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A constructive version of the Lov ász Local Lemma

A constructive version of the Lov ász Local Lemma. Robin Moser, ETH, Zürich Gábor Tardos, Rényi Institute, Budapest and Simon Fraser University, Vancouver. Triviality: A 1 , A 2 , ..., A n bad events in a prob. space , mutually independent , Pr [ A i ] < 1

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A constructive version of the Lov ász Local Lemma

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  1. A constructive version ofthe Lovász Local Lemma Robin Moser, ETH, Zürich Gábor Tardos, Rényi Institute, Budapest and Simon Fraser University, Vancouver

  2. Triviality: A1, A2, ..., An bad events in a prob. space, • mutually independent, • Pr [Ai] < 1 then all of them can be avoided: Pr [∩Ai] > 0

  3. Triviality: A1, A2, ..., An bad events in a prob. space, • mutually independent, • Pr [Ai] < 1 then all of them can be avoided: Pr [∩Ai] > 0 Lovász Local Lemma: • relaxed independence • smaller bound on probability • same conclusion n arbitrarily high

  4. Lovász Local Lemma size of G is arbitrary A simple form: A1, A2, ..., An bad events in a prob. space: G: graph on the vertex set {A1, A2, ..., An} • each Ai independent from set of non-neighbors • d = max degree in G • (d+1)Pr[Ai]<e-1 Pr[∩Ai] > 0

  5. Lovász Local Lemma size of G is arbitrary A simple form: A1, A2, ..., An bad events in a prob. space: G: graph on the vertex set {A1, A2, ..., An} • each Ai independent from set of non-neighbors • d = max degree in G • (d+1)Pr[Ai]<e-1 Pr[∩Ai] > 0 Simplest application: • =k-CNF: all clauses contain exactly k literals • any one clause intersects less than d = 2k/e-1 other clauses  CNF is satisfiable Eg: (xyz)(xtz)(yuw)(tuw) size of formula is arbitrary

  6. Lovász Local Lemma Asymptotically tight Shearer A simple form: A1, A2, ..., An bad events in a prob. space: G: graph on the vertex set {A1, A2, ..., An} • each Ai independent from set of non-neighbors • d = max degree in G • (d+1)Pr[Ai]<e-1 Pr[∩Ai] > 0 Simplest application: • =k-CNF: all clauses contain exactly k literals • any one clause intersects less than d = 2k/e-1 other clauses  CNF is satisfiable Eg: (xyz)(xtz)(yuw)(tuw)

  7. Lovász Local Lemma Asymptotically tight Shearer A simple form: A1, A2, ..., An bad events in a prob. space: G: graph on the vertex set {A1, A2, ..., An} • each Ai independent from set of non-neighbors • d = max degree in G • (d+1)Pr[Ai]<e-1 Pr[∩Ai] > 0 Simplest application: • =k-CNF: all clauses contain exactly k literals • any one clause intersects less than d = 2k/e-1 other clauses  CNF is satisfiable Eg: (xyz)(xtz)(yuw)(tuw) Recent: also tight Gebauer, Szabó, T.

  8. Lovász Local Lemma A simple form: A1, A2, ..., An bad events in a prob. space: G: graph on the vertex set {A1, A2, ..., An} • each Ai independent from set of non-neighbors • d = max degree in G • (d+1)Pr[Ai]<e-1 Pr[∩Ai] > 0 Simplest application: • =k-CNF: all clauses contain exactly k literals • any one clause intersects less than d = 2k/e-1 other clauses  CNF is satisfiable Eg: (xyz)(xtz)(yuw)(tuw) Original proof non-constructive. Find point in ∩Ai . Find satisfying assignment.

  9. Lovász Local Lemma A very general form: A1, A2, ..., An bad events in a prob. space: G: graph on the vertex set {A1, A2, ..., An} • each Ai independent from set of non-neighbors • x1, x2, …, xn (0,1) • Pr [Ai]≤xi  (1-xm) i~m Pr [∩Ai] > 0 A combinatorial version: • V = {v1, v2, …, vz} independent random variables • Each Ai determined by a subset vbl(Ai)  V. • Ai and Am connected in G iff vbl(Ai)∩vbl(Am)0

  10. Lovász Local Lemma A very general form: A1, A2, ..., An bad events in a prob. space: G: graph on the vertex set {A1, A2, ..., An} • each Ai independent from set of non-neighbors • x1, x2, …, xn (0,1) • Pr [Ai]≤xi  (1-xm) i~m Pr [∩Ai] > 0 A combinatorial version: • V = {v1, v2, …, vz} independent random variables • Each Ai determined by a subset vbl(Ai)  V. • Ai and Am connected in G iff vbl(Ai)∩vbl(Am)0 Original proof non-constructive. Find assignment in ∩Ai .

  11. History of constructive local lemma Finding satisfying assignment for =k-CNF, each clause intersecting at most d other • Beck 1991 d < 2k/48 • Alon d < 2k/8 • Molloy, Reed (general random variables) • Czumaj, Scheideler (uneven version of LLL) • Srinivasan d < 2k/4 • Moser d < 2k/2 , d < 2k/32 • This result:d≤ 2k/e-1 General random variables, uneven version. Applies every time the LLL applies. Simplest algorithm (randomized).

  12. History of constructive local lemma Finding satisfying assignment for =k-CNF, each clause intersecting at most d other • Beck 1991 d < 2k/48 • Alon d < 2k/8 • Molloy, Reed (general random variables) • Czumaj, Scheideler (uneven version of LLL) • Srinivasan d < 2k/4 • Moser d < 2k/2 , d < 2k/32 • This result:d≤ 2k/e-1 General random variables, uneven version. Applies every time LLL applies. Simplest algorithm (randomized).

  13. The simplest algorithm there is to find satisfying assignment orassignment avoiding bad events • Evaluate variables randomly Most clauses satisfied / most bad events avoidedbut some are not. • Re-evaluate randomly all variables involved in unsatisfied clauses / bad events not avoided. • Repeat till needed. Hope it stops fast. This algorithm was suggested by Molloy/Reed + others. Still open if works.

  14. Finding assignment avoiding bad eventsAlmost as simple – and works • Evaluate variables randomly. • Find an arbitrary single bad event not avoided. • Re-evaluate randomly all involved variables. • Repeat, till good assignment is found. Bad events: A1, A2, ..., An;reals: x1, x2, …, xn (0,1)Pr [Ai]≤xi  (1-xm)i~m THEOREMEx[# times Ai is picked] ≤ xi/(1-xi) Tight (only if Ai is isolated)

  15. Proof ideas bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D B C A D F E Variable sets of bad events

  16. Proof ideas bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? B C A D F E Variable sets of bad events

  17. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A D F E Variable sets of bad events

  18. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A D F E Variable sets of bad events

  19. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A D F E Variable sets of bad events

  20. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A B D F E Variable sets of bad events

  21. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A B D B F E Variable sets of bad events

  22. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A B D B F E C Variable sets of bad events

  23. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A B D B F E C Variable sets of bad events

  24. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A B F D B F E C Variable sets of bad events

  25. Building a witness tree bad events: A, B, C, D, E, F re-sampled in step 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 E, F, E, C, B, B, C, E, A, D accounting:why is A re-sampled in step 9? Witness-tree will explain. B C A A B F D B E F E C Variable sets of bad events

  26. Chance of a witness tree EASY: The probability of this exact witness tree to be built is ≤ Pr[C]Pr[B]Pr[E]Pr[B]Pr[F]Pr[A] B C A A B F D B E F E C Variable sets of bad events

  27. Each re-sampling of A generates different witness tree.  Ex[# times A picked for re-sampling] =  Pr [T appears as witness tree] root of T is labeled A Need (weighted) counting of labeled trees

  28. The miracle For a simple multi-type Galton-Watson process output = labeled trees with root labeled A. Pr [T is output by G-W process] ≥ Pr [T appears as witness tree] 1-xi xi

  29. The miracle For a simple multi-type Galton-Watson process output = labeled trees with root labeled A. Pr [T is output by G-W process] ≥ Pr [T appears as witness tree] T ≤ 1  T ≤ Q.E.D. 1-xi xi xi 1-xi

  30. Extensionsparallel – deterministic - lopsided

  31. Extensionsparallel – deterministic - lopsided • Evaluate variables randomly. • Find a maximal independent set of bad events not avoided. • Re-evaluate randomly all involved variables. • Repeat, till satisfying assignment is found. Bad events: A1, A2, ..., An;reals: x1, x2, …, xn (0,1)Pr[Ai]≤(1-)xi  (1-xm)i~m THEOREMEx [# cycles] = O( -1logixi /(1-xi)) (looks like logarithmic time but is)O(log2) parallel steps

  32. Extensionsparallel – deterministic - lopsided • Evaluate variables randomly. • Find a maximal independent set of bad events not avoided. • Re-evaluate randomly all involved variables. • Repeat, till satisfying assignment is found. Bad events: A1, A2, ..., An;reals: x1, x2, …, xn (0,1)Pr[Ai]≤(1-)xi  (1-xm)i~m THEOREMEx[# cycles] = O( -1logixi /(1-xi)) (looks like logarithmic time but is)O(log2) parallel steps

  33. Extensionsparallel – deterministic - lopsided Deterministic poly time derandomization if • Pr [Ai]≤(1-)xi  (1-xm)i~m • Pr [Ai | partial evaluation] computable inP • Dependency graph has constant maximum degree

  34. Extensionsparallel – deterministic - lopsided Deterministic poly time derandomization if • Pr [Ai]≤ (xi  (1-xm))1+i~m • Pr [Ai | partial evaluation] computable inP • Dependency graph has constant maximum degree Goyal, Haeupler

  35. Extensionsparallel – deterministic - lopsided Lopsided local lemma: Positive correlations don’t matter E.g.: Want to satisfy an CNF formula Two clauses are lopsidependent if one contains a variable in positive form, the other contains same variable negated. (xyz) and (xuv) are NOT lopsidependent Lovász local lemma still works. Our algorithm still works.

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