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Lecture 9: PSPACE

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Lecture 9: PSPACE Lecture 9: PSPACE PSPACE = DSPACE[nO(1)] = NSPACE[nO(1)] = ATIME[nO(1)] • PSPACE consists of what we could compute with a feasible amount of hardware, but with no time limit. • PSPACE is a large and very robust complexity class. • With polynomially many bits of memory, we can search any implicitly-defined graph of exponential size. This leads to complete problems such as reachability on exponentially-large graphs. • We can search the game tree of any board game whose configurations are describable with polynomially- many bits and which lasts at most polynomially many moves. This leads to complete problems concerning winning strategies. 1 PSPACE-Complete Problems Recall PSPACE = ATIME[nO(1)] Recall QSAT, the quantified satisfiability problem. Proposition 9.1 QSAT is PSPACE-complete. Proof: We’ve already seen that QSAT 2 ATIME[n] ⊆ PSPACE. QSAT is hard for ATIME[nk]: Let M be an ATIME[nk] TM, w an input, n = jwj k Let M write down its n alternating choices, c1 c2 :::c nk . Deterministic TM D evaluates the answer, i.e., for all inputs w, M(w) = 1 , 9c1 8c2 · · · 9cnk (D(c; w) = 1) By Cook’s Theorem 9 reduction f : L(D) ≤ SAT: D(c; w) = 1 , f(c; w) 2 SAT Let the new boolean variables in f(c; w) be d1 : : : dt(n). M(w) = 1 , “9c1 8c2 · · · 9cnk d1 : : : dt(n) (f(c; w))” 2 QSAT 2 Geography is a two-person game. 1. E “chooses” the start vertex v1. 2. A chooses v2, having an edge from v1 3. E chooses v3, having an edge from v2, etc. No vertex may be chosen twice. Whoever moves last wins. Texas Selma Amherst Africa Tulsa 3 Let GEOGRAPHY be the set of positions in geography games s.t. 9 has a winning strategy. Proposition 9.2 GEOGRAPHY is PSPACE-complete. Proof: GEOGRAPHY 2 PSPACE: search the polynomial-depth game tree. A polynomial-size stack suffices. Show: QSAT ≤ GEOGRAPHY a a Given formula, ', build graph G' s.t. 9 chooses existential variables; 8 chooses (a v b) universal variables. ' ≡ 9a 8b 9c b [(a _ b) ^ b (b v c) (¯b _ c) ^ (b _ c¯)] c (b v c) c 4 Definition 9.3 A succinct representation of a graph is G(n; C; s; t) = (V; E; s; t) where C is a boolean circuit with 2n inputs and n V = w w 2 f0; 1g 0 0 E = (w; w ) C(w; w ) = 1 SUCCINCT REACH = (n; C; s; t) G(n; C; s; t) 2 REACH Proposition 9.4 SUCCINCT REACH 2 PSPACE Why? Remember Savitch’s Thm: REACH 2 NSPACE[log n] ⊆ DSPACE[(log n)2] SUCCINCT REACH 2 NSPACE[n] ⊆ DSPACE[n2] ⊆ PSPACE Proposition 9.5 SUCCINCT REACH is PSPACE-complete. Proof: Let M be a DSPACE[nk] TM, input w, n = jwj M(w) = 1 $ CompGraph(M; w) 2 REACH CompGraph(n; w) = (V; E; s; t) k V = ID = hq; h; pi q 2 States(N); h ≤ n; jpj ≤ c n E = (ID1; ID2) ID1(w) −! ID2(w) M s = initial ID t = accepting ID 5 Succinct Representation of CompGraph(n; w): k V = ID = hq; h; pi q 2 States(N); h ≤ n; jpj ≤ c n E = (ID1; ID2) ID1(w) −! ID2(w) M Let V = f0; 1gc0nk Build circuit Cw: on input u; v 2 V , accept iff u −! v. M 0 k M(w) = 1 , G(c n ;Cw; s; t) 2 SUCCINCT REACH 6 The vertices of an alternating graph, G = (V;A , E), are split into: existential vertices and universal vertices. Def: vertex t is reachable from vertex s in G iff 1. s = t, or 2. s is existential and for some edge, hs; ai 2 E, t is reachable from a, or, 3. s is universal and there is an edge leaving s and for all edges, hs; ai 2 E, t is reachable from a. t s E A 7 Def: Let AREACH = G = (V;A; E; s; t ) t is reachable from s Prop: AREACH is P complete. Proof: Think about it! (Very similar to the proof that REACH is NL complete.) Cor: CVP and MCVP are P complete. Proof: Easy to see that AREACH ≤ MCVP ≤ CVP. 8 co-r.e. complete Arithmetic Hierarchy FO(N) r.e. complete Halt Halt FO8(N) FO9(N) co-r.e. r.e. Recursive Primitive Recursive SO[2nO(1) ] EXPTIME QSAT PSPACE complete FO[2nO(1) ] SO[nO(1)] PSPACE co-NP complete PTIME Hierarchy SO NP complete SAT SAT SO9 co-NP SO8 NP NP \ co-NP FO[nO(1)] P complete Horn- P FO(LFP) SAT FO[logO(1) n] “truly NC FO[log n] feasible” AC1 FO(CFL) sAC1 FO(TC) 2SAT NL comp. NL FO(DTC) 2COLOR L comp. L FO(REGULAR) NC1 FO(COUNT) ThC0 FO LOGTIME Hierarchy AC0 9.
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