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 of the lov sz local lemma

A constructive version ofthe Lovász Local Lemma

Robin Moser, ETH, Zürich

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


A constructive version of the lov sz local lemma

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


A constructive version of the lov sz local lemma

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


Lov sz local lemma

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


Lov sz local lemma1

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


Lov sz local lemma2

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)


Lov sz local lemma3

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.


Lov sz local lemma4

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.


Lov sz local lemma5

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


Lov sz local lemma6

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

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 lemma1

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 or assignment avoiding bad events

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 events almost as simple and 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

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 ideas1

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

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 tree1

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 tree2

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 tree3

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 tree4

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 tree5

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 tree6

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 tree7

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 tree8

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

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


A constructive version of the lov sz local lemma

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

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 miracle1

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


Extensions parallel deterministic lopsided

Extensionsparallel – deterministic - lopsided


Extensions parallel deterministic lopsided1

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


Extensions parallel deterministic lopsided2

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


Extensions parallel deterministic lopsided3

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


Extensions parallel deterministic lopsided4

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


Extensions parallel deterministic lopsided5

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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