Mosers Algorithmic LLL Presented by Zhiwei Zhang Sep 6th, 2018 - - PowerPoint PPT Presentation

Moser’s Algorithmic LLL

Presented by Zhiwei Zhang Sep 6th, 2018

k-CNF

• conjunctive normal form:

k-CNF φ = C1 ∧ C2 ∧ · · · ∧ Cm

• m clauses, n Boolean variables
• each clause Ci = li1 ∨ lv2 ∨ · · · ∨ lik

is a disjunction of exact k literals

• degree d: each clause shares variables

with at most d other clauses

Lovasz Local Lemma (symmetric version)

“bad events”: B1, B2, . . . , Bn

each one happens with probability p and is dependent on at most d other events if

4pd ≤ 1

then

Pr[B1 ∧ B2 ∧ · · · ∧ Bn > 0]

“All bad events do not happen is possible!”

k-CNF

• conjunctive normal form:

k-CNF φ = C1 ∧ C2 ∧ · · · ∧ Cm

• m clauses, n Boolean variables
• each clause Ci = li1 ∨ lv2 ∨ · · · ∨ lik

is a disjunction of exact k literals

• degree d: each clause shares variables

with at most d other clauses What if ?

d ≤ 2k−2

Ci = li1 ∨ lv2 ∨ · · · ∨ lik

for each clause

p = Pr[Ci is unsatisfied] = 2−k

Bad event: Ci is unsatisfied

What if ?

d ≤ 2k−2

4pd ≤ 1

d ≤ 2k−2

p = Pr[Ci is unsatisfied] = 2−k

Bad event: Ci is unsatisfied

Ci = li1 ∨ lv2 ∨ · · · ∨ lik

for each clause

Pr[φ is satified] > 0

(LLL condition) What if ?

d ≤ 2k−2

4pd ≤ 1

(LLL condition)

d ≤ 2k−2

p = Pr[Ci is unsatisfied] = 2−k

Bad event: Ci is unsatisfied

satisfying assignment for φ

∃

Ci = li1 ∨ lv2 ∨ · · · ∨ lik

for each clause

Pr[φ is satified] > 0

What if ?

d ≤ 2k−2

Algorithmic LLL

: k-CNF of max degree d with m clauses on n variables

φ

Theorem

d ≤ 2k−2

satisfying assignment for φ

∃

Algorithmic LLL

: k-CNF of max degree d with m clauses on n variables

φ

Theorem (Moser, 2009)

d ≤ 2k−3

satisfying assignment can be found in w.h.p.

O(n + kmlogm)

Theorem

d ≤ 2k−2

satisfying assignment for φ

∃

: k-CNF of max degree d with m clauses on n variables

φ

Solve( )

φ

pick a random assignment

x1, x2, . . . , xn;

while exists unsatisfied clause C

Fix(C) Fix(C)

replace variables in C with random values while exists unsatisfied clause D overlapping with C

Fix(D)

: k-CNF of max degree d with m clauses on n variables

φ

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

Observation: In Solve(), for every clause C, Fix(C) can be called for

• nly once.

: k-CNF of max degree d with m clauses on n variables

φ

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

Observation: In Solve(), for every clause C, Fix(C) can be called for

• nly once.

# of top-level calls to Fix(C):

≤

(# of clauses)

total # of calls to Fix(C) (including recursive calls): m t

: k-CNF of max degree d with m clauses on n variables

φ

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

the recursion trees can be encoded to:

Encode the recursion trees/running log

for each recursion tree: root:

dlog2me

bits

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

the recursion trees can be encoded to: each internal node: bits

Encode the recursion trees/running log

for each recursion tree: root:

dlog2me

bits

log2d + O(1)

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

the recursion trees can be encoded to: each internal node: bits

Encode the recursion trees/running log

for each recursion tree: root:

dlog2me

bits

+

log2d + O(1)

( )

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

the sequence of random bits can be recovered from: fjnal assignment: bits + recursion trees /running log: bits

n

≤

t

dlog2me

+

log2d + O(1)

( )

Recover the sequence of random bits

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

the sequence of random bits can be recovered from: fjnal assignment: bits + recursion trees /running log: bits

n

≤

t

dlog2me

+

log2d + O(1)

( )

Recover the sequence of random bits

Observation: Fix(C) is called assignment of C is uniquely determined

Demonstration

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

φ = A ∧ B ∧ C

A = ¬x1 ∨ x2 ∨ x3

B = ¬x2 ∨ x3 ∨ x4

C = x1 ∨ x2 ∨ x4

φ(0, 1, 0, 1) = 1

Demonstration

Log start Fix(A) Fix(B) Fix(C) end

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

Solution 0 1 0 1

φ = A ∧ B ∧ C

A = ¬x1 ∨ x2 ∨ x3

B = ¬x2 ∨ x3 ∨ x4

C = x1 ∨ x2 ∨ x4

φ(0, 1, 0, 1) = 1

Random Bits Assigned in Fix()

Demonstration

Log start Fix(A) Fix(B) Fix(C) end

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

Current Assignment Random Bits Assigned in Fix() 0 1 0 1

φ = A ∧ B ∧ C

A = ¬x1 ∨ x2 ∨ x3

B = ¬x2 ∨ x3 ∨ x4

C = x1 ∨ x2 ∨ x4

φ(0, 1, 0, 1) = 1

011

Demonstration

Log start Fix(A) Fix(B) Fix(C) end

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

0 1 0 1

φ = A ∧ B ∧ C

A = ¬x1 ∨ x2 ∨ x3

B = ¬x2 ∨ x3 ∨ x4

C = x1 ∨ x2 ∨ x4

φ(0, 1, 0, 1) = 1

011 0 1 0 0 Current Assignment Random Bits Assigned in Fix()

Demonstration

Log start Fix(A) Fix(B) Fix(C) end

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

0 1 0 1

φ = A ∧ B ∧ C

A = ¬x1 ∨ x2 ∨ x3

B = ¬x2 ∨ x3 ∨ x4

C = x1 ∨ x2 ∨ x4

φ(0, 1, 0, 1) = 1

011 0 1 0 0 100 Current Assignment Random Bits Assigned in Fix()

Demonstration

Log start Fix(A) Fix(B) Fix(C) end

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

0 1 0 1

φ = A ∧ B ∧ C

A = ¬x1 ∨ x2 ∨ x3

B = ¬x2 ∨ x3 ∨ x4

C = x1 ∨ x2 ∨ x4

φ(0, 1, 0, 1) = 1

011 0 0 0 0 000 0 1 0 0 Current Assignment Random Bits Assigned in Fix()

Demonstration

Log start Fix(A) Fix(B) Fix(C) end

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

0 1 0 1

φ = A ∧ B ∧ C

A = ¬x1 ∨ x2 ∨ x3

B = ¬x2 ∨ x3 ∨ x4

C = x1 ∨ x2 ∨ x4

φ(0, 1, 0, 1) = 1

011 0 0 0 0 000 0 1 0 0 1 0 0 0 010 Current Assignment Random Bits Assigned in Fix()

Log start Fix(A) Fix(B) Fix(C) end 0 1 0 1 011 000 1 0 0 0 010 Log+ Final Solution All random bits used in Solve() and Fix()

Demonstration

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

Recover

Log start Fix(A) Fix(B) Fix(C) end 0 1 0 1 011 000 1 0 0 0 010 Log+ Final Solution All random bits used in Solve() and Fix()

Demonstration

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

Recover Encode

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

is encoded to: + bits

n

mdlog2me + t(log2d + O(1))

Log+ Final Solution: All random bits used in Solve() and Fix()

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

the sequence of random bits is encoded to: + bits

n

mdlog2me + t(log2d + O(1))

Incompressibility Theorem (Kolmogorov)

N uniform random bits cannot be encoded to less than N-l bits with probability

1 − O(2−l)

m t

≤

recursion trees total # nodes: total # of random bits:

n + tk

(assigned bits)

. . . ... . . . ...

the sequence of random bits is encoded to: bits

≤

w.h.p. +

n

mdlog2me + t(log2d + O(1))

m t

≤

recursion trees total # nodes: total # of random bits: n + tk

(assigned bits)

. . . ... . . . ...

the sequence of random bits is encoded to: bits +

n

mdlog2me + t(log2d + O(1))

+

n

mdlog2me + t(log2d + O(1))

n + tk

≤

w.h.p. when d < 2k−c

t ≤ mlog2m c − O(1)

m t

≤

recursion trees total # nodes: total # of random bits: (assigned bits)

. . . ... . . . ...

the sequence of random bits is encoded to: bits +

n

mdlog2me + t(log2d + O(1))

+

n

mdlog2me + t(log2d + O(1))

n + tk

≤

w.h.p. when d < 2k−c

t ≤ mlog2m c − O(1)

total running time(total assigned random bits):

n + tk

= O(n + kmlog2m)

Value Axis 12.5 25 37.5 50

t

+

n

mdlog2me + t(log2d + O(1))

n + tk

length of assignment random bits length of running log

n

n + mlog2m

length of bits

length of initial random assignment

Value Axis 12.5 25 37.5 50

t

+

n

mdlog2me + t(log2d + O(1))

n + tk

length of assignment random bits length of running log

n

n + mlog2m

length of bits

length of initial random assignment

almost impossible to happen! w.h.p

Value Axis 12.5 25 37.5 50

t

+

n

mdlog2me + t(log2d + O(1))

n + tk

length of assignment random bits length of running log

n

n + mlog2m

length of bits

length of initial random assignment

almost impossible to happen! w.h.p

O(n + kmlog2m)

Algorithmic LLL

: k-CNF of max degree d with m clauses on n variables

φ

Theorem (Moser, 2009)

d ≤ 2k−3

satisfying assignment can be found in w.h.p.

O(n + kmlogm)

x1, x2, . . . , xn;

φ

Solve( )

pick a random assignment

while exists unsatisfied clause C

Fix(C)

Fix(C)

replace variables in C with random values

while exists unsatisfied clause D overlapping with C

Fix(D)

Acknowledgement

• Prof. Yitong Yin, Nanjing University

Reference

A majority of pages of this slides are made using reference

• f Prof. Yin slides and materials in class

[1]A constructive Proof of the Lovasz Local Lemma, Robin A.Moser [2]https://terrytao.wordpress.com/2009/08/05/ mosers-entropy-compression-argument/ [3]http://tcs.nju.edu.cn/slides/aa2017/LLL.pdf