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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 ofthe Lovász Local Lemma

Robin Moser, ETH, Zürich

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

A1, A2, ..., An bad events in a prob. space,

• mutually independent,

• Pr [Ai] < 1

then all of them can be avoided:

Pr [∩Ai] > 0

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

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

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

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)

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.

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.

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

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 .

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).

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.