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I know that CNF SAT is in NP (and also NP-complete), because SAT is in NP and NP-complete. But what I don't understand is why? Is there anyone that can explain this?

Raphael
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user2795095
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CNF-SAT is in NP since you can verify a satisfying assignment in polynomial time. CNF-SAT is NP-hard since SAT is a special case of CNF-SAT, and so we can reduce the NP-hard problem SAT to the CNF-SAT. Since it is both in NP and NP-hard, we conclude that CNF-SAT is NP-complete.

Yuval Filmus
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  • Thank you. But to prove that CNF-SAT is in NP, it's necessary to construct a non-deterministic TM that solves it in polynomial time? Why do I have to use a non-det TM? Is it because I have to non-deterministically choose the valuation of the literals in CNF-SAT? (I have just started learning time-complexity, so forgive my question) – user2795095 Apr 02 '14 at 17:38
  • You use non-determinism to guess the satisfying assignment. We don't expect CNF-SAT to be solvable in polynomial time (this is the P vs. NP question). – Yuval Filmus Apr 02 '14 at 18:21
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    CNF-SAT is in NP owing to the fact that, given a solution, you can verify it in polynomial time. NP can also be viewed as the complexity class of the problems whose solutions can be verified in deterministic polynomial time. – Massimo Cafaro Apr 03 '14 at 17:40
  • CNF-SAT is a special case of SAT, not the other way around, as you state. Specifically, CNF-SAT is the special case of Boolean satisfiability (SAT) for formulas in conjunctive normal form (CNF). Also, the normal proof of NP-completeness for SAT is not to reduce to CNF-SAT, since converting a formula to CNF naively takes exponential time. – David Richerby Jun 30 '14 at 14:46
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Try reading up on the Cook-Levin theorem. SAT is basically the first problem proven NP-complete. High level sketch of the proof: simulate a nondeterministic (NP-time, nondeterministic polynomial time) TM computation using a cascading circuit that computes the TM iterations, which can be converted to SAT. Loosely speaking, the "size" of the circuit is "polynomial".

Juho
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vzn
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Note: This is a repeat of an answer for a previous question, but the answer works better here than it did there.

There were two independent proofs that SAT is NP-hard, one by Stephen Cook in 1971 [1] and the other by Leonid Levin in 1973. We now know it as the Cook-Levin theorem. You can read the paper, or consult the Wikipedia article [2] for details, but I'll give a brief outline of the basic idea here.

Let's look at the recogniser problem. There is a language $L$, and a nondeterministic TM $M$ which recognises $L$ in polynomial time. Let $w$ be a string. The idea is to construct a boolean formula $A(w)$ in conjunctive normal form, where the number of formulas and the number of propositions is polynomial in the length of $w$ and the size of $M$ and the size of the alphabet, which is true if $M$ accepts $w$ and false if $M$ rejects $w$.

The proof depends on the fact that $M$ takes polynomial time in the length of the string. Suppose that the maximum number of steps that $M$ can take for a string of length $n$ is $Q(n)$. Then this is also an upper bound on the amount of tape that $M$ can use. We can trivially modify $M$ so that all computations take at least this time. We could, for example, modify $M$ so that it loops forever in an accepting or rejecting state, let it run for $Q(n)$ time and then see which state it was in.

We introduce the following proposition symbols:

  • $P_{i,s,t}$ is true if and only if the tape square $s$ contains symbol $i$ at time step $t$.
  • $Q_{i,t}$ is true if and only if the machine is in state $i$ at time step $t$.
  • $S_{s,t}$ is true if and only if symbol $s$ is scanned by the tape head at time step $t$.

Next, we construct formulas which model the actions of $M$ and test whether or not $w$ is accepted. We can do this using only the above proposition symbols, and in conjunctive normal form.

I encourage you to think through the details yourself by working out what the formulas might look like. The ones that Cook used are:

  • At each time step $t$, one and only one square is scanned.
  • At each time step $t$ and tape square $s$, there is one and only one symbol.
  • At each time step $t$, $M$ is in one and only one state.
  • At time step $1$, $M$ is in its start state and the tape contains exactly $w$ followed by "blank" symbols.
  • At each time step transition, the $P$, $Q$ and $S$ propositions are updated correctly, according to the transition function of $M$. Remember $M$ is nondeterministic, so you need to include all possible transitions. (If you're playing along at home, use three formulas for this.)

And the final, most important, formula states that:

  • $M$ enters the "accept" state at some time.

Then the conjunction of all of these formulas is true if and only if $M$ accepts $w$. Solve using your favourite SAT solver, and you're done.

  1. The Complexity of Theorem-Proving Procedures by Stephen A. Cook (1971).
  2. Cook-Levin theorem on Wikipedia.
Juho
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Pseudonym
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