Foundations of Complexity Lesson 19: The Immerman-Szelepcsenyi 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 c^s for some constant c. Let t=c^s. 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 m_i be the number of configurations reachable from I in at most i steps. We have m₀=1 and m_t=m. We show how to compute m_i+1 from m_i. Then starting at m₀ we compute m₁ then m₂ all the way up to m_t=m and then run the algorithm above.

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

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<m_i 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 m_i to get m_i+1 we can run the whole algorithm in space O(s).