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INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011

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QUIZ 6 B is NP-Complete if: (a) B ∈ NP (b) ∀ A ∈ NP, A ≤ P B Reduction from HAMPATH to KPATH: ƒ( 〈 G,s,t 〉 ) = 〈 G,s,t,|V|-1 〉 nifty proof that 〈 [20, 46, 43, 50, 36, 45, 14, 52], 99 〉 ∈ SSUM: e.g. 20+43+36= 99 Nifty Proof that 〈 G 〉 ∈ 3COLOR:

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Extended Office Hours posted at: http://cs4011.org/ FINAL EXAM THURSDAY, DECEMBER 18 10:30AM – 12:30PM. Reminder: one page (8.5×11”) “cheat sheet”

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1*(0 ∪ 01) 0 q0q0 q2q2 1 q1q1 0 1 {0 k 1 k | k ≥ 0} R → LRL | b L → a | b S → R | ε { ww | w ∈ {0,1}* } MODELS

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0 → 0, R readwritemove → , R q accept 0 → 0, R → , R 0 → 0, R → , L TURING MACHINES UNBOUNDED TAPE 0 q0q0 q0q0 q1q1 q2q2 q1q1 q2q2 0

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A TM = { M,w | M is a TM that accepts string w } A TM is undecidable:(proof by contradiction) Assume there is a program accepts to decide A TM. accepts( M ,w) = true if M(w) accepts false if M(w) does not. Construct a new TM LLPF that on input M , runs accepts( M , M ) and “does the opposite”: LLPF(PROG ) =if (accepts( PROG, PROG )) then reject; else accept. LLPF

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REDUCTIONS A m B if there is a computable ƒ so that w A ƒ(w) B ƒ is called a reduction from A to B A P B if there is a poly-time computable ƒ so that w A ƒ(w) B

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COMPLEXITY THEORY P = Problems where it is easy to find the answer. NP = Problems where it’s easy to check the answer. If P = NP then generation is as easy as recognition. Is there a fast program for this problem? PSPACE = Problems that can be solved in polynomial space. If P = PSPACE then TIME is as powerful as SPACE.

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COMPLETE PROBLEMS NP:3SAT,SUBSET-SUM,HAMPATH, VERTEX COVER, 3-COLOR,CLIQUE,… PSPACE:TQBF,FG,GEOGRAPHY,… If C is a class of languages and B is a language, then B is C-Complete if: 1. B ∈ C. 2. ∀ A ∈ C, A ≤ P B (i.e. B is C-Hard)

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SPACE COMPLEXITY

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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. Definition: SPACE(s(n)) = { L | L is a language decided by a O(s(n)) space deterministic Turing Machine }

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PSPACE = SPACE(n k ) k N 3SAT PSPACE PSPACE = NPSPACE P NP PSPACE EXPTIME

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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) HARDEST PROBLEMS IN PSPACE

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QUANTIFIED BOOLEAN FORMULAS x y x y [ ] x [ x x ] x [ x ] x y [ (x y) ( x y) ]

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Definition: A fully quantified Boolean formula (or a sentence) is a Boolean formula in which every variable is quantified x y x y [ ] x [ x x ] x [ x ] x y [ (x y) ( x y) ]

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TQBF = { | is a true fully quantified Boolean formula} Theorem: TQBF is PSPACE-complete

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TQBF PSPACE is_true( ): 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.

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Claim: Every language A in PSPACE is polynomial time reducible to TQBF We build a poly-time reduction from A to TQBF The reduction turns a string w into a fully quantified Boolean formula that is true iff M has an accepting computation on w. Let M be a deterministic TM that decides A in space n k

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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, c, h, where h = 2 n k+1 startaccept

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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 equals d” says that each for each i, c i = d i. “d follows from c in a single step of M” can be expressed as we did with SAT

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If t > 1, we can build c,d,t recursively: c,d,t = m [ c,m,t/2 m,d,t/2 ] m 1 m 2 … ∃ m l How long is this formula? O(t) = 2 O(n k ) c,d,t = m a,b[ [(a,b)=(c,m) (a,b)=(m,d)] [ a,b,t/2 ] ] Since O(t) is too long, we modify the formula to be:

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THE FORMULA GAME …is played between two players, E and A Given a fully quantified Boolean formula 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 y x [ (x y) ( x y) ]

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FG = { | Player E can force a win in } Theorem: FG is PSPACE-Complete Proof: FG = TQBF x y [ (x y) ( x y) ] x y x y [ ]

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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 Austin Nashua Albany York Cities cannot be repeated Whoever cannot name any more cities loses

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GENERALIZED GEOGRAPHY b a e c d f g i h

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GG = { 〈 G,b 〉 | Player I has a winning strategy for the generalized geography game played on graph G starting at node b } Theorem: GG is PSPACE-Complete

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GG PSPACE def current_player_wins(G, s): G’ = remove_node(s,G) for t ∈ neighbors(s,G): else: return False if neighbors(s,G) = Ø: return if not(current_player_wins(G’,t)): return True. False Maximum recursion depth: |G| Space complexity: O(|G|²)

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We show that FG P GG GG IS PSPACE-HARD We convert a formula into 〈 G,b 〉 such that: Player E has winning strategy in if and only if Player I has winning strategy in 〈 G,b 〉 For simplicity we assume is of the form: = x 1 x 2 x 3 … x k [ ] where is in cnf

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b c TRUEFALSE x1x1 x2x2 xkxk c1c1 c2c2 cncn x1x1 (x 1 x 1 x 2 ) ( x 1 x 2 x 2 ) … x2x2 x2x2 x1x1 x1x1 x2x2

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b TRUEFALSE x1x1 c1c1 x1x1 x1x1 x1x1 c x 1 [ (x 1 x 1 x 1 ) ]

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n x n GO, chess and checkers can be shown to be PSPACE-hard Question: Is Chess PSPACE complete? No, because determining whether Player I has a winning strategy takes constant time and space

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