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When I said a succinct representation of a graph of n nodes, I meant a Boolean circuit C of 2*b input gates (where b = |n| and |n| is the binary string length of n), such that for every b-bits integers i and j, then C accepts the input i and j if and only if (i, j) is an edge of the graph and the size of C is O(b^{k}), that is polylogarithmic in relation to n.

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    $\begingroup$ The answer is no. I have already see it in this post: cstheory.stackexchange.com/questions/41/… $\endgroup$ – Frank Vega Aug 15 '16 at 15:40
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    $\begingroup$ No. An obvious information-theoretic argument tells you that random graphs whp cannot be described using less than $\binom n2$ bits, hence they need exponential circuit size $\Omega(n^2/\log n)=\Omega(2^{2b}/b)$. $\endgroup$ – Emil Jeřábek Aug 15 '16 at 15:40
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With n vertices, there are $O(2^{n \choose 2})$ possible labelled graphs (based on which edges are present). So you need ${n\choose 2}$ bits to store.

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