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Anonymous asked in Computers & InternetProgramming & Design · 2 months ago

Simplify a Boolean expression using K-maps?

!A!B+!AB!C+!(A+!C)

I don't get it, aren't K-Maps for simplifying expressions when all the terms have the same variables? Every term has a different number or type of variable, how am I supposed to use K-maps here?

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  • 2 months ago
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    No, you don't need every variable in every term in order to use a Karnaugh map.  You just need to write out the truth table for the boolean expression.  That truth table is effectively an equivalent representation of your expression as a sum of minterms. (Each 1 or T in the final column corresponds to a minterm.)

    If you don't want to do the truth table, you can use boolean algebra to get those entries to plug into a K-map grid.  Pardon me if I use A' for "not A" instead of the C-like !A.  First, use de Morgan to get rid of parentheses:

        A' B' + A' B C' + (A + C')' = A' B' + A' B C' + A' C

    Then add (x + x') factors in all terms with a missing variable x.  (This is the magic step to make every term a minterm.)

                                     = A' B' (C + C') + A' B C' + A' C (B + B')

    That's valid because C + C' = 1 and B + B' = 1.  Now use the distributive property to get rid of parentheses:

        = A' B' C + A' B' C' + A' B C' + A' B C + A' B' C

    Finally, remove duplicate minterms if there are any.  This is easier to do if you reorder the term in "index order".  You find that A' B' C occurs twice:

        = A' B' C'  + A' B' C + A' B C + A' B' C'

    In indexed minterm "m_x for the minterm with index x" form, that's:

        = m_0 + m_1 + m_2 + m_3

    Those are the index spots in your K-map grid that get ones.  All the rest are zeroes.

    If you've done more than a half dozen or so K-maps (or had any practice with boolean algebra), it's pretty obvious now that the answer will be simply A'.   Or !A in your notation.

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