Chapter 3 Simplification of Switching Functions. Karnaugh Maps (K-Map) A K-Map is a graphical...

Chapter 3

Simplification of Switching Functions

Karnaugh Maps (K-Map)

A K-Map is a graphical representation of a logic function’s truth table

Relationship to Venn Diagrams

a b

ab

ab abab

Relationship to Venn Diagrams

a b

0m

2m 1m3m

Relationship to Venn Diagrams

0m

1m

2m

3mb

a

Relationship to Venn Diagrams

0

1

2

3b

a

Relationship to Venn Diagrams

0 1

0

1

ab0

1

2

3

Two-Variable K-Map

0 1

0

1

ab

Three-Variable K-Map

abc 00 01 11 10

0

1

0m

1m

2m

3m

6m

7m

4m

5m

Three-Variable K-Map

abc 00 01 11 10

0

1

Three-Variable K-Map

abc 00 01 11 10

0

1

Edges are adjacent

Four-variable K-Map

abcd 00 01 11 10

00

01

11

10

0m

1m

2m

3m

6m

7m

4m

5m

12m

13m

14m

15m

10m

11m

8m

9m

Four-variable K-Mapab

cd 00 01 11 10

00

01

11

10

Four-variable K-Mapab

cd 00 01 11 10

00

01

11

10

Edges are adjacent

Edg

es

are

ad

jace

nt

Plotting Functions on the K-map

SOP Form

Canonical SOP Form

Three Variable Example

F ABC ABC ABC ABC

using shorthand notation

6 3 1 5F m m m m

, , 1,3,5,6F A B C m

Three-Variable K-Map Example

abc 00 01 11 10

0

1

, , 1,3,5,6F a b c m

Plot 1’s (minterms) of switching function

1 1 1

1

Three-Variable K-Map Example

abc 00 01 11 10

0

1

, ,F a b c ab bc

Plot 1’s (minterms) of switching function

1 1 1

1 abbc

Four-variable K-Map Exampleab

cd 00 01 11 10

00

01

11

10

, , , 0, 2,9,12,14F a b c d m

1

1

1

1

1

Karnaugh Maps (K-Map)

Simplification of Switching Functionsusing K-MAPS

Terminology/Definition Literal

A variable or its complement Logically adjacent terms

Two minterms are logically adjacent if they differ in only one variable position

Ex:

abc abcand

m6 and m2 are logically adjacent

Note: abc abc a a bc bc Or, logically adjacent terms can be combined

Terminology/Definition

Implicant Product term that could be used to cover

minterms of a function Prime Implicant

An implicant that is not part of another implicant

Essential Prime Implicant An implicant that covers at least one minterm

that is not contained in another prime implicant

Cover A minterm that has been used in at least one

group

Guidelines for Simplifying Functions

Each square on a K-map of n variables has n logically adjacent squares. (i.e. differing in exactly one variable)

When combing squares, always group in powers of 2m , where m=0,1,2,….

In general, grouping 2m variables eliminates m variables.

Guidelines for Simplifying Functions

Group as many squares as possible. This eliminates the most variables.

Make as few groups as possible. Each group represents a separate product term.

You must cover each minterm at least once. However, it may be covered more than once.

K-map Simplification Procedure

Plot the K-map Circle all prime implicants on the K-

map Identify and select all essential

prime implicants for the cover. Select a minimum subset of the

remaining prime implicants to complete the cover.

Read the K-map

Example

Use a K-Map to simplify the following Boolean expression

, , 1, 2,3,5,6F a b c m

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 1: Plot the K-map

1 1 1

1

, , 1, 2,3,5,6F a b c m

1

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 2: Circle ALL Prime Implicants

1 1 1

1

, , 1, 2,3,5,6F a b c m

1

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 3: Identify Essential Prime Implicants

1 1 1

1

, , 1, 2,3,5,6F a b c m

1

EPI

EPI

PI

PI

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 4: Select minimum subset of remaining Prime Implicants to complete the cover.

1 1 1

1

, , 1, 2,3,5,6F a b c m

1

EPIPI

EPI

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 5: Read the map.

1 1 1

1

, , 1, 2,3,5,6F a b c m

1

bcab

bc

Solution

, ,F a b c ab bc bc ab b c

Example

Use a K-Map to simplify the following Boolean expression

, , 2,3,6,7F a b c m

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 1: Plot the K-map

11

11

, , 2, 4,5,7F a b c m

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 2: Circle Prime Implicants

11

11

, , 2,3,6,7F a b c m

Wrong!!We reallyshould drawA circle aroundall four 1’s

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 3: Identify Essential Prime Implicants

EPIEPI

, , 2,3,6,7F a b c m

11

11Wrong!!We reallyshould drawA circle aroundall four 1’s

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 4: Select Remaining Prime Implicants to complete the cover.

EPIEPI

11

11

, , 2,3,6,7F a b c m

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 5: Read the map.

abab

11

11

, , 2,3,6,7F a b c m

Solution

, ,F a b c ab ab b

Since we can still simplify the functionthis means we did not use the largestpossible groupings.

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 2: Circle Prime Implicants

11

11

, , 2,3,6,7F a b c m

Right!

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 3: Identify Essential Prime Implicants

EPI

, , 2,3,6,7F a b c m

11

11

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 5: Read the map.

b

11

11

, , 2,3,6,7F a b c m

Solution

, ,F a b c b

Special Cases

Three-Variable K-Map Example

abc 00 01 11 10

0

1 1 1 1

1

, , 1F a b c

11

1

1

Three-Variable K-Map Example

abc 00 01 11 10

0

1

, , 0F a b c

Three-Variable K-Map Example

abc 00 01 11 10

0

1 1

, ,F a b c a b c

1

1

1

Four Variable Examples

Example

Use a K-Map to simplify the following Boolean expression

, , , 0, 2,3,6,8,12,13,15F a b c d m

Four-variable K-Mapab

cd 00 01 11 10

00

01

11

10

, , , 0, 2,3,6,8,12,13,15F a b c d m

1

1

1

1

11

1

1

Four-variable K-Mapab

cd 00 01 11 10

00

01

11

10

0,2,3,6,8,12,13,15F m

1

1

1

1

11

1

1

Four-variable K-Mapab

cd 00 01 11 10

00

01

11

10

F abd abc acd abd acd

1

1

1

1

11

1

1

Example

Use a K-Map to simplify the following Boolean expression

, , , 0, 2,6,8,12,13,15

3,9,10

F a b c d m

d

D=Don’t care (i.e. either 1 or 0)

Four-variable K-Mapab

cd 00 01 11 10

00

01

11

10

1

1

d

1

11

1

1

, , , 0, 2,6,8,12,13,15 3,4,9F a b c d m d

d

d

Four-variable K-Mapab

cd 00 01 11 10

00

01

11

10

1

1

d

1

11

1

1

F ac ad abd

d

d

Five Variable K-Maps

, , , ,F a b c d e

Five variable K-map

A=1

A=0

Use two four variable K-maps

Use Two Four-variable K-Maps

bcde 00 01 11 10

00

01

11

10

bcde 00 01 11 10

00

01

11

10

A=0 map A=1 map

Five variable example

, , , , 5,7,13,15,21,23,29,31F a b c d e m

Use Two Four-variable K-Maps

bcde 00 01 11 10

00

01

11

10

bcde 00 01 11 10

00

01

11

10

A=0 map A=1 map

, , , , 5,7,13,15,21,23,29,31F a b c d e m

1

1

1

1

1

1

1

1

Use Two Four-variable K-Maps

bcde 00 01 11 10

00

01

11

10

bcde 00 01 11 10

00

01

11

10

A=0 map A=1 map

1

1

1

1

1

1

1

1

1F a ce 2F a ce

Five variable example

1 2F F F a ce a ce ce

Plotting POS Functions

K-map Simplification Procedure

Plot the K-map for the function F Circle all prime implicants on the K-map Identify and select all essential prime

implicants for the cover. Select a minimum subset of the remaining

prime implicants to complete the cover. Read the K-map Use DeMorgan’s theorem to convert F to F

in POS form

Example

Use a K-Map to simplify the following Boolean expression

, , 1, 2,3,5,6F a b c M

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 1: Plot the K-map of F

1 1 1

11

, , 1, 2,3,5,6F a b c M

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 2: Circle ALL Prime Implicants

1 1 1

11

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 3: Identify Essential Prime Implicants

1 1 1

11

EPI

EPI

PI

PI

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 4: Select minimum subset of remaining Prime Implicants to complete the cover.

1 1 1

11

EPIPI

EPI

Three-Variable K-Map Example

abc 00 01 11 10

0

1

Step 5: Read the map.

1 1 1

11

bcab

bc

Solution

F ab bc bc

F ab bc bc

a b b c b c

, , 1, 2,3,5,6F a b c M

TPS Quiz