Karnaugh Map (K-map)

The Karnaugh The Karnaugh MapMap

Contents

• Introduction• The Map Methods•Rules for Grouping•Solution of Sop and POS expression with K-map •Don’t Care Conditions

dEPARTMENT OF cOMPUTER SCIENCE

KARNAUGH MAP• INTRODUCED BY MAURICE KARNAUGH IN 1953• ALSO KNOWN AS THE K-MAP• IT IS A METHOD TO SIMPLIFY BOOLEAN ALGEBRA

EXPRESSIONS.• IT REDUCES THE NEED FOR EXTENSIVE CALCULATIONS.• MODIFIED FORM OF TRUTH TABLE.• USE A 2-DIMENSIONAL GRID• USED TO PRODUCE THE SIMPLEST SOP OR POS EXPRESSION

POSSIBLE, KNOWN AS THE MINIMUM EXPRESSION.• DIFFICULT FOR MORE THAN 5 VARIABLES• LESS TIME AND SPACE CONSUMING

THE KARNAUGH MAPTHE KARNAUGH MAP

• NUMBER OF SQUARES = NUMBER OF COMBINATIONS

• EACH SQUARE REPRESENTS A MINTERM

• N VARIABLES 2^N SQUARES

• 2 VARIABLES 2^2=4 SQUARES



STEPS TO SOLVE A BOOLEAN EXPRESSION USING K-MAP

Following are the steps to solve a boolean expression using K-MAP.Step 1: InitiateStep 2: Mapping A Standard Expression.Step 3: Form GroupsStep 4: Determining The Minimum Expression

Express the given expression in its SOP or POS form.

Consider the adjacent 'ones' (or ‘zeroes’) in the K-map cells and group them.

Enter the value of 'one' for each product-term into the K-map cell, while filling others with zeros.

The 2 Variable K-Map• THERE ARE 4 CELLS AS SHOWN:

AB 0

0

1

1 ABAB’

A’BA’B’m1m0

m2 m2

The 3 Variable K-Map

• THERE ARE 8 CELLS AS SHOWN:

CCABAB 00 11

0000

0101

1111

1010

CBA CBA

CBA BCA

CAB ABC

CBA CBA

m2

m1

m6

m3

m4

m7

m5

m0

The 4-Variable K-Map

CDCDABAB 0000 0101 1111 1010

0000

0101

1111

1010 DCBA

DCAB

DCBA

DCBA

DCBA

DCAB

DCBA

DCBA

CDBA

ABCD

BCDA

CDBA

DCBA

DABC

DBCA

DCBAm0

m4

m2

m7m5

m12

m6

m13 m15 m14

m3m1

m10m11m9m8

TYPES OF MINIMIZATIONS

1.SOP MinimizationOR

Sum of products Minimization

2. POS MinimizationOR

Product od sums Minimization

1. K-Map SOP Minimization

• A BOOLEAN EXPRESSION MUST BE IN STANDARD FORM BEFORE YOU USE A K-MAP. THERE ARE TWO METHODS FOR THIS PURPOSE:

• 1.ALGEBRIC MANIPULATION

• 2. NUMERICAL EXPANSION

STEP:1Initiate

Step 1: Write the binary value of the two variables and attach a 0 for the missing variable.

Step 2: Write the binary value of the two variables and attach a 1 for the missing variable.

The two resulting binary numbers are the values of the standard SOP terms.

If the function is not in standard fom then we first convert into standard form as given

below:• F = A‘C + A‘B + AB‘C +BC

• 001+011+010+011+101+011+111 (NUMERICAL EXPANSION)

• =A‘C(B+B‘)+A’B(C+C’)+AB’C+BC(A+A’) (SIMPLIFICATION USING RULES)

• =A’BC+A’B’C+A’BC+A’BC’+AB’C+ABC+A’BC

• =A’BC+A’B’C+A’BC’+AB’C+ABC

• 011 001 010 101 111

• =M1+M2+M3+M5+M7

Mapping a Standard SOP Expression

For an SOP expression in standard form:A 1 is placed on the

K-map for each product term in the expression.

Each 1 is placed in a cell corresponding to the value of a product term.

CBA

CCABAB 00 11

0000

0101

1111

1010

CBA CBA

CBA BCA

CAB ABC

CBA CBA1

STEP:2

CCABAB 00 11

0000

0101

1111

1010

Mapping a Standard SOP Expression (full example)

The expression: CBACABCBACBA

000 001 110 100

1 1

1

1

• F = A‘C + A‘B + AB‘C +BC

• 001+011+010+011+101+011+111 (NUMERICAL EXPANSION)

• =A‘C(B+B‘)+A’B(C+C’)+AB’C+BC(A+A’) (SIMPLIFICATION USING RULES)

• =A’BC+A’B’C+A’BC+A’BC’+AB’C+ABC+A’BC

• =A’BC+A’B’C+A’BC’+AB’C+ABC

• 011 001 010 101 111

• =M1+M2+M3+M5+M7

CCAABB

00 11

0000010111111010

Mapping a Non-Standard SOP Expression

1

1

1

1

1

GROUPS AND THEIR OUT PUT IN VARIABLE

16

For a 3-variable map:1. A 1-cell group yields a 3-

variable product term2. A 2-cell group yields a 2-

variable product term3. A 4-cell group yields a 1-

variable product term4. An 8-cell group yields a

value of 1 for the expression.

• For a 4-variable map:1. A 1-cell group yields a 4-variable

product term2. A 2-cell group yields a 3-variable


product term4. An 8-cell group yields a a 1-

variable product term5. A 16-cell group yields a value of 1

for the expression.

• For a 2-variable map:1. A 1-cell group yields a 2-variable


product term3. A 4-cell group yields a value of 1

for the expression.

Cell Adjacency

CDCDABAB

0000 0101 1111 1010

0000010111111010

Adjacency in k map is defined one variable change

Grouping The Karnaugh map uses the following rules for the

simplification of expressions by grouping together adjacent cells containing ones.

1.Groups may not include any cell containing a zero.

2.Groups may be horizontal or vertical, but not diagonal.

STEP:3

RULES

3. Groups must contain 1, 2, 4, 8, or in general 2n cells.

That is if n = 1, a group will contain two 1's since 21 = 2.

If n = 2, a group will contain four 1's since 22 = 4.

4. Each group should be as large as possible.

5.Each cell containing a one must be in at least one group.

6.Groups may overlap.

7.Groups may wrap around the table. The leftmost cell in a row may be grouped with the

rightmost cell and the top cell in a column may be grouped with the bottom cell.

8. There should be as few groups as possible, as long as this does not contradict any of the previous rules.

Prime Implicants

• A PRIME IMPLICANT IS A PRODUCT TERM OBTAINED BY COMBINING THE MAXIMUM POSSIBLE NUMBER OF

• ADJACENT SQUARES IN THE MAP.• IF A MINTERM IN A SQUARE IS COVERED BY ONLY

ONE PRIME IMPLICANT, THAT PRIME IMPLICANT IS SAID TO BE ESSENTIAL.

• THE PRIME IMPLICANTS OF A FUNCTION CAN BE OBTAINED FROM THE MAP BY COMBINING ALL POSSIBLE MAXIMUM NUMBERS OF SQUARES

Grouping the 1s (example)

CCABAB 00 11

0000 11

0101 11

1111 11 11

1010

CCABAB 00 11

0000 11 11

0101 11

1111 11

1010 11 11

Grouping the 1s (example)CDCD

ABAB 0000 0101 1111 1010

0000 11 11

0101 11 11 11 11

1111

1010 11 11

CDCDABAB 0000 0101 1111 1010

0000 11 11

0101 11 11 11

1111 11 11 11

1010 11 11 11

Determining the Minimum SOP Expression from the Map• THE FOLLOWING RULES ARE APPLIED TO FIND THE

MINIMUM PRODUCT TERMS AND THE MINIMUM SOP EXPRESSION:

1. AFTER GROUPING EACH GROUP VARIABLES THAT OCCUR BOTH COMPLEMENTED AND UNCOMPLEMENTED WITHIN THE GROUP ARE ELIMINATED CALLED CONTRADICTORY VARIABLES.

STEP:4

Determining the Minimum SOP Expression from the Map (example)

CDCDABAB 0000 0101 1111 1010

0000 11 110101 11 11 11 111111 11 11 11 111010 11

BCA

DCA

DCACAB

Determining the Minimum SOP Expression from the Map (exercises)

ACCAB

CCABAB 00 11

0000 11 11

0101 11

1111 11

1010 11 11

CDCDABAB 0000 0101 1111 1010

0000 11 11

0101 11 11 11

1111 11 11 11

1010 11 11 11

CBCBAD

Y=Σ(0,1,3)

=A’(B+B’)=A’.1=A’

=(A+A’)B=1.B=B

=(A+A’)(B’C’+BC’)=1.C’(B+B’)=C’.1=C’

=A(B’C’+B’C+BC+BC’)=A(B’(C+C’)+B(C+C’))=A(B+B’)=A

Y=Σ(0,2,4,5,6,7)

=(A’B’+AB’)(C’D’+CD’)=B’(A+A’)D’(C+C’)=B’D’

=(A’B+AB)+(C’D+CD)=B(A+A’)+D(C+C’)=BD

2. K-Map POS Minimization

• The approaches are much the same (as SOP) except that with POS expression, 0’s representing the standard sum terms are placed on the K-map instead of 1s.

CCABAB 00 11

0000

0101

1111

1010

Mapping a Standard POS Expression The expression:

))()()(( CBACBACBACBA

000 010 110

000

0

F= π(0,2,5,6)

Simplify the following Boolean function into (a) sum-of-products form and(b) product-of-sums form:F (A, B, C, D) = (0, 1, 2, 5, 8, 9, 10)Combining the squares with 1’s gives the simplified function in sum-of-products form:

(a) F = B’D’ + B’C’ + A’C’If the squares marked with 0’s are combined, as shown in the diagram, we obtainthe simplified complemented function:

F’ = AB + CD + BD’Applying DeMorgan’s theorem

(b) F = (A’ + B’) (C’ + D’) (B’ + D)

K-map Simplification of POS Expression

))()()()(( CBACBACBACBACBA

CCABAB 00 11

0000

0101

1111

1010 BA

0 0

0 00 AC1

11

ACBAF

F=π(0,1,2,3,6)

SOP Form

POS Form

F’=A’+BC’

F=A(B’+C)

Don’t Care Conditions

• A DON’T CARE CONDITION, MARKED BY (X) IN THE TRUTH TABLE, INDICATES A CONDITION WHERE THE DESIGN DOESN’T CARE IF THE OUTPUT IS A (0) OR A (1).

• A DON’T CARE CONDITION CAN BE TREATED AS A (0) OR A (1) IN A K-MAP.

• TREATING A DON’T CARE AS A (0) MEANS THAT YOU DO NOT NEED TO GROUP IT.

• TREATING A DON’T CARE AS A (1) ALLOWS YOU TO MAKE A GROUPING LARGER, RESULTING IN A SIMPLER TERM IN THE SOP EQUATION.

SIMPLIFY THE BOOLEAN FUNCTIONF (W, X, Y, Z) = (1, 3, 7, 11, 15)

WHICH HAS THE DON’T-CARE CONDITIONSD (W, X, Y, Z) = (0, 2, 5)

Karnaugh Map (K-map)

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