A constructive version of the lov sz local lemma
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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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A constructive version ofthe Lovász Local Lemma

Robin Moser, ETH, Zürich

Gábor Tardos, Rényi Institute, Budapest and Simon Fraser University, Vancouver


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


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


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


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


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)


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.


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.


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


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 .


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).


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).


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.


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)


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


Extensionsparallel – deterministic - lopsided


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


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


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


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


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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