New answers tagged boolean-algebra
3
Yes, you are right. Having $x \leftrightarrow (x \rightarrow z)$ one can deduce that $x\equiv 1$ and $z \equiv 1$.
In the paper you cite (the link is to another one btw.), rule 6 allows only to deduce that $x\equiv 1$ holds.
But then you have the triplet $(1,1,z)$ and you can apply rule 4 $\frac{(x,1,z)}{x/z}$ to deduce $z\equiv x$, i.e., $z \equiv 1$ holds, ...
1
The relationship is that a short SOP expression implies a low Kolmogorov complexity, but not vice versa. If you have a short SOP expression, then that is a concise way to compress the string, hence Kolmogorov complexity is low. However, there might be strings that can be compressed in another way even though the SOP expression is long, and those strings ...
1
It's 4 because you're allowed to "wrap around" from the top edge to the bottom edge or from the right edge to the left edge (those moves still represent 1-bit changes because the Gray code is cyclic).
So there is one PI of 3 minterms in the first column, one in the fourth column, one in the third row, and one of 2 minterms in the second row.
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