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# Class Note for CMPSCI 711 at UMass(6)

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Date Created: 02/06/15

CMPSCI 711 Really Advanced Algorithms Lecture 8 Probabilistic Method and Lova39sz Local Lemma Andrew McGregor Last Compiied February 25 2009 Outline Probabilistic Method Probabilistic Method It39s obvious that l X 2 r gt O or E X 2 r implies that the event X 2 r can happen But it39s very powerful Probabilistic Method It39s obvious that l X 2 r gt O or E X 2 r implies that the event X 2 r can happen But it39s very powerful Examples we39ve already seen gt Any graph has a cut of size at least m2 gt For any collection of m clauses it is possible to satisfy 17 2 km of the clauses if each clause has at least k literals gt For any collection of subsets A17 An C it is possible to partition n into A and B such that maxllAi BlilAi Cll 4annn i6n k SAT gt Input A CNF formula consisting of m clauses in n Boolean variables X17 7Xn eg X1 VXQ V23 A A X9 VX10 VX21 where 217 X gt Problem Is there a satisfying assignment of the formula k SAT gt Input A CNF formula consisting of m clauses in n Boolean variables X17 7Xn eg X1 VXQV3 AAX9VX10VX21 where 217 X gt Problem Is there a satisfying assignment of the formula Theorem If each clause contains exactly k 2 3 literals and each variable appears complemented or un complemented in at most 2quot50 clauses then the formula can always be satisfied Outline Lova39sz Local Lemma Warm up gt Suppose we have n bad events 31 Bn Warm up gt Suppose we have n bad events 31 B gt To show that it is possible for no bad events to occur it is sufficient to find some random process where P ielnl i gt 0 Warm up gt Suppose we have n bad events 31 B gt To show that it is possible for no bad events to occur it is sufficient to find some random process where P ielnl i gt 0 gt Eg if B are independent and maxiP B lt 1 we are good Warm up V Suppose we have n bad events 31 B V To show that it is possible for no bad events to occur it is sufficient to find some random process where P ielnl i gt 0 V Eg if B are independent and maxiP B lt 1 we are good Eg if ZiPB lt 1 we are good V Warm up V Suppose we have n bad events 31 B V To show that it is possible for no bad events to occur it is sufficient to find some random process where P ielnl i gt 0 V Eg if B are independent and maxiP B lt 1 we are good Eg if ZiPB lt 1 we are good What if probabilities aren39t tiny and events not independent V V Lov sz Local Lemma Theorem Consider events 81 B with B independent ofBj j 39i Suppose that there exist X E 07 1 fori E n such that WBIKXI H 1939 jeri Then 1 ielani 2 Hidnl ixi Where G Bi Lov sz Local Lemma Theorem Consider events 81 B with B independent ofBj j 39i Suppose that there exist X E 07 1 fori E n such that WEI SXI H 1939 jeri Then 1 ielani 2 Hidnl ixi Where G 3 Corollary Let 817Bn be events With 1PB p and B independent of all but at most d other events then P ielani gt O ifepd 1 g 1 Lov sz Local Lemma Theorem Consider events 81 B with B independent ofBj j 39i Suppose that there exist X E 07 1 fori E n such that LPlBil SXI H 1939 jeri Then l ielanil 2 Hidn l ixi Where G 3 Corollary Let 817Bn be events With lP B p and B independent of all but at most d other events then l ielanil gt O ifepd 1 g 1 Proof Use Lova39sz Local Lemma with X 1d 1 for all i E D Proof of Lov sz Local Lemma gt Sufficient to prove 1PB jes Gj X for any 5 C n7 139 5 E i6nGi 1P311P3251Dm1 13HmidnaBID Proof of Lov sz Local Lemma gt Sufficient to prove 1PBi jes Gj X for any 5 C n7 139 5 E i6nGii 1P311P32i51 1 1 Bni misinai Bii gt Proof by induction on k 5 Base case k O is immediate Proof of Lov sz Local Lemma gt Sufficient to prove 1PBi jes Gj X for any 5 C n7 139 5 ED i6nGii 1 P311P32i61i41 Bni iqnq Bii gt Proof by induction on k 5 Base case k O is immediate gt Inductive Step Let 1 j E 5 j E 39i and 2 1 LPBi N 165151 i miss 51 P 3quot 3965 6quot P in65161 i 1652 Gj Proof of Lovasz Local Lemma gt Sufficient to prove 1PBi jes Gj X for any 5 C n7 139 5 E i6nGii 1P311P32i51 1 1 Bni lanai Bii gt Proof by induction on k 5 Base case k O is immediate gt Inductive Step Let 1 j E 5 j E 39i and 2 1 LPBi N 165151 i 1652 51 E jesl G i 1652 61 gt Numerator By independence assumptions LPiiBi N 165151 i 1652 51 S LP Bii 1652 51 LP Bi 1P Bii jes 5 Proof of Lovasz Local Lemma gt Sufficient to prove 1PBi jes Gj X for any 5 C n7 139 5 E i6nGii 1P311P32i51 1 1 Bni lanai Bii gt Proof by induction on k 5 Base case k O is immediate gt Inductive Step Let 1 j E 5 j E 39i and 2 1 LPBi mes Gquot i 39es 639 1PBi m 5 G m K J Pimjesl G i 1652 61 gt Numerator By independence assumptions LPiiBi N 165151 i 1652 51 S LP Bii 1652 51 LP Bi gt Denominator Let 1 j1j amp Tk 52 U j1 jk r71 PGj1 Gjri jesg H 7 PBjk1i jeTk k0 2 H 1939 EpiBiiXi jeri Return to k SAT Theorem If each clause contains exactly k 2 3 literals and each variable appears complemented or un complemented in at most 2quot50 clauses then the formula can always be satisfied Return to k SAT Theorem If each clause contains exactly k 2 3 literals and each variable appears complemented or un complemented in at most 2quot50 clauses then the formula can always be satisfied Proof gt Pick each X value uniformly and independently from 01 Return to k SAT Theorem If each clause contains exactly k 2 3 literals and each variable appears complemented or un complemented in at most 2quot50 clauses then the formula can always be satisfied Proof gt Pick each X value uniformly and independently from 01 gt Let BJ be the event that the j th clause is unsatisfied Return to k SAT Theorem If each clause contains exactly k 2 3 literals and each variable appears complemented or un complemented in at most 2quot50 clauses then the formula can always be satisfied Proof gt Pick each X value uniformly and independently from 01 gt Let BJ be the event that the j th clause is unsatisfied gt By previous analysis p lP BJ 24 Return to k SAT Theorem If each clause contains exactly k 2 3 literals and each variable appears complemented or un complemented in at most 2quot50 clauses then the formula can always be satisfied Proof gt Pick each X value uniformly and independently from 01 gt Let BJ be the event that the j th clause is unsatisfied gt By previous analysis p lP BJ 24 gt BJ is independent of all but at most d k2k50 71 other events Return to k SAT Theorem If each clause contains exactly k 2 3 literals and each variable appears complemented or un complemented in at most 2quot50 clauses then the formula can always be satisfied Proof gt Pick each X value uniformly and independently from 01 gt Let BJ be the event that the j th clause is unsatisfied gt By previous analysis p lP BJ 24 gt BJ is independent of all but at most d k2k50 71 other events gt Since epd l 1 g 1 for k 2 3 using LLL l1 ielani gt O

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