1 CS 20 Lecture 14 Karnaugh Maps Professor CK Cheng CSE Dept. UC San Diego.

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CS 20 Lecture 14Karnaugh Maps

Professor CK Cheng

CSE Dept.

UC San Diego

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1. Introduction2. The Maps3. Boolean Optimization

Karnaugh Maps

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Definitions:Literal: xi or xi’Product Term: x2x1’x0

Sum Term: x2 + x1’ + x0

Minterm of n variables: A product of n literals in which every variable appears exactly once. f(a,b,c,d): ab’cd’, a’bc’d’ Maxterm of n variables: A sum of n literals in which every variable appears exactly once. f(a,b,c,d): (a’+b+c+d), (a’+b’+c+d)

Introduction

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Introduction

Input: Boolean expression of n binary variablesGoal: Simplification of the expression.E.g. we want to minimize # terms and # literals.Applications:Logic: rule reductionHardware Design: cost and performance optimization. Cost (wires, gates): # literals, product terms,

sum terms Performance: speed, reliability

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Introduction

ID A B f(A,B) minterm

0 0 0 0

1 0 1 1 A’B

2 1 0 1 AB’

3 1 1 1 AB

An example of 2-variable function f(A,B)=A+B

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Function can be represented by sum of minterms:f(A,B) = A’B+AB’+AB

This is not minimal however! We want to minimize the number of literals and terms.We factor out common terms –

A’B+AB’+AB= A’B+AB’+AB+AB=(A’+A)B+A(B’+B)=B+A

Hence, we havef(A,B) = A+B

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K-Map: Truth Table in 2 Dimensions

A = 0 A = 1

B = 0

B = 1

0 2

1 3

0 1

1 1

A’B

AB’

AB

f(A,B) = A + B

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ID A B f(A,B) minterm

0 0 0 0

1 0 1 1 A’B

2 1 0 0

3 1 1 1 AB

Another Example f(A,B)=B

f(A,B)=A’B+AB=(A’+A)B=B

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On the K-map:

A = 0 A= 1

B= 0

B = 1

0 2

1 3

0 0

1 1

A’B AB

f(A,B)=B

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ID A B f(A,B) Maxterm

0 0 0 0 A+B

1 0 1 1

2 1 0 0 A’+B

3 1 1 1

Using Maxterms

f(A,B)=(A+B)(A’+B)=(AA’)+B=0+B=B

The Maps: Representation of k-Variable Functions

• Boolean Expression• Truth Table• Cube• K Map• Binary Decision

Diagram

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(1,1,1)(1,1,0)

(0,0,0) (0,0,1)

(0,1,0) (0,1,1)

(1,0,1)

A cube of 3 variables: (A,B,C)

C

B

A

Representation of k-Variable Func.

• Boolean Expression• Truth Table• Cube• K Map• Binary Decision

Diagram

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(0,1,1,1)(0,1,1,0)

(0,0,0,0) (0,0,0,1) (1,0,0,1)

(1,1,1,1)

(1,1,0,1)(1,0,0,0)

(0,0,1,0)

(1,1,1,0)

(0,0,1,1) (1,0,1,1)

(0,1,0,1)

(1,0,1,0)

A cube of 4 variables: (A,B,C,D)

D

C

B

A

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Three-Variable K-Map

Id a b c f (a,b,c)

0 0 0 0 f(0,0,0)

1 0 0 1 f(0,0,1)2 0 1 0 f(0,1,0)3 0 1 1 f(0,1,1)4 1 0 0 f(1,0,0)5 1 0 1 f(1,0,1)6 1 1 0 f(1,1,0)7 1 1 1 f(1,1,1)

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Three-Variable K-Map

Id a b c f (a,b,c)

0 0 0 0 1

1 0 0 1 02 0 1 0 13 0 1 1 04 1 0 0 15 1 0 1 06 1 1 0 17 1 1 1 0

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Corresponding K-map

0 2 6 4

1 3 7 5

b = 1

c = 1

a = 1

1 1 1 1

0 0 0 0

(0,0) (0,1) (1,1) (1,0)

c = 0

Gray code

f(a,b,c) = c’

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Karnaugh Maps (K-Maps)• Boolean expressions can be minimized by combining

terms• K-maps minimize equations graphically

C 00 01

0

1

Y

11 10AB

1

1

0

0

0

0

0

0

C 00 01

0

1

Y

11 10AB

ABC

ABC

ABC

ABC

ABC

ABC

ABC

ABC

B C0 00 11 01 1

A0000

0 00 11 01 1

1111

11000000

Y

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• Find rectangles to cover 1’s in adjacent entries.• Rectangles can overlap but should not include 0’s.• Use the rectangle that corresponds to a product

term.

K-map

C 00 01

0

1

Y

11 10AB

1

0

0

0

0

0

0

1

B C0 00 11 01 1

A0000

0 00 11 01 1

1111

11000000

Y

y(A,B)=A’B’C’+A’B’C= A’B’(C’+C)=A’B’

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Another 3-Input example

Id a b c f (a,b,c)

0 0 0 0 0

1 0 0 1 02 0 1 0 13 0 1 1 04 1 0 0 15 1 0 1 16 1 1 0 -7 1 1 1 1

Don’t Care Entry: “-” means the entry is not relevant either at input or output.In other words, we are free to assign either 0 or 1 to reduce the Boolean expression.

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Corresponding K-map

0 2 6 4

1 3 7 5

b = 1

c = 1

a = 1

0 1 - 1

0 0 1 1

(0,0) (0,1) (1,1) (1,0)

c = 0

f(a,b,c) = a + bc’

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Yet another example

Id a b c f (a,b,c,d)

0 0 0 0 1

1 0 0 1 12 0 1 0 -3 0 1 1 04 1 0 0 15 1 0 1 16 1 1 0 07 1 1 1 0

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Corresponding K-map

0 2 6 4

1 3 7 5

b = 1

c = 1

a = 1

1 - 0 1

1 0 0 1

(0,0) (0,1) (1,1) (1,0)

c = 0

f(a,b,c) = b’

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Quiz

Given Boolean function, f(a,b,c)= a’b’+a’c’+bc’+ab+b’c1.Write the truth table2.Use Karnaugh map to derive the function in a minimal expression of sum of product form.

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4-input K-map

01 11

01

11

10

00

00

10AB

CD

Y

0

C D0 00 11 01 1

B0000

0 00 11 01 1

1111

1

110111

YA00000000

0 00 11 01 1

0000

0 00 11 01 1

1111

11111111

11

100000

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4-input K-map

01 11

1

0

0

1

0

0

1

101

1

1

1

1

0

0

0

1

11

10

00

00

10AB

CD

Y

0

C D0 00 11 01 1

B0000

0 00 11 01 1

1111

1

110111

YA00000000

0 00 11 01 1

0000

0 00 11 01 1

1111

11111111

11

100000

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4-input K-map

01 11

1

0

0

1

0

0

1

101

1

1

1

1

0

0

0

1

11

10

00

00

10AB

CD

Y

Y = AC + ABD + ABC + BD

0

C D0 00 11 01 1

B0000

0 00 11 01 1

1111

1

110111

YA00000000

0 00 11 01 1

0000

0 00 11 01 1

1111

11111111

11

100000

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K-maps with Don’t Cares

0

C D0 00 11 01 1

B0000

0 00 11 01 1

1111

1

110X11

YA00000000

0 00 11 01 1

0000

0 00 11 01 1

1111

11111111

11

XXXXXX

01 11

01

11

10

00

00

10AB

CD

Y

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K-maps with Don’t Cares

0

C D0 00 11 01 1

B0000

0 00 11 01 1

1111

1

110X11

YA00000000

0 00 11 01 1

0000

0 00 11 01 1

1111

11111111

11

XXXXXX

01 11

1

0

0

X

X

X

1

101

1

1

1

1

X

X

X

X

11

10

00

00

10AB

CD

Y

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K-maps with Don’t Cares

0

C D0 00 11 01 1

B0000

0 00 11 01 1

1111

1

110X11

YA00000000

0 00 11 01 1

0000

0 00 11 01 1

1111

11111111

11

XXXXXX

01 11

1

0

0

X

X

X

1

101

1

1

1

1

X

X

X

X

11

10

00

00

10AB

CD

Y

Y = A + BD + C