Key Observation

Adjacencies in the K-Map

000

001

010

01 1

1 10

1 1 1

100

101

00 01 11 10

0

1

BC A

B

C

011

010

000

001

100

110

101

111

B

C

A

Any two adjacant cells in the K-map differ in exactly one variable.

The uniting theorem can be applied when adjacent cells contain (function =1)to eliminat the changing variable.

Design ExamplesTwo Bit Comparator

F 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1

F 2 0 1 1 1 0 0 1 1 0 0 0 1 0 0 0 0

F 3 0 0 0 0 1 0 0 0 1 1 0 0 1 1 1 0

D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

B 0 1 0 1

A 0 0 1 1

=, >, < F 1 A B = C D F 2 A B < C D F 3 A B > C D

A B

C D

N 1

N 2

Block Diagram

Truth Table

Design Steps:1- Simplify to reduce cost (three 4-variable K-maps, one for each output)2- Implement using a suitable design style (e.g. 2-level AND-OR, NAND-NAND, or multilevel techniques)

Design ExamplesTwo Bit Comparator (continued)

F1 = A' B' C' D' + A' B C' D + A B C D + A B' C D'

F2 = A' B' D + A' C

F3 = B C' D' + A C' + A B D'

AB 00 01 11 10

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

00

01

11

10 C

CD

A

D

B K-map for F 1

AB 00 01 11 10

0 0 0 0

1 0 0 0

1 1 0 1

1 1 0 0

00

01

11

10 C

CD

A

D

B K-map for F 2

AB 00 01 11 10

0 1 1 1

0 0 1 1

0 0 0 0

0 0 1 0

00

01

11

10 C

CD

A

D

B K-map for F 3

f1 = AC(BD+bd) + ac(bd+BD) = (AC+ac) (bd+BD) = (a c) (b d)

Design ExamplesTwo-Bit Adder

Truth Table

+ N 3

X 0 0 0 0 0 0 0 1 0 0 1 1 0 1 1 1

Y 0 0 1 1 0 1 1 0 1 1 0 0 1 0 0 1

Z 0 1 0 1 1 0 1 0 0 1 0 1 1 0 1 0

D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

B 0 1 0 1

A 0 0 1 1

A B

C D

N 1

N 2

X Y Z

Block Diagram

Design ExampleTwo-Bit Adder (Continued)

X = A C + B C D + A B D

Z = B D' + B' D = B xor D

Y = A' B' C + A B' C' + A' B C' D + A' B C D' + A B C' D' + A B C D

AB 00 01 11 10

0 0 0 0

0 0 1 0

0 1 1 1

0 0 1 1

00

01

11

10 C

CD

A

D

B K-map for X

AB 00 01 11 10

0 1 1 0

1 0 0 1

1 0 0 1

0 1 1 0

00

01

11

10 C

CD

A

D

B K-map for Z

AB 00 01 11 10

0 0 1 1

0 1 0 1

1 0 1 0

1 1 0 0

00

01

11

10 C

CD

A

D

B K-map for Y

5-Variable K-MapsConstructed from two 4-variable K-Maps.

E

C C

100 101 111 110

B ABC

000 001 011 010

00

01

11

10 D

DE

B

1

1

1

1

1 1 1 1

1 1 1 1

1

1

100 101 111 110

B

C

ABC 000 001 011 010

00

01

11

10 D

DE

B

C

E

1

1

1

1

1

1

1

1

1 1

f = bCDE + Bd

g = ABe + Bce + bE

A = 0 A = 1

5-Variable K-Maps

ƒ(A,B,C,D,E) = m(2,5,7,8,10,13,15,17,19,21,23,24,29 31)

BC DE

BC DE

A =0

A =1

00 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

Another View

ƒ(A,B,C,D,E) = C E + A B' E + B C' D' E' + A' C' D E'

BC DE 00 01 11 10

00

01

11

10

A=0

BC DE 00 01 11 10

00

01

11

10

A=1

1 1

1

1

1 1

1

1

1 1 1

1 1 1

6-Variable K-Maps

ƒ(A,B,C,D,E,F) =m(2,8,10,18,24,

26,34,37,42,45,50,53,58,61)

CD EF

CD EF

AB =00

AB =01

00 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

CD EF

AB =11

00 01 11 10 00

01

11

10

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

CD EF

AB =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

CD EF

CD EF

AB =00

AB =01

00 01 11 10 00

01 11

10

00 01 11 10 00

01 11

10

CD EF

AB =11 00 01 11 10

00 01

11 10

CD EF

AB =10 00 01 11 10

00 01

11 10

1

1 1

1

1 1

1 1

1 1

1 1

1 1

6-Variable K-Maps

ƒ(A,B,C,D,E,F) =m(2,8,10,18,24,

26,34,37,42,45,50,53,58,61)

= D' E F' + A D E' F+ A' C D' F'

CD EF

CD EF

AB =00

AB =01

00 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

CD EF

AB =11

00 01 11 10 00

01

11

10

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

CD EF

AB =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

CD EF

CD EF

AB =00

AB =01

00 01 11 10 00

01 11

10

00 01 11 10 00

01 11

10

CD EF

AB =11 00 01 11 10

00

01 11

10

CD EF

AB =10 00 01 11 10

00 01

11 10

1

1 1

1

1 1

1 1

1 1

1 1

1 1

Definitions

Implicant: Single 1 entry or any group of 1’s that can be combined together in a K-map to form a product term.

(Dual): Single 0 entry or any group of 0’s that can be combined together in a K-map to form a sum term.

Prime Implicant: an implicant that cannot be combined with another implicant to eliminate a term

Essential Prime Implicant: a prime implicant that contains a 1 entry not covered by any other implicant.

6 Prime Implicants:A' B C’ , C D, A‘ D , B C' D’ , A C , AB D'

essential

Minimum cover = A’ D + A C + B C ‘ D ’

CD 00 01 11 10

0 1 1 0

1 1 1 0

1 0 1 1

0 0 1 1

00

01

11

10 A

AB

C

B

D

Example

Definitions

Implicant: Single 1 entry or any group of 1’s that can be combined together in a K-map to form a product term.

(Dual): Single 0 entry or any group of 0’s that can be combined together in a K-map to form a sum term.

Prime Implicant: an implicant that cannot be combined with another implicant to eliminate a term

Essential Prime Implicant: a prime implicant that contains a 1 entry not covered by any other implicant.

CD 00 01 11 10

0 1 1 0

1 1 1 0

1 0 1 1

0 0 1 1

00

01

11

10 A

AB

C

B

D

Example

Illustrating the Definitions

Prime Implicants:B D, C D, A C, B' C

essential

Essential primes form the minimum cover

AB 00 01 11 10

0 0 0 0

0 1 1 0

1 1 1 1

1 0 1 1

00

01

11

10 C

CD

A

D

B

5 Prime Implicants:

B D, A B C', A C D, A' B C, A' C' D

essential

Essential implicants form minimum cover

AB 00 01 11 10

0 0 1 0

1 1 1 0

0 1 1 1

0 1 0 0

00

01

11

10 C

CD

A

D

B

Illustrating the Definitions

Prime Implicants:A D’, AB, A C, BD

essential

Essential primes form the minimum cover

CD 00 01 11 10

0 0 0 0

0 1 1 0

1 1 1 1

1 0 1 1

00

01

11

10 A

AB

C

B

D

5 Prime Implicants:

B D, A B C', A C D, A' B C, A' C' D

essential

Essential implicants form minimum cover

AB 00 01 11 10

0 0 1 0

1 1 1 0

0 1 1 1

0 1 0 0

00

01

11

10 C

CD

A

D

B

Try the other axis

Quine-McCluskey (Tabular) Method systematically finds all prime implicants

Implication Table

Column I 0000 0100 1000 0101 0110 1001 1010 0111 1101

1111

Example: Simplify ƒ(A,B,C,D) = m(4,5,6,8,9,10,13) + d(0,7,15)

Step 1: List minterms and don’t cares using their binary representation, and group according to number of 1’s

Remarks:Only terms in adjacent groups have to becompared with one another. Terms in non-adjacent groups differ in morethan one variable. Thus can not be combined

Principle:Combine terms in adjacent groups which differ in a single variable, and eliminate thechanging variable.

Q & M Method

Implication Table

Column I Column II 0000 ¦ 0-00 -000 0100 ¦ 1000 ¦ 010- 01-0 0101 ¦ 100- 0110 ¦ 10-0 1001 ¦ 1010 ¦ 01-1 -101 0111 ¦ 011- 1101 ¦ 1-01

1111 ¦ -111 11-1

ƒ(A,B,C,D) = m(4,5,6,8,9,10,13) + d(0,7,15)

Step 2:

Compare elements of a group with k 1's against those with k+1 1's. If they differ by one bit, eliminate changing variable and place reduced term in next column.

E.g., 0000 vs. 0100 yields 0-00 0000 vs. 1000 yields -000

When a term is used in a combination, mark that term with a check. If cannot be combined, mark term with a star. These are the prime implicants.

Implication Table

Column I Column II Column III 0000 ¦ 0-00 * 01-- * -000 * 0100 ¦ -1-1 * 1000 ¦ 010- ¦ 01-0 ¦ 0101 ¦ 100- * 0110 ¦ 10-0 * 1001 ¦ 1010 ¦ 01-1 ¦ -101 ¦ 0111 ¦ 011- ¦ 1101 ¦ 1-01 *

1111 ¦ -111 ¦ 11-1 ¦

Q & M MethodStep 2 Continues:

Compare elements of a group in Col. II with

k 1's against those with k+1 1's. If they differ by one bit, eliminate changing variable and place reduced term in next column.

E.g., 010- vs. 011- yields 01- - -101 vs. -111 yields -1-1

Q & M - Prime Implicant Chart

rows = prime implicantscolumns = mintermsplace an "X" if minterm iscovered by the prime implicant

If column has a single X, than theimplicant associated with that rowis essential. It must appear inminimum cover

5,7,13,15 (-1-1)

4,5,6,7(01--)

9,13(1-01)

8,10(10-0)

8,9(10-0)

0,8(-000)

0,4(0-00) 4 5 6 8 9 10 13x

x xx

x

xxx

x

xx

x

x

4 5 6 8 9 10 13x

x xx

x

xxx

x

xx

x

x

5,7,13,15 (-1-1)

4,5,6,7(01--)

9,13(1-01)

8,10(10-0)

8,9(10-0)

0,8(-000)

0,4(0-00)

Gate Logic

• Review

Description Z = 1 if X and Y are both 1

Truth Table SwitchesY 0 1 0 1

X 0 0 1 1

Z 0 0 0 1

X Y

X • Y

false

true

NOT

AND

OR

Description If X = 0 then X ' = 1 If X = 1 then X ' = 0

Switches

X

X 0 1

X 1 0

T ruth T able

T rue

False X

Description Z = 1 if X or Y (or both) are 1

T ruth T able Switches

X 0 0 1 1

Y 0 1 0 1

Z 0 1 1 1

X Y

X + Y

False

T rue

Gate realization

Switch realizationa

e’

c’

d

w

1 s

Example : Logic circuit & its Switch realization

s = a (d+w)(c’+ e’)

c

e

dwa

s

Logic Functions: NAND

NANDDescription Z = 1 if X is 0 or Y is 0

T ruth T able Switches

X Y

X • Y

False

T rue X 0 0 1 1

Y 0 1 0 1

Z 1 1 1 0

Description Z = 1 if X and Y are both 1

Truth Table SwitchesY 0 1 0 1

X 0 0 1 1

Z 0 0 0 1

X Y

X • Y

false

true

AND

Note that switch circuit of the NAND gate is the complementof the switch circuit for the AND gate.

Logic Functions: NAND, NOR, XOR, XNOR

NORDescription Z = 1 if both X and Y are 0

T ruth T able Switches

X 0 0 1 1

Y 0 1 0 1

Z 1 0 0 0

X Y

X + Y False

T rue

OR

Description Z = 1 if X or Y (or both) are 1

T ruth T able Switches

X 0 0 1 1

Y 0 1 0 1

Z 0 1 1 1

X Y

X + Y

False

T rue

Note that switch circuit of the NOR gate is the complementof the switch circuit for the OR gate.