COMPLEXITY THEORY. PSPACE = SPACE(n k ) k N. NPSPACE = NSPACE(n k ) 10/30/2012. Space Complexity: Savitch's Theorem and PSPACE- Completeness

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1 COMPLEXITY THEORY Space Complexity: Savitch's Theorem and PSPACE- Completeness October 30,2012 MEASURING SPACE COMPLEXITY FINITE STATE CONTROL I N P U T We measure space complexity by looking at the furthest tape cell reached during the computation Let M = deterministic TM that halts on all inputs. Definition: The space complexity of M is the function s : N N, where s(n) is the furthest tape cell reached by M on any input of length n. Let N be a non-deterministic TM that halts on all inputs in all of its possible branches. Definition: The space complexity of N is the function s : N N, where s(n) is the furthest tape cell reached by M, on any branch if its computation, on any input of length n. Definition: SPACE(s(n)) = { L L is a language decided by a O(s(n)) space deterministic Turing Machine } Definition: NSPACE(s(n)) = { L L is a language decided by a O(s(n)) space non-deterministic Turing Machine } PSPACE = SPACE(n k ) k N k N NPSPACE = NSPACE(n k ) 1

2 3SAT SPACE(n) PSPACE # x y # x 0 y 0 Assume a deterministic Turing machine that halts on all inputs runs in space s(n) Question: What s an upper bound on the number of time steps for this machine? A configuration gives a head position, state, and tape contents. Number of configurations is at most: # x 0 y 1 # x 1 y 0 s(n) Q Γ s(n) = 2 O(s(n)) Recall: Q is the set of states; Γ, the set to tape symbols MORAL: Space S computations can be simulated in at most 2 O(S) time steps PSPACE EXPTIME k N n k EXPTIME = TIME(2 ) Is NTIME(t(n)) TIME(t(n))? Is NTIME(t(n)) TIME(t(n) k ) for some k > 1? We don t know in general! If the answer is yes, then P = NP What about the space-bounded setting? s(n) n Is NTIME(t(n)) TIME(t(n))? Is NTIME(t(n)) TIME(t(n) k ) for some k > 1? We don t know in general! If the answer is yes, then P = NP What about the space-bounded setting? therefore NPSPACE PSPACE Is NTIME(t(n)) TIME(t(n))? Is NTIME(t(n)) TIME(t(n) k ) for some k > 1? We don t know in general! If the answer is yes, then P = NP What about the space-bounded setting? therefore PSPACE = NPSPACE 2

3 Theorem: For functions s(n) where s(n) n Proof Try: Let N be a non-deterministic TM with space complexity s(n) Construct a deterministic machine M that tries every possible branch of N Since each branch of N uses space at most s(n), then M uses space at most s(n)? Theorem: For functions s(n) where s(n) n Proof Try: Let N be a non-deterministic TM with space complexity s(n) Construct a deterministic machine M that tries every possible branch of N Since each branch of N uses space at most s(n), then M uses space at most s(n)? There are 2^(O(2^O(s))) branches to keep track of! IDEA: Given two configurations C 1 and C 2 of an s(n) space machine N, and a number t, determine if N can get from C 1 to C 2 within t steps We need to simulate a non-deterministic computation and save as much space as possible Procedure CANYIELD(C 1, C 2, t): If t = 0 then accept iff C 1 = C 2 If t = 1 then accept iff C 1 yields C 2 within one step. Use transition map of N to check [uses space O(s(n)) ] If t > 1, then Accept if and only if for some configuration C m of size s(n), both CANYIELD(C 1,C m,t/2) and CANYIELD(C m,c 2, t/2) accept CANYIELD(C 1, C 2, t) has log(t) levels of recursion. Each level of recursion uses O(s(n)) additional space to store C m. So CANYIELD(C 1, C 2, t) uses O(s(n) log(t)) space. IDEA: Given two configurations C 1 and C 2 of an s(n) space machine N, and a number t, determine if N can get from C 1 to C 2 within t steps Procedure CANYIELD(C 1, C 2, t): If t = 0 then accept iff C 1 = C 2 If t = 1 then accept iff C 1 yields C 2 within one step. Use transition map of N to check [uses space O(s(n)) ] If t > 1, then Accept if and only if for some configuration C m of size s(n), both CANYIELD(C 1,C m,t/2) and CANYIELD(C m,c 2, t/2) accept M: On input w, Output the result of CANYIELD(c start, c accept, 2 ds(n) ) Here d > 0 is chosen so that 2 d s w ) upper bounds the number of configurations of N(w) IDEA: Given two configurations C 1 and C 2 of an s(n) space machine N, and a number t, determine if N can get from C 1 to C 2 within t steps Procedure CANYIELD(C 1, C 2, t): If t = 0 then accept iff C 1 = C 2 If t = 1 then accept iff C 1 yields C 2 within one step. Use transition map of N to check [uses space O(s(n)) ] If t > 1, then Accept if and only if for some configuration C m of size s(n), both CANYIELD(C 1,C m,t/2) and CANYIELD(C m,c 2, t/2) accept M: On input w, Output the result of CANYIELD(c start, c accept, 2 ds(n) ) CANYIELD(C 1, C 2, 2 ds(n) ) uses O(s(n) log(2 ds(n) )) space. 3

4 IDEA: Given two configurations C 1 and C 2 of an s(n) space machine N, and a number t, determine if N can get from C 1 to C 2 within t steps Procedure CANYIELD(C 1, C 2, t): If t = 0 then accept iff C 1 = C 2 If t = 1 then accept iff C 1 yields C 2 within one step. Use transition map of N to check [uses space O(s(n)) ] If t > 1, then Accept if and only if for some configuration C m of size s(n), both CANYIELD(C 1,C m,t/2) and CANYIELD(C m,c 2, t/2) accept M: On input w, Output the result of CANYIELD(c start, c accept, 2 ds(n) ) CANYIELD(c start, c accept, 2 ds(n) ) uses O(s(n) 2 )) space. Theorem: For a function s where s(n) n Proof: Let N be a nondeterministic TM using s(n) space Modify N so that when it accepts, it goes to a special state q s, clears its tape, and moves its head to the leftmost cell N has a UNIQUE accepting configuration: C acc = q s Construct a deterministic M that on input w, runs CANYIELD(C 0, C acc, 2 ds( w ) ) Here d > 0 is chosen so that 2 d s( w ) upper bounds the number of configurations of N(w) => 2 d s w ) is an upper bound on the running time of N(w). Theorem: For a function s where s(n) n Proof: Let N be a nondeterministic TM using s(n) space Modify N so that when it accepts, it goes to a special state q s, clears its tape, and moves its head to the leftmost cell N has a UNIQUE accepting configuration: C acc = q s Construct a deterministic M that on input w, runs CANYIELD(C 0, C acc, 2 ds( w ) ) Uses log(2 d s( w ) ) recursions. Each level of recursion uses O(s(n)) extra space. Therefore uses O(s(n) 2 ) space! PSPACE = NPSPACE P NP PSPACE EXPTIME P NP PSPACE NPSPACE P EXPTIME TIME HIERARCHY THEOREM EXPTIME 4

5 TIME HIERARCHY THEOREM Intuition: If you have more TIME to work with, then you can solve strictly more problems! Theorem: For functions f, g where g(n)/(f(n)) 2 infinity TIME(g(n)) ( TIME(f(n)) So, for all k, since 2 n /n 2k infinity TIME(2 n ) TIME(n k ) Therefore, TIME(2 n ) P TIME HIERARCHY THEOREM Intuition: If you have more TIME to work with, then you can solve strictly more problems! Theorem: For functions f, g where g(n)/(f(n)) 2 infinity TIME(g(n)) ( TIME(f(n)) Proof IDEA: Diagonalization Make a machine M that works in g(n) time and does the opposite of all f(n) time machines on at least one input So L(M) is in TIME(g(n)) but not TIME(f(n)) TIME HIERARCHY THEOREM Intuition: If you have more TIME to work with, then you can solve strictly more problems! Theorem: For functions f, g where g(n)/(f(n)) 2 infinity TIME(g(n)) ( TIME(f(n)) Proof IDEA: Diagonalization Need g(n) >> f(n) 2 to ensure that you can simulate an arbitrary machine running in f(n) time with a single machine that runs in g(n) time. Definition: Language B is PSPACE-complete if: 1. B PSPACE 2. Every A in PSPACE is poly-time reducible to B (i.e. B is PSPACE-hard) So L(M) is in TIME(g(n)) but not TIME(f(n)) QUANTIFIED BOOLEAN FORMULAS (in prenex normal form) x y [ x y ] x [x x x] x [ x ] x y [ (x y) ( x y) ] Definition: A fully quantified Boolean formula is a Boolean formula where every variable is quantified x y [ x y ] x [x x x] x [ x ] x y [ (x y) ( x y) ] x y [ (x 0) ( x y) ] 5

6 TQBF PSPACE TQBF = { is a true fully quantified Boolean formula} Theorem: TQBF is PSPACE-complete T( ): 1. If has no quantifiers, then it is an expression with only constants. Evaluate. Accept iff evaluates to If = x, recursively call T on, first with x = 0 and then with x = 1. Accept iff either one of the calls accepts. 3. If = x, recursively call T on, first with x = 0 and then with x = 1. Accept iff both of the calls accept. Claim: Every language A in PSPACE is polynomial time reducible to TQBF We build a poly-time reduction from A to TQBF A tableau for M on w is an table whose rows are the configurations of the computation of M on input w # q 0 w 1 w 2 w n # # # The reduction turns a string w into a fully quantified Boolean formula that simulates the PSPACE machine for A on w Let M be a deterministic TM that decides A in space n k How do we know M exists? 2 O(nk ) # # n k We design to encode a simulation of M on w will be true if and only if M accepts w Given two collections of variables denoted c and d representing two configurations and t > 0, we construct a formula c,d,t If we assign c and d to actual configurations, c,d,t will be true if and only if M can go from c to d in t steps We let = c, c, h, where h = 2 e s(n) for a start accept constant e chosen so that M has less than 2 e s(n) possible configurations on an input of length n Here s(n) = n k We design to encode a simulation of M on w will be true if and only if M accepts w Given two collections of variables denoted c and d representing two configurations and t > 0, we construct a formula c,d,t If we assign c and d to actual configurations, c,d,t will say: there exists a configuration m such that c,m,t/2 is true and m,d,t/2 is true We let = c, c, h, where h = 2 e s(n) for a start accept constant e chosen so that M has less than 2 e s(n) possible configurations on an input of length n Here s(n) = n k 6

7 HIGH-LEVEL IDEA: Encode the Algorithm of Savitch s Theorem with a Quantified Boolean Formula If M uses n k space, then the QBF will have size O(n 2k ) If we assign c and d to actual configurations, c,d,t will say: there exists a configuration m such that c,m,t/2 is true and m,d,t/2 is true We let = c, c, h, where h = 2 e s(n) for a start accept constant e chosen so that M has less than 2 e s(n) possible configurations on an input of length n Here s(n) = n k To construct c,d,t use ideas of Cook-Levin plus Savitch: Each cell in a configuration is associated with variables representing possible tape symbols and states. Each config has n k cells so and is encoded by O(n k ) variables. If t = 0 or 1, we can easily construct c,d,t : c,d,t = c equals d OR d follows from c in a single step of M How do we express c equals d? Wit Write a Boolean formula saying that t each of the variables representing c is equal to the corresponding one in d d follows from c in a single step of M? Use 2 x 3 windows as in the Cook-Levin theorem, and write a CNF formula If t > 1, we construct c,d,t recursively: c,d,t = m [ c,m,t/2 m,d,t/2 ] x 1 x 2 x L L= O(n k ) But how long is this formula? Every level of the recursion cuts t in half but roughly doubles the size of the formula. So, we modify the formula to be: c,d,t = m a,b[ [(a,b)=(c,m) (a,b)=(m,d)] => [ a,b,t/2 ] ] This folds the 2 recursive sub-formulas into 1 c,d,t = m a,b[ [(a,b)=(c,m) (a,b)=(m,d)] =>[ a,b,t/2 ] ] Set = c, c, h where h = 2 e f(n) start accept Each recursive step adds a portion that is linear in the size of the configurations, so has size O(f(n)) Number of levels of recursion is log h = O(f(n)) Hence, the size of is O(f(n) 2 ) PSPACE is often called the class of games Formalizations of many popular games are PSPACE-Complete 7

8 THE FORMULA GAME (FG) is played between two players, E and A Given a fully quantified Boolean formula y x [ (x y) ( x y) ] E chooses values for variables quantified by A chooses values for variables quantified by Start at the leftmost quantifier E wins if the resulting formula is true A wins otherwise x y [ (x y) ( x y) ] x y [ x y] FG = { Player E has a winning strategy in } Theorem: FG is PSPACE-Complete Proof: FG = TQBF GEOGRAPHY GENERALIZED GEOGRAPHY Two players take turns naming cities from anywhere in the world d Each city chosen must begin with the same letter that the previous city ended with Cities cannot be repeated a b e g i Austin Nashua Albany York c h Whoever cannot name any more cities loses f GG = { (G, b) Player 1 has a winning strategy for generalized geography played on graph G starting at node b } Theorem: GG is PSPACE-Complete C GG PSPACE WANT: Machine M that accepts (G,b) Player 1 has a winning strategy on (G, b) M(G, b): If b has no outgoing edges, reject. 1. Remove node b and all edges touching it to get to a new graph G 1 2. For each of the nodes b 1, b 2,, b k that b originally pointed at, recursively call M(G 1, b i ) 3. If all of these accept, Player 2 has a winning strategy, so reject. Otherwise, accept. 8

9 GG IS PSPACE-HARD We show that FG P GG TRUE x 1 T b F FALSE x 1 x 2 x k (x 1 x 1 x 2 ) ( x 1 x 2 x 2 ) x 1 We convert a formula into (G, b) such that: x 1 c 1 Player E has winning strategy in if and only if Player 1 has winning strategy in (G, b) x 2 T F x 2 c 2 For simplicity we assume is of the form: x 1 c = x 1 x 2 x 3 x k [ ] x 2 where is in cnf. (Quantifiers alternate, and the last move is E s) x k T F x 2 c n x 1 [ (x 1 x 1 x 1 ) ] TRUE b FALSE x 1 T F x 1 x1 c 1 GG = { (G, b) Player 1 has a winning strategy for generalized geography played on graph G starting at node b } Theorem: GG is PSPACE-Complete C x 1 c Question: Is Chess a PSPACE complete problem? No, because determining whether a player has a winning strategy takes CONSTANT time and space (OK, the constant is large ) But n x n GO, Chess and Checkers can be shown to be PSPACE-hard 9