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INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011

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QUIZ 5 The language L TIME(t(n)) if: L can be decided by a TM in time O(t(n)) The language L NP if: L can be decided by a NTM in time O(n c ) L has a polynomial time verifier 3SAT is the language: { ϕ | ϕ is a satisfiable 3cnf formula}

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QUIZ 5 Give a satisfying assignment to x 1 = 0, x 2 = 0 Asymptotic upper bounds for: (a) 4n n log 3 n = O(n 2.1 ) (b) 4 (log n)/3 + 8n(n/2) 1/2 = O(n 3/2 )

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TIME COMPLEXITY Definition: Let M be a TM that halts on all inputs. The running time or time-complexity of M is the function f : N N, where f(n) is the maximum number of steps that M uses on any input of length n. Definition: TIME(t(n)) = { L | L is a language decided by a O(t(n)) time Turing Machine }

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P = TIME(n c ) c N IMPORTANT COMPLEXITY CLASSES Problems in P can be efficiently solved. Problems in NP can be efficiently verified. NP = NTIME(n c ) c N L P iff there is an n c -time TM that decides L. L NP iff there is an n c -time NTM that decides L iff there is an n c -time verifier for L

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Definition: A language B is NP-complete if: 1. B NP 2. Every A in NP is poly-time reducible to B (i.e. B is NP-hard) HARDEST PROBLEMS IN NP Theorem: if B is NP-Complete and B P, then NP=P. Corollary: if B is NP-Complete and P NP, then there is no fast algorithm for B. Theorem: if A is NP-Hard and A P B, then B is NP-Hard.

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P NP B C HARDEST PROBLEMS IN NP

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Theorem (Cook-Levin): 3SAT is NP-complete Proof: (1) 3SAT NP (2) Every language A in NP is polynomial time reducible to 3SAT Our proof of (2) has three steps. (a) CIRCUIT-SAT is NP-Hard (b) CIRCUIT-SAT P CNF-SAT (c) CNF-SAT P 3SAT Corollary: 3SAT P if and only if P = NP

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A CIRCUIT … is a collection of (boolean) gates and inputs connected by wires. ¬ x0x0 x1x1 x2x2 … is satisfiable if some setting of inputs makes it output 1. … has arbitrary fan-in and fan-out CIRCUIT-SAT = { C | C is a satisfiable circuit }

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Given TM M and time bound t, we create a circuit that takes n input bits and runs up to t steps of M. The circuit will have t rows, where the i th row represents the configuration of M after i steps: q00q q11q q10q q01q q00q q11q11 010q1q1 010qaqa t t a tableau

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q0q0 Rows are made up of cells. Each cell has a light for every state and every tape symbol. q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 Each light has a circuit that turns it on or off based on the previous row. q0q0

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q0q0 EXAMPLE q0q0 q1q1 qaqa 0 0, R 0, R 1 1, R 1, R, L q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 0 1

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The lights in the first row are connected to the circuit inputs and the tape head is hardwired in: x1x1 x2x2 x3x3 … x n-1 xnxn q0q0 The circuit should output 1 iff M ends in q accept. qaqa qaqa qaqa qaqa qaqa qaqa qaqa qaqa

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CNF-SAT = { | is a satisfiable CNF formula } Theorem. CIRCUIT-SAT P CNF-SAT. Proof. Given a circuit C, we will output a CNF formula that is satisfiable iff C is. x1x1 x2x2 (x 1 x 2 ) ((x 1 x 2 ) (x 1 x 2 ))

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x1x1 x2x2 For every gate in the circuit C, we introduce a new variable g i and force g i to satisfy the gate. g 1 = (x 1 x 2 ) g 2 = (x 1 x 2 ) g 3 = (g 1 g 2 ) g 4 = (g 1 g 2 ) g 5 = (g 3 g 4 ) g 1 (x 1 x 2 ) (x 1 x 2 ) g 1 g 2 (x 1 x 2 ) (x 1 x 2 ) g 2 g 3 (g 1 g 2 ) (g 1 g 2 ) g 3 g 4 (g 1 g 2 ) (g 1 g 2 ) g 4 g 5 (g 3 g 4 ) (g 3 g 4 ) g 5 g5g5

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We can convert to CNF using the fact that (p q) is equivalent to (¬p q) so: g ( 1 2 … n ) (¬g 1 2 … n ) ( 1 2 … n ) g (¬ 1 ¬ 2 … ¬ n g) g ( 1 2 … n ) (¬g 1 ) (¬g 2 ) … (¬g n ) ( 1 2 … n ) g (¬ 1 g) (¬ 2 g) … (¬ n g) The final output is the CNF with all the clauses produced from gates in this way, plus the clause (g i ), where g i corresponds to Cs output gate.

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If C has k inputs, m gates, n wires, how long is ? O(m+n+k) How long to compute ? O(m+n+k) Suppose an assignment to x 1 …x k and g 1 …g m satisfies. Then C outputs 1 on input x 1 …x k. Suppose C(x 1 …x k )=1. Then setting each g i to the value of the corresponding gate satisfies.

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Theorem. CNF-SAT P 3SAT. Proof. We show how to create an equivalent 3cnf for any cnf formula. cnf formula is not in 3cnf iff there is a clause C with less than 3 literals or more than 3 literals. if C = ( i ), replace C with ( i _ i _ i ) if C = ( i _ j ), replace C with ( i _ i _ j )

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We can reduce the number of literals in any clause by one if we introduce a new variable and clause: ( 1 2 … n ) (z 3 … n ) (z ( 1 2 )) (z 3 … n ) (¬z 1 2 )) if we satisfy the first clause by setting z=1, then to satisfy the second, we must set ( 1 2 )=1. Repeat until every clause has three literals.

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MORE SATISFIABILITY PROBLEMS Let NAESAT = { | is a 3cnf and x. (x)=1 and each clause has 1 false literal} Theorem. NAESAT is NP-Complete. Let DSAT = { | is a cnf with 2 satisfying assignments} Theorem. DSAT is NP-Complete. Let 2-in-4-SAT = { | is a 4cnf and x: each clause has exactly two true literals.} Theorem. 2-in-4-SAT is NP-Complete.

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SATISFYING CONSTRAINTS A 2csp is a list of constraints on pairs of variables Each constraint C(x,y) is a list of values for (x,y). An assignment satisfies a constraint if (x,y) C(x,y). An assignment satisfies a 2csp if it satisfies all constraints. e.g. Scheduling a project; seating at a wedding… 2CSP = { C | C is a satisfiable 2csp } Theorem. 2CSP is NP-Complete.

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Proof. 1. 2CSP NP. Given an assignment to the variables we can check that all constraints are satisfied in linear time. 2. 3SAT P 2CSP. Idea: the main difference is that a 2csp constraint should have only two variables. Add variables to the 2csp that represent pairs of variables in the 3cnf, and constraints to enforce consistency with the 3cnf variables.

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Map 3cnf ϕ with n variables and m clauses to a 2csp C ϕ with 2n+m variables and 3m+n constraints: for each variable x ϕ : add (v x,v ¬x ) {(0,1),(1,0)} for each clause (x y z) ϕ : add v xy {00,01,10,11} add (v xy,v z ) (00,0) add (v xy,v x ) {(00,0), (01,0), (10,1), (11,1)} add (v xy,v y ) {(00,0), (01,1), (10,0), (11,1)} Claim. C ϕ 2CSP ϕ 3SAT: Proof. A satisfying assignment to ϕ can be mapped to a satisfying assignment to C ϕ by assigning each v x = x and each v xy = 2v x +v y because… An assignment to C ϕ maps to an assignment to ϕ by setting each x = v x, because…

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2SAT 2SAT = { | is in 2cnf and is satisfiable} Theorem. 2SAT P! Idea: a 2SAT clause (x y) is equivalent to (¬x y) and (¬y x). If there is a chain (x z 1 ) (z 1 z 2 ) … (z k ¬x) and (¬xy 1 ) (y 1 y 2 ) … (y k x) then: x ¬x If not, the formula is consistent, so satisfiable. e.g. (x x) (¬x y) (¬y ¬x)

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def is_satisfiable_2cnf(F): V = Ø, E = Ø, G = (V,E) for l literals(F): V = V {l, ¬l} for (x y) clauses(F): E = E {(¬x,y), (¬y,x)} for x vars(F): if has_path(G,x,¬x) and has_path(G,¬x,x): return False return True (xx)(xx̅)(x̅x) x x x x x x

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3SAT P DSAT, NAESAT, 2CSP… We will use 3SAT to prove other problems are NP-Complete or NP-Hard. Examples include 3SAT P CLIQUE 3SAT P 0/1-ILP 3SAT P HAMPATH 3SAT P 3COLOR 3SAT P GRADUATION 3SAT P VERTEX-COVER

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