A Simple, Greedy Approximation Algorithm for MAX SAT

Transcription

1 A Simple, Greedy Approximation Algorithm for MAX SAT David P. Williamson Joint work with Matthias Poloczek (Frankfurt, Cornell) and Anke van Zuylen (William & Mary)

2 Greedy algorithms Greedy Greed, for algorithms lack of a work. better word, Alan is Hoffman, good. Greed IBM is right. Greed works. Gordon Gekko, Wall Street

3 Another reason When I interviewed at Watson, half of my talk was about maximum satisfiability, the other half about the max cut SDP result. I thought, Oh no, I have to talk about Hardness of approximation in front of Madhu Sudan, Randomized rounding in front of Prabhakar Raghavan, And eigenvalue bounds in front of Alan Hoffman. Today I revisit the first part of that talk.

4 Maximum Satisfiability Input: n Boolean variables x 1,, x n m clauses C 1,, C m with weights w j 0 each clause is a disjunction of literals, e.g. C 1 = x 1 x 2 x 3 Goal: truth assignment to the variables that maximizes the weight of the satisfied clauses

5 Approximation Algorithms An α-approximation algorithm runs in polynomial time and returns a solution of at least α times the optimal. For a randomized algorithm, we ask that the expected value is at least α times the optimal.

6 A ½-approximation algorithm Set each x i to true with probability ½. Then if l j is the number of literals in clause j

7 What about a deterministic algorithm? Use the method of conditional expectations (Erdős and Selfridge 73, Spencer 87) If E W x 1 true E W x 1 false then set x 1 true, otherwise false. Similarly, if X i 1 is event of how first i 1 variables are set, then if E W X i 1, x i true E W X i 1, x i false, set x i true. Show inductively that E[W X i ] E W 1 2 OPT.

8 An LP relaxation

9 Randomized rounding Pick any function fsuch that 1 4 x f x f(y i ), where y is an optimal LP solution. 4 x 1. Set x i true with probability

10 Analysis

11 Integrality gap The result is tight since LP solution z 1 = z 2 = z 3 = z 4 = 1 and y 1 = y 2 = 1 feasible for instance above, but OPT = 3. 2

12 Current status NP-hard to approximate better than (Håstad 01) Combinatorial approximation algorithms Johnson s algorithm (1974): Simple ½-approximation algorithm (Greedy version of the randomized algorithm) Improved analysis of Johnson s algorithm: 2 / 3 -approx. guarantee [Chen-Friesen-Zheng 99, Engebretsen 04] Randomizing variable order improves guarantee slightly [Costello-Shapira-Tetali 11] Algorithms using Linear or Semidefinite Programming Yannakakis 94, Goemans-W 94: ¾-approximation algorithms Best guarantee [Avidor-Berkovitch-Zwick 05] Question [W 98]: Is it possible to obtain a 3/4-approximation algorithm without solving a linear program?

13 (Selected) recent results Poloczek-Schnitger 11: randomized Johnson combinatorial ¾- approximation algorithm Van Zuylen 11: Simplification of randomized Johnson probabilities and analysis Derandomization using Linear Programming Buchbinder, Feldman, Naor, and Schwartz 12: Another ¾-approximation algorithm for MAX SAT as a special case of submodular function maximization We show MAX SAT alg is equivalent to van Zuylen 11.

14 (Selected) recent results Poloczek-Schnitger 11 Van Zuylen 11 Buchbinder, Feldman, Naor and Schwartz 12 Common properties: iteratively set the variables in an online fashion, the probability of setting x i to true depends on clauses containing x i or x i that are still undecided.

15 Today Give textbook version of Buchbinder et al. s algorithm with an even simpler analysis

16 Buchbinder et al. s approach Keep two bounds on the solution Lower bound LB = weight of clauses already satisfied Upper bound UB = weight of clauses not yet unsatisfied Greedy can focus on two things: maximize LB, maximize UB, but either choice has bad examples Key idea: make choices to increase B = ½ (LB+UB)

17 LB 0 (= 0) B 0 = ½(LB 0 +UB 0 ) UB 0 (= w j )

18 Weight of undecided clauses satisfied by x 1 = true Weight of undecided clauses unsatisfied by x 1 = true LB 0 LB 1 B 0 = ½(LB 0 +UB 0 ) UB 1 UB 0 Set x 1 to true

19 Weight of undecided clauses satisfied by x 1 = true B 1 Weight of undecided clauses unsatisfied by x 1 = true LB 0 LB 1 B 0 UB 1 UB 0 Set x 1 to true

20 Weight of undecided clauses satisfied by x 1 = true B 1 Weight of undecided clauses unsatisfied by x 1 = true LB 0 LB 1 LB 1 B 0 UB 1 UB 1 UB 0 Set x 1 to true or Set x 1 to false

21 Weight of undecided clauses satisfied by x 1 = true B 1 B 1 Weight of undecided clauses unsatisfied by x 1 = true LB 0 LB 1 LB 1 B 0 UB 1 UB 1 UB 0 Set x 1 to true or Set x 1 to false Guaranteed that (B 1 -B 0 )+(B 1 -B 0 ) 0 t 1 f 1

22 Weight of undecided clauses satisfied by x i = true Remark: This is the algorithm proposed independently by BFNS 12 and vz 11 B i B i Weight of undecided clauses unsatisfied by x i = true B i-1 LB i-1 LB i LB i UB i UB i UB i-1 Algorithm: if t i < 0, set x i to false if f < 0, set x i i to true else, set x i to true with probability t i t i +f i (B i -B i-1 )+(B i -B i-1 ) 0 t i f i

23 Example Initalize: LB = 0 UB = 6 Step 1: Clause Weight x 1 2 x 1 x 2 1 x 2 x 3 3 t 1 = 1 2 LB + UB = ( 2) = 1 2 f 1 = 1 2 LB + UB = = 1 Set x 1 to false

24 Example Step 2: Clause Weight x 1 2 x 1 x 2 1 x 2 x 3 3 t 2 = 1 2 LB + UB = = 1 2 f 2 = 1 2 LB + UB = ( 1) = 1 Set x 2 to true with probability 1/3 and to false with probability 2/3

25 Example Clause x 1 2 x 1 x 2 1 x 2 x 3 3 Algorithm s solution: x 1 = false x 2 = true w.p. 1/3 and false w.p. 2/3 x 3 = true Expected weight of satisfied clauses: Weight

26 Different Languages Bill, Baruch, and I would say: Alan would say: Let G be a graph... Let A be a matrix... And we would be talking about the same thing!

27 Relating Algorithm to Optimum Let x 1, x 2,, x n be an optimal truth assignment Let OPT i = weight of clauses satisfied if setting x 1,, x i as the algorithm does, and x i+1 =,, x n = x n x i+1 Key Lemma: E B i B i 1 E[OPT i 1 OPT i ]

28 Let x 1, x 2,, x n an optimal truth assignment OPT Let OPT i = weight of clauses B 1 satisfied if setting OPT 1 x 1,, x n as the algorithm does, and x i+1 =,, x n = x n LB 0 B 0 UB 0 x i+1 Key Lemma: E B i B i 1 E[OPT i 1 OPT i ]

29 Let an optimal truth assignment OPT n = B n = weight of ALG s solution OPT Let = weight of clauses satisfied B 1 if setting OPT as the 1 algorithm does, and LB 0 B 0 UB 0 Key Lemma: B 0 ½ OPT ½ (OPT-B 0 ) Conclusion: expected weight of ALG s solution is E B n B OPT B 0 = 1 2 OPT + B OPT

30 Relating Algorithm to Optimum Weight of undecided clauses satisfied by x i = true B i B i Weight of undecided clauses unsatisfied by x i = true B i-1 LB i-1 LB i LB i UB i UB i UB i-1 Suppose x i = true If algorithm sets x i to true, B i B i 1 = t i OPT i 1 OPT i = 0 If algorithm sets x i to false, B i B i 1 = f i Want to show: OPT i 1 OPT i LB i LB i 1 + UB i UB i 1 = 2 B i B i 1 = 2t i Key Lemma: E B i B i 1 E[OPT i 1 OPT i ]

31 Relating Algorithm to Optimum Want to show: Key Lemma: E B i B i 1 E[OPT i 1 OPT i ] Know: If algorithm sets x i to true, B i B i 1 = t i OPT i 1 OPT i = 0 If algorithm sets x i to false, B i B i 1 = f i OPT i 1 OPT i 2t i Case 1: f i < 0 (algorithm sets x i to true): E B i B i 1 = t i > 0 = E OPT i 1 OPT i Case 2: t i < 0 (algorithm sets x i to false): E B i B i 1 = f i > 0 > 2t i E OPT i 1 OPT i

32 Relating Algorithm to Optimum Want to show: Key Lemma: E B i B i 1 E[OPT i 1 OPT i ] Know: If algorithm sets x i to true, B i B i 1 = t i OPT i 1 OPT i = 0 If algorithm sets x i to false, Equal to B i B i 1 = f (t i i f OPT i ) 2 +2t i 1 i f OPT i i 2t i Case 3: t i 0, f i 0 (algorithm sets x i to true w.p. t i ti +f i ): E B i B i 1 = t i t i t i +f i + f i E OPT i 1 OPT i 0 f i = 1 t i +f i t i t i + f i + 2t i (t t i +f i2 + f i2 ) i f i = 1 (2t t i + f i t i + f i f i ) i

33 Hi David, After seeing your , the very next thing I did this morning was to read a paper I'd earmarked from the end of the day yesterday: Walter Gander, Gene H. Golub, Urs von Matt "A constrained eigenvalue problem" Linear Algebra and its Applications, vol , March April 1989, Pages "Special Issue Dedicated to Alan J. Hoffman On The Occasion Of His 65th Birthday" The table of contents of that special issue: Citations for papers in this issue:.. Johan Ugander

34 Question Is there a simple combinatorial deterministic ¾-approximation algorithm?

35 Deterministic variant?? Greedily maximizing B i is not good enough: Clause Weight x 1 1 x1 x 2 2+ x 2 1 x 2 x x n 1 1 x n 1 x n 2+ Optimal assignment sets all variables to true OPT = (n-1)(3+ ) Greedily increasing B i sets variables x 1,, x n 1 to false GREEDY= (n-1)(2+ )

36 A negative result Poloczek 11: No deterministic priority algorithm can be a ¾ -approximation algorithm, using scheme introduced by Borodin, Nielsen, and Rackoff 03. Algorithm makes one pass over the variables and sets them. Only looks at weights of clauses in which current variable appears positively and negatively (not at the other variables in such clauses). Restricted in information used to choose next variable to set.

37 But It is possible with a two-pass algorithm (Joint work with Ola Svensson). First pass: Set variables x i fractionally (i.e. probability that x i true), so that E W 3 OPT. 4 Second pass: Use method of conditional expectations to get deterministic solution of value at least as much.

38 Buchbinder et al. s approach expected Keep two bounds on the fractional solution Lower bound LB = weight of clauses already satisfied Upper bound UB = weight of clauses not yet unsatisfied expected Greedy can focus on two things: maximize LB, maximize UB, but either choice has bad examples expected Key idea: make choices to increase B = ½ (LB+UB)

39 As before Let t i be (expected) increase in bound B i 1 if we set x i true; f i be (expected) increase in bound if we set x i false. Algorithm: For i 1 to n if t i < 0, set x i to 0 if f < 0, set x i i to 1 else, set x i to t i t i +f i For i 1 to n If E W X i 1, x i true E W X i 1, x i false, set x i true Else set x i false

40 Analysis Proof that after the first pass E W 3 OPT 4 is identical to before. Proof that final solution output has value at least E W 3 4 conditional expectation. OPT is via method of

41 Conclusion We show this two-pass idea works for other problems as well (e.g. deterministic ½- approximation algorithm for MAX DICUT). Can we characterize the problems for which it does work?

42 Thank you for your attention and Happy Birthday Alan!