# Simplifying Boolean Function with Karnaugh Maps

Given the boolean function

f(x,y,z) = xyz + xyz' + xy'z + xy'z' + x'yz + x'y'z + x'y'z' (where x' = not x)


In a three variable Karnaugh Map:

   yz   yz'  y'z'  y'z
x  1    1    1     1
x' 1         1     1


The goal is to group the adjacent units and simplifying using the distributive law since y+y' would equal one. This is all good, but when it comes to the above Karnaugh map, which one do I group together? The textbook says, it should be the biggest block but I am a bit confused in terms of what that means.

The final answer after simplification would yield:

x + y' + z

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I've marked the groups on the image below. As usual, the value of each group is the variable that remains constant in the group.

Red = x

Blue = y'

Green = z

So the answer is x + y' + z

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Thanks for the answer. The problem I am having is know which to include inside the circles. How come the blue circle not include the first column too? Wouldn't such grouping making it a bigger block? – user1234440 Nov 13 '12 at 23:30
The groups should always be rectangles, and the sides should be powers of two. So, in this case, we have groups that are 1x4, 2x2 and 2x2. If the blue group included the first column, it would be 2x3, which is not permitted. – Ricbit Nov 13 '12 at 23:31

K-Maps must be grouped in either 1,2,4,8 basically powers of 2.

This K map can be grouped in to 3 groups that have 4 in each group. Then look fro the variables that don't change.

The groups would look like this:

Group 1:

y' z' x'
y' z' x
y' z  x'
y' z  x


This simplifies to y'.

Group 2:

y  z  x
y  z' x
y' z' x
y' z  x


This simplifies to x.

Group 3:

y  z  x'
y' z  x'
y' z  x
y' z  x


This simplifies to z.

So the boolean function is: y' + x + z

This is how the final answer is resolved.

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