﻿ 3 var Kmap minimization example.
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3 variable K-map, Example -2

Minimize following

F(x,y,z) =     (0,1,4,5,6,7)

Above is a common format of representing the K-map problems. The numbers 0,1,4,5,6,7 are the location of cells in the 3-var k-map table. Discussed below.

3 variable K– map with 1 and 0 values assigned to cells.

00

01

4

5

11

10

7

6

x’y’z’ = 1

x’y’z = 1

x’yz = 0

x’yz’ = 0

xy’z’ = 1

xy’z = 1

xyz = 1

xyz’ = 1

0

1

x

yz

The K-map for 3 variables is plotted above. You will notice the column for 11 and 10 is inter-changed. This is done to allow only one variable to change across adjacent cells. This adjustment in columns allows in minimization of logic mapped into tables.

Any adjacent 1, 2, 4 or 8 cells can be grouped to find a minimized logic value.
Following plot will show grouping of adjacent cells.

00

01

4

5

11

10

7

6

x’y’z’ = 1

x’y’z = 1

x’yz = 0

x’yz’ = 0

xy’z’ = 1

xy’z = 1

xyz = 1

xyz’ = 1

0

1

x

yz

The two step minimization equation is shown below.
With reference to the table above the cells under the dotted box’s can be combined to come up with following reduced  equation.

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