if a non-deterministic program executes only lg(n) decisions on each branch of the computation tree, then the problem this program solves is in P? That means, there is a deterministic algorithm that computes the solution in a polynomial time.

The answer yes looks intuitive to me when I'm looking the tree, but I don't know how to actually "proof" that

a binary tree representing the computation tree

  • 2
    $\begingroup$ One idea is $2^{C\log n} = n^C$. $\endgroup$ – Yuval Filmus Nov 8 '20 at 0:05

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