Today well talk about NL completeness NLcoNL this differs from what we think is

# Today well talk about nl completeness nlconl this

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Today we’ll talk about NL-completeness NL=coNL (this differs from what we think is true for NP) 127
Lecture 19 Notes on Theory of Computation Recall that L=SPACE(log ? ) and NL=NSPACE(log ? ). We have a nice hierarchy: L NL P NP PSPACE . We don’t know whether these containments are proper. We can show that PSPACE and NL are different (and will eventually do so), so not everything in the picture collapses down. Most people believe that these spaces are all different; however, we don’t know adjacent inclusions are proper. However, NL=coNL shows that surprising things do happen, and we do have unexpected collapses. First let’s review a theorem from last time. Theorem (Theorem 18.6) : NL P. Proof. For a NL-machine ? , a configuration of ? on ? is ( ?, ? 1 , ? 2 , ? ). The number of configurations of ? on ? is polynomial in ? where ? = | ? | ( ? is fixed). The computation graph is the graph where nodes are configurations, and edges show how ? can move. Here is a polynomial time algorithm that simulates ? . “On input ? , 1. Construct the computation graph. 2. Test if there is a path from start to accept (using any polynomial time algorithm for PATH). 3. Accept if yes and reject if no.” 128
Lecture 19 Notes on Theory of Computation S 1 L vs. NL Now we turn our attention to L vs. NL. We’ll show that the situation is analogous to the situation of P vs. NP. How much space deterministically do we actually need for a NL problem? We can do it with polynomial space but that’s pretty crude. We can do much better. We have using Savitch’s Theorem that NL = NSPACE(log ? ) SPACE(log 2 ? ) We stated Savitch’s Theorem for space bounds ? ; with space bounds of log ? the same argument goes through. No one knows whether we can reduce the exponent, or whether L=NL. (We will show that SPACE(log ? ) is provably different fron SPACE(log 2 ? ), using the hierarchy theorem 20.1. When we increase the amount of space/time, we actually get new stuff. But maybe some other argument could show NL SPACE(log ? ).) We will show that there are NL-complete problems, an example of which is PATH. If you can solve PATH or any other NL-complete problems in deterministic log space, then it brings down everything with it to L. We’ll show everything in NL is reducible to the PATH problem. This shouldn’t be a surprise because it’s what we did in the previous theorem: whether a machine accepts is equivalent to whether there’s a path. We’ll just need to define NL-completeness in the appropriate way and then we’ll be done by the argument given in the NL P theorem. Definition 19.1: ? is NL-complete if 1. ? NL 2. Every NL-problem is log-space reducible to ? : for every ? NL, ? ? ? . We need to define what it means to be log-space reducible. We have to be careful because the input is roughly ? , and the output is roughly ? . we don’t want to count the output of machine in the space bound. The input and output should be kept seprate.