05-AdditionalKmaps - 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

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CS231 Boolean Algebra 1 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, if it’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) 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 f(x,y,z) = x’y’z + xy’z xyz’ xyz
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CS231 Boolean Algebra 2 Unsimplifying expressions You can also convert the expression to a sum of minterms with Boolean algebra. 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 combine them together with the K- map. xy + y’z + xz = (xy 1) + (y’z 1) + (xz 1) = (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
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CS231 Boolean Algebra 3 Making the example K-map Next up is drawing and filling in the K-map. Put 1s in the map for each minterm, and 0s in the other squares. You can use either the minterm products or the shorthand to show you where the 1s and 0s belong. In our example, we can write f(x,y,z) in two equivalent ways. In either case, the resulting K-map is shown below. Y 0 1 0 0 X 0 1 1 1 Z Y x y z x y z x yz x yz X xy z xyz Z f(x,y,z) = x’y’z + xy’z xyz’ Y m 0 m 1 m 3 m 2 X m 4 m 5 m 7 m 6 Z
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CS231 Boolean Algebra 4 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 i of the K-map. 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) 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 Y 0 1 0 0 X 0 1 1 1 Z
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CS231 Boolean Algebra 5 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.
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This note was uploaded on 04/15/2008 for the course CS 231 taught by Professor - during the Spring '08 term at University of Illinois at Urbana–Champaign.

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05-AdditionalKmaps - 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

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