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 ½ P SP ASE so we conclude NL ½ DSP ACE(n) ½ DSP ACE(n3) ½ 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 di®er. 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 2 NL by Immerman-Szelepsc¶enyi Theorem (NL = vo ¡ NL) we know that A 2 NL. As C = A \ s ¡ t ¡ conn and s ¡ t ¡ conn 2 NL we get that A 2 NL. 3. If L 2 NICE, then taking \quit" as reject we get an NP machine (when x2 = L we always reject) for L, and by taking \quit" as accept we get coNP machine for L (if x 2 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 de¯ne 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 2 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 BP Ppath 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 unsatis¯able equation the probability of returning 'NO' is 1 (thus > 3 ). If the equation is satis¯able 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 . 1.