What are the consequences of saying a non-deterministic Turing Machine can solve NP in polynomial time?

Question

these days I have been studying about NP problems, computational complexity and theory. I believe I have finally grasped the concepts of Turing Machine, but I have a couple of doubts.

I can accept that a non-deterministic turing machine has several options of what to do for a given state and symbol being read and that it will always pick the best option, as stated by wikipedia

How does the NTM "know" which of these actions it should take? There are two ways of looking at it. One is to say that the machine is the "luckiest possible guesser"; it always picks the transition which eventually leads to an accepting state, if there is such a transition. The other is to imagine that the machine "branches" into many copies, each of which follows one of the possible transitions. Whereas a DTM has a single "computation path" that it follows, an NTM has a "computation tree". If any branch of the tree halts with an "accept" condition, we say that the NTM accepts the input.

What I can not understand is, since this is an imaginary machine, what do we gain from saying that it can solve NP problems in polynomial time? I mean, I could also theorize of a magical machine that solves NP problems in O(1), what do I gain from that if it may never exist?

Thanks in advance.

Answer 1

A non-deterministic Turing machine is a tricky concept to grasp. Try some other viewpoints:

1. Instead of running a magical Turing machine that is the luckiest possible guesser, run an even more magical meta-machine that sets up an infinite number of randomly guessing independent Turing machines in parallel universes. Every possible sequence of guesses is made in some universe. If in at least one of the universes the machine halts and accepts the input, that's enough: the problem instance is accepted by the meta-machine that set up these parallel universes. If in all universes the machine rejects or fails to halt, the meta-machine rejects the instance.

2. Instead of any kind of guessing or branching, think of one person trying to convince another person that the instance should be accepted. The first person provides the set of choices to be made by the non-deterministic Turing machine, and the second person checks whether the machine accepts the input with those choices. If it does, the second person is convinced; if it does not, the first person has failed (which may be either because the instance cannot be accepted with any sequence of choices, or because the first person chose a poor sequence of choices).

3. Forget Turing machines. A problem is in NP if it can be described by a formula in existential second-order logic. That is, you take plain-vanilla propositional logic, allow any quantifiers over propositional variables, and allow tacking at the beginning existential quantifiers over sets, relations, and functions. For example, graph three-colorability can be described by a formula that starts with existential quantification over colors (sets of nodes):

   ∃ R ∃ G ∃ B

   Every node must be colored:

   ∃ R ∃ G ∃ B (∀ x (R(x) ∨ G(x) ∨ B(x)))

   and no two adjacent nodes may have the same color – call the edge relation E:

   ∃ R ∃ G ∃ B (∀ x (R(x) ∨ G(x) ∨ B(x))) ∧ (∀ x,y ¬ (E(x,y) ∧ ((R(x) ∧ R(y)) ∨ (G(x) ∧ G(y)) ∨ (B(x) ∧ B(y)))))

   The existential quantification over second-order variables is like a non-deterministic Turing machine making perfect guesses. If you want to convince someone that a formula ∃ X (...) is true, you can start by giving the value of X. That polynomial-time NTMs and these formulas not just "like" but actually equivalent is Fagin's theorem, which started the field of descriptive complexity: complexity classes characterized not by Turing machines but by classes of logical formulas.

You also said

I could also theorize of a magical machine that solves NP problems in O(1)

Yes, you can. These are called oracle machines (no relation to the DBMS) and they have yielded interesting results in complexity theory. For example, the Baker–Gill–Solovay theorem states that there are oracles A and B such that for Turing machines that have access to A, P=NP, but for Turing machines that have access to B, P≠NP. (A is a very powerful oracle that makes non-determinism irrelevant; the definition of B is a bit complicated and involves a diagonalization trick.) This is a kind of a meta-result: any proof solving the P vs NP question must be sensitive enough to the definition of a Turing machine that it fails when you add some kinds of oracles.

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The value of non-deterministic Turing machines is that they offer a comparatively simple, computational characterization of the complexity class NP (and others): instead of computation trees or second-order logical formulas, you can think of an almost-ordinary computer that has been (comparatively) slightly modified so that it can make perfect guesses.

Answer 2

What you gain from that is that you can prove that a problem is in NP by proving that it can be solved by an NTM in polynomial time.

In other words you can use NTMs to find out whether a given problem is in NP or not.