[My Computational Complexity Web Log] Foundations of Complexity Lesson 19: The Immerman-Szelepcsényi Theorem
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. rejectIf 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+bThe 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).
Posted by Lance Fortnow to My Computational Complexity Web Log at 6/3/2003 2:09:50 PM
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