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� simply called the K-map

� a systematic method of reducing the complexity

KarnaughMaps

� a systematic method of reducing the complexity of algebraic expressions

� guarantees the simplest Boolean expression using very straightforward procedures

� can be presented in different ways basing on

the number of variables the Boolean function has

Combinational Logic (Part 2) * Property of STIPage 1 of 35

� 22 = 4 minterms

� a map with four quadrants

Two-VariableMap

� a map with four quadrants

Example 4:

Obtain the K-Map of the function: F = ΣΣΣΣ m(0,3)

Two-VariableMap

Obtain the K-Map of the function: F = ΣΣΣΣ m(0,3)

Solution:

� there are two minterms that has a value of 1(m0, m3)

� put a 1 on the appropriate squares of the two-

Truth Table:

� put a 1 on the appropriate squares of the two-variable K-map

� blank squares automatically have 0 values (you

may not write the 0 in the squares)

Example 5:

Obtain the K-Map of the function: F = XZ

Two-VariableMap

Obtain the K-Map of the function: F = XZ

Solution:

Truth Table

� the minterm that has a value of 1 is m3

� 23 = 8 minterms

� have eight quadrants containing a minterm

Three-VariableMap

� have eight quadrants containing a minterm each

� does not follow the normal binary count sequence like that of the two-variable map

� the way the minterms are assigned to each of

the squares differ by only one variable

� the fourth column of this map wraps around to the first column

Three-VariableMap

� count from the outside to the middle:

Example 6:

Obtain the K-Map of the function: F = ΣΣΣΣ

Three-VariableMap

Obtain the K-Map of the function: F = ΣΣΣΣm(1,3,5,7)

Solution:

� this is a three-variable map since its highest value is 7

� four minterms (1,3,5,7) has a value of 1

Truth Table:

Example 7:

Obtain the K-Map of the function:

Three-VariableMap

F = X’YZ’ + X’YZ + XY’Z + XY’Z’

Solution:

Truth Table:

Example 8:

Three-VariableMap

F = X + X’Z + Y’Z

Solution:

Truth Table:

� 24 = 16 minterms

� have sixteen quadrants containing a minterm

Four-VariableMap

� have sixteen quadrants containing a minterm each

� an extension of the three-variable map

� the assignment of columns and rows are similar to that of the three-variable map

Example 9:

The Boolean function: F = ΣΣΣΣ m(0,3,4,15) needs

Four-VariableMap

The Boolean function: F = ΣΣΣΣ m(0,3,4,15) needs to be represented in a K-Map.

Solution:

Minterms: m0, m3, m4, m15

Truth Table:

Example 10:

The Boolean function:

Four-VariableMap

The Boolean function:

F (A, B, C, D) = ΣΣΣΣ m(0,1,5,7)

needs to be represented in a K-Map.

Solution:

Minterms: m0, m1, m5, m7

Truth Table:

� remember: K-map is directly related to the minterms (and therefore, its truth table)

K-Map Simplification

K-Map for F (X, Y) = 1

K-Map for F (X, Y) = ΣΣΣΣ m(0,1,2,3)

F (X, Y) = m0+m1+m2+m3 = X’Y’+X’Y+XY’+XY

= X’Y’+X’Y+XY’+XY

= X’(Y’+Y) + X(Y’+Y) Distributive

= X’(1) + X(1) Postulate 6

� What happens when we take out one minterm from the equation below?:

F (X, Y) = ΣΣΣΣ m(0,1,2,3)

E (X, Y) = ΣΣΣΣ m(1,2,3)

E (X, Y) = m1+m2+m3 = X’Y+XY’+XY

Simplify using Boolean algebra:

E = X’Y + XY’ + XY

= X’Y + X(Y’+Y) Distributive Prop.

= X’Y + X(1) Postulate 6

= X’Y + X

= X’Y + (X+XY) Theorem 6

= X’Y + X + XY

= X + Y(X’+X) Distributive Prop.

= X + Y(1) Postulate 6

= X + Y

Step 1: Draw the K-Map

Step 2: Group into clusters

� Take note of all the adjacent squares that have

� Take note of all the adjacent squares that have a value of “1”

Rules in Grouping:

1. must be of size 1,2,4,8,16....2n

2. all the 1’s that can be grouped must be included in a cluster of maximum size

3. the minimum size for the cluster is two

4. combine squares that are adjacent

Step 3: Get the simplified expression from the clustering

Step 3 (cont.):

TIP: Get only the variables that appear in all

TIP: Get only the variables that appear in all the squares included in the cluster

� the two values (X,Y) are the terms included for the sum of products expression

E (X, Y) = ΣΣΣΣ m(1,2,3) = X’Y+XY’+XY

E = X + Y

Example 11:

Draw the K-map of the truth table shown here

Draw the K-map of the truth table shown here and minimize the resulting expression.

Solution:

Solution (cont.):

Step 3: Write the equivalent expression

Example 12:

Design a logic circuit to implement the following

Design a logic circuit to implement the following truth table:

Solution:

Solution (cont.):

F = A’B’ + BC’ + B’C’

= A’B’ + C’(B + B’) Distributive

= A’B’ + C’(1) Postulate 6

F = A’B’ + C’

The Wrapping-Around Method:

Back to Step 2:Group into clusters - wrap the

leftmost column with the rightmost column to get the

maximum number of groupings

Solution (cont.):

Step 4: Draw the logic diagram

F = A’B’ + C’

Three- and Four-Variable K-Maps

with Wrap-around

Summary forWrapping Around

with Wrap-around

Example 13:

Design a logic circuit to implement the following

Design a logic circuit to implement the following truth table:

Solution:

Solution (cont.):

Step 3: Write down the terms and the final

Step 3: Write down the terms and the final expression

upper-left square cluster: B’C’

outer two-four cluster: D’

F = B’C’ + D’

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