Satisfiability threshold for 2-SAT in new model - work in...
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Satisfiability threshold for 2− SAT in newmodel - work in progress
Mikołaj Pudo
Jagiellonian UniversityMathematics and Computer Science Faculty
Foundations of Computer Science Department
CLA, 2009
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Table of contents
1 Introductionk − SAT problemSatisfiability threshold conjecture
2 Satisfiability threshold conjecture - new approachNew modelAttempt of attack
3 Some more detailsGraph constructionCounting unsatisfiable formulasThe quest for E [Xn,m]
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Introduction
Table of contents
1 Introductionk − SAT problemSatisfiability threshold conjecture
2 Satisfiability threshold conjecture - new approachNew modelAttempt of attack
3 Some more detailsGraph constructionCounting unsatisfiable formulasThe quest for E [Xn,m]
![Page 4: Satisfiability threshold for 2-SAT in new model - work in ...perso.ens-lyon.fr/pierre.lescanne/CLA/Talks/cla09-MPudo.pdf · Introduction Table of contents 1 Introduction k SAT problem](https://reader035.fdocuments.us/reader035/viewer/2022070810/5f08d3807e708231d423e87f/html5/thumbnails/4.jpg)
Introduction
k − SAT problem
k − SAT problem
Vm = x1, x2, . . . , xm - set of boolean variables,C = ±xi1 ∨ . . . ∨ ±xik - k -clause,F = (±x11 ∨ . . . ∨ ±x1k ) ∧ . . . ∧ (±xn1 ∨ . . . ∨ ±xnk ) -k − CNF formula,Ωk (n,m) - set of k − CNF formulas with n clauses over setof m variables,
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Introduction
k − SAT problem
k − SAT problem
Vm = x1, x2, . . . , xm - set of boolean variables,C = ±xi1 ∨ . . . ∨ ±xik - k -clause,F = (±x11 ∨ . . . ∨ ±x1k ) ∧ . . . ∧ (±xn1 ∨ . . . ∨ ±xnk ) -k − CNF formula,Ωk (n,m) - set of k − CNF formulas with n clauses over setof m variables,
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Introduction
k − SAT problem
Satisfiability probability
A : Vm → 0,1 - valuation,A(φ) - set of all solutions for formula φ,c = n
m - clause to variables density,Sk (m, c) = Pr [φ ∈ Ωk (mc,m) satisfiable],
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Introduction
k − SAT problem
Satisfiability probability
A : Vm → 0,1 - valuation,A(φ) - set of all solutions for formula φ,c = n
m - clause to variables density,Sk (m, c) = Pr [φ ∈ Ωk (mc,m) satisfiable],
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Introduction
Satisfiability threshold conjecture
Satisfiability threshold conjecture
ConjectureFor each k ≥ 2 there is ck such that for all ε > 0
limm→∞
Sk (m, ck − ε) = 1
andlim
m→∞Sk (m, ck + ε) = 0.
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Introduction
Satisfiability threshold conjecture
The state of the art
For k = 2 it is known that c2 = 1,For k ≥ 3 we don’t know. . .Experiments show that: c3 ≈ 4.25± 0.05,Lower and upper bounds for interval containig ck (if itexists), eg. 3.42 < c3 < 4.506.
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Introduction
Satisfiability threshold conjecture
The state of the art
For k = 2 it is known that c2 = 1,For k ≥ 3 we don’t know. . .Experiments show that: c3 ≈ 4.25± 0.05,Lower and upper bounds for interval containig ck (if itexists), eg. 3.42 < c3 < 4.506.
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Satisfiability threshold conjecture - new approach
Table of contents
1 Introductionk − SAT problemSatisfiability threshold conjecture
2 Satisfiability threshold conjecture - new approachNew modelAttempt of attack
3 Some more detailsGraph constructionCounting unsatisfiable formulasThe quest for E [Xn,m]
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Satisfiability threshold conjecture - new approach
New model
Variables permutation
Consider two formulas over V = x , y , z:
(x ∨ y) ∧ (y ∨ z)
and(z ∨ x) ∧ (x ∨ y)
So far they were considered different, but actually they don’tdiffer that much.Take permutation σ : V → V such that:σ(x) = z, σ(y) = x , σ(z) = y
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Satisfiability threshold conjecture - new approach
New model
Variables permutation
Consider two formulas over V = x , y , z:
(x ∨ y) ∧ (y ∨ z)
and(z ∨ x) ∧ (x ∨ y)
So far they were considered different, but actually they don’tdiffer that much.Take permutation σ : V → V such that:σ(x) = z, σ(y) = x , σ(z) = y
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Satisfiability threshold conjecture - new approach
New model
Φk (n,m) - set of k − CNF formulas with n clauses over setof m variables,Sk (m, c) = Pr [φ ∈ Φk (mc,m) satisfiable],For each k ≥ 2 find ck such that for all ε > 0
limm→∞
Sk (m, ck − ε) = 1
andlim
m→∞Sk (m, ck + ε) = 0
Since 2− SAT is always the easiest part of the game. . .
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Satisfiability threshold conjecture - new approach
New model
Φk (n,m) - set of k − CNF formulas with n clauses over setof m variables,Sk (m, c) = Pr [φ ∈ Φk (mc,m) satisfiable],For each k ≥ 2 find ck such that for all ε > 0
limm→∞
Sk (m, ck − ε) = 1
andlim
m→∞Sk (m, ck + ε) = 0
Since 2− SAT is always the easiest part of the game. . .
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Satisfiability threshold conjecture - new approach
Attempt of attack
Sketch of proof
Construct a graph from each 2− CNF formula,Describe all graphs constructed from unsatisfiableformulas,Count those graphs and show that the number tends tozero.
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Some more details
Table of contents
1 Introductionk − SAT problemSatisfiability threshold conjecture
2 Satisfiability threshold conjecture - new approachNew modelAttempt of attack
3 Some more detailsGraph constructionCounting unsatisfiable formulasThe quest for E [Xn,m]
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Some more details
Graph construction
Graph construction
Every formula φ ∈ Φ(n,m) can be transformed into a directedgraph with coloring function Gφ = (V ,E , α), where:
V - vertex set, |V | = 4 · nE - edge set,α : V → 1 . . .m - coloring function.
Gφ is made of 4-element gadgets.
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Some more details
Graph construction
Gadgets
Consider formula φ. Every clause a = (x ∨ y) ∈ φ gives one4-element gadget with vertices: ax ,ax ,ay ,ay , and two gadgetedges: ax → ay ,ay → ax .
Example
(x ∨ y)
qqq
qI ax
ax
ay
ay
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Some more details
Graph construction
Mid-gadget edges
For all a,b ∈ φ, if y ∈ a ∧ y ∈ b, then take gadgets constructedfrom a and b and join them with mid-gadget edges: ay → byand by → ay
Example
φ = a ∧ ba = (x ∨ y)b = (y ∨ z)
qqI
ax
ax
ay
ay
q q-
by
by
bz
bz
-
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Some more details
Graph construction
Graph coloring
Construct α : V → 1 . . .m such that for all a,b ∈ φ if(y ∈ a ∨ y ∈ a) and (y ∈ b ∨ y ∈ b), thenα(ay ) = α(ay ) = α(by ) = α(by ).
Example(x1 ∨ x2)∧(x2 ∨ x3)∧(x3 ∨ x1)∧(x4 ∨ x5)
x1 → 1x2 → 2x3 → 3x4 → 4x5 → 5
R
?6
?6
R
w
~
>1 2
1
1
1
222
34
5
3
3 34
5
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Some more details
Graph construction
Contradictory cycle
DefinitionA cycle in Gφ is a contradictory cycle if it contains at least twovertices marked with the same color, but with opposit signs.Such vertices are called contradictory vertices. Mid-gadgetedges connecting contradictory vertices are calledcontradictory edges.
Lemma
Formula φ ∈ Φ(n,m) is unsatisfiable if and only if Gφ containscontradictory cycle.
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Some more details
Graph construction
Example
(x1 ∨ x2)∧(x2 ∨ x3)∧(x3 ∨ x2)∧(x2 ∨ x1)∧
x1 → 1x2 → 2x3 → 3
qq R
qq ?
6 qq
?
6
qqR
y
N
q
M
1 2
1
2
2
2
22
3
2
1
3
3 32
1
:
?
-
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Some more details
Graph construction
Example
(x1 ∨ x2)∧(x2 ∨ x3)∧(x3 ∨ x4)∧(x4 ∨ x1)∧
x1 → 1x2 → 2x3 → 3x4 → 4 q
q I
qqq
qIq
q R
qq R y
N
)
z
O
1 2
1
4
4
2
22
3
4
1
3
3 34
1
:
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Some more details
Counting unsatisfiable formulas
Random variable
DefinitionLet Xn,m : Φ(n,m)→ N be a random variable such that for aformula φ ∈ Φ(n,m), Xn,m(φ) is equal to the number ofcontradictory cycles in graph Gφ.
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Some more details
Counting unsatisfiable formulas
Markov’s inequality
Since the random variable Xn,m(φ) gives the number ofcontradictory cycles in random formula φ, by Lemma only thoseformulas for which Xn,m(φ) ≥ 1 holds are unsatisfiable.We can use Markov’s inequality:
Pr (φ ∈ Φ(n,m)|φ - unsatisfiable) =
= Pr (φ ∈ Φ(n,m)|Xn,m(φ) ≥ 1) ≤ E [Xn,m],
where:E [Xn,m] - expected value of random variable Xn,m.
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Some more details
Counting unsatisfiable formulas
Markov’s inequality
Since the random variable Xn,m(φ) gives the number ofcontradictory cycles in random formula φ, by Lemma only thoseformulas for which Xn,m(φ) ≥ 1 holds are unsatisfiable.We can use Markov’s inequality:
Pr (φ ∈ Φ(n,m)|φ - unsatisfiable) =
= Pr (φ ∈ Φ(n,m)|Xn,m(φ) ≥ 1) ≤ E [Xn,m],
where:E [Xn,m] - expected value of random variable Xn,m.
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Some more details
The quest for E [Xn,m ]
All we have to do is:Find E [Xn,m],Show that E [Xn,m] = o(1), assuming n = c ·m, and c < 1.
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Some more details
The quest for E [Xn,m ]
E [Xn,m] =∑
φ∈Φ(n,m)
Xn,m(φ) · Pr (φ ∈ Φ (n,m)) =
=
∑φ∈Φ(n,m) Xn,m(φ)
|Φ(n,m)|
The denominator of E [Xn,m] is easy:
|Φ(n,m)| =
2nm
22n,
where:2nm
- Stirling number of the second kind.
The numerator requires more perspiration. . .
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Some more details
The quest for E [Xn,m ]
E [Xn,m] =∑
φ∈Φ(n,m)
Xn,m(φ) · Pr (φ ∈ Φ (n,m)) =
=
∑φ∈Φ(n,m) Xn,m(φ)
|Φ(n,m)|
The denominator of E [Xn,m] is easy:
|Φ(n,m)| =
2nm
22n,
where:2nm
- Stirling number of the second kind.
The numerator requires more perspiration. . .
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Some more details
The quest for E [Xn,m ]
E [Xn,m] =∑
φ∈Φ(n,m)
Xn,m(φ) · Pr (φ ∈ Φ (n,m)) =
=
∑φ∈Φ(n,m) Xn,m(φ)
|Φ(n,m)|
The denominator of E [Xn,m] is easy:
|Φ(n,m)| =
2nm
22n,
where:2nm
- Stirling number of the second kind.
The numerator requires more perspiration. . .
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Some more details
The quest for E [Xn,m ]
Numerator of E [Xn,m]
Choose clauses for thecontradictory cycle.Count all contradictory cycleswhich can be built on thoseclauses
?66
?
6
??66
Contradictory cycle
Ordinary clauses
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Some more details
The quest for E [Xn,m ]
Numerator of E [Xn,m]
Color ordinary clauses withsome variables and signs.
?66
?
6
??66
Contradictory cycle
Ordinary clauses
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Some more details
The quest for E [Xn,m ]
Numerator of E [Xn,m]
Some variables from ordinaryclauses can be joined withvariables from contradictorycycle.
?66
?
6
??66
Contradictory cycle
Ordinary clauses
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Some more details
The quest for E [Xn,m ]
Numerator of E [Xn,m]
Remember to count allpossible choices of clauses forcontradictory cycle, and alllengths of those cycles.
?66
?
6
??66
Contradictory cycle
Ordinary clauses
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Some more details
The quest for E [Xn,m ]
A scary formula
∑φ∈Φ(n,m)
Xn,m(φ) =
n∑a=2
b a2 c∑
b=1
a−b∑r=0
na · a(2b) · (m − r)!
(m − a + b)!
(a− b
r
)2n − 2am − r
22n−2b,
where:na = n!
(n−a)! - falling factorial power
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Some more details
The quest for E [Xn,m ]
Even more scary formula
E [Xn,m] =
∑na=2
∑b a2 c
b=1∑a−b
r=0na·a(2b)·(m−r)!a·4b·(m−a+b)!
(a−br
)2n−2am−r
2nm
,
where:nm
- Stirling number of the second kind,
na = n!(n−a)!
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Some more details
The quest for E [Xn,m ]
Since n = mc
E [Xmc,m] =
=mc∑a=2
b a2 c∑
b=1
a−b∑r=0
(mc)a · a(2b) · (m − r)!
a · 4b · (m − a + b)!
(a− b
r
)2mc−2am−r
2mcm
![Page 39: Satisfiability threshold for 2-SAT in new model - work in ...perso.ens-lyon.fr/pierre.lescanne/CLA/Talks/cla09-MPudo.pdf · Introduction Table of contents 1 Introduction k SAT problem](https://reader035.fdocuments.us/reader035/viewer/2022070810/5f08d3807e708231d423e87f/html5/thumbnails/39.jpg)
Some more details
The quest for E [Xn,m ]
What next?
What next?1 Find upper bound for c2
2 Find upper and lower bound for ck , k ≥ 33 Attack conjecture
![Page 40: Satisfiability threshold for 2-SAT in new model - work in ...perso.ens-lyon.fr/pierre.lescanne/CLA/Talks/cla09-MPudo.pdf · Introduction Table of contents 1 Introduction k SAT problem](https://reader035.fdocuments.us/reader035/viewer/2022070810/5f08d3807e708231d423e87f/html5/thumbnails/40.jpg)
Thank you.