SAT (The Cook-Levin Theorem)

An overview of the Cook-Leven theorem and the Satisfiability Problem, covering how the problem is NP-Complete without reducing any other problem to it

Transcript

1. S A T The First NP-Complete Problem
3. Stephen Cook Professor, University of Toronto 72 Years Old Leonid Levin Professor, Boston University 63 Years Old
4. “The Complexity of Theorem- Proving Procedures” 1971
5. What is SAT? Given a formula S of boolean variables X = {x1 , x2 … xm } is there an assignment of variables X such that S is true? S1 = (a) ! (b)" S2 = (a ! b ! -c) # (-a ! b ! d)" S3 = (a ! b) # (-a ! c)"
6. Easy to show SAT in NP If given a certificate C of assignments to variables X, it is easy to evaluate the formula and see if it is true. X= { x1 , x2 , x3 }" S = (x1 ! x2 ! -x3 ) # (-x1 ! x2 ! x3 )" C = { x1 =T, x2 =F, x3 =T}" Note: Many different C’s would work. x2 being true satisfies S = (T ! F ! -T) # (-T ! F ! T)" = (T ! F ! F) # ( F ! F ! T)" = T" ✓
7. Is SAT NP-Complete? Reduce another NP-Complete problem to SAT.
8. Is SAT NP-Complete? Must be in NP ✔ Everything else in NP must be reducible to SAT in polynomial time
9. Can everything in NP be reduced to SAT? If a problem/language A is in NP, there’s a Nondeterministic Turing Machine NTMA that recognizes it in polynomial time Can we simulate NTMA using only boolean expressions? Boolean expressions are made up of: ANDs ORs NOTs
10. Terms A – a language in NP NTMA - Nondeterministic Turing Machine that accepts A w - input to NTMA
11. Assumptions One tape Head does not move left of start position Input w is of length n" NTMA runs in p(n) steps" If NTMA accepts w, it does so through a sequence of p(n) configurations The config itself is p(n) in length
12. NTMA Configuration Snapshot (running on w) # q0" w1" w2" …" wn" _" …" _" # state input extra space p(n)"
13. w on NTMA Tableau # # # # # q0" w1" w2" …" wn" _" …" _" # p(n)" if any config (r0w) accepts, the entire tableau accepts # #
14. “Does the tableau for w on NTMA accept?” “Does NTMA accept w?” Non-determinism: if any configuration/row accepts, the whole machine does
15. Where’s this going, again? We can represent NTMA on w as a tableau Solve $ using SAT! Use tableau to construct a boolean expression $ such that: $ is satisfied iff NTMA accepts w"
16. C0,0" C0,1" …" C0,p(n)" C1,0" C1,1" …" C1,p(n)" ." ." ." ." ." ." …" ." ." ." Cp(n),0" Cp(n),1" …" Cp(n),p(n)" There are only so many symbols that could be in Ci,j" " For each symbol s, xi,j,s is true when Ci,j has s in it. We will form $ from x’s
17. We’re now representing the entire tableau as boolean variables. Our expression will be made up of four sub- expressions: " •  $unique "(Exactly one symbol in each cell) •  $start "(The start configuration is valid) •  $move "(All of the transitions are valid) •  $accept "(One of the configurations accepts) $ = $unique # $start # $move # $accept
18. $unique •  For all i,j < p(n) (each cell Ci,j )" – For all possible symbols Y, Z (Y ≠ Z)" •  Take the AND of all (-xi,j,Y ! -xi,j,Z )" In other words, for all pairs of symbols, it’s true that any given cell fails to have one of them (aka: each cell only has one symbol)
19. $start x0,0,# # x0,1,q0 # x0,2,w1 # x0,3,w2 # … " " "# x0,n+1,wn # x0,n+2,_ # …" " "# x0,p(n)-1,_ # x0,p(n),# " " " " " " Ensure that the start state looks exactly as it’s supposed to
20. $unique •  For all i,j < p(n) (each cell Ci,j )" – Take the OR of all xi,j,qaccept" Check each cell to see if it’s the unique symbol for acceptance
21. $move Trickiest one Look at all of the different 2-row “chunks” of the tableau and make sure the top one can lead the bottom one
22. # q0" w1" w2" …" wn" _" ... _ # # # # # window
23. Suppose NTMA has a transition function & that includes: &(q,A) è (p,B,<-) " If machine in state q reads an A, write a B, move left, enter state p"
24. &(q,A) è (p,B,<-) C" B" A" C" B" A" C" q" A" p" C" B" C" B" A" C" B" p" q" A" B" C" B" B" A" B" C" B" B" C"
25. Constant number of window arrangements to check for any given transition Constant number of transitions for NTMA" " We can check if a window is legal in constant time " $move = The AND of all windows with top center cell Ci,j being legal
26. " $ = $unique # $start # $move # $accept Reduced NP Language A to Boolean Satisfiability Problem
27. What is the complexity of this reduction?
28. $unique "- O(p(n)2) symbols The possible pairs of symbols is independent of n, so it’s a constant. The number of cells is p(n) * p(n)" " $start " "- O(p(n)) symbols Simply checking every symbol in a single row of the tableau " $move " "- O(p(n)2) symbols Need to check every “window”, of which there are almost as many as there are cells, p(n) * p(n) " $accept "- O(p(n)2) symbols Need to check each cell for accept symbol, of which there are p(n) * p(n).
29. Total number of variables: " O(p(n)2 + p(n) + p(n)2 + p(n)2)" " O(p(n)2) The reduction can be done in polynomial time.
30. In polynomial time, we can reduce any instance of an NP problem into an instance of SAT Success! SAT is NP-Complete!
31. “[SAT] is a good candidate for an interesting set not in [P], and I feel it is worth spending considerable effort trying to prove this conjecture.” “Such a proof would be a major breakthrough in complexity theory”
32. questions Easy only ?