusingratto
obtain
rationalapproxim
ations
topow
ersofσ.Replace
.02by.01or
.10.Youshould
findthat.02worksbest.
(c)Modifypianoexto
alsouse
therationalapproxim
ationsproducedbyrat.Can
you
detectanydifference
inthesoundgeneratedbypianoexiftheseratapproxi-
mationsareincorporated.
20.8
Musicalscore.Ourpianoexisable
toprocess
aMatlabfunctionthatrepre-
sents
amusicalscore.Thescore
isacellarraywithtw
ocolumns.
Thefirstcolumn
containsnote
numbersor
chord
vectors.Thesecondcolumncontainsdurations.
Ifthesecondcolumnis
notpresent,
allthenoteshavethesame,
default,duration.
For
example,hereis
aC-m
ajorscale.
s=
{0
24
57
91112}’
pianoex(s)
Amore
comprehensiveexample
istheportionof
Vivaldi’s“FourSeasons”
inthe
EXM
functionvivaldi.
typevivaldi
pianoex(vivaldi)
Frankly,Idonotfindthisattem
ptto
express
musicin
amacine-readable
form
very
satisfactory.Canyou
create
somethingbetter?
