k-maps-11,12,13

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Function Minimization:

Karnaugh Maps

Maher al omari

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Introduction

• Karnaugh maps provide a systematic method toobtain simplified Boolean functions.

• Objective: Fewest possible terms/literals.

• Advantage: Easy with visual aid.

• Disadvantage: Limited to 5 variables.

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2-variable K-mapsF (a,b) = a b + a' b'

equivalent to:

ab

0 1

0

1

11

01

10

00

ba

1

0

0

1

F

= m0 + m3

1

1

0

0

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2-variable K-maps

• Equivalent labelling:

a

b

equivalent to:

ab

0 1

0

1

b

a

equivalent to:

ba

1 0

0

1

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2-variable K-maps

a'b' a'b

ab' aba

b

m 0 m 1

m 2 m 3 a

b

OR

Karnaugh-map (K-map) is organised as a matrix of squares, where

each square represents a minterm

adjacent squares always differ by just

one literal (so that the unifying

theorem may apply: a + a' = 1)

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2-variable K-maps

• The K-map for a function is specified by putting – a ‘1’ in the square corresponding to a minterm

– a ‘0’ otherwise

• For example:

0 0

0 1a

b

0 1

1 0a

b

C = ab S = ab' + a'b

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3-variable K-maps

• There are 8 minterms for 3 variables (a, b, c).

Therefore, there are 8 cells in a 3-variable K-map.

ab'c' ab'ca

b

abc abc'

a'b'c' a'b'c a'bc a'bc'

0

1

00 01 11 10

c

abc

ORm 4 m 5 a

b

m 7 m 6

m 0 m 1 m 3 m 2 0

1

00 01 11 10

c

a

bc

Note Gray code sequenceAbove arrangement ensures that mintermsof adjacent cells differ by only ONE literal .

(Other arrangements which satisfy thiscriterion may also be used.)

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3-variable K-maps

• There is wrap-around in the K-map:

– a'b'c' (m0) is adjacent to a'bc' (m2) – ab'c' ( m4) is adjacent to abc' (m6 )

m 4 m 5 m 7 m 6

m 0 m 1 m 3 m 2 0

1

00 01 11 10abc

Each cell in a 3-variable K-map has 3 adjacent neighbours.In general, each cell in an n -variable K-map has n adjacent

neighbours.

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4-variable K-maps

m 4 m 5

w

y

m 7 m 6

m 0 m 1 m 3 m 2

z

w xyz

m 1 2 m 1 3 m 1 5 m 1 4

m 8 m 9 m 1 1 m 1 0

x

16 cellsEvery cell has 4

neighbours

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Simplification Using K-maps• Example:

F (w,x,y,z) = w'xy'z' + w'xy'z + wx'yz'

+ wx'yz + wxyz' + wxyz

z

1 1

w

y

00

01

11

10

00 01 11 10w x

yz

1 1

1 1

x

(cells with ‘0’ are notshown for clarity)

= Σ m(4, 5, 10, 11, 14, 15)

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Simplification Using K-maps

• Each group of adjacent minterms corresponds to a possible

product term of the given function.

1 1

w

00

01

11

10

00 01 11 10

z

w xyz

1 1

1 1

x

A

B

y

F (w,x,y,z) =Σ

m(4, 5,) +Σ

m(10, 11, 14, 15,)

= A + B

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Simplification Using K-maps• There are 2 groups of minterms: A and B, where:

A = w'xy'z' + w'xy'z

= w'xy'(z' + z)

= w'xy'

1 1

w

00

01

11

10

00 01 11 10

z

w x

yz

1 1

1 1

x

A

B

y

B = wx'yz' + wx'yz + wxyz' + wxyz

= wx'y(z' + z) + wxy(z' + z)

= wx'y + wxy

= w(x'+x)y

= wy

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Simplification Using K-maps

1 1

w

00

01

11

10

00 01 11 10

z

w x

yz

1 1

1 1

x

A

B

y

w'xy'

wy

•Each product term of agroup, represents the sum

of minterms in that group.

F (w,x,y,z) = A + B = w'xy' + wy

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Simplification Using K-

maps• Based on the Unifying Theorem:

A + A' = 1• Each group of adjacent cells must have size in

powers of twos: 1, 2, 4, 8, …)

• Grouping 2 adjacent squares eliminates 1 variable,grouping 4 squares eliminates 2 variables,grouping 8 squares eliminates 3 variables, and so

on. In general, grouping 2n squares eliminates nvariables.

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Simplification Using K-

maps• Group as many squares as possible.

– The larger the group is, the fewer the number of literalsin the resulting term.

• Select as few groups as possible to cover all the

squares (minterms/maxterms) of the function. – The fewer the groups, the fewer the number of terms in

the minimized function.

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Simplification Using K-maps• Other possible valid groupings of a 4-variable K-map

include:

1

11

1

1

1

1

11

11

1 1

111

1

11

1

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Simplification Using K-maps• Larger groups correspond to product terms of

fewer literals. In the case of a 4-variable K-map:

1 cell = 4 literals, e.g.: wxyz, w'xy'z

2 cells = 3 literals, e.g.: wxy, wy'z'

4 cells = 2 literals, e.g.: wx, x'y

8 cells = 1 literal, e.g.: w, y', z

16 cells = no literal, e.g.: 1

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Simplification Using K-maps• Groups of minterms must be

(1) rectangular, and(2) have size in powers of 2’s.

Otherwise they are invalid groups. Some

examples of invalid groups:1

11

1 1

111

1

1

1

1

1

1

1

1

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Drawing the K-map

1

1

C

A

00

01

11

10

00 01 11 10

B

CD

AB

D

1

or

= A(C'D')(B'+D') + BC + CC' + A'CD

= AB'C'D' + AC'D' + BC + A'CD

=AB'C'D' + ABC'D' + ABCD + ABCD' +

A'BCD + A'BCD' + A'B'CD

f(A,B,C,D) = A(C+D)'(B'+D') +C(B+C'+A'D)

1

1

11

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prime implicant • A prime implicant is a product term obtained by

combining the maximum possible number of minterms from adjacent squares in the map.

• Use bigger groupings (prime implicants) where

possible.11 1

111

11 1

111

Sec 3-2

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prime implicants 1

All prime implicants

1

1

C

A

00

01

11

10

00 01 11 10

B

CD

AB

1

1

1

1

D

1

1

1

Sec 3-2

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prime implicants 2


1

1

C

A

00

01

11

10

00 01 11 10

B

CD

AB

1

1

1

D

1

1

is not fully contained in any one other implicant .

Sec 3-2

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Example: find all prime implicats for

F = y’z’ + wyz + w’xz

Y

Y Z

Z

11

W

00

01

11

10

00 01 11 10

X

W X

1

1

1

1 1

1

Sec 3-2

y’z’

w’xzw’xy’

WyZxyZ

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NOT Essential

prime implicant

essential prime implicant

1

1*

C

A

00

01

11

10

00 01 11 10

B

CD

AB

1

1

1*

D

1*

1

* Essential prime implicant

includes at least one 1 that is not included

in any other prime implicant.Sec 3-2

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NOT Essential

prime implicant1

Y

W

00

01

11

10

00 01 11 10

X

Y Z

W X

1

1

1

1*

Z

1*

1

* Essential prime implicant

includes at least one 1 that is not included

in any other prime implicant.

1

Example: find all essential prime implicants for

F = y’z’ + wyz + w’xz

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Simplest SOP Expressions – bigger groupings of minterms ( prime

implicants)

– no redundant groupings (look for

essential prime implicants)

–minimum number of

literals per product term

–minimum number of productterms

Leads to

Leads to

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no redundant groupings

1

1

1

11

1

1

1

1

1

1

11

1 1

1

redundant

group

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Simplest SOP Expressions• Algorithm

1. Identify & Circle all essential prime implicants onthe K-map.

.

2. Select a minimum prime implicants to cover those mintermsnot covered by the essential prime implicants.

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1

1

C

A

00

01

11

10

00 01 11 10

B

CD

AB

1

1

1

1

D

1

1

1

Essential prime

implicants

Minimum cover

• Example f(A,B,C,D) = ∑ m(2,3,4,5,7,8,10,13,15)

BD

AB'D'A'BC'

A'B'C

f(A,B,C,D) = BD + A'B'C + AB'D' + A'BC'

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Getting POS Expressions

• This gives the SOP of F' to be:

F' = BD' + AB

• To get POS of F, we have:

F = (BD' + AB)'

= (BD')'(AB)' DeMorgan

= (B'+D)(A'+B') DeMorgan

0

0

C

A

00

01

11

10

00 01 11 10

B

CD

AB

0

1

0

0

D

0

0

0 0

01

11

1 1

A

CD

1

1

C

00

01

11

10

00 01 11 10

B

AB

1

0

1

1

D

1

1

1 1

10

00

0 0K-map

of F

K-map

of F'

Sec 3-4

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Getting POS Expressions• Simplified POS expression can be obtained by

grouping the maxterms (i.e. 0s) of given function.

• Example:

Given F=∑m(0,1,2,3,5,7,8,9,10,11), we first

draw the K-map, then group the maxtermstogether:

1

1

C

A

00

01

11

10

00 01 11 10

B

CD

AB

1

0

1

1

D

1

1

1 1

10

00

0 0(B'+D) (A'+B')

Sec 3-4

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No. A B C D P

0 0 0 0 0 1

1 0 0 0 1 0

2 0 0 1 0 0

3 0 0 1 1 1

4 0 1 0 0 0

5 0 1 0 1 1

6 0 1 1 0 1

7 0 1 1 1 0

8 1 0 0 0 0

9 1 0 0 1 1

10 1 0 1 0 X

11 1 0 1 1 X

12 1 1 0 0 X

13 1 1 0 1 X

14 1 1 1 0 X

15 1 1 1 1 X

Don’t-care Conditions• In certain problems, some

outputs are not specified.

• These outputs can be either ‘1’or ‘0’.

• They are called don’t-careconditions, denoted by X.

• Example: An odd parity

generator for BCD code whichhas 6 unused combinations.

Sec 3-5

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Don’t-care Conditions

• Don’t-care conditions can be used to help simplifyBoolean expression further in K-maps.

• They could be chosen to be either ‘1’ or ‘0’,

depending on which gives the simpler expression.

Sec 3-5

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Don’t-care Conditions SOP

WITH Don’t-cares:

P =

1

A

C

00

01

11

10

00 01 11 10

D

AB

CD

1

B

1

1

1

1

A

C

00

01

11

10

00 01 11 10

D

AB

CD

1

B

1

1

1

X X

XXXX

Can you find Pin simplest POS?

WITHOUT Don’t-cares:

P =

No. A B C D P

0 0 0 0 0 1

1 0 0 0 1 0

2 0 0 1 0 0

3 0 0 1 1 1

4 0 1 0 0 0

5 0 1 0 1 16 0 1 1 0 1

7 0 1 1 1 0

8 1 0 0 0 0

9 1 0 0 1 1

10 1 0 1 0 X

11 1 0 1 1 X

12 1 1 0 0 X

13 1 1 0 1 X

14 1 1 1 0 X

15 1 1 1 1 X

Sec 3-5

1

A

C

00

01

11

10

00 01 11 10

D

AB

CD

1

B

1

11

X X

XXXX

00

00

0

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5-variable K-maps• Organised as two 4-variable K-maps:

Corresponding squares of each map are adjacent.Can visualise this as being one 4-variable map on TOP of the

other 4-variable map .

m 2 0 m 2 1

w

y

m 2 3 m 2 2

m 1 6 m 1 7 m 1 9 m 18 00

01

11

10

00 01 11 10

z

w x

yz

m 2 8 m 2 9 m 3 1 m 30

m 2 4 m 2 5 m 2 7 m 26

x

m 4 m 5

w

y

m 7 m 6

m 0 m 1 m 3 m 2 00

01

11

10

00 01 11 10

z

w x

yz

m 1 2 m 13 m 1 5 m 1 4

m 8 m 9 m 1 1 m 1 0

x

v ' v

Sec 3-3

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5-variable K-maps

1

w

y

1 1

1 1 00

01

11

10

00 01 11 10

z

w x

yz

1 1 1

1 1

x

1

w

y

1 1

1 1 00

01

11

10

00 01 11 10

z

w x

yz

1 1 1

1 1

x

v ' v

Sec 3-3

k-maps-11,12,13

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