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Show that if $SAT \in P/klog(n)$ then $SAT \in P$

Assuming that there is a a constant $k \in \mathbb{N}$ such that $SAT \in P/klog(n)$, I need to prove that $SAT \in P$.

Since $SAT \in P/klog(n)$, then there is a sequence of advices $\{a_n\}_{n \in \mathbb{N}}$ of length $|a_n \le klog(n)$ and a turing machine $M$ that runs for at most $n^c$ steps (for some $c \in \mathbb{N}$ such that:

$$\phi \in SAT \Leftrightarrow M(\phi, a_{|\phi|}) = 1 $$

I tried to show a Turing machine $U$ that runs in a polynomial time and solves $SAT$.

$U$ on an input $\phi$:

  1. calculate the size $|\phi|$
  2. write $a_{|\phi|}$ on start of the working tape
  3. simulate $ M(\phi, a_{|\phi|})$ in a way that when $M$ tries to read $a_{|\phi|}$, it will read it from the start of the working tape, and the working tape of $M$ will start from offset $|a_{|\phi|}|$.
  4. $U$ will return the same answer as $M$

Is my answer correct? It seems like I'm missing something

Help would be appreciated.

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    $\begingroup$ Your approach assumes that for any $|\phi|$ you know $\alpha_{|\phi|}$. You can't encodeall possible advices (for all input lengths) in your algorithm. Hint: how many different advices of length $k \log n$ exist? $\endgroup$ – Dmitry Dec 5 '20 at 18:52
  • $\begingroup$ Okay, I understand. Since there are $n^k$ advices of length $klogn$ then I can check for every one of them, and it will still be polynomial. $\endgroup$ – Gabi G Dec 5 '20 at 19:11
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You need to combine two ideas here:

  1. There are only polynomially many possible advice strings.
  2. If you can solve the decision version of SAT, then you can solve the search version of SAT.

There is a slight annoyance that you need the search-to-decision reduction to involve only constantly many different input lengths. Assuming for simplicity that you can perform everything using a single input length $n$, here is what you do: you go over all possible advices for length $n$, and for each one, either determine that the SAT instance is unsatisfiable, or attempt to find a satisfying assignment. If you find a satisfying assignment, then the SAT instance is satisfiable, otherwise it isn't.

I'll let you fill in the details.

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