Lecture 10: Lovász Local Lemma. Lovász Local Lemma

Transcription

1 Lecture 10:

2 Introduction Let A 1,..., A n be indicator variables for bad events in an experiment Suppose P [A i ] p We want to avoid all the bad events If P [ A 1 A n ] > 0, then there exists a way to avoid all the bad events simultaneously Suppose, the event A i is independent of all other events Then, it is easy to see that: P [ A 1 A n ] (1 p) n > 0 will help us conclude the same even in presence of limited independence

3 The Statement Theorem () Let (A 1,..., A n ) be a set of bad event. For each A i, where i [n], we have P [A i ] p and each event A i depends on at most d other bad events. If ep() 1, then ( P [ A 1 A n ] 1 1 ) n > 0 The condition is also stated sometimes as 4pd 1, instead of ep() 1.

4 Application: k-sat I Let Φ be a k-sat formula such that each variable occurs in at most 2 k 2 /k different clauses Experiment: X i be an independent uniform random variable that assigns the variable x i a value from {true, false} Bad Event: For the j-th clause we have the bad event A j that is the indicator variable for the bad event: The j-th clause is not satisfied Probability of Bad Event: For any j, note that P [ A j ] 1 2 k, Because there is at most one assignment of variables to make the clause false.

5 Application: k-sat II Dependence: Note that the j-th clause has k literals, and each variable of the literal occurs in 2 k 2 /k different clauses. So, the clause A j can depend on at most d = 2 k 2 different bad events Conclusion: Note that 4pd = 1, so implies that there exists an assignment that satisfies all the clauses in the formula simultaneously Observation: The probability p of each bad events does not depend on the overall problem instant size (i.e., the number of variables).

6 Application: Vertex Coloring Let G be a graph with degree at most Experiment: X v be the random variable that represents the color of the vertex v. Let X v be a independent and uniformly random over the set {1,..., C} Bad Event: For every edge e, we have a bad event A e that is the indicator variable for both its vertices receiving identical color Probability of the Bad Event: Note that P [A e ] = 1 C Dependence: Note that the event A e does not depend on any other event A e if the edges do not share a vertex. So, the event A e depends on at most 2( 1) other bad events Conclusion: A valid coloring exists if 4pd 1, i.e., C 8( 1)

7 Application: Vertex Coloring (Bad Bound) Let G be a graph with degree at most Experiment: X v be the random variable that represents the color of the vertex v. Let X v be a independent and uniformly random over the set {1,..., C} Bad Event: For every edge v, we have a bad event A v that is the indicator variable for one of the neighbors of v receiving the same color as v Probability of the Bad Event: Note that ( ) P [A v ] C Dependence: Note that the event A v does not depend on any other event A v if {v} N(v) does not intersect with {v } N(v ). So, the event A e depends on at most + ( 1) = 2 other bad events Conclusion: A valid coloring exists if 4pd 1, i.e., C???

8 Proof of Claim Let S {1,..., n}, then we have: P A i A k 1 k S Assuming this claim, it is easy to prove the. n n P A i = P A i A k i=1 i=1 n i=1 k<i ( 1 1 ) = ( 1 1 ) n > 0

9 Proof of the Claim I We will proceed by induction on S Base Case: If S = 0, then the claim holds, because: P A i A k 1 = P [A i ] p e() 1 k S Assume that for all S S < t, the claim holds We will prove the claim for S = t. Suppose D i be the set of all j such that the bad event A i depends on the bad event A j Easy Case. Suppose S D i =. This case is easy, because P A i A k 1 = P [A i ] p e() 1 k S

10 Proof of the Claim II Remaining Case. Suppose S D i. P A i A k = P A i A k, A k k S k D i k S\D i P [A i, ] A k Di k A k S\Di k = [ ] P A k Di k A k S\Di k ] P [A i A k S\Di k [ ] P A k Di k A k S\Di k P [A i ] = [ ] P A k Di k A k S\Di k

11 Proof of the Claim III Suppose D i = {i 1,..., i z } Using chain rule, we can write the denominator P A k A k k D i k S\D i as follows z P A il l=1 A k, k S\D i A k k {i 1,...,i l 1 }

12 Proof of the Claim Note that each probability term is condition on < t bad events. So, we can apply the induction hypothesis. We get P z ( A k A k 1 1 ) k D i k S\D i l=1 ( = 1 1 ) z ( 1 1 ) d 1 e Now, let us return to our original expression P A i A k P [A i ] [ ] P A k Di k A k S\Di k k S ep [A i ] 1 IV

13 Proof of the Claim V This completes the proof by induction We will prove a more general result in the next lecture