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I know how to minimize a Boolean function with X as outputs (using K-map).

I have encountered an exercise in which I am asked to write the Normal Disjunctive Canonical Form (NDCF)/ Sum of Products (SoP) but i have X as inputs.

Does it mean I will not take them into consideration when writing the minterms for where the functions has output high (1)?

For example:

A B C | Y
0 X 0 | 1
0 X 1 | 0
1 0 0 | 1
1 0 1 | 0
1 1 0 | 0
1 1 1 | 1

Will Y = A'C' + AB'C' + ABC ?

Also can this be done directly from the truth table, no need for a K-map unless we have to minimize?

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  • \$\begingroup\$ You can use Boolean logic for distribution with A(XOR{B,C}) but may not help much. \$\endgroup\$
    – Tony Stewart EE75
    Feb 13, 2019 at 15:27

2 Answers 2

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Yes, x means the input is ignored. Technically it's not really ignored

A B C | Y
0 X 0 | 1

could be expanded to

A B C | Y
0 0 0 | 1
0 1 0 | 1
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  • \$\begingroup\$ @Mr.Hondo Should you choose to expand it, you would (typically) not include that term in your final Boolean expression as it makes no contribution. In this case, !A||!C is same as !A!B!C || !AB!C i.e. optimization. \$\endgroup\$
    – CapnJJ
    Feb 13, 2019 at 18:09
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Actually, since X is a don't care condition, it can be used to simplify the expression.

You state:

Will \$Y = \bar A \bar C + A \bar B \bar C + ABC\$ ?

\$Y = \bar A \bar C (1 + \bar B) + A \bar B \bar C + ABC\$

\$Y = \bar A \bar C + \bar A \bar B \bar C + A \bar B \bar C + ABC\$

\$Y = \bar A \bar C + \bar B \bar C (\bar A + A) + ABC\$

The minimal expression is \$Y = \bar A \bar C + \bar B \bar C + ABC\$

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