NP-Completeness. A language B is NP-complete iff B NP. This property means B is NP hard
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1 NP-Completeness A language B is NP-complete iff B NP A NP A P B This property means B is NP hard 1
2 3SAT is NP-complete 2
3 Result Idea: B is known to be NP complete Use it to prove NP-Completeness of C IF C is NP-hard B is NP-complete and B P C and C NP THEN C is NP-complete 3
4 Result CLIQUE is NP-hard 3SAT is NP-complete and 3SAT P CLIQUE and CLIQUE NP THEREFORE CLIQUE is NP-complete 4
5 Result VERTEX-COVER is NP-hard 3SAT is NP-complete and 3SAT P VERTEX-COVER and VERTEX-COVER NP THEREFORE VERTEX-COVER is NP-complete 5
6 Result IF B is NP-complete and B P THEN P = NP 6
7 SAT P P = NP 3SAT P P = NP CLIQUE P P = NP VERTEX-COVER P P = NP 7
8 SAT = { Φ Φ is a satisfiable Boolean formula} SAT NP
9 Cook-Levin Theorem SAT is NP-complete In other words SAT NP A NP A P SAT 9
10 To prove A NP A P SAT define polynomial time computable function f: Σ * Σ * w Σ * w A f(w) SAT w Σ * w A Φ SAT 10
11 Consider any A NP NTM N that decides A in polytime n k For any input w Σ * valid tableau of configurations 11
12 Properties of a Valid Tableau There is exactly one symbol in each cell The first row is the ( legal ) start configuration Every subsequent row is generated legally 12
13 For any input w A there is an accepting (valid) tableau of configurations 13
14 Properties of an Accepting Tableau There is exactly one symbol in each cell The first row is the ( legal ) start configuration Every subsequent row is generated legally One of these rows is an accepting configuration 14
15 Proof Idea Given N and w construct a Boolean formula that is satisfiable exactly when N has an accepting tableau on input w 15
16 Constructing the formula Define Boolean formula with variables x ijs for 1 i n k 1 j n k s State Set Tape Alphabet Delimiter Want following semantics: x ijs is T iff cell (i, j) contains symbol s for some valid accepting tableau 16
17 x ijs is T iff cell (i, j) contains symbol s x 12q_0 is T whereas x 12 is F 17
18 Properties of an Accepting Tableau There is exactly one symbol in each cell The first row is the ( legal ) start configuration Every subsequent row is generated legally One of these rows is an accepting configuration 18
19 represent valid tableau with a Boolean formula with components Φ cell exactly one symbol per cell for any pair (i,j) the cell contains at least one symbol the cell contains at most one symbol 19
20 Properties of an Accepting Tableau There is exactly one symbol in each cell The first row is the ( legal ) start configuration Every subsequent row is generated legally One of these rows is an accepting configuration 20
21 represent valid tableau with a Boolean formula with components Φ cell exactly one symbol per cell Φ start legal starting configuration 21
22 Properties of an Accepting Tableau There is exactly one symbol in each cell The first row is the ( legal ) start configuration Every subsequent row is generated legally One of these rows is an accepting configuration 22
23 represent valid tableau with a Boolean formula with components Φcell exactly one symbol per cell Φstart legal starting configuration Φmove legal moves 23
24 represent valid tableau with a Boolean formula with components Φ move legal moves transition function N q1 b c, L q2 a b, R b a, R # a b q1 b c # a q2 b c c # a b q1 b c # a b a q2 c 24
25 represent valid tableau with a Boolean formula with components Φ move legal moves represented by legal windows transition function N q1 b c, L q2 a b, R b a, R 25
26 represent valid tableau with a Boolean formula with components Φ move legal moves represented by legal windows 26
27 Properties of an Accepting Tableau There is exactly one symbol in each cell The first row is the ( legal ) start configuration Every subsequent row is generated legally One of these rows is an accepting configuration 27
28 represent valid accepting tableau with a Boolean formula with components Φ cell exactly one symbol per cell Φ start legal starting configuration Φ move legal moves Φ accept legal accepting configuration 28
29 represent valid accepting tableau with Boolean formula Φcell Φstart Φmove Φaccept 29
30 SAT is NP-complete w Σ * w A there is a valid accepting tableau constructed formula is SATISFIABLE Corollary 7.42: 3SAT is NP-complete 30
31 HALTTM = { M, w : M is a TM that halts on input w} HALTTM is NP-hard In other words A NP A P HALTTM 31
32 3SAT P HALTTM define polynomial time computable function f: Σ * Σ * w Σ * w 3SAT f(w) HALT TM 32
33 Define TM M as follows M: On input w: 3SAT P HALTTM If w is not a valid 3CNF encoding then loop If w is a valid 3CNF encoding then check if w evaluates to True for any possible assignment to variables in w If yes then accept else loop w Σ * w 3SAT M, w HALT TM 33
34 HALTTM = { M, w : M is a TM that halts on input w} Is HALTTM NP-complete? 34
35 NP-hard NP-complete NP P All languages 35
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