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EECS 465: Digital Systems

Lecture Notes # 2

Two-Level Minimization Using Karnaugh Maps

SHANTANU DUTT

Department of Electrical & Computer Engineering

University of Illinois, ChicagoPhone: (312) 355-1314: e-mail: dutt@ece.uic.edu

URL: http://www.ece.uic.edu/~dutt

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Concepts in 2-Level Minimization

Defn: An implicantg of a function is aproduct term(e.g., g=xyz) s.t.

when g=1f=1

Defn: A mintermis an implicant whose product term representation

consists of all n variables of an n-var. function (each in either compl.or

uncompl. form)

Defn: Let g, and h be 2 implicants of a function f. g is said to coverh

if h=1g=1 (e.g., g = xz, h = xyz)Defn: Let G = {g1, .., gk} be a set of implicants of func f, and let h be

another implicant of f. Then G is coversh if h=1g1+ .. + gk=1

Defn: An SOP or POS expression is also called a 2-level expression

Defn: A 2-level AND-OR (OR-AND) gate realizationof an SOP(POS) expression is one in which all product terms (OR

terms in POS) in the SOP (POS) expression are implemented

by multiple input AND (OR) gates & the ORing (ANDing)

of the product (OR) terms is realized by a multiple input OR

(AND) gate.

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E.g. f1= AB + BC + ACD --- 2-level. (SOP)

f2= ( B+C )( A+D )( C+D ) --- 2-level (POS)

f3=AB + AC( B+D ) --- not SOP or POS

level 1 I/Ps --- not 2-level

--- 3 level ( can be realized

directly by 3 levels of gates)A

B

A

C

B

D

f3

level 1

gates

level 0

I/Ps

level 2

gates

level 3

gate

level 2

I/Ps

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AB + BC + ACDA

B

B

C

ACD

f1

Level 2

Level 1

(B + C)(A + D)(C + D)B

C

A

D

C

D

f2

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Defn. A gate in a logic circuit is a level-1 gate if all its inputs are

primary inputs (i.e., literals, A, A, B, B, etc.)

A gate is a level-i gate i > 1 if the highest-level gate whose

output feeds the (level-i) gate is a level-(i-1) gate.

NOTE: If a given expr. is not 2-level, we can convert it to a

2-level one using the distributive law.

The goal of 2-level minimization is to minimize the number of

literals (a literal is a var. in complemented or uncompl. form)

in a 2-level logic expr..

This approx. reduces the total # of inputs over all gates in the

circuit in a 2-level gate realization.

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E.g. f = ABC + ABC + ABC (non-minimized 9 literals)

A

BCA

BC

ABC

f

(12 gate I/Ps)

= Circuit complexity

f = AB + AC (minimized 4 literals)

A

B

A

C

f(6 gate I/Ps)

= Circuit complexity

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Note: Minimizing # of literals + # of product terms (approx) minimizing total # ofgate inputs

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Multilevel Minimization (Example):

f = x1x2x3+ x2x3x4+ x1x5+ x4x5 (2-level minimized)

Apply distributive law (factoring)

f = x2x3(x1+ x4) + x5(x1+ x4)

= (x2x3+ x5)(x1+ x4) -- Multilevel expr.

Gate Impl.:

x2x3

x5

x1x4

f

Multilevel

circuit

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Defn. Two implicants (or product terms) of a function f are said to

be adjacent (logically) if they have the same literals

except in one variable xi

which occurs in uncomplemented for

(xi) in one implicant and in complemented form (xi) in the

other. The 2 implicants are said to differ in xi.

E.g. ABC, ABC (adjacent implicants, differ in B)

ABD, ABC (not adjacent)

NOTE: Adjacent implicants can be combined into one implicant

by the combining theorem (ABC + ABC = AC)

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Karnaugh Maps

2-variable TT outputs

A B Z1 Z2

0 0 0 0

0 1 1 1

1 0 0 1

1 1 1 0

Binary

place-value

ordering

(Binary

ordering)

Physically adjacent

but not logically

adjacent.

logically adjacent but not physically adjacent

AB + AB

B

Another ordering of inputs ( Gray-code ordering)A B Z1

0 0 0

0 1 1

1 1 1

1 0 0

Z1= BPhysically as well as

logically adjacent.

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n-variable Gray-code ordering

G1 = 0 Base.1

G2 = 0 G1

1 rev(G1)=

0 0

0 1

1 1

1 0

G3 = 0 G2

1 rev(G2)=

A B C

0 0 0

0 0 1

0 1 10 1 0

1 1 0

1 1 1

1 0 11 0 0

Gn = 0 Gn-1

1 rev(Gn-1)

Convert to a 2-dimensional

Gray-code ordered TT

ABC 00 01 11 10

0

1

x

y

001

110

Throw away variable (A)

changing across the 2 squares.

We thus obtain BC.

1 1

Defn. Gi= i-bit Gray-code ordering

rev(Gi) = reverse order of Gi

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Example ( 3-var. K-Map)

Function f:AB

C 00 01 11 10

0

1

1 1 1

1 1f = AB + AB + AC + BC

Consensus = BC

Four Prime Implicants AB, BC, AC, AB formed. Not all these PIs

are needed in the expression. Only a minimum set that covers all

minterms is needed.

Defn. A Prime Implicant (PI) of a function f is an implicant of f that

is not covered by any other implicant of f.

Defn. An Essential PI is a PI that covers/includes at least one minterm

that is not covered by any other PI.

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In the above example, AB & AB are essential PIs. These will need

to be present in any SOP expr. of the function. To form a minimal

set, the PI BC can be used. Thus f = AB + AB + BC is a minimized

expression.

Larger than 2-minterm PIs can also exist:

E.g.,AB

C 00 01 11 10

0

1

1 1

1 1

AB ABB

B

AB

C 00 01 11 10

0

1 1 1 1 1

C

2-minterm implicants (AB, AB in the 1st example above) can

be merged if they are logically adjacent to form a 4-minterm

implicant and so on.

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When an implicant can not be grown any further, then it is a PI.

Defn. In a K-map, 2 implicants can be logically adjacent if they

are symmetric,i.e., if : (1) They are disjoint (no common minterms).

(2) They cover the same # of minterms.

(3) Each minterm in one implicant is adjacent to a

unique minterm in the other implicant.More Examples:

AB

C

AB

C

00 01 11 10

0

1

0

11 1

1 1 1

1 1

1 1 1

Symmetric ( logically adjacent)

Not symmetric

2-minterm implicants : Can be merged to form

a 4-minterm implicant, which will be a PI.

Redrawn

Both PIs

(A, BC)

are

essential.Thus, f = A + BC

is minimized.

00 01 11 10

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4-variable K-Map:

AB

C

D00 01 11 10

00

01

11

10

14

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Example:AB

CD

AB

CD

00 01 11 10

00

01

11

10

00

01

11

10

00 01 11 10

1 1 1 1

1 1

1 1 1 1

1 1

1 1

1 1 1

1

AC

f = AC + DInstead of starting with 2-minterm

implicants & growing them, you

can form larger implicants directlyby grouping power of 2 (2, 4, 8,

etc.) # of minterms so as to form

a rectangle or square ( a convex

region).Convex

Concave

Valid

groupings

Invalid grouping

(region formed is concave)

f = AB + AC + BCD

D

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Another Example:

A B C D Z0 0 0 0 1

0 0 0 1 0

0 0 1 0 1

0 0 1 1 1

0 1 0 0 0

0 1 0 1 10 1 1 0 1

0 1 1 1 1

1 0 0 0 1

1 0 0 1 0

1 0 1 0 1

1 0 1 1 11 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 1

TT:

Binary

order

ABCD 00 01 11 10

00

01

11

10

1 1

1

1 1 1 1

1 1 1 1

ABD

BD

C

f = C + BD + ABD

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Dont Cares in K-Maps

Many times certain input combinations are invalid for a function

(E.g. BCD to 7-segment display functions; see pp. 212-214 inKatz text )

For such combinations, we do not care what the output is, and

we put an x in the o/p column for the TT

A B C f

0 0 0 1

0 0 1 0

0 1 0 1

0 1 1 x

1 0 0 0

1 0 1 1

1 1 0 x

1 1 1 x

ABCD 00 01 11 10

0

1

x x 1 0

0 1 1 0

f = BC + AB

by considering allxs as 0s

f = B The xs can, however, be profitably

used to make larger PIs & thus

further simplify the function.

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by selectively

considering the 010cells x as 1

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STEPS IN K-MAP MINIMIZATION (sop EXPR.)

STEP1: Form all PIs by including the Xs with the 1s to form larger

PIs. (Do not form PIs covering only Xs !)

STEP2: Identify all essential PIs by visiting each minterm (1s) &

checking if it is covered by only one PI. If so, then that

PI is an essential PI.

STEP3: Identify all 1s not covered by essential PIs. Select a minimal-

cost set of PIs S to cover them. This requires some trial-and error (in

fact, this is a hard computational problem).

STEP4: Minimal SOP Expr. for function f:

PI Piis in set S

f = essential PIs + PiPi S

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Mi i l POS E f K M

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Minimal POS Expr. from a K-Map :

METHOD 1: Obtain minimal SOP expression for the complement

function f, and complement this SOP expression to get a minimal

POS expr. for f (using De-Morgans Law)

Example:

F (A,B,C,D) = M(0,2,4,8,10,11,14,15) D(6,12,13)

Big M notation

0s of FXs of F

ABCD

ABCD 00 01 11 10

00

01

11

10

00

01

11

10

0 4 12 8

1 5 13 9

3 1 7 1 15 0 11 0

2 0 6 x 14 0 10 0

0 0 x 0

1 1 x 1

Complement 1 1 x 1

0 0 x 0

0 0 1 1

1 x 1 1AC

D

F =D + AC F = D + AC = D(A +C)

F F

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METHOD2 Di t M th d 21

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METHOD2 : Direct Method :

(1) Obtain all prime implicates(PTs) (largest groups of 0s & Xs

of size 2 , i 0, forming a convex region, that can not be grown

any further).(2) Identify all essential PTs (those that cover at least one 0 not

covered by any other PT).

(3) Choose a minimal set T of PTs to cover the 0s not covered bythe set of essential PTs.

(4) The expression for a PT is an OR term obtained by discarding

all changing variables, & keeping variables complemented if they

are constant at 1 or uncomplemented if they are constant at 0 &

taking the OR (sum) of these literals.

(5) Minimal POS Expr.:

f = (essential PTs) qiqi T

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i

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Example: (Direct Method)

ABCD 00 01 11 10

00

01

11

10

0 0 x 0

1 1 x 1

1 1 0 0

0 x 0 0

(A + C)

D

(OR terms)

Prime Implicate

( Groups of 0s that cannot

be grown any further)

Both D and A + C areEssential Prime Implicates.

F = D( A + C )

( Product of Prime Implicates )

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5-variable K-map: f(A,B,C,D,E)---Juxtapose two 4-var. K-submaps

one for the MSB A=0 and the other for A=1:BC

DE

BCDE00 01 11 10

00

01

11

10

00 01 11 10

00

01

11

10

BCDE

BC

DE00 01 11 10

00

01

11

10

00 01 11 10

00

01

11

10

A=0 A=1

A=0 A=1

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

0 1 4 12 1 8

1 5 13 1 9

3 1 7 1 15 x 11

2 6 14 x 10

16 20 28 x 24

17 21 29 1 25

19 x 23 1 31 x 27

18 22 1 30 1 26

4-var. K-submap 4-var. K-submapAdjacencies of a square: Adjacent squares of the 4-var. K-submap

plus the corresponding or mirror square in the other 4-var.

K-submap.

Example:

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EF EF00 01 11 10

00

01

11

10

00 01 11 10

00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

6-variable K-map: f(A,B,C,D,E,F)--Juxtapose two 5-var. K-submaps

one for the MSB A=0 and other for A=1

B=0 B=1

A=05-var.

K-submap

EF EF00 01 11 10

00

01

11

10

00 01 11 10

00

01

11

10

32 36 44 40

33 37 45 41

35 39 47 43

34 38 46 42

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

CD CD

CD CD

A=1

5-var.

K-submap

Adjacencies of a square: Adjacent squares of the 5-var.

K-submap plus the corresponding or mirror squarein the other 5-var. K-submap.

B=0 B=1

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i bl f( ) l 25

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EF EF00 01 11 10

00

01

11

10

00 01 11 10

00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

B=0 B=1

A=0

EF EF00 01 11 10

00

01

11

10

00 01 11 10

00

01

11

10

32 36 44 40

33 37 45 41

35 39 47 43

34 38 46 42

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

CD CD

A=1

B=1

6-variable K-map: f(A,B,C,D,E,F,)---Example:

CD CD

B=0

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