Valeriy Balabanov NTU GIEE AlCom lab Outline Basic definitions Keyfacts about resolution proofs Intractability of resolution Heuristics for proof minimization Resolution in firstorder logic ID: 653083
Download Presentation The PPT/PDF document "Resolution proof system Presenter" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
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 definitionsSlide4Slide5
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:Slide25Slide26
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!!!