Resolution proof system Presenter - PowerPoint Presentation

Slide1

Resolution proof system

Presenter

Valeriy

Balabanov

NTU, GIEE,

AlCom

labSlide2

Outline

Basic definitions

Key-facts about resolution proofs

Intractability of resolution

Heuristics for proof minimization

Resolution in first-order logic

Conclusion and future work

ReferencesSlide3

Basic definitionsSlide4
Slide5

Resolution is a deductive rule in a form:

where

a, b, c

are some distinct logical facts

“

a

” is called

pivot (b or c) is called resolventA Resolution refutation proof for F is a sequence of clauses R = (C1, ..,Ct), where Ct = ∅;Ci ∈ F or Ci is derived from two previous clauses by the resolution ruleSlide6

The length of the proof =

#

of clauses in the derivation

Resolution proof

can also be seen as

DAG

,

where the nodes represent clauses, and edges represent resolution steps; the single sink node is an empty clauseTree-like resolution is a resolution, with special property – each parent node has exactly one child (in other words each clause in a proof is resolved only once)Note: Tree-like resolution can be derived from DAG resolution by splitting multiply used nodes into separate nodesSlide7

Key-facts about resolution proofsFor 2SAT it is possible to find the shortest resolution proof in polynomial time (

2SAT∈ P

)

For

HornSAT

polynomial resolution proof exists (

HornSAT

∈ P-complete), but finding the shortest proof is NP-hardGenerally, finding the shortest resolution proof is NP-hard (generally, as we will see the shortest proof can be exponential in number of clauses)Slide8

Intractability of resolution

Resolution is complete and sound

Proof:

Soundness: every clause, resolved from the formula is implied by that formula, thus, if resolved clause is empty – formula is UNSAT

Completeness:

elimination of variable “

a

” from CNF, is a procedure, when we make all possible resolutions using “a” as a pivot, and then eliminating all the clauses containing “a” from the original formulaSlide9

Completeness(continued):Let F be UNSAT CNF with m-variables

a

1

,a

2

…a

m

Let Si be the set of clauses, which are left after elimination of i variables from F; S0 is the original formula F; Sm has at most the empty clause.Let’s prove by induction on i, that every truth assignment to variables in F will make some clause in Si to be falseFor i=0 S0 is UNSAT, and thus has false clause for every assignmentAssume for Sk it is also true, and for some assignment

V

, the false clause is

θ

, then if

θ

doesn’t contain variable

a

k+1

, then

θ

also will be present after elimination of

a

k+1

;Slide10

Completeness(continued):now, if θ has variable

a

k+1

,

let

W

be the truth assignment, same as

V, but with different assignment to variable ak+1; let β be the clause which is false for W; if β doesn’t contain variable ak+1, then β will be in Sk+1; if it does – then the resolvent of β and θ will be present in Sk+1 and obviously will be false for V(also W); thus for every truth assignment, S

i

must contain a clause which will be false under it

Thus,

S

m

should contain the empty clause, and by the construction of

S

m

it was derived by resolutionSlide11

Pigeonhole principle:Let A be a sequence of n=sr

+ 1

distinct numbers. Then either A has:

an increasing subsequence of s + 1 terms or

a decreasing subsequence of r + 1 terms (or both).

Consequence:

Suppose we have

n=s+1 pigeons (r=1)If we put them in at most s holes, then there definitely will be at least 2 pigeons in the same holeIn other words it is impossible to put every pigeon to it’s own holeSlide12

Proof:Every number in sequence a

i

has score

(x

i

,

yi).xi is the longest increasing subsequence ending at aiyi is the longest decreasing subsequence starting at ai(xi, yi) ≠ (xj,

y

j

)

whenever

i

≠ j

.

Assume

i

< j, then

:

if

a

i

<

a

j

→ x

i

<

x

j

if

a

i

>

a

j

→

y

i

>

y

j

Thus we have

rs+1

points on a plane, and there is

ai

with coordinate

(x

i

,

y

i

)

outside the

rs

-square.

So, for that

a

i

we will have x

i

≥ s+1 or

y

i

≥ r+1Slide13

Formalizing PHP to CNF formulaxi,j

- pigeon

i

sits in hole

j

(type 1): xi,1 ∨ xi,2 ∨ .. ∨ xi,n−1 for i = 1..n (every pigeon sits in at least one hole)(type 2): (¬xi,k ∨ ¬xj,k) for 1 ≤ i ≠ j ≤ n ; 1 ≤ k ≤ n − 1 (no two pigeons sit in the same hole)From pigeonhole principle conjunction of above clauses is UNSATExample:

Note: deleting any clause will lead to

SATSlide14

Haken’s super-polynomial lower bound Original proof shows the bound for n>200

We present modified proof:

Ω

(2

√n/32

)

Definition:

A critical assignment is a one-to-one mapping of n − 1 pigeons to n − 1 holes, with one pigeon unset. Having i-th pigeon unset defines a i-critical assignment.Presenting the assignments of the xi,j as a matrix, the critical assignments would look like this: Example of 9-critical assignment for PHP with n=9Slide15

Let R be the proof of unsatisfiability of PHP

n

Replace

x

i,j

’ in all clauses C by:

Definition:

The resulting sequence of positive clauses R+ = (C1+ , ..,Ct+ ) is a positive pseudo-proof of PHPnLemma: C+(α) = C(α

) for any critical

α

Proof:

Suppose

∃C

+

(

α

) ≠ C(

α

)

⇒

∃

xi,j’ ∈ C s.t. Ci,j(α) ≠ xi,j’(α) ⇔(x1,j ∨ .. ∨ xn,j)(α) ≠ xi,j’(α). This is impossible, since α is critical, therefore has exactly one 1 in the column j.Slide16

We will show now, that t ≥ 2n/32.

For a contradiction, assume

t < 2

n/32

, t is the number of clauses in R

+

.Definition: A long clause has at least n2/8 variables. (more than 1/8 of all possible n(n − 1) variables). l is the number of long clauses in R. l ≤ t < 2n/32By the pigeonhole principle, there exists a variable xi,j, which occurs in at least l/8 of the long clauses.Set the special variable xi,j to 1.Set all xi,j’, x

i’,j

for

j’

≠

j

,

i’

≠

i

to 0.

Clauses containing

x

i,j

are set to 1 and therefore disappear from the proof.

The variables set to 0 disappear from all clauses.Slide17

We are left with a pseudo-proof of PHPn−1 with at most l(1 − 1/8) long clauses.

Doing this d = 8log(l) times, we will eliminate all long clauses, since

We are left now with a pseudo-proof of

PHP

m

with no long clauses (of length more than n

2/8).Since m = n – d, and from assumption l < 2n/32, we can obtainSlide18

Lemma: Any positive pseudo-proof of

PHP

m

must have a clause with at least 2m

2

/9 variables.

Proof: let R’ be a positive pseudo-proof of PHPmDefinition: ∀C∈R’, W is a witness of C if W is a set of clauses from PHPm, whose conjunction implies C for critical assignments. (∀ critical α: α satisfies all ω∈W → α satisfies C). The weight of C = # clauses in minimal witness. Note: for any C there exist witness W

Clauses of (type 2) are not the part of a minimal witness

Clauses of (type 1) have weight 1

The weight of the final clause is m

The weight of a clause is at most the sum of the two clauses its been derived from

There exists a clause C∈R’ of weight s, m/3 ≤ s ≤ 2m/3.Slide19

Let S is a set of indices of witness clauses for CW = {C

i

|i

∈

S}, |S| = s,

C

i

= xi,1 ∨ xi,2.. ∨ xi,m−1; Ci ∈ PHPm∧Ci → CAlso leti ∈ Sα is i-critical assignment with C(α)=0

j ∉ S;

α

’ is j-critical

α

’ is obtained from

α

, by swapping

row

i

and

row

j

:

If

α maps pigeonj to holek, then α’ maps pigeoni to holekSlide20

Since j ∉ S α’ satisfies all C

i

∈ W, so C(

α

’)=1

From the construction

α

differs from α’ only in xi,k, xj,kThis implies xi,k ∈ CWe can run this argument for current i-critical assignment under all (m − s) different choices for j ∉ SThus C contains the variables xi,k1, xi,k2, .., xi,km−s

And by repeating this for all

i

∈ S, we conclude that C contains at least

(m-s)s

different variables

Since m/3 ≤ s ≤ 2m/3, we have

(m-s)s ≥ 2m

2

/9,

concluding the proof for lemma

We reached a contradiction to our assumption that

t ≤ 2

n/32Slide21

Thus we conclude, that pigeonhole family of clauses requires super-polynomial minimal proofs for large nPeople have also found many exponentially hard examples for resolution using graph theory

Definition:

extended resolution

, is a regular resolution, but with additional property: any definition can be added to original formula, if it doesn’t change its

satisfiability

Example: if x is not in original formula, we can add

Extended resolution can find polynomial proofs for pigeonhole formulas

Extended resolution is one of the strongest known proof systemsSlide22

Heuristics for proof minimization

Resolution proofs are useful for

Extracting

unsatisfiable

cores

Extracting

interpolants

Detecting useful clauses for incremental SAT-solvingRun-till-fix and Trim-till-fixUse SAT-solver repeatedly to minimize UNSAT-coreUse incremental SAT-solver to analyze the structure of the proof and restructure itRunning time is usually large, since we need to rerun SAT-solver again and againSlide23

Recycling learned unit clausesIf (x) is a unit clause that was learned by the SAT solver, it can be used for simplifying resolution inferences that used x as the pivot prior to learning this clause

May lead to circular reasoning, so must be applied carefully

Let

P – is a resolution proof of the empty clause

For a given node n in P:

n.C

- is the clause represented by n

n.L and n.R are parents of nn.piv – is the variable used to resolve n.C from n.L.C and n.R.CSlide24

Example:Slide25
Slide26

Example:

It is easy to see, that recycling units will only make proof stronger

The size of the proof also will be reduced

The time complexity is quadratic in size of the proof, and no SAT-solving is usedSlide27

Recycling PivotsObservation:

along each path from root to sink in a proof graph there is no need for resolving on the same variable more than once

Proof:

Key point here is: why do we want to use resolution?

We use current resolution step to eliminate variable “x”

If in few steps variable “x” will reappear again – then what was the purpose of first resolution?

The proof with above mentioned property is called RegularThe shortest proof for a given problem must be regularThe Reconstruct-Proof algorithm will be the same as that for Recycling UnitsRuntime of Recycling Pivots is linear in proof sizeSlide28

Example:Slide29

Experimental results

Run-till-fix finds the smallest UNSAT core (# of roots), but it increases the proof-size

Recycle Units and Pivots significantly simplify the proof, but cannot make UNSAT core small enoughSlide30

Resolution in first-order logicPropositional logic vs. First-order logic

Example

Universal reduction

Example

but

Q-resolution combines resolution and universal reductionSlide31

Example:

Red lines

: universal reduction

Green lines

: exist. resolutionSlide32

Q-resolution is both complete and soundSoundness: if the empty clause was generated, as in SAT, QBF obviously evaluates to 0

Completeness:

Induction on number of quantifiers:

For single ∃-variable it is just a usual resolution

For single ∀-variable, falsity of formula->there is at least one non-tautological clause, which can be universally reduced

Induction step for ∀-variable (a) will choose the value of a, which leads to UNSAT, and use the same resolution steps;

For ∃-variable (a) both assignments to a lead to a conflict; we use Q-res steps for those assignments; if in one of them a (a’) was not present – we are done; if both present – we resolve resulting clauses on a, and thus get the conflict clause Slide33

As QBF is a general case of SAT, Q-resolution is also intractableMore definitions:

∃-unit clause is clause with only one ∃-variable

Q-unit resolution is a Q-resolution where one of the clauses is a positive ∃-unit clause

Horn clause is a clause with only one positive literal

Extended quantified Horn formula has every clause’s existential part to be a Horn clauseSlide34

Theorem: Q-unit resolution is complete and sound for extended quantified Horn formulas

Proof:

look into [7]

Theorem:

For every

t>0

there exists a quantified extended Horn formula of length

18t+1 which is FALSE, and the refutation to the empty clause requires at leas 2t Q-resolution stepsProof: look into [7]Q-resolution can’t simulate usual resolutionExample can’t conclude xSlide35

Conclusion and future work

Resolution is simplest, but yet efficient proof system

Resolution is intractable

Existence of exponential lower bounds

Resolution proofs are used in model checking

Shorter proofs can be produced using some heuristics

Q-resolution is an extension of resolution in first-order logicSlide36

Other proof systemsExchange of the nodes in the resolution graphDifferent heuristics for proof-length reduction

Interpolants

in first-order logic

Q-resolution vs. QBF’s certificatesSlide37

References

“The relative efficiency for propositional proofs”, Stephen A. Cook and Robert A.

Reckhov

“Hard examples for Resolution”, Alasdair Urquhart

“On the complexity of derivation in propositional calculus”, G.S.

Tseitin

“Optimal length tree-like resolution refutations for 2SAT formulas”, K.

Subramani“The intractability of resolution”, Armin Haken“Reducing the size of resolution proofs in linear time”, O.B.Ilan, O. Fuhrmann, S. Hoory, O. Shacham, O.Strichman

“Resolution for Quantified Boolean Formulas”,

H.Buning

,

M.

Karpinski

, A.

FlogelSlide38

Thank you very much!!!