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Tuesday, June 03, 2003

Foundations of Complexity
Lesson 19: The Immerman-Szelepcsenyi Theorem

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In this lesson we will prove the Immerman-Szelepcsényi Theorem.

Theorem (Immerman-Szelepcsényi): For reasonable s(n)≥ log n, NSPACE(s(n))=co-NSPACE(s(n)).

Let M be a nondeterministic machine using s(n) space. We will create a nondeterministic machine N such that for all inputs x, N(x) accepts if and only if M(x) rejects.

Fix an input x and let s=s(|x|). The total number of configurations of M(x) can be at most cs for some constant c. Let t=cs. We can also bound the running time of M(x) by t because any computation path of length more than t must repeat a configuration and thus could be shortened.

Let I be the initial configuration of M(x). Let m be the number of possible configurations reachable from I on some nondeterministic path. Suppose we knew the value of m. We now show how N(x) can correctly determine that M(x) does not accept.

Let r=0
For all nonaccepting configurations C of M(x)
  Try to guess a computation path from I to C
  If found let r=r+1
If r=m then accept o.w. reject
If M(x) accepts then there is some accepting configuration reachable from I so there must be less than m non-accepting configurations reachable from I so N(x) cannot accept. If M(x) rejects then there is no accepting configurations reachable from I so N(x) on some nondeterministic path will find all m nonaccepting paths and accept. The total space is at most O(s) since we are looking only at one configuration at a time.

Of course we cannot assume that we know m. To get m we use an idea called inductive counting. Let mi be the number of configurations reachable from I in at most i steps. We have m0=1 and mt=m. We show how to compute mi+1 from mi. Then starting at m0 we compute m1 then m2 all the way up to mt=m and then run the algorithm above.

Here is the algorithm to nondeterministically compute mi+1 from mi.

Let mi+1=0
For all configurations C
  Let b=0, r=0
  For all configurations D
    Guess a path from I to D in at most i steps
    If found
      Let r=r+1
      If D=C or D goes to C in 1 step
        Let b=1
  If r<mi halt and reject
  Let mi+1=mi+1+b
The test that r<mi guarantees that we have looked at all of the configurations D reachable from I in i steps. If we pass the test each time then we will have correctly computed b to be equal to 1 if C is reachable from I in at most i+1 steps and b equals 0 otherwise.

We are only remembering a constant number of configurations and variables so again the space is bounded by O(s). Since we only need to remember mi to get mi+1 we can run the whole algorithm in space O(s).

8 comments:

  1. I am not sure I understand this algorithm. Why can't we just define the following non-deterministic turing machine N(x):

    1. Non-deterministically guess the accepting configuration of M(x)
    2. Verify the accepting configuration
    3. If successful, then reject, otherwise accept

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  2. M(x) may have different nondeterministic choices, some lead to accept and others leading to reject. In this case both M(x) and N(x) will have (different) choices that lead to acceptance.

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  3. Let me try to modify the algorithm just a little:

    Let r=0
    For all accepting configurations C of M(x)
      Try to guess a computation path from I to C
      If found let r=r+1
    If r>0 then reject o.w. accept

    We are now iterating through all accepting configurations of M(x). The existence of a single reachable, accepting configuration is enough to reject the computation.

    What is wrong with this algorithm? Any space-bounds I am violating? Over-using the power of nondeterminism? Misunderstanding complementation?

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  4. How is it even possible that we compute the set of all non-accepting configurations? Wouldn't that require exponential space?

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  5. This comment has been removed by the author.

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  6. This is my suggestion for the second algorithm. Why can't we just use nondeterminism to "guess" the path to n+1 ?

    Let mi+1=0
    For all configurations C
      Guess a path from I to C in at most i+1 steps
      If found
        Let mi+1=mi+1+b

    I think I might be misunderstanding nondeterminism. Maybe it is related to the fact that the nondeterministic turing machine needs to be able to provide a certificate, which can then be verified by the deterministic machine.

    In my algorithm, the certificate would be the set of all reachable nodes. Maybe I am violating the space bound with that? I am confused..

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  7. We need to make sure m_{i+1} is correct on every nondeterministic path that doesn't reject. Your algorithm could undercount m_{i+1} causing problems later.

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  8. You are right. Thank you for the replies!

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