Ex. 8 Complexity 1. by Savich Theorem NL ⊆ DSPACE (Log2n)

Ex. 8 Complexity

1. By Savich theorem NL ⊆ DSP ACE(log2n) ⊆ DSP ASE(n). we saw in one of the previous ex. that DSP ACE(n2f(n)) 6= DSP ACE(f(n)) we also saw NL ⊂ P SP ASE so we conclude NL ⊂ DSP ACE(n) ⊂ DSP ACE(n3) ⊂ P SP ACE but DSP ACE(n) 6= DSP ACE(n3) as needed.

2. (a) We emulate the running of the s − t − conn algorithm in parallel for two paths (that is, we maintain two pointers, both starts at s and guess separately the next step for each of them). We raise a flag as soon as the two pointers differ. We stop updating each of the two pointers as soon as it gets to t. If both got to t and the flag is raised, we return true. (b) As A ∈ NL by Immerman-Szelepscényi Theorem (NL = vo − NL) we know that A ∈ NL. As C = A ∩ s − t − conn and s − t − conn ∈ NL we get that A ∈ NL.

3. If L ∈ NICE, then taking “quit” as reject we get an NP machine (when x∈ / L we always reject) for L, and by taking “quit” as accept we get coNP machine for L (if x ∈ L we always reject), and so NICE ⊆ NP ∩ coNP .

For the other direction, if L is in NP ∩coNP , then there is one NP machine M1(x, y) solving L, and another NP machine M2(x, z) solving coL. We can therefore deﬁne a new machine that guesses the guess y for M1, and the guess z for M2. If M1(x, y) accept we accept, if M2(x, z) accept we reject (and notice that it is impossible that both accept. Why?) otherwise we answer quit. Now, prove that this machine is a nice machine for L.

4. ZPP ⊆ RP as one can replace the ’quit’ state (recall ZPP 1 of previous question) with ’reject’ (the probability of mistake can be made arbitrary small by repetition). Similarly ZPP ⊆ coRP , hence ZPP ⊆ RP ∩ coRP . If L ∈ RP ∩ coRP , we run both algorithm (each has a one side error) for deciding L. If they agree - return their answer, otherwise return ’quit’.

5. We solve SAT by a BPPpath machine M. M generates a random assignment to the n variables, and check it. If it doesn’t satisfy the equation then M returns ’NO’. If it does satisfy the equation then M flips another 2n coins and returns ’YES’. 2 For every unsatisfiable equation the probability of returning ’NO’ is 1 (thus > 3 ). If the equation is satisfiable then at most 2n − 1 of the computation paths yield ’NO’, and at least 2n 2 2 yield ’YES’. Therefore the probability of acceptance is clearly larger than 3 .