quantified formulas

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Quantified Formulas

Acknowledgement: QBF slides borrowed from S. Malik

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Why do we need quantifiers ?

As always: more modeling power Examples of quantifiers usage:

“Everyone in the room has a friend” “There is a person in the room that all of his cars are red” “There is not more than one person in the room that earns

more than $1M”

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Quantifiers in Math…

For any integer x there is a smaller integer y

8x2Z 9y2Z. y < x X Reverse claim: There exists an integer y such that any

integer x is greater than y

8x2Z 9y2Z. y < x £

(Bertrand’s postulate) For any natural number greater than 1 there is a prime number p such that n < p < 2n

8n2 N. 9p2 N. n >1 ! (isprime(p) Æ n < p < 2n)

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Actually…

Satisfiability of (x1,,xn) = does there exist an interpretation of x1,,xn that satisfies

Validity of (x1,,xn) = does it hold that all interpretation of x1,,xn satisfy

Conclusion: what we did so far (satisfiability, validity) is non-alternating quantification.

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Prenex Normal-Form (PNF)

Formulas in PNF look like this:

’: Q[n]V[n]. .Q[1]V[1]. Quantifier-free formula

where Q[i] 2 {8,9} and V[i] is a variable.

Every quantified formula can be transformed to PNF while preserving validity. How ?

prefix

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Prenex Normal Form (PNF)

1. Eliminate ! and $ (transform to Ç Æ :)

2. Push negations inside using:: 8 x. $ 9 x. : : 9 x. $ 8 x. :

3. If there are name conflicts across scopes, solve with renaming.

4. Move quantifiers out by using recursively rules such as:1. Q1 x. 1(x) Æ Q2 y.

2(y) $ Q1 x. Q2 y. (1(x) Æ 2(y)) Qi2{8,9}

2. Q1 x. 1(x) Ç Q2 y. 2(y) $ Q1 x. Q2 y. (1(x) Ç 2(y))Qi2{8,9}

3. 1 Æ 9 x. 2(x) $ 9 x. (1 Æ 2(x)) where x does not appear in 1

4. 1 Æ 8 x. 2(x) $ 8 x. (1 Æ 2(x)) where x does not appear in 1

5. 8 x. 1(x) Æ 8 x.

2(x) $ 8 x. (

1(x) Æ

2(x))

6. 9 x. 1(x) Ç 9 x.

2(x) $ 9 x. (

1(x) Ç

2(x))

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Prenex Normal Form (PNF): example

:9 x. : (9 y. ((y ! x) Æ (: x Ç y)) Æ :8 y. ((y Æ x) Ç (: x Æ : y)))

1,2. Eliminate !, push negations inside:

8 x. (9 y. ((:y Ç x) Æ (: x Ç y)) Æ 9y. ((:y Ç : x) Æ (x Ç y)))

3. Renaming:

8 x. (9y1. ((:y1 Ç x) Æ (: x Ç y1)) Æ 9y2. ((:y2 Ç : x) Æ (x Ç y2)))

4. Move quantifiers to front:

8 x. 9y1. 9y2. (x Ç :y1) Æ (: x Ç y1) Æ (:y2 Ç : x) Æ (x Ç y2)

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Why eliminating 9x. ÆiLi is enough

A procedure for eliminating an existential quantifier applied to a conjunction of literals is enough, because: Given a formula , write it in DNF. Use the fact that

Eliminate universal quantifiers using the fact

8x. $ :9x. :

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Quantifier Elimination

Examples first, generalization later. Example #1: Quantified Boolean Formulas (QBF) Example #2: Quantified Linear Arithmetic (QLA)

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Example #1: QBF

Examples of Quantified Boolean Formula

: ue.(uÇ :e)(:uÇ e)

: e4e5u1u2u3e1e2e3. f(e1,e2,e3,e4,e5,u1,u2,u3)

QBF Problem: is valid? P-Space Complete, theoretically harder than NP-Complete

problems such as SAT.

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Motivations

QBF has practical applications: AI Planning Sequential circuit verification …

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a Ç b Ç g Ç h’Ç fa Ç b Ç g Ç h’

Solving QBF with projection: 9

Eliminate 9x. by projecting x on variables in higher quantification levels (their scope includes x’s scope).

In Propositional Logic projection can be done with Resolution.

Resolution example:

a Ç b Ç c’ Ç f g Ç h’ Ç c Ç f

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Solving QBF with projection: 8

)'')(')('( 13211113121 euueueuueuu

))'')(')('(( 13211113121 euueueuueuu

))'()'()'(( 121111121 eueueueuu

Transform 8 to 9 via: (8x. $ (:9x. : CNF is easier than general formulas:

))'()'()'(( 13211113121 euueueuueuu Suffix is DNF

))'()'()'( 121111121 eueueueuu Back to CNF

Shortcut for CNF formulas: simply erase universally quantified variables!

))).('()'()'(( 13321111121 euuueueueuu Replace with true

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)'( 2121 uuuu

)')('')(')('( 311323111133121 eeeuueeueueueuu

)')()('')('')('( 312113232111233121 eeeeeuueeueueeueuu

Resolution Based QBF Algorithm

)')(')('( 121111121 eueueueuu

false

)'')(')('( 13211113121 euueueuueuu

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Solving QLA with projection

Eliminate 9x. by projecting x. In Linear Arithmetic over R projection can be done

with Fourier-Motzkin elimination.

Fourier-Motzkin method to eliminate a variable xn:- for each pair of constraints: i=1..n-1ai’xi < xn < i=1..n-1aixi

add a constraint i=1..n-1ai’xi < i=1..n-1aixi

- in the end remove all constraints involving xn.

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x + 1 · 3z + 3

2y · 2z + 4

Solving QLA with projection

Fourier Motzkin: example.

Eliminate y:

x + 1 · z + 2 Æ

x + 1 · y Æ y · 3z + 3 Æ

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Quantifier elimination - example

8x.9y.9z. (y+1 · x Æ z+1 · y Æ 2x+1 · z)

8x.9y. (y+1 · x Æ 2x+1 · y-1 )

8x. (2x+2 · x-1) // transform to 9

:9x. : (2x+2 · x-1)

:9x. x > -3

:true

false

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Quantifier elimination by projection: summary

Given a PNF formula f = Q[n]V[n]Q[1]V[1]

For i = 1 .. n {

if Q[i] = 9 then = project(,V[i])

else = :project(:,V[i])

}

Return

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More about QBF

Example of using QBF (the diameter problem) A search-based procedure for QBF.

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The State Space Diameter Problem

S0

S2S1

S3

S5 S4 diameter = 3

Start from the initial states, the minimum number of steps needed to visit every reachable state

S0

initial state: S0S2S1

step 1: S1, S2

S4

S3step 2: S3, S4

S5

step 3: S5

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Why is the Diameter Problem important?

Bounded model checking (BMC): search for a ‘bad’ state up to k steps from an initial step.

BMC can be formulated as SAT. Increasing k makes is harder. Q: how deep should we go ? A: as deep as the diameter The diameter can be found by solving a QBF problem

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Circuit Constructed for the Diameter Problem

CombinationalLogic

I1

O1

1s0s CombinationalLogic

In

On

ns1ns CombinationalLogic

In+1

On+1

1ns

CombinationalLogic

I1’

O1’

'1s'0s CombinationalLogic

In’

On’

'ns'1ns

The idea: prove that for every state reachable in k+1 steps, there exists inputs that drive the model to this state earlier.

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Some Terminology for the Formulations

CombinationalLogic

I1

O1


In

On


In+1

On+1

1ns

CombinationalLogic

I1’

O1’


In’

On’

'ns'1nsVariables: V1

Circuit consistency condition: C(V1)

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Some Terminology for the Formulations

CombinationalLogic

I1

O1


In

On


In+1

On+1

1ns

CombinationalLogic

I1’

O1’


In’

On’

'ns'1ns

Variables: V2

Circuit consistency condition: C(V2)

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QBF Formulation

CombinationalLogic

I1

O1


In

On


In+1

On+1

1ns

CombinationalLogic

I1’

O1’


In’

On’

'ns'1ns

C(V1)

C(V2)

)1()()()(.),\( '10212

1

11121 i

SSVCVCVIVIII nni

n

iin

Other V1variables

V1 inputs V2 variables

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Another way to project Boolean variables

Shannon expansion:9x. = |x=0 Ç |x=1

8x. = |x=0 Æ |x=1 // can be derived from 8x. = :9x.:

The same applies for all finite-range variables. Applying to CNF $ resolution But: does not need to be in CNF, and there is no

need to transform the formula to DNF.

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Projection for non-CNF formulas: example

9y 8z 9x. (y Ç (x Æ z))

9y 8z. (y Ç (x Æ z))|x=0 Ç (y Ç (x Æ z))|x=1

9y 8z. ((y) Ç (y Ç z))

9y :9z. (:y Æ :z)

9y. : ((:y Æ :z)|z=0 Ç (:y Æ :z)|z=1)

9y. : (:y)

True

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Search Based QBF Algorithms

Work by gradually assigning variables A partial assignment

[KGS98] M. Cadoli, A. Giovanardi, M. Schaerf. An Algorithm to Evaluate Quantified Boolean Formulae. In Proc. of 16th National Conference on Artificial Intelligence (AAAI-98)

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Undetermined Continue search


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Undetermined Conflict

Backtrack Record the reason


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Undetermined Conflict Satisfied

Backtrack Determine the covered satisfying space


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Undetermined Conflict Satisfied

The majority of QBF solvers are search based, the DPLL algorithm is an example of this

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Basic DPLL Flow for QBF

eu (e Ç u)(:e Ç :u)

Unknown

True (1)

False(0)

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e = 0

Unknown

True (1)

False(0)


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e = 0

u = 1

Unknown

True (1)

False(0)

Satisfying Node

eu (e Ç u)(:e Ç :u) Universal quantification

Existential quantification

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e = 0

u = 1

Unknown

True (1)

False(0)

Backtrack


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e = 0

u = 1 u = 0

Unknown

True (1)

False(0)


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e = 0

u = 1 u = 0

Unknown

True (1)

False(0)


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e = 1

Unknown

True (1)

False(0)


e = 0

u = 1 u = 0

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e = 1

u = 1

Unknown

True (1)

False(0)


e = 0

u = 1 u = 0

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e = 1

u = 1

Unknown

True (1)

False(0)

Conflicting Node


e = 0

u = 1 u = 0

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e = 1

u = 1

Unknown

True (1)

False(0)


e = 0

u = 1 u = 0

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e = 1

u = 1

e = 0

u = 1 u = 0

False

Unknown

True (1)

False(0)


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ue (u Ç e)(:u Ç :e)

Unknown

True (1)

False(0)

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Unknown

True (1)

False(0)

u = 1


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Unknown

True (1)

False(0)

u = 1

e = 1


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Unknown

True (1)

False(0)

u = 1

e = 1 e = 0


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Unknown

True (1)

False(0)

u = 1

e = 1 e = 0


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Unknown

True (1)

False(0)

u = 1

e = 1 e = 0

u = 0

e = 1


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Unknown

True (1)

False(0)

u = 1

e = 1 e = 0

u = 0

e = 1

Trueue (u Ç e)(:u Ç :e)

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Naïve DPLL Based Approach

Works on a CNF database Learning and non-chronological backtracking is

much harder – requires a change! Modern QBF solvers do not work with CNF, rather

with a combination of CNF with Cubes. This lets them apply learning efficiently.

quantified formulas

Documents

x x y2why

x x y24

x x y3

integer x

projecting x

x y1 9y2

y 8x2z 9y2z

y xbertrands postulate