schöning's random walk algorithm for k-sat -...

Schöning's Random Walk Algorithm for k-SAT

time：2016.12 group：LLWJJ

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The Boolean Satisfiability K-SAT Probabilistic Algorithm Analysis

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PART ONE PART THREE

The Boolean Satisfiability

𝐴𝐴 OR 𝐵𝐵

A B 0 1

0

1

0 1

1 1

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PART ONE PART THREE


𝐴𝐴 OR 𝐵𝐵

A B 0 1

0

1

0 1

1 1 Satisfiable

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PART ONE PART THREE


¬𝐴𝐴 AND 𝐴𝐴

1

0

1

0

0

A

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PART ONE PART THREE


¬𝐴𝐴 AND 𝐴𝐴

1

0

1

0

0

A

Unsatisfiable

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PART ONE PART THREE


(𝑨𝑨𝟏𝟏 AND 𝑨𝑨𝟐𝟐) XOR (𝐴𝐴3 AND ¬𝐴𝐴4) … 𝐴𝐴50

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PART ONE PART THREE

K-SAT

Literal: (𝑥𝑥, ¬𝑥𝑥) Clause: (𝑥𝑥1 ∨¬𝑥𝑥2∨𝑥𝑥3) Conjunctive Normal Form:

• 𝑥𝑥1∨¬𝑥𝑥2 ∧ ¬𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧¬𝑥𝑥1 • …∨ … ∧ …∨ … ∧...∧ …∨ …

K ≥ 3

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PART ONE PART THREE

K-SAT

Example: 3-SAT • 𝑥𝑥1∨ 𝑥𝑥3 ∧ 𝑥𝑥2∨ 𝑥𝑥4 ∧ 𝑥𝑥1∨ 𝑥𝑥4∨ 𝑥𝑥2 ∧ 𝑥𝑥1∨ 𝑥𝑥2

2 2 3 2

1

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PART ONE

Probabilistic Algorithm

1 input: a formula in k-CNF with n variables Guess an initial assignment a∈｛0,1｝n，uniformly at

random Repeat 3n times: If the formula is satisfied by the actual assignment: stop and accept Let C be some clause not being satisfied by the actual assignment Pick one of the ≤k literals in the clause at random and flip its value in the current assignment

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state j)

1 for 1 (true) for 0 (false)

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state j) a(1) = 1000

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state j) a(1) = 1000 a(2) = 1001

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state j) a(1) = 1000 a(2) = 1001 a(3) = 1000

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state j) a(1) = 1000 a(2) = 1001 a(3) = 1000

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state j) a(1) = 1000 a(2) = 1001 a(3) = 1000 a(4) = 1100, state 0, reached √

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NP-complete Problem all known algorithms require superpolynomial time search for an "almost" optimal solution Use randomness to get a faster average running

time, and allow the algorithm to fail with some small probability

Analysis

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Suppose: existing satisfying assignment α* Success probability p

Expected independent repetitions 1𝑝𝑝

The probability that we don’t get a satisfying assignment after t repetitions is (1 − 𝑝𝑝)𝑡𝑡≤ 𝑒𝑒−𝑝𝑝𝑡𝑡

Analysis

acceptable

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j := number of different bits between α and α * := Hamming distance between α and α *

Analysis

Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state j) Reach state 0 after 3j steps a(1) = 1000 a(2) = 1001 a(3) = 1000 a(4) = 1100, state 0, reached √

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Analysis – Intuition in 3-SAT


j =2

j =1 j =2 j =1 j =0

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the initial assignment α is ramdomly selected by j P 𝑋𝑋 = 𝑗𝑗 = 𝑛𝑛

𝑗𝑗 2−𝑛𝑛


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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) Initial State: a = 0000 (state 2)


2 3 1

𝑥𝑥1, 𝑥𝑥2

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𝑥𝑥3

13

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) a(1) = 1000 (state 1)


1 2 0

𝑥𝑥2

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Example: 𝑥𝑥1∨𝑥𝑥2∨𝑥𝑥3 ∧ 𝑥𝑥2∨𝑥𝑥3 ∨𝑥𝑥4 ∧ ¬𝑥𝑥1∨¬𝑥𝑥4∨𝑥𝑥2 a* = 1100 (state 0) a(2) = 1001 (state 2)


2 3 1

𝑥𝑥2

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A clause 𝑥𝑥a∨𝑥𝑥𝑏𝑏∨𝑥𝑥c not satisfied by the assignment α At least one literal needs to be flipped Pr 𝑖𝑖 → 𝑖𝑖 − 1 ≥ 1

3, Pr (𝑖𝑖 → 𝑖𝑖 + 1) ≤ 2

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i i+1 i-1

13

13

13

13

23

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23

23

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𝑃𝑃𝑖𝑖 := possibility of reaching 0 from i, walking ∞ long 𝑃𝑃𝑖𝑖 = 1

3𝑃𝑃𝑖𝑖−1 + 2

3𝑃𝑃𝑖𝑖+1 , 𝑃𝑃0 = 1


i i+1 i-1

13

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13

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𝑃𝑃𝑖𝑖 = 13𝑃𝑃𝑖𝑖−1 + 2

3𝑃𝑃𝑖𝑖+1 (*)

𝑃𝑃𝑖𝑖 = 1 would be a solution to (*), but sadly it’s not true


i i+1 i-1

13

13

13

13

23

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23

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3𝑃𝑃𝑖𝑖+1 (*)

𝑃𝑃𝑖𝑖 = 2−𝑖𝑖 is also a solution to (*)


i i+1 i-1

13

13

13

13

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3𝑃𝑃𝑖𝑖+1 (*)

𝑃𝑃𝑖𝑖 = 2−𝑖𝑖 is a solution to (*) Proof. By Mathematical Induction, when 𝑖𝑖 ≥ 3, 𝑃𝑃𝑖𝑖 = 2−𝑖𝑖, 𝑃𝑃𝑖𝑖−1 = 2−(𝑖𝑖−1), we can get 𝑃𝑃𝑖𝑖+1 = 2−(𝑖𝑖+1) from 𝑃𝑃𝑖𝑖 = 1

3𝑃𝑃𝑖𝑖−1 + 2

3𝑃𝑃𝑖𝑖+1.

When 𝑖𝑖 = 1,2, 𝑃𝑃𝑖𝑖 = 2−𝑖𝑖 also exists. ∎


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3𝑃𝑃𝑖𝑖+1 (*)

𝑃𝑃𝑖𝑖 = 2−𝑖𝑖 is a solution to (*) ∎ In fact, that “usually” happens in 3i steps


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PART ONE

Analysis – in k-SAT

To-do: calculate 𝑃𝑃𝑗𝑗 in k-SAT “Wrong” direction: i step “Right” direction: i+j step (i≤j)

j j+1 j-1 Right Right Right Right

Wrong Wrong Wrong Wrong

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PART ONE PART THREE


To-do: calculate 𝑃𝑃𝑗𝑗 in k-SAT At least one literal needs to be flipped in a clause Pr(go wrong direction) ≤ 𝑘𝑘−1

𝑘𝑘

Pr(go right direction) ≥ 1𝑘𝑘

j j+1 j-1 ≥

1𝑘𝑘 ≤

𝑘𝑘 − 1𝑘𝑘

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PART ONE PART THREE


To-do: calculate 𝑃𝑃𝑗𝑗 in k-SAT Using binominal coefficient

𝑃𝑃𝑗𝑗 ≥ ∑ 𝑗𝑗+2𝑖𝑖𝑖𝑖

𝑗𝑗𝑖𝑖=0 ∙ 𝑖𝑖

𝑗𝑗+2𝑖𝑖∙ 1

𝑘𝑘

𝑖𝑖+𝑗𝑗∙ 𝑘𝑘−1

𝑘𝑘

𝑖𝑖

~ 1𝑘𝑘−1

𝑗𝑗

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PART ONE PART THREE


To-do: calculate overall success possiblity 𝑝𝑝 in k-SAT

P 𝑋𝑋 = 𝑗𝑗 = 𝑛𝑛𝑗𝑗 2−𝑛𝑛

𝑝𝑝 = ∑ P 𝑋𝑋 = 𝑗𝑗 ∗ 𝑝𝑝𝑗𝑗𝑛𝑛𝑗𝑗=0

≥ 12

𝑛𝑛∑ 𝑛𝑛

𝑗𝑗𝑛𝑛𝑗𝑗=0

1𝑘𝑘−1

𝑗𝑗

= 12

1 + 1𝑘𝑘−1

𝑛𝑛

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PART ONE PART THREE


To-do: calculate overall success possiblity 𝑝𝑝 in k-SAT

𝑝𝑝 ≥ 12

1 + 1𝑘𝑘−1

𝑛𝑛 , especially 𝑝𝑝 ≥ 3

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𝑛𝑛 in 3-SAT

Complexity of k-SAT is within a polynomial factor of

2 1 − 1𝑘𝑘

𝑛𝑛, especially (4

3)𝑛𝑛 in 3-SAT

THANKS Q & A

schöning's random walk algorithm for k-sat -...

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