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A constructive version of the Lov ász Local Lemma

Discover the Lovász Local Lemma, a powerful tool for probability theory and its constructive versions, history, and algorithmic applications. Learn how this lemma proves existence of satisfying solutions even when bad events happen. Find out about recent developments and application in CNF satisfiability.

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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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