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# [Solved]: Why IS is not in NL, only in NP?

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Problem Detail:

All we need to do is guess $k$ vertices. We look at vertex $v_1$, and make sure $v_1$ is not connected to $v_2...v_k$. Then, we "throw" $v_1$, and look at $v_2$. We do this to all vertices.

Meaning that we only need to guess $k$ vertices, and in our working memory (which determinates NL or NP) we only need to keep 2 vertices.

Where is my mistake?

First, $NL\stackrel{?}{=} NP$ is still open, so $IS\in NL$ is possible (within our current knowledge, though we believe that they are different).

The problem with your proof, is that when you "guess $k$ vertices", you have to write them in your memory (when we say "guess", we mean that we write a vertex in the working tape, where at each step our machine writes either $0$ or $1$). Keeping $k$ vertices simultaneously (where $k$ is part of the input), requires more than logarithmic space.

Question Source : http://cs.stackexchange.com/questions/56818

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