CS 683: Advanced Design and Analysis of Algorithms

Transcription

1 CS 683: Advaced Desig ad Aalysis of Algorithms Lecture 6, February 1, 2008 Lecturer: Joh Hopcroft Scribes: Shaomei Wu, Etha Feldma February 7, Threshold for k CNF Satisfiability I the previous lecture, we discussed the threshold for 3 CNF problem ad claimed that for ay k CNF, there is a threshold r k = 2 k l 2 for formula satisfiability. Today s lecture gives a more formal proof for this statemet. Theorem 1 Give a k CNF formula with c clauses(c is a fixed umber) ad variables, assume c = r k, the r k = 2 k l 2 is a threshold for the give formula to be satisfiable. Proof: (1) Pick a radom assigmet S for variables. (2) For ay clause (with k literals), the probability that it is false uder assigmet S is 1 2 k, hece P rob(clause is true uder S) = k. (3) Give c clauses, the formula is true with probability P rob(formula is true) = (1 1 2 k )c (*The clauses are cosidered to be geerated idepedetly.) Whe r k = 2 k l 2 ad c = r k, P rob(formula is true) = (1 1 2 k )c = (1 1 2 k )2k l 2. The secod part of this lecture o graph coectivity was re-stated ad corrected i February 3 s lecture, hece we reduced the amout of cotet i that part. Please refer to later otes for more clear proof ad explaatios 1

2 For large k, (1 1 2 k ) 2k e 1, hece P rob(formula is true) = (1 1 2 k )2k l 2 e (l 2) = 1 2. I other words, the probability that the give formula is satisfied by a radom assigmet of variables is 1 2. Let s use a set of idicator radom variables I 1, I 2,..., I 2 to represet the evets of 1st, 2d,..., ad 2 t h assigmets satisfyig the give formula, respectively. Kowig that each assigmet has the same probability of makig the formula true, the expected values for all I i s are same as E(I 1 ) = E(I 2 ) =... = E(I 2 ) = = 1 2. Therefore, the expected umber of assigmets that satisfy the give formula would be E(I) = E(I 1 ) + E(I 2 ) E(I 2 ) = = 1 Figure 1: Probability of k-cnf beig satisfied by a radom assigmet. As proved i previous lectures, the umber of satisfyig assigemets for the give formula is a mootoe property. Icreasig c will itroduce a decrease of the probability that the formula is true uder a radom assigmet, which meas, whe c is greater tha r k, the expected umber of satisfyig assigmets will be less tha 1. As show i Figure 1, r k = 2 k l 2 is the place where the upperboud of threshold for k-cnf appears. 2

3 For k = 3, Chao ad Fraco [1] preseted a heuristic algorithm based o Uit-Clause rule ad proved that: whe r k 2k k, the probability of the heuristic fidig a solutio is approachig 1; while whe r k 2 k l 2, the probability is approachig 0. For example, assumig there are 10 variables appearig i a 3-CNF formula: if the umber of clauses is below 26, the heuristic algorithm ca almost always fid the satisfyig assigmet; if the umber of clauses is above 55, it would be early impossible for the heuristic algorithm to geerate a solutio. However, the situatio betwee r k = 2k k ad r k = 2 k l 2 is still uclear to us, ad it is possible that there is a secod-momet threshold existig i this rage. Oe guess ca be made by observig the chage of solutio space accordig to r k : startig from r k = 0, the solutio space is a fully-coected graph; alog with the icrease of r k, the solutio space will fall apart ad form a set of coected compoets; after passig the threshold of r k = 2 k l 2, the solutio space becomes very sparse, cosists of isolated vertices. Ivestigatig the satisfiability of k-cnf from the solutio space view brigs us back to the coectivity problem of graphs, which will be discussed i the followig sectio. 2 Whe does a graph become coected? Give a radom G(, p), whe does it become coected? states that G(, p) ca be: 1. G(, p) is coected; 2. G(, p) cosists of isolated vertices; There are three 3. G(, p) has o isolated vertex, but is ot coected either (It is ulikely for this case to happe). 2.1 Disappearace of isolated vertices Specific vertex is isolated with probability (1 p) 1. Threshold at p = log +c Note: P rob(o isolated vertices) was wrog for this lecture because the P rob(isolated vertex) is ot idepedet across all vertices. See lecture otes for 01/25/2008. If isolated vertices are the last to disappear before the graph becomes coected, the the probability that the graph is coected is e e c. 2.2 Disappearace of compoets of size 2 Give 2 vertices u, v coected by a edge. Threshold at p = log +c. 3

4 2 {}}{ u v P rob(compoet of size 2) = (1 p) 2 4 ( 1 log + c 2(log +c) = e = e 2log e 2c = e 2c 2 ) 2 } {{ } 2 Number of coected vertices log. Note: P rob(o compoets of size 2) was wrog for this lecture because the P rob(compoet of size 2) is ot idepedet across coected odes. See lecture otes for 02/04/ Graphs with Giat Compoets p = d, d > 1 Algorithm for traversig coected compoets: 1. Select a vertex v 2. Mark all vertices as udiscovered ad uexplored 3. Mark v as discovered 4. While discovered but uexplored vertex, select oe ad 5. add all adjacet vertices to discovered ad mark the selected vertex explored Let d = 2 for this example. S = total # of discovered vertices. t = iteratio #. F = expected # of vertices discovered per iteratio (frotier). Graph: 4

5 F = S t # of discovered vertices o each iteratio time t Iitially, the algorithm discovers 2 vertex per iteratio (expected degree of each vertex = 2, so o each iteratio we expect to fid 2 more vertices), but also t is icremeted so iitially we expect the lie F to have a slope 1. F is a expected value, so there will be some variatio from the give curve (idicated by the lighter lies above.) If F = 0, the frotier is empty so we just foud the last vertex of a coected compoet, so there exists a coected compoet of size t. As the graph shows, we expect F to be ear 0 oly durig the first few iteratios ad the ear (the giat coected compoet). Differetial Equatio: Solvig for S: ds (1 dt = d s ) S = (1 e d t) Refereces [1] M. T. Chao ad J. Fraco. Probabilistic aalysis of a geeralizatio of the uit clause selectio heuristic for the k-satisfiability problem. Iformatio Scieces, 51: ,