1 Agenda Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and.

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Agenda

Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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Conflict clauses and Resolution

Binary-resolution is a sound inference rule:

Example:

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Consider the following example:

Conflict clause: c5: (x2 Ç :x4 Ç x10)

We show that c5 is inferred by resolution from c1,…,c4

Conflict clauses and resolution

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Conflict clause: c5: (x2 Ç :x4 Ç x10)

BCP order: x4,x5,x6,x7 T1 = Res(c4,c3,x7) = (:x5 Ç :x6) T2 = Res(T1, c2, x6) = (:x4 Ç :x5 Ç X10 ) T3 = Res(T2,c1,x5) = (x2 Ç :x4 Ç x10 )

Conflict clauses and resolution

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Applied to our example:

Finding the conflict clause:

cl is asserting the first UIP

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Conflict clauses and resolution

c2

c3

c4

c5

c1

This will be part of the (Hyper )Resolution Graph:

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The resolution graph

What is it good for ? Example: for computing an Unsatisfiable core

[Picture Borrowed from Zhang, Malik SAT’03]

Minimizing the core

The proof is not unique. Different proofs / different cores.

Can we find a minimum / minimal / smaller cores/proofs?

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Minimizing the core

Core compression [ZM03,…]… Min-core-biased search [NRS’13]

Proof compression:

Exponential-time transformations [GKS’06].

Linear time transformations “Recycle units” [BFHSS’08]

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Core compression (smaller core)

A basic approach: run until reaching a fixpoint [chaff]

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SAT solver

core== last_core ?

corecore

last_core = core

yes

no

initially last_core = ;

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Remove an unmarked clause c 2

SAT() ?

mark c

yes

:= core

no

Initially ’s clauses are unmarkedReturn

All marked

Core compression (minimal core)

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Proof compression Example of a linear-time transformation / “Recycle-units” Observation:

When learning (resolving) a new clause in SAT, The resolving clauses are not satisfied Hence, the resolution-variable is unassigned

l l1 :l l2

l1 l2

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Example (cont’d)

Suppose that the pivot’s constant value is learned later on.

We will use it to simplify the resolution proof.

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Recycle-units / easy case

1 3 -1 2 5

2 3 5 1 -2

1 3 5

-1 4 -1 -4

-1

3 5

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Recycle-units / easy case

1 3 -1 2 5

2 3 5 1 -2

1 3 5

-1 4 -1 -4

-1

3 5

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Recycle-units

1 3

3 1 -2

1 3 5

-1 4 -1 -4

-1

3 5

3

3

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Recycle-units

1 3

3

-1 4 -1 -4

-1

Reduced proof by 4 clauses Reduced core by 2 clauses

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Recycle-units / beware of cycles

1 3

2 3 5 1 -2

1 3 5

-1 4 -1 -4

-1

By making this connection we created

cyclic reasoning

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Recycle-units / beware of cycles

Solution: mark antecedents of units apply only to unmarked nodes

1 3

2 3 5 1 -2

1 3 5

-1 4 -1 -4

-1

-1 2 5

3 5

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Agenda

Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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4. Periodically, all the counters are divided by a constant.

3. The unassigned variable with the highest counter is chosen.

2. When a clause is added, the counters are updated.

1. Each variable in each polarity has a counter initialized to 0.

)Implemented in Chaff(

Decision heuristicsVSIDS (Variable State Independent Decaying Sum)

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• Chaff holds a list of unassigned variables sorted by the counter value.

• Updates are needed only when adding conflict clauses.

• Thus - decision is made in constant time.

Decision heuristicsVSIDS (cont’d)

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VSIDS is a ‘quasi-static’ strategy:

- static because it doesn’t depend on current assignment

- dynamic because it gradually changes. Variables that appear in recent conflicts have higher priority.

This strategy is a conflict-driven decision strategy.

“..employing this strategy dramatically )i.e. an orderof magnitude( improved performance ... “

Decision heuristicsVSIDS (cont’d)

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Decision Heuristics - Berkmin

Keep conflict clauses in a stack Choose the first unresolved clause in the stack

If there is no such clause, use VSIDS

Choose from this clause a variable + value according to some scoring (e.g. VSIDS)

This gives absolute priority to conflicts.

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Berkmin heuristic

tail-first conflict clause

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Deduction)( allocates new implied variables and conflicts. How can this be done efficiently ?

Observation: More than 90% of the time SAT solvers perform Deduction)(.

More engineering aspects of SAT solvers

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Hold 2 counters for each clause :

val1( ) - # of negative literals assigned 0 in +

# of positive literals assigned 1 in .

val0(( - # of negative literals assigned 1 in +

# of positive literals assigned 0 in .

Grasp implements Deduction)( with counters

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Every assignment to a variable x results in updating the counters for all the clauses that contain x.

Grasp implements Deduction)( with counters

is satisfied iff val1)( > 0

is unsatisfied iff val0)( = ||

is unit iff val1)( = 0 val0)( = || - 1

is unresolved iff val1)( = 0 val0)( < || - 1..

Backtracking: Same complexity.

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Observation: during Deduction(), we are only interested in newly implied variables and conflicts.

These occur only when the number of literals in with value ‘false’ is greater than || - 2

Conclusion: no need to visit a clause unless (val0() > || - 2)

How can this be implemented ?

Chaff implements Deduction)( with 2 observers

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For each clause ¼, define two ‘observers’: O1(), O2().

O1() and O2(): two literals which are not ‘false’.

is unit if updating one observer leads to O1() = O2().

Visit clause only if O1() or O2() become ‘false’.

Chaff implements Deduction)( with 2 observers

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1 2 3 4 5

V[1]=0

1 2 3 4 5

V[5]=0, v[4]= 0 1 2 3 4 5 Unit clause

Backtrackv[4] = v[5]= Xv[1] = 1

1 2 3 4 5

Chaff implements Deduction)( with 2 observers

1 2 3 4 5

V[2]=0

O1 O2

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Complexity of Backtracking: Observers of a unit clause are on the highest decision level

present in the clause. Hence backtracking will un-assign them first.

Conclusion: when backtracking, observers stay in place.

Chaff implements Deduction() with 2 observers

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Chaff implements Deduction() with 2 observers

Which observing literals? Strategy: the least frequently updated variables

The observers method has a learning curve Initial observers: chosen arbitrarily. Then: shifted from variables that were recently updated

These variables will most probably be reassigned soon.

In our example: the next time v[5] is updated, it will point to a significantly smaller set of clauses.

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Agenda

Modeling problems in Propositional Logic SAT basics Decision heuristics Non-chronological Backtracking Learning with Conflict Clauses SAT and resolution More techniques: decision heuristics, deduction. Stochastic SAT solvers: the GSAT approach

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GSAT: stochastic SAT solving

for i = 1 to max_tries {

T := randomly generated truth assignment

for j = 1 to max_flips {

if T satisfies return TRUE

choose v s.t. flipping v’s value gives largest increase in

the # of satisfied clauses )break ties randomly(.

T := T with v’s assignment flipped. } }

Given a CNF formula , choose max_tries and max_flips

Many alternative heuristics

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Numerous progressing heuristics

Hill-climbing Tabu-list Simulated-annealing Random-Walk Min-conflicts ...

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Improvement # 1: clause weights

Initial weight of each clause: 1 Increase by k the weight of unsatisfied clauses. Choose v according to maximum increase in weight

Clause weights is another example of conflict-driven decision strategy.

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Improvement # 2: Averaging-in

Q: Can we reuse information gathered in previous tries in order to speed up the search ?

A: Yes!

Rather than choosing T randomly each time, repeat ‘good assignments’ and choose randomly the rest.

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Let X1, X2 and X3 be equally wide bit vectors.

Define a function bit_average : X1 X2 X3 as follows:

b1i b1

i = b2i

random otherwise

(where bji is the i-th bit in Xj, j {1,2,3})

b3i :=

Improvement # 2: Averaging-in

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Notation:

Tiinit

: The initial assignment T in cycle i.

Tibest

: The assignment with highest # of satisfied clauses in

cycle i.

T1init := random assignment.

T2init := random assignment.

i > 2, Tiinit := bit_average(Ti-1

best, Ti-2best)

Improvement # 2: Averaging-in