k-map simplification - kfupm · k-map method provides a straightforward procedure for simplifying...

K-Map Simplification

COE 202

Digital Logic Design

Dr. Abdulaziz Tabbakh

College of Computer Sciences and Engineering

King Fahd University of Petroleum and Minerals

COE 202– Digital Logic Design – KFUPM slide 2

Outline

K-Map Method

Two-, Three-, and Four-variable K-Map

Prime Implicants and Essential Prime Implicants

SOP Simplification

POS Simplification

Don’t Care Conditions

Five-variables K-maps

Six-variables K-Maps


Introduction

The design task of finding an optimal gate-level

implementation of the Boolean functions.


The Map Method The complexity of the digital circuit that implements a Boolean

function is related to the complexity of the algebraic expression from

which the circuit is implemented.

Although Boolean expressions can be simplified by algebraic

manipulation, such an approach lacks clear regular rules for each

succeeding step and it is difficult to determine whether the simplest

expression has been achieved.

Each Boolean function has:

Unique truth table

Many different, but equivalent, algebraic expression

The map is pictorial form of a truth table

It is known as the Karnaugh map or K-map

K-map method provides a straightforward procedure for simplifying

Boolean functions.


K-map

A K-Map is a diagram made up of squares, with each

square representing one minterm of the function to be

minimized.

Adjacent minterms are combined to form simpler terms.

The map presents a visual diagram of all possible ways

a function may be expressed in standard form.

The distance between two binary codes is the number of

bit positions in which the two code have different values.

0011 and 0100 have a distance of 3 (called Hamming distance)

Any 2 adjacent squares in the k-map should have a

distance of exactly 1


K-map

The simplified expressions produced either SOP or POS

The simplest algebraic expression means:

Minimum number of terms ( minimum number of gates)

Smallest number of literals in each term ( less inputs to gates)

the simplest expression is not unique


2-Variable K-Maps

Two K-map squares are considered adjacent if the input

codes they represent have a Hamming distance of 1.

A K-map square with a function value of 1 will be

referred to as a 1-Square.

A K-map square with a function value of 0 will be

referred to as a 0-Square.

The simplification procedure is summarized below:

Step 1: Draw the map according to the number of input

variables of the function.

Step 2: Fill “1’s” in the squares for which the function is true.


2-Variable K-Maps Step 3: Form as big group of adjacent 1-squares as possible.

There are some rules for this which you will learn with bigger

maps.

Step 4: Find the common literals for each group and write the

simplified expression in SOP.

In an n-variable map each square is adjacent to “n” other

squares, e.g., in a 2-variable map each square is

adjacent to two other squares.


2-variable K-map

(a) F=m3 = xy(b) F=m1+m2+m3

= x’y+xy’+xy

= x+x’y

= (x+x’)(x+y)

= x+y


3-variable K-map

Three-variable map consists of 8 squares (8 minterms)

To maintain adjacent columns physically adjacent on the map, the

column coordinates do not follow the binary count sequence. This

choice yields unit distance between codes of one column to the next

(00 – 01—11 – 10), like Grey Code.

There are cases where two squares in the map are considered to be

adjacent even though they do not physically touch each other.

m0 is adjacent to m2 and m4 is adjacent to m6 because the minterms differ by only

one variable.


3-variable K-map

The number of adjacent squares that may be combined

must always represent a number that is a power of two

More adjacent squares product term with fewer

literals

One square represents one minterm (a term with three literals.)

Two adjacent squares represent a term with two literals

Four adjacent squares represent a term with one literal

Eight adjacent squares cover the entire map and produce a

function F=1.

Members of a group must have a closed loop adjacency,

i.e., L-Shaped 4 squares do not form a valid group.


Example

Simplify

F(x,y,z)= Ʃ(2,3,4,5)

F = xy’+x’y

Simplify :F(x,y,z)= Ʃ(3,4,6,7)


Example

Simplify

F(x,y,z)= Ʃ(3,4,6,7)

F = xz’+yz

yz

x 00 01 11 10

0 1

1 1 1 1

ZY

X


Example

Simplify

F(x,y,z)= Ʃ(0,2,4,5,6)

F = z’+xy’

yz

x 00 01 11 10

0 1 1

1 1 1 1

Z

Y

X


Example

Let the Boolean function

F = AB + A’C + A’BC’

Express in Sum-Of-minterms

Minimize.

Convert to Canonical form

F = ABC + ABC’ + A’BC + A’B’C + A’BC’

Fill in the map

F= (1,2,3,6,7)

= B + A’C

Solve Example 3.4 page 95

bc

a 00 01 11 10

0 1 1 1

1 1 1

c

b

a


4-variable K-map One square represents one minterm, giving a term with four literals.

Two adjacent squares represent a term with three literals.

Four adjacent squares represent a term with two literals.

Eight adjacent squares represent a term with one literal.

Sixteen adjacent squares produce a function that is always equal to

1


Example

Simplify

F (w, x, y, z) = Ʃ(0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14)

F = y’+w’z’+xz’

yz

wx 00 01 11 10

00 1 1 1

01 1 1 1

11 1 1 1

10 1 1

Z

Y

X

W


Example

F = A’B’C’ + B’CD’ + A’BCD’ + AB’C’ = B’D’ + B’C’ + A’CD’


Prime Implicants

In choosing adjacent squares in a map, we must ensure

that:

All minterms of the function are covered when we combine

squares

The number of terms in the expression is minimized

No redundant terms

Sometimes there might be two or more expressions

that satisfy the simplification criteria.


Prime Implicant

The procedure for combining squares in the map may be

made more systematic if we understand the meaning of

two special types of terms

A product term of a function is said to be an implicant.

A prime implicant is a product term obtained by

combining the maximum possible number of adjacent 1-

squares in a map.

If a minterm is covered by only one prime implicant, that

prime implicant is said to be essential prime implicant.

The prime implicants of a function can be obtained from

the map by combining all possible maximum numbers of

squares (i.e. we cannot remove any of its literals)


SOP Simplification procedure

1. Identify all prime implicants covering 1’s

Example: For a function of 3 variables, group all possible groups

of 4, then groups of 2 that are not contained in groups of 4, then

minterms that are not contained in a group of 4 or 2.

2. Identify all essential prime implicants and select them.

3. Check all minterms (1’s) covered by essential prime

implicants

4. Repeat until all minterms (1’s) are covered:

Select the prime implicant covering the largest uncovered

minterms (1’s).


Example

F(A, B, C, D) = ∑(0, 2, 3, 5, 7, 8, 9, 10, 11, 13, 15)

cd

ab 00 01 11 10

00 1 1 1

01 1 1

11 1 1

10 1 1 1 1

d

c

b

a


Example


Example

The simplified expression is obtained from the logical

sum of the two essential prime implicants and any two

prime implicants that cover minterms m3, m9, and m11

The simplified expression is obtained from the logical sum of

all the essential prime implicants, plus other prime

implicants that may be needed to cover any remaining

minterms not covered by the essential prime implicants.


Product of Sums Simplification

When combining minterms, we get SOP out of the K-

MAP simplification process.

If we combine maxterms (0 squares) we get the

compliment of F (F’).

Using DeMorgan’s Theorems, we can get F as a product

of sums (POS)


Example

F(A, B, C, D) = Σ(0, 1, 2, 5, 8, 9, 10)

F(A, B, C, D) = B’D’ + B’C’ + A’C’D

F’ = AB + CD + BD’

F’’= F = (A’ + B’)(C’ + D’)

(B’ + D)


Gate-Level Implementation

the implementation of a function in a standard form is

said to be a two-level implementation. The two-level

implementation may not be practical, depending on the

number of inputs to the gates.


The 1’s of the function represent the minterms and the

0’s represent the maxterms. The map for this function

For the sum of products,

we combine the 1’s to obtain:

F= x’z + xz’

For the product of sums,

we combine the 0’s to obtain:

F’=xz+x’z’ F = (x’+z’)(x+z)

Convert PO Maxterms to K-map


K-map from POS

To enter a function expressed in product-of-sums form

into the map, use the complement of the function to find

the squares that are to be marked by 0’s.

Marking 0’s in the squares representing the minterms of

F’. The remaining squares are marked with 1’s.


Don’t Care Conditions

Functions that have unspecified outputs for some input

combinations are called incompletely specified functions.

We DON’T CARE what value is assumed by the function

for the unspecified minterm.

call the unspecified minterms of a function don’t-care

conditions.

can be used on a map to provide further simplification of the

Boolean expression.

Don’t cares cannot be marked with ‘1’ or ‘0’ so they are

marked with ‘x’

Don’t cares can be assumed ‘1’ or ‘0’ to simplify the

function.


Example

Simplify the Boolean Function

F(w, x, y, z) = (1, 3, 7, 11, 15)

Which has the don’t care conditions: d(w, x, y, z) = (0, 2, 5)

Sol: F = yz + w’x’

Or F = yz + w’z

yz

wx 00 01 11 10

00 x 1 1 x

01 x 1

11 1

10 1

Z

Y

X

W


Simplification with Don’t Cares

To get the simplified expression in sum-of-products form,

we must include all 1’s in the map, but we may or may

not include any of the X’s

The two functions obtained are not algebraically equal,

but both cover the specified minterms but different

don’t care conditions (Both are VALID SOLUTIONS)!

It is also possible to obtain a simplified product-of-sums

expression.


SOP Simplification procedure using Don’t Cares

1. Identify all prime implicants covering 1’s & X’s

Each prime implicant must contain at least a single 1

2. Identify all essential prime implicants and select them.

An essential prime implicant must be the only implicant covering

at least a 1.

3. Check all 1’s covered by essential prime implicants

4. Repeat until all 1’s are covered:

Select the prime implicant covering the largest uncovered 1’s.


Five-Variable K-Maps

There are 32 minterms (squares) for a Boolean function

with five-variables.

It consists of 2 four-variable maps. Variable A

distinguishes between the two maps. The left-hand four-

variable map represents the 16 squares where A=0, and

the other four-variable map represents the squares

where A=1.


Five-Variable K-Maps

Minterms 0 through 15 belong to the four-variable map

with A=0 and minterms 16 through 31 belong to the four-

variable map with A=1.

Each four-variable map retains the previously defined

adjacency when taken separately.

In addition, each square in the A=0 map is adjacent to

the corresponding square in the A=1 map.

For example, minterm 4 is adjacent to minterm 20 and minterm

15 to 31.

The best way to visualize this new rule for adjacent

squares is to consider the two half maps as being one on

top of the other. Any two squares that fall one over the

other are considered adjacent.


Five-Variable K-Map Example


Six-Variable K-Maps

There are 64 minterms (squares)

for a Boolean function with six-

variables.

By following the procedure used

for the five-variable map, it is

possible to construct a six-

variable map with 4 four-variable

maps to obtain the required 64

squares.


Six-Variable K-Map Example

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