Čapek, Turing, von Neumann, and the 20th Century Evolution ...
Theory of computing, part 1. Von Neumann Turing machine Finite state machines NP complete problems...
-
Upload
bernard-joshua-watts -
Category
Documents
-
view
218 -
download
2
Transcript of Theory of computing, part 1. Von Neumann Turing machine Finite state machines NP complete problems...
Theory of computing, part 1
Von NeumannTuring machineFinite state machinesNP complete problems-maximum clique-travelling salesman problem-colour graph
1 Introduction
2 Theoretical background Biochemistry/molecular biology
3 Theoretical background computer science
4 History of the field
5 Splicing systems
6 P systems
7 Hairpins
8 Detection techniques
9 Micro technology introduction
10 Microchips and fluidics
11 Self assembly
12 Regulatory networks
13 Molecular motors
14 DNA nanowires
15 Protein computers
16 DNA computing - summery
17 Presentation of essay and discussion
Course outline
Old computers
Abacus
Abacus
Born December 26, 1791 in Teignmouth, Devonshire UK,
Died 1871, London; Known to some as the Father of
Computing for his contributions to the basic design of
the computer through his Analytical machine..
The difference engine (1832)
The ENIAC machine occupied a room 30x50 feet. The
controls are at the left, and a small part of the
output device is seen at the right.
ENIAC (1946)
The IBM 360 was a revolutionary advance in
computer system architecture, enabling a family
of computers covering a wide range of price and
performance.
ENIAC (1946)
Books
Books
Introduction
Finite state machines (automata)
Pattern recognition
Simple circuits (e.g. elevators, sliding doors)
Automata with stack memory (pushdown autom.)
Parsing computer languages
Automata with limited tape memory
Automata with infinite tape memory
Called `Turing machines’,
Most powerful model possible
Capable of solving anything that is solvable
Models of computing
Regular grammars
Context free grammars
Context sensitive grammars
Unrestricted grammars
Chomsky hierarchy of grammars
Computers can be made to recognize, or accept, the
strings of a language.
There is a correspondence between the power of the
computing model and the complexity of languages
that it can recognize!
Finite automata only accept regular grammars.
Push down automata can also accept context free
grammars.
Turing machines can accept all grammars.
Computers can recognise languages
Languages
A set is a collection of things called its elements. If
x is an element of set S, we can write this:
x S A set can be represented by naming all its elements,
for example:
S = {x, y, z}
There is no particular order or arrangement of the
elements, and it doesn’t matter if some appear more
than once. These are all the same set:
{x, y, z}={y, x, z}={y, y, x, z, z}
Sets
A set with no elements is called the empty set, or
a null set. It is denoted by ={}. If every element of set A is also an element of
set B, then A is called a subset of B :
A B The union of two sets is the set with all elements
which appear in either set:
C = A B The intersection of two sets is the set with all
the elements which appear in both sets:
C = A B
Combining sets
A string a sequence of symbols that are placed
next to each other in juxtaposition
The set of symbols which make up a string are
taken from a finite set called an alphabet
E.g. {a, b, c} is the alphabet for the string
abbacb.
A string with no elements is called an empty
string and is denoted . If Σ is an alphabet, the infinite set of all
strings made up from Σ is denoted Σ*.
E.g., if Σ ={a}, then Σ *={, a, aa, aaa, …}
Alphabet and strings
A language is a set of strings.
If Σ is an alphabet, then a language over Σ
is a collection of strings whose components
come from Σ.
So Σ* is the biggest possible language over
Σ, and every other language over Σ is a
subset of Σ*.
Languages
Four simple examples of languages over an
alphabet Σ are the sets , {}, Σ, and Σ *.
For example, if Σ={a} then these four simple
languages over Σ are
, {}, {a}, and {, a, aa, aaa, …}.
Recall {} is the empty string while is the empty set. Σ* is an infinite set.
Examples of languages
The alphabet is Σ = {a,b,c,d,e…x,y,z}
The English language is made of strings
formed from Σ: e.g. fun, excitement.
We could define the English Language as the
set of strings over Σ which appear in the
Oxford English dictionary (but it is clearly
not a unique definition).
Example: English
The natural operation of concatenation
of strings places two strings in
juxtaposition.
For example, if then the concatenation
of the two strings aab and ba is the
string aabba.
Use the name cat to denote this
operation. cat(aab, ba) = aabba.
Concatenation
Languages are sets of strings, so they
can be combined by the usual set
operations of union, intersection,
difference, and complement.
Also we can combine two languages L
and M by forming the set of all
concatenations of strings in L with
strings in M.
Combining languages
This new language is called the product of L
and M and is denoted by L M.
A formal definition can be given as follows:
L M = {cat(s, t) | s L and t M}
For example, if L = {ab, ac} and M = {a, bc,
abc}, then the product L•M is the language
L M = {aba, abbc, ababc, aca, acbc, acabc}
Products of languages
The following simple properties hold for any
language L:
L {} = {} L = L
L = L =
The product is not commutative. In other
words, we can find two languages L and M such
that
L M M L
The product is associative. In other words,
if L, M, and N are languages, then
L (M N) = (L M) N
Properties of products
If L is a language, then the product
L L is denoted by L2.
The language product Ln for every n {0, 1, 2, …} is as follows:
L0 = {}
Ln = L Ln-1 if n > 0
Powers of languages
For example, if L = { a, bb} then the first
few powers of L are
L0 = {}
L1 = L = {a, bb}
L2 = L L = {aa, abb, bba, bbbb}
L3 = L L2 = {aaa, aabb, abba, abbbb,
bbaa, bbabb, bbbba, bbbbbb}
Example
If L is a language over Σ (i.e. L A*) then the closure of L is the language
denoted by L* and is defined as follows:
L* = L0 L1 L2 …
The positive closure of L is the language
denoted by L+ and defined as follows:
L+ = L1 L2 L3 …
Closure of a language
It follows that L* = L+ {}. But it’s not necessarily true that
L+ = L* - {}
For example, if we let our alphabet
be Σ = {a} and our language be L =
{, a}, then
L+ = L*
Can you find a condition on a
language L such that
L+ = L* - {}?
L* vs. L+
The closure of Σ coincides with our
definition of Σ* as the set of all
strings over Σ. In other words, we have
a nice representation of Σ* as follows:
A* = A0 A1 A2 ….
Closure of an alphabet
Let L and M be languages over the
alphabet Σ. Then:
a) {}* = * = {}
b) L* = L* L* = (L*)*
c) L if and only if L+ = L*
d) (L* M*)* = (L* M*)* = (L M)*
e) L (M L)* = (L M)* L
Properties of closure
Grammars
A grammar is a set of rules used to
define the structure of the strings in a
language.
If L is a language over an alphabet Σ,
then a grammar for L consists of a set of
grammar rules of the following form:
where and denote strings of symbols taken from Σ and from a set of grammar
symbols (non-terminals) that is disjoint
from Σ
Grammars
A grammar rule is often called a production, and it can be read in any of
several ways as follows:
replace by
produces
rewrites to
reduces to
Productions
Every grammar has a special grammar symbol
called a start symbol, and there must be at
least one production with left side
consisting of only the start symbol.
For example, if S is the start symbol for a
grammar, then there must be at least one
production of the form
S
Where to begin ……
1. An alphabet N of grammar symbols called
non-terminals. (Usually upper case
letters.)
2. An alphabet T of symbols called
terminals. (Identical to the alphabet of
the resulting language.)
3. A specific non-terminal called the start
symbol. (Usually S. )
4. A finite set of productions of the form , where and are strings over the alphabet N T
The 4 parts of a grammar
Let Σ = {a, b, c}. Then a grammar for the
language Σ* can be described by the
following four productions:
S
S aS
S bS
S cS
Or in shorthand:
S | aS | bS | cS
S can be replaced by either , or aS, or bS, or cS.
Example
A string made up of grammar symbols
and language strings is called a
sentential form.
If x and y are sentential forms and is a production, then the
replacement of by in xy is
called a derivation, and we denote it
by writing
xy xy
String
S | aS | bS | cS,
S aS
S aS aaS.
S aS aaS aacS aacbS..
S aS aaS aacS aacbS aacb
= aacb
A short hand way of showing a derivation
exists:
S * aacb
Sample derivation
The following three symbols with their
associated meanings are used quite often in
discussing derivations:
derives in one step,
+ derives in one or more steps,
* derives in zero or more steps.
Other shorthand
S AB
A | aA
B | bB
We can deduce that the grammar non-terminal
symbols are S, A, and B, the start symbol is
S, and the language alphabet includes , a, and b.
A more complex grammar
Let's consider the string aab.
The statement S + aab means that there
exists a derivation of aab that takes one or
more steps.
For example, we have
S AB aAB aaAB aaB aabB aab
Another derivation
If G is a grammar, then the language of G
is the set of language strings derived
from the start symbol of G. The language
of G is denoted by:
L(G)
Grammar specifies the language
If G is a grammar with start symbol S and
set of language strings T, then the
language of G is the following set:
L(G) = {s | s T* and S + s}
Grammar specifies the language
If the language is finite, then a grammar can
consist of all productions of the form
S w
for each string w in the language.
For example, the language {a, ba} can be
described by the grammar S a | ab.
Finite languages
If the language is infinite, then some
production or sequence of productions must be
used repeatedly to construct the derivations.
Notice that there is no bound on the length of
strings in an infinite language.
Therefore there is no bound on the number of
derivation steps used to derive the strings.
If the grammar has n productions, then any
derivation consisting of n + 1 steps must use
some production twice
Infinite languages
For example, the infinite language {anb | n
0 } can be described by the grammar,
S b | aS
To derive the string anb, use the production
S aS
repeatedly --n times to be exact-- and then
stop the derivation by using the production
S b
The production S aS allows us to say
If S derives w, then it also derives aw
Finite languages
A production is called recursive if its
left side occurs on its right side.
For example, the production S aS is
recursive.
A production S is indirectly
recursive if S derives (in two or more
steps) a sentential form that contains
S.
Recursion
For example, suppose we have the
following grammar:
S b | aA
A c | bS
The productions S aA and A bS are
both indirectly recursive
S aA abS
A bS baA
Indirect recursion
A grammar is recursive if it contains
either a recursive production or an
indirectly recursive production.
A grammar for an infinite language must
be recursive!
However, a given language can have many
grammars which could produce it.
Recursion
Suppose M and N are languages. We can
describe them with grammars have disjoint
sets of nonterminals.
Assign the start symbols for the grammars of
M and N to be A and B, respectively:
M : A N : B
… …
Then we have the following rules for creating
new languages and grammars:
Combining grammars
The union of the two languages, M N, starts with the two productions
S A | B
followed by the grammars of M and N
A
…
B
…
Combining grammars, union rule
Similarly, the language M N starts with the production
S AB
followed by, as above,
A
…
B
…
Combining grammars, product rule
Finally, the grammar for the closure of a
language, M*, starts with the production
S AS |
followed by the grammar of M
A
…
Combining grammars, closure rule
Suppose we want to write a grammar for the
following language:
L = {, a, b, aa, bb, ..., an, bn, ..}
L is the union of the two languages
M = { an | n N}
N = { bn | n N}
Example, union
Thus we can write a grammar for L as follows:
S A | B union rule,
A | aA grammar for M,
B | bB grammar for N.
Example, union
Suppose we want to write a grammar for the
following language:
L = { ambn | m,n N}
L is the product of the two languages
M = {am | m N}
N = {bn | n N}
Example, product
Thus we can write a grammar for L
as follows:
S AB
product rule,
A | aA grammar for M
B | bB grammar for N
Example, product
Suppose we want to construct the
language L of all possible strings made
up from zero or more occurrences of aa
or bb.
L = {aa, bb}* = M*
M = {aa, bb}
Example, closure
So we can write a grammar for L as follows:
S AS | closure rule,A aa | bb grammar for {aa, bb}.
Example, closure
We can simplify this grammar:
Replace the occurrence of A in S
AS by the right side of A aa to
obtain the production S aaS.
Replace A in S AS by the right
side of A bb to obtain the
production S bbS.
This allows us to write the the
grammar in simplified form as:
S aaS | bbS |
An equivalent grammar
Language Grammar
{a, ab, abb, abbb} S a | ab | abb | abbb
{, a, aa, aaa, …} S aS |
{b, bbb, bbbbb, … b2n+1} S bbS | b
{b, abc, aabcc, …, anbcn} S aSc | b
{ac, abc, abbc, …, abnc} S aBc
B bB |
Some simple grammars
Regular languages
There are many possible ways of describing the
regular languages:
Languages that are accepted by some finite
automaton
Languages that are inductively formed from
combining very simple languages
Those described by a regular expression
Any language produced by a grammar with a
special, very restricted form
What is a regular language?
We start with a very simple basis of
languages and build more complex ones by
combining them in particular ways:
Basis: , {} and {a} are regular
languages for all a Σ.
Induction: If L and M are regular
languages, then the following languages
are also regular:
L M, L M and L*
Building a regular language
For example, the basis of the definition
gives us the following four regular
languages over the alphabet Σ = {a,b}:
, {}, {a}, {b}
Sample building blocks
Regular languages over {a, b}.
Language {, b}
We can write it as the union of the two
regular languages {} and {b}:
{,b} = {} {b}
Example 1
Language {a, ab }
We can write it as the product of the
two regular languages {a} and {, b}:
{a, ab} = {a} {, b}
Example 2
Language {, b, bb, …, bn,…}
It's just the closure of the regular
language {b}:
{b}* = {, b, bb, ..., bn ,...}
Example 3
{a, ab, abb, ..., abn, ...}
= {a} {, b, bb, ..., bn, ... }
= {a} {b}*
Example 4
{, a, b, aa, bb, ..., an, …, bm, ...}
= {a}* {b}*
Example 5
A regular expression is basically a
shorthand way of showing how a regular
language is built from the basis
The symbols are nearly identical to
those used to construct the languages,
and any given expression has a language
closely associated with it
For each regular expression E there is
a regular language L(E)
Regular expressions
The symbols of the regular expressions are
distinct from those of the languages
Regular expression Language
L () =
L () = {}
a L {a} = {a}
Basis of regular Basis of regular
expressions languages
Regular expressions versus languages
There are two binary operations on regular expressions
(+ and ) and one unary operator (*)
Regular expression Language
R + S L (R + S ) = L (R ) L (S )
R S , R S L (R S ) = L (R ) L (S )
R* L (R* ) = L (R )*
These are closely associated with the union, product and
closure operations on the corresponding languages
Operators on regular expressions
Like the languages they represent, regular
expressions can be manipulated inductively to
form new regular expressions
Basis: , and a are regular expressions for all a Σ.
Induction: If R and S are regular expressions,
then the following expressions are also regular:
(R), R + S, R S and R*
Building regular expressions
For example, here are a few of the infinitely
many regular expressions over the alphabet Σ
= {a, b }:
, , a, b
+ b, b*, a + (b a), (a + b) a,
a b*, a* + b*
Regular expressions
To avoid using too many parentheses, we
assume that the operations have the
following hierarchy:
* highest (do it first)
+ lowest (do it last)
For example, the regular expression
a + b a*
can be written in fully parenthesized form
as
(a + (b (a*)))
Order of operations
Use juxtaposition instead of whenever no confusion arises. For example, we can
write the preceding expression as
a + ba*
This expression is basically shorthand
for the regular language
{a} ({b} ({a}*))
So you can see why it is useful to write
an expression instead!
Implicit products
Find the language of the regular expression
a + bc*
L(a + bc*) = L(a) L(bc*)
= L(a) (L(b) L(c*))
= L(a) (L(b) L(c)*)
= {a} ({b} {c}*)
= {a} ({b) {, c, c2, ., cn,…})
= {a} {b, bc, bc2, .. .... bcn,…}
= {a, b, bc, bc2, ..., bcn, ...}.
Example
Many infinite languages are easily seen to
be regular. For example, the language
{a, aa, aaa, ..., an, ... }
is regular because it can be written as the
regular language {a} {a}*, which is
represented by the regular expression aa*.
Regular language
The slightly more complicated language
{, a, b, ab, abb, abbb, ..., abn, ...}
is also regular because it can be represented
by the regular expression
+ b + ab*
However, not all infinite languages are
regular!
Regular language
Distinct regular expressions do not always
represent distinct languages.
For example, the regular expressions a + b
and b + a are different, but they both
represent the same language,
L(a + b) = L(b + a) = {a, b}
Regular language
We say that regular expressions R and
S are equal if
L(R) = L(S)
and we denote this equality by
writing the following familiar
relation:
R = S
Regular language
For example, we know that
L (a + b) = {a, b} = {b, a} = L (b + a)
Therefore we can write
a + b = b + a
We also have the equality
(a + b) + (a + b) = a + b
Regular language
Properties of regular language
Additive (+) properties:
R + T = T + R
R + = + R= R
R + R = R
(R +S) +T = R + (S+ T)
These follow simply from the
properties of the union of sets
Properties of regular language
Product (•) properties
R = R =
R = R = R
(RS)T =R(ST)
Distributive properties
R(S + T) = RS + RT
(S + T)R = SR +TR
* = * =
R* = R*R* = (R*)* = R+R*
RR* = R*R
R(SR)* = (RS)* R
(R+S)* = (R*S*)* = (R* + S*)* = R*(SR*)*
Closure properties
Show that ( + a + b)* = a*(ba*)*
( + a + b)* = (a + b)* (+ property)
= a*(ba*)* (closure property)
Example
Regular grammars
A regular grammar is one where each
production takes one of the following
forms:
S
S w
S T
S wT
where the capital letters are non-
terminals and w is a non-empty string
of terminals
Regular grammar
Only one nonterminal can appear on the
right side of a production. It must
appear at the right end of the right
side.
Therefore the productions
A aBc and S TU
are not part of a regular grammar, but
the production A abcA is.
Regular grammar