Tuesday, April 05, 2011

A New Proof of the Nondeterministic Time Hierarchy


A nondeterministic time hierarchy was first proved by Cook and in the strongest form by Seiferas, Fischer and Meyer.  Zàk gave a simple proof that we sketched in this post. Here is another.

Theorem: If t1 and t2 are time-constructible functions and t1(n+1)=o(t2(n)) then NTIME(t1(n)) is strictly contained in NTIME(t2(n)).

Proof: 

Let M1,… be an enumeration of nondeterministic Turing machines. We define a nondeterministic machine M that acts as follows on input w=1i01m0y
  • If |y|<t1(i+m+2) then accept if Mi accepts both inputs 1i01m0y0 and 1i01m0y1 in t2(|w|) steps.
  • If |y|=t1(i+m+2) then accept if Mrejects input 1i01m0 on the computation path described by y.
This machine uses time O(t2(n)). If NTIME(t1(n))=NTIME(t2(n)) then there is an equivalent machine Mi using time O(t1(n)).
Since t1(n+1)=o(t2(n)) we have for sufficiently large m,



1i01m0 in L(M) ⇔ 1i01m0y in L(M) for |y|=1⇔ … ⇔ 1i01m01y in L(M) for |y|=t1(i+m+2)⇔ M(1i01m0) rejects on all computation paths y
a contradiction. QED

The advantage over Zàk is that you only need t1 steps instead of exponential in t1. On the other hand Zàk can give you a unary language and can be generalized to a broader set of complexity measures.

Rahul Santhanam and I needed and discovered this proof for our recent paper. The proof came out of a failed attempt at an oracle to show that no such relativized proof would be possible.

15 comments:

  1. I believe that t_2(|w|) in the line of the first bullet is t_1(|w|).

    Bin Fu

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  2. Bin Fu: I'm not sure it would make any difference. Maybe the error there is that "|y|=t_1(m)" should read "|y| = t_1(m+i+2)" ?

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  3. Now I feel Lance's proof is fine.
    It does not need any revision.
    It is a very nice proof for this
    important theorem.

    Bin

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  4. Hi Lance,
    Could you explain why does the machine M run in non-deterministic time t_2?

    I am probably missing something, but don't you need need time of at least 2^t_1(n) in order to emulate M_i on all possible computation paths?

    Thanks!
    Or

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  5. What "broader set of complexity measures" are you referring to?

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  6. The proof is amazingly compact. It's only 10 lines long, which gives it elegance, sort of a Sudoku square stripped to its essential clues from which the full solution can be expanded. I'm stuck (I didn't get today's Sudoku either though), here's where. The proof defines a machine M. There are two clauses in the definition: M accepts if condition 1, and M accepts if condition 2. So I provisionally believe that M is a machine which can never reject anything, as that's not in its nature. But later in the proof there's a clause "M rejects", at least I think it's M and not some other indexed nondeterministic Turing machine, which threatens my provisional belief. Back to square 1 !

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  7. OK, I see it now. Thanks to Arnab Bhattacharyya for a patient explanation.

    Very nice proof!

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  8. Michaël: Good catch. Fixed.

    Or: In the second bullet, M only simulates M_i on the single path described by y.

    Anon 5: Zàk's proof automatically works for any measure that has universal simulation (like Sigma_5-Time). It's much harder to generalize our techniques.

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  9. Btw, I really wonder if my previous comment goes through moderation - on one hand, it is not offensive at all, on the other, it does not add much, so it might seem immodest if you let it appear. Meta: Will this comment go through?

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  10. We publish all comments good and bad. We moderate to stop the ugly.

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  11. I'm trying to give this proof in class, but am missing something. Consider the reverse direction. Say $1^i 0 1^m 0 \not\in L(M)$. Then we know that M_i rejects on all computation path. But M_i runs in time O(t_1(n)), not t_1(n). So maybe all its computation paths have length 2t_1(n), say. (And I don't think you can use the speedup theorem here, without allowing y to be non-binary. Isn't that right?) So then M_i rejects on computation path y for all y of length 2t_1(n). But then the argument stops.

    What am I missing?

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  12. I guess machine M actually needs to check if |y|=t_2(i+m+2). That seems to work. (Note: the error, if it is one, appears to be in the paper as well.)

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  13. Dear Lance, it looks like the proof also works for deterministic machines. So, we should have DTIME(t1(n)) strictly contained in DTIME(t2(n)) as soon as t1(n+1) = o(t2(n)) and both t1, t2 time constructible. But then where is the log factor that usually appears in deterministic time hierarchy?

    --Martin & Stephan from Munich

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    Replies
    1. For deterministic computation you need an extra log factor for the simulation, technically a log factor to reduce from k-tapes to 2-tapes. For nondeterministic there is a way to get from k-tapes to 2-tapes without an extra log overhead.

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