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Course: CSCI 4011, Fall 2008
School: Minnesota
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LIMITATIONS INHERENT OF COMPUTER PROGRAMS CSci 4011 SPACE COMPLEXITY Definition: Let M be a deterministic TM that halts on all inputs. The space complexity of M is the function f : N N, where f(n) is the rightmost tape position that M reaches on any input of length n. PSPACE = SPACE(n ) k kN 3SAT PSPACE Definition: SPACE(t(n)) = { L | L is a language decided by a O(t(n)) space deterministic Turing...

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LIMITATIONS INHERENT OF COMPUTER PROGRAMS CSci 4011 SPACE COMPLEXITY Definition: Let M be a deterministic TM that halts on all inputs. The space complexity of M is the function f : N N, where f(n) is the rightmost tape position that M reaches on any input of length n. PSPACE = SPACE(n ) k kN 3SAT PSPACE Definition: SPACE(t(n)) = { L | L is a language decided by a O(t(n)) space deterministic Turing Machine } PSPACE = NPSPACE P NP PSPACE EXPTIME 1 HARDEST PROBLEMS IN PSPACE 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) QUANTIFIED BOOLEAN FORMULAS xy [ x y ] x [ x x ] x [ x ] xy [ (x y) (x y) ] Definition: A fully quantified Boolean formula (or a sentence) is a Boolean formula in which every variable is quantified xy [ x y ] x [ x x ] x [ x ] xy [ (x y) (x y) ] TQBF = { | is a true fully quantified Boolean formula} Theorem: TQBF is PSPACE-complete 2 TQBF PSPACE isTrue(): 1. If contains no quantifiers, then it is an expression with only constants, so simply evaluate 2. If = x , recursively call isTrue on , first with x = 0 and then with x = 1. Accept if either one of them is true. 3. If = x , recursively call isTrue on , first with x = 0 and then with x = 1. Accept if both of them are true. Claim: Every language A in PSPACE is polynomial time reducible to TQBF We build a poly-time reduction from A to TQBF Let M be a deterministic TM that decides A in space nk The reduction turns a string w into a fully quantified Boolean formula that is true iff M has an accepting computation on w. We now design so that a satisfying assignment to the variables corresponds to M accepting w Given two collections of variables denoted c and d representing two configurations of M and a number 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 start , c accept , h, where h = 2nk+1 If t = 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" "c d" equals says that each for each i, ci = di. "d follows from c in a single step of M" can be expressed as we did with SAT 3 If t > 1, we can build c,d,t recursively: c,d,t = m [c,m,t/2 m,d,t/2 ] m1m2 ... ml How long is this formula? PSPACE is frequently called the class of games because many popular games are PSPACE-Complete c,d,t = ma,b[ [(a,b)=(c,m) (a,b)=(m,d)] [ a,b,t/2 ] ] THE FORMULA GAME ...is played between two players, E and A Given a fully quantified Boolean formula xy [ (x y) (x y) ] xy [ x y ] FG = { | Player E can force a win in } Theorem: FG is PSPACE-Complete Proof: yx [ (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 4 GEOGRAPHY Two players take turns naming cities from anywhere in the world Each city chosen must begin with the same letter that the previous city ended with Cities cannot be repeated Austin Nashua Albany York Whoever cannot name any more cities loses GENERALIZED GEOGRAPHY d b a c e h f g i GG PSPACE currentPlayerWins(G, s): GG = { (G,b) | Player I has a winning strategy for the generalized geography game played on graph G starting ...

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