CS 172 : Automata , Computability and Complexity Handout 5

Sections 4 and 5 are " bonus " material, not part of the syllabus. These notes do not include the proof of PSPACE-completeness of the TQBF problem, which is well explained in Sipser's book. The problem called PATH in Siper's book is called STCONN here. L.T. A machine solves a problem using space s(·) if, for every input x, the machine outputs the correct answer and uses only the first s(|x|) cells of the tape. For a standard Turing machine, we can't do better than linear space since x itself must be on the tape. So we will often consider a machine with multiple tapes: a read-only " input " tape, a read/write " work " or " memory " tape, and possibly a write-once " output " tape. Then we can say the machine uses space s if for input x, it uses only the first s(|x|) cells of the work tape. We denote by L the set of decision problems solvable in O(log n) space. We denote by PSPACE the set of decision problems solvable in polynomial space. A first observation is that a space-efficient machine is, to a certain extent, also a time-efficient one. In general we denote by SPACE(s(n)) the set of decision problems that can be solved using space at most s(n) on inputs of length n. Theorem 1 If a machine always halts, and uses s(·) space, with s(n) ≥ log n, then it runs in time 2 O(s(n)). Proof: Call the " configuration " of a machine M on input x a description of the state of M , the position of the input tape, and the contents of the work tape at a given time. Write down c 1 , c 2 ,. .. , c t where c i is the configuration at time i and t is the running time of M (x). No two c i can be equal, or else the machine would be in a loop, since the c i completely describes the present, and therefore the future, of the computation. Now, the number of possible configurations is simply the product of the number of states, the number of positions on the input tape, and the number of possible contents of the work tape (which itself depends on the number of allowable positions on the input tape). This is O(1) · n · |Σ| s(n) = 2 …