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Karnaugh Map Technique for Simplifying Boolean Expressions, Study notes of Computer Architecture and Organization

University of Illinois - Urbana-ChampaignComputer Architecture and Organization

Karnaugh maps as a graphical technique for simplifying boolean expressions into a minimal sum of products (msp) form. The concept of minterms, sum of minterms form, and the process of simplifying expressions using karnaugh maps. It also includes examples of simplifying two-variable and three-variable expressions.

Typology: Study notes

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February 2, 2004 CS 231 1
Karnaugh maps
•Last time we saw applications of Boolean logic to circuit design.
–The basic Boolean operations are AND, OR and NOT.
–These operations can be combined to form complex expressions,
which can also be directly translated into a hardware circuit.
–Boolean algebra helps us simplify expressions and circuits.
•Today we’ll look at a graphical technique for simplifying an expression
into a minimal sum of products (MSP) form:
–There are a minimal number of product terms in the expression.
–Each term has a minimal number of literals.
•Circuit-wise, this leads to a
minimal
two-level implementation.
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February 2, 2004

CS 231

Karnaugh maps

•^

Last time we saw applications of Boolean logic to circuit design.^ –

The basic Boolean operations are AND, OR and NOT.

–^

These operations can be combined to form complex expressions,which can also be directly translated into a hardware circuit.

–^

Boolean algebra helps us simplify expressions and circuits.

•^

Today we’ll look at a graphical technique for simplifying an expressioninto a minimal sum of products (MSP) form:^ –

There are a minimal number of product terms in the expression.

–^

Each term has a minimal number of literals.

•^

Circuit-wise, this leads to a

minimal two-level implementation.

February 2, 2004

CS 231

Review: Standard forms of expressions

•^

We can write expressions in many ways, but some ways are more usefulthan others

•^

A sum of products (SOP) expression contains:^ –

Only OR (sum) operations at the “outermost” level

–^

Each term that is summed must be a product of literals

•^

The advantage is that any sum of products expression can beimplemented using a two-level circuit^ –

literals and their complements at the “0th” level

–^

AND gates at the first level

–^

a single OR gate at the second level

•^

This diagram uses some shorthands…^ –

NOT gates are implicit

–^

literals are reused

–^

this is

not okay in LogicWorks!

f(x,y,z) = y’ + x’yz’ + xz

February 2, 2004

CS 231

Terminology: Sum of minterms form

•^

Every function can be written as a sum of minterms, which is a specialkind of sum of products form

•^

The sum of minterms form for any function is

unique

•^

If you have a truth table for a function, you can write a sum ofminterms expression just by picking out the rows of the table wherethe function output is 1.

x^

y^

z^

f(x,y,z)

f’(x,y,z)

f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’

= m

0

+ m

+ m 1

2

+ m

3

+ m

6

=^

Σm(0,1,2,3,6)

f’ = xy’z’ + xy’z + xyz

= m

4

+ m

5

+ m

7

=^

Σm(4,5,7)

f’ contains all the minterms not in f

February 2, 2004

CS 231

Re-arranging the truth table

•^

A two-variable function has four possible minterms. We can re-arrangethese minterms into a Karnaugh map.

•^

Now we can easily see which minterms contain common literals.^ –

Minterms on the left and right sides contain y’ and y respectively.

–^

Minterms in the top and bottom rows contain x’ and x respectively.

x^

y^

minterm

x’y’

x’y

xy’

xy

Y

x’y’

x’y

X

xy’

xy

Y

x’y’

x’y

X

xy’

xy

Y’

Y

X’

x’y’

x’y

X

xy’

xy

February 2, 2004

CS 231

More two-variable examples

•^

Another example expression is x’y + xy.^ –

Both minterms appear in the right side, where y is uncomplemented.

–^

Thus, we can reduce x’y + xy to just y.

•^

How about x’y’ + x’y + xy?^ –

We have x’y’ + x’y in the top row, corresponding to x’.

–^

There’s also x’y + xy in the right side, corresponding to y.

–^

This whole expression can be reduced to x’ + y.

Y

x’y’

x’y

X

xy’

xy^ Y

x’y’

x’y

X

xy’

xy

February 2, 2004

CS 231

A three-variable Karnaugh map

•^

For a three-variable expression with inputs x, y, z, the arrangement ofminterms is more tricky:

•^

Another way to label the K-map (use whichever you like):

Y

x’y’z’

x’y’z

x’yz

x’yz’

X^

xy’z’

xy’z

xyz

xyz’

Z

Y

m

0

m

1

m

3

m

2

X^

m

4

m

5

m

7

m

6

Z

YZ

x’y’z’

x’y’z

x’yz

x’yz’

X

xy’z’

xy’z

xyz

xyz’

YZ

m

0

m

1

m

3

m

2

X

m

4

m

5

m

7

m

6

February 2, 2004

CS 231

Example K-map simplification

•^

Let’s consider simplifying f(x,y,z) = xy + y’z + xz.

•^

First, you should convert the expression into a sum of minterms form, ifit’s not already.^ –

The easiest way to do this is to make a truth table for the function,and then read off the minterms.

–^

You can either write out the literals or use the minterm shorthand.

•^

Here is the truth table and sum of minterms for our example:

x^

y^

z^

f(x,y,z)

f(x,y,z) = x’y’z + xy’z + xyz’ + xyz

=^

m

1

+^

m

5

+^

m

6

+^

m

7

February 2, 2004

CS 231

Unsimplifying expressions

•^

You can also convert the expression to a sum of minterms with Booleanalgebra.–

Apply the distributive law in reverse to add in missing variables.

–^

Very few people actually do this, but it’s occasionally useful.

•^

In both cases, we’re actually “unsimplifying” our example expression.–

The resulting expression is larger than the original one!

–^

But having all the individual minterms makes it easy to combinethem together with the K-map.

xy + y’z + xz = (xy

1) + (y’z

1) + (xz

= (xy

(z’ + z)) + (y’z

(x’ + x)) + (xz

(y’ + y))

= (xyz’ + xyz) + (x’y’z + xy’z) + (xy’z + xyz)= xyz’ + xyz + x’y’z + xy’z

February 2, 2004

CS 231

K-maps from truth tables

•^

You can also fill in the K-map directly from a truth table.^ –

The output in row

i of the table goes into square

m

of the K-map.i

–^

Remember that the rightmost columns of the K-map are “switched.”

Y

m

0

m

1

m

3

m

2

X^

m

4

m

5

m

7

m

6

Z

x^

y^

z^

f(x,y,z)

Y

X

Z

February 2, 2004

CS 231

Grouping the minterms together

•^

The most difficult step is grouping together all the 1s in the K-map.^ –

Make rectangles around groups of one, two, four or eight 1s.

–^

All of the 1s in the map should be included in at least one rectangle.

–^

Do

not include any of the 0s.

•^

Each group corresponds to one product term. For the simplest result:^ –

Make as few rectangles as possible, to minimize the number ofproducts in the final expression.

–^

Make each rectangle as large as possible, to minimize the number ofliterals in each term.

–^

It’s all right for rectangles to overlap, if that makes them larger.

Y

X^

Z

February 2, 2004

CS 231

Practice K-map 1

•^

Simplify the sum of minterms m

+ m 1

3

+ m

5

+ m

.^6

Y

X

Z

Y

m

0

m

1

m

3

m

2

X

m

4

m

5

m

7

m

6

Z

February 2, 2004

CS 231

Solutions for practice K-map 1

•^

Here is the filled in K-map, with all groups shown.^ –

The magenta and green groups overlap, which makes each of themas large as possible.

–^

Minterm m

6

is in a group all by its lonesome.

•^

The final MSP here is x’z + y’z + xyz’.

Y

X

Z

February 2, 2004

CS 231

Example: Simplify m

+m 0

+m 2

+m 5

+m 8

+m

•^

The expression is already a sum of minterms, so here’s the K-map:

•^

We can make the following groups, resulting in the MSP x’z’ + xy’z.

Y

X

W

Z

Y

m

0

m

1

m

3

m

2

m

4

m

5

m

7

m

6

m

12

m

13

m

15

m

14

X

W

m

8

m

9

m

11

m

10

Z

Y

X

W

Z

Y

w’x’y’z’

w’x’y’z

w’x’yz

w’x’yz’

w’xy’z’

w’xy’z

w’xyz

w’xyz’

wxy’z’

wxy’z

wxyz

wxyz’

X

W

wx’y’z’

wx’y’z

wx’yz

wx’yz’

Z

February 2, 2004

CS 231

K-maps can be tricky!

•^

There may not necessarily be a

unique MSP. The K-map below yields two

valid and equivalent MSPs, because there are two possible ways toinclude minterm m

•^

Remember that overlapping groups is possible, as shown above.

Y

X

Z

y’z + yz’ + xy

y’z + yz’ + xz

Y

X

Z

Y

X

Z