1. Theory of Computational Complexity Probability and Computing Chapter 6.6-6.9 Hikaru Inada Iwama and Ito lab M1

2. The conditional Expectation Inequality Theorem 6.10 Let, where each is a 0-1 random variable. Then need not be independent

3. The conditional Expectation Inequality Proof If X>0 Otherwise If X>0 Otherwise XY is Bernoulli random variable. Then

4. The conditional Expectation Inequality

5. Use Jensen’s inequality (chapter 2)

6. Theorem 6.8 In suppose that, where Then, for any and for sufficiently large n, the probability that a random graph chosen from has a clique of four or more vertices is less than. Similarly, if then, for sufficiently large n, the probability that a random graph chosen from does not have a clique with four or more vertices is less than. The conditional Expectation Inequality We can use Theorem 6.10 to give an alternate proof of Theorem 6.8

7. The conditional Expectation Inequality Let, where is 1 if is a 4-clique and 0 otherwise. Theorem 6.10 applies, Let be an enumeration of all the subsets of four vertices in G. For each, We have

8. The conditional Expectation Inequality (1) (2) (3) (4)

9. The conditional Expectation Inequality (1) There are sets of vertices. Each is 1 with probability. (2) There are sets of vertices. Each is 1 with probability.

10. The conditional Expectation Inequality (3) There are sets of vertices. Each is 1 with probability. (4) There are sets of vertices. Each is 1 with probability.

11. The conditional Expectation Inequality When, this value approaches 1 as n grows large.

12. The Lovasz Local Lemma Let be a set of (bad) events in some probability space. We want to show that there is an element in the sample space that is not included in any of (bad) events. If the events are mutually independent then so are.(chapter 1) If for all then But mutually independence is too much to ask for many arguments.

13. The Lovasz Local Lemma The Lovasz local lemma considers the case where the n events are not mutually independent but the dependency is limited. We say that an event is mutually independent of the events if, for any subset, The dependency between events can be represented in terms of a dependency graph.

14. The Lovasz Local Lemma Definition 6.1 A dependency graph for a set of events is a graph such that and, for, event is mutually independent of the events 3 2 1 4 5 6 7 8 is mutually independent of

15. The Lovasz Local Lemma We discuss first a special case, the symmetric version of the Lovasz local lemma. Theorem 6.10[Lovasz local lemma] Let be a set of events, and assume that the following hold: 1.For all ; 2.The degree of the dependency graph given by is bounded by d ; 3. ; Then,

16. The Lovasz Local Lemma proof. We prove by induction on that, if, then for all we have 1) To perform the inductive step, we show that.

17. Proof of a)s=1 b)s>1 without loss of generality. Then (induction hypothesis)

18. The Lovasz Local Lemma 2) s>0 let and a) If is mutually independent of the events b) the case similarly define

19. The Lovasz Local Lemma Numerator: Denominator: Then

20. Numerator: Denominator:

21. Recall that And the degree of the dependency graph is bounded by d. Because we can apply the induction hypothesis

23. Theorem 6.10[Lovasz local lemma] Let be a set of events, and assume that the following hold: 1.For all ; 2.The degree of the dependency graph given by is bounded by d; 3. ; Then,

24. Application: Edge-Disjoint Paths Given: n pairs of users network Each pair can choose a path from a collection of m paths. We want to show that, if the possible paths do not share too many edges, there is a way to choose n edge-disjoint paths connecting the n pairs.

25. Application: Edge-Disjoint Paths 1 1 2 2 33 4 4 5 5

26. Theorem 6.12 If any path in shares edges with no more than k paths in where and, then there is a way to choose n edge-disjoint paths connecting the n pairs. proof Consider the probability space defined by each pair choosing a path independently and uniformly at random from its set of m paths. : the event that the paths chosen by pairs and share at least one edge.

27. Application: Edge-Disjoint Paths Consider the dependency graph Event is independent of all events when and.Then the degree of dependency graph is the number of events and All of the conditions of Lovasz local lemma are satisfied

28. Application: Edge-Disjoint Paths Hence, there is a choice of paths such that the n paths are edge-disjoint.

29. Application: Satisfiability k-SAT problem, the formula is restricted so that each clause has exactly k literals. We assume that no clause contains both a literal and its negation, as these clauses are trivial. We prove that any k-SAT formula in which no variable appears in too many clauses has a satisfying assignment. Theorem 6.13 If no variable in a k-SAT formula appears in more than clauses, then the formula has a satisfying assignment.

30. Application: Satisfiability Proof Consider the probability space defined by giving a random assignment to the variables. : the number of clauses : the event that the ith clause is not satisfied by the random assignment. Each of the k variables in clause can appear in no more than clauses.

31. So we can conclude that Hence there is a satisfying assignment for the formula Application: Satisfiability

32. Explicit Construction Using Local Lemma The Local lemma says But this probability might be too small for an algorithm that is based on simple sampling. The existential result of the Lovasz local lemma can be used to derive efficient construction algorithm. All the known algorithms are based on a common two- phase scheme.

33. Explicit Construction Using Local Lemma First phase: a subset of the variables of the problem are assigned random values. The subset of variables is chosen so that: 1) Using the local lemma, one can show that the random partial solution fixed in the first phase can be extended to a full of solution of the problem without modifying any of the variables fixed in the first phase. 2) The dependency graph H between events defined by the variables deferred to the second phase has, with high probability, only small connected components.

34. Explicit Construction Using Local Lemma second phase: an exhaustive search for remaining variables

35. Application: A Satisfiability Algorithm Consider a k-SAT formula F, with k an even constant such that each variable appears in no more than clauses for some constant determined in the proof. We can obtain a polynomial time algorithm with this condition. F has variables,,and m clauses. We define a dangerous clause. We call a clause dangerous if both the following conditions hold 1. literals of the clause have been fixed 2. is not yet satisfied

36. Application: A Satisfiability Algorithm Phase 1: Consider the variable sequentially. If is not in a dangerous clause, assign it independently and uniformly at random a value in {0,1} Phase 2: We use exhaustive search in order to assign value to the deferred variables deferred variable: the variable that was not assigned a value in the first phase

37. Application: A Satisfiability Algorithm We want to show that: 1.the partial solution computed in phase 1 can be extended to a full satisfying assignment of F 2.with high probability, the exhaustive search in phase 2 is completed in time that is polynomial in m

38. Application: A Satisfiability Algorithm Lemma 6.14: There is an assignment of value to the deferred variables such that all the surviving clauses are satisfied surviving clause: the clause that was not satisfied in phase 1 Proof: : a graph on m nodes, if and only if is the dependency graph for the original problem. : such that (a): if and only if is surviving clause (b): if and only if and share a deferred variable

39. Application: A Satisfiability Algorithm Consider the probability space defined by assigning a random value in {0,1} independently to each deferred variable. By phase 1 and this, we assign all the variables. : the event surviving clause is not satisfied by this assignment The graph is the dependency graph for this set of events.

40. Application: A Satisfiability Algorithm (surviving clause has at least k/2 deferred variables) (a variable appears no more than T clauses) For a sufficiently small constant α > 0 By the Lovasz local lemma, there is an assignment for the deferred variables that satisfies the formula.

41. Application: A Satisfiability Algorithm We want to show that: 1.the partial solution computed in phase 1 can be extended to a full satisfying assignment of F 2.with high probability, the exhaustive search in phase 2 is completed in time that is polynomial in m

42. Application: A Satisfiability Algorithm second phase: an exhaustive search for remaining variables We show that the size of each component in H’ is Then, the number of deferred variables in each subformula is If we can show that, an exhaustive search can be done in polynomial time.

43. Application: A Satisfiability Algorithm Lemma 6.15: all connected components in H’ are of size with probability R R has r vertices

44. is dangerous clause A surviving clause is either a dangerous clause or it share at least one deferred variable with a dangerous clause. The probability that a given clause is dangerous is The probability that a given clause survives is bounded by

45. We define a 4-tree S of a connected component R in H as follows: 1.S is a rooted tree 2.Any two nodes in S are at distance at least 4 in H 3.There can be an edge in S only between two nodes with distance exactly 4 between them in H 4.Any node of R is either in S or is at distance 3 or less from a node in S

46. R S S

47. R S The event that a node u in a 4- tree survives and the event that another node v in a 4- tree survive are actually independent u v

48. Application: A Satisfiability Algorithm For any 4-tree S, the probability that the nodes in the 4-tree survive is at most We consider a maximal 4-tree S of a connected component. Since the degree of the dependency graph is bounded by d, there are no more than nodes at distance 3 or less from any given vertex.

49. Application: A Satisfiability Algorithm Therefore a maximal 4-tree of R must have at least Otherwise there must be a vertex of distance at least 4 from all vertices in S. Next, we calculate the probability of the event that 4-tree of H of size survives in H’. Since a surviving connected component R would have a maximal 4-tree of size, the absence of such a 4-tree implies the absence of such a component.

50. Application: A Satisfiability Algorithm Count the number of 4-trees of size in H. we can choose the root of 4-tree in m ways A tree with root v is uniquely defined by an Eulerian tour that starts and ends at v and traverses each edge of the tree twice, once in each direction.

51. S v Length 4 in H Each vertex can choose nodes at each step. Therefore the number of 4-tree of size is

52. Application: A Satisfiability Algorithm The probability that a 4-tree survives in H’ is at most The probability that at least one 4-tree of size survives in H’ is at most

53. Application: A Satisfiability Algorithm Theorem 6.16: Consider a k-SAT formula with m clause, where k is an even constant and each variable appears in up to clauses for a sufficiently small constant.Then there is an algorithm that find a satisfying assignment for the formula in expected time that is polynomial in m.

54. Theorem 6.10 Let be a set of events, and assume that the following hold: 1.For all ; 2.The degree of the dependency graph given by is bounded by d; 3. ; Then, Theorem 6.17: Let be a set of events in arbitrary probability space and let be the dependency graph for these events. Assume there exist such that, for all Then,

55. . We prove by induction on that, if, then for all we have 1) To perform the inductive step, we show that. Lovasz Local Lemma: The General Case proof

56. Lovasz Local Lemma: The General Case 2) s>0 let and a) If is mutually independent of the events b) the case similarly define

57. Lovasz Local Lemma: The General Case Numerator: Denominator: Then

58. Numerator: Denominator:

60. Lovasz Local Lemma: The General Case