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In my homework I came across a situation where I had a Karnaugh map that only contained don't cares and trues. Since there are no false outputs possible, it seems like the equation would just be f(x,y,z) = TRUE. Am I right or am I missing something?

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    $\begingroup$ If there are k "don't care"s in the map, there are 2^k possibilities. But I'd use the simplest one, f = true. $\endgroup$ – Charles Apr 26 '11 at 17:08
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That sounds about right to me - your specification is basically for a circuit that outputs 1 or anything, so outputting 1 is correct.

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