1. Mining backbone literals in incremental SAT
A new kind of incremental data
Alexander Ivrii
IBM Haifa
Vadim Ryvchin
Intel Haifa
Ofer Strichman
Technion, Haifa
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2. Backbone literals
l is a backbone literal of  if all models of  satisfy l.
Checking whether a literal is a backbone is NP-C [1].
In this work:
We show that many backbone literals (BL) can be found in P- time in the incremental SAT setting.
Based on analyzing the proof of the last unsat instance.
[1] Janota, Lynce, Marques-Silva. Algorithms for computing backbones of propositional formulae. AI Communication 2015.

3. Observation 1
Let π be a refutation. Then every vertex cut in π represents an inconsistent set of clauses [2] ?
[2] A. Nadel. Understanding and Improving a Modern SAT Solver. PhD thesis, 2009.

4. Observation 2 Suppose our next instance is ’ = ¼ n cone(c) Let ® ² ’
Then ® ² :c
’ ) :c c2 ? c c3 c1 ’
(this observation is the basis of ‘redundancy removal’ in MUC extraction)

5. Observation 2 Suppose our next instance is ’ = ¼ n cone(c) Let ® ² ’
Then ® ² :c1
’ ) :c1 c2 ? c c3 c1 ’

6. Observation 2 Suppose our next instance is ’ = ¼ n cone(c) Let ® ² ’
Then ® ² :c2 Ç :c3
’ ) :c2 Ç :c3 c2 ? c c3 c1 ’

7. Observation 2 Suppose our next instance is ’ = ¼ n cone(c) Let ® ² ’
Then ® ² :c Æ :c1 Æ (:c2 Ç :c3)
’ ) :c Æ :c1 Æ (:c2 Ç :c3) c2 ? c c3 c1 ’
So what ?

8. In general… Let Cuts be the set of vertex cuts in cone(c)
Then (¼ n cone(c)) )
So we can add the redundant constraints à ’ Ã
So what ?

9. In general Adding Ã: Two problems:
Not clear that adding such redundancy helps
Exploring all cuts is ineffective
A Solution preview:
Find in P-time literals BL that are implied by Ã
Since ’ ) Ã ) BL
…then we can check ’ Æ BL
But how?

10. Finding those backbone literals: Example
Suppose c2,c3 contain a mutual literal l
e.g. c2 = (l Ç X1), c3 = (l Ç X2)
Then l ² c2 Æ c3
But since we saw that
à ² (:c2 Ç :c3)
then
à ² :l.
Conclusion: ’ ) :l. c2 ? c c3 c1 ’

11. Observation 3
’ ² :l if l appears in every clause along some cut in cone(c)
Luckily, it can be done in P-time
But exploring
all cuts is
exponential…

12. Mining BL literals in P-time: example
Literals on all paths from root to here
{1 2 -3}
{ }
{ }
{ }
{ }
Each of these literals satisfies a cut in cone(c)

13. Let’s keep mining… So far we ignored the state of the solver….
Suppose at decision level 0, ’ implies the literal -7
Denote such literals by cons.
{1 2 -3}
{ }
{ }
{ }
{ }
{ }

14. Let’s keep mining… We can create a feedback loop:
…and activate it every time there is an increase in cons (decision level 0)
(future work): … or at higher decision levels,
Using an interface similar to SMT, where MBL is the theory.
cons
BCP
MBL BL

15. Other optimizations… Cutoff values
When the span of cone(c) exceeds a threshold, stop.
Likely to take too much time
Not likely to produce many literals because cuts are long

16. A major problem The more BL literals there are… Without a proof of …
… the better the chance the proof will rely on them…
… and hence be a proof of :BL, rather than of .
Without a proof of …
We cannot repeat the process.
In MUC extraction: we cannot apply clause-set refinement (extract a core).
Hence, can only remove one clause.
A known problem for simpler techniques [1][2]
We actually have an idea how to extract a core (see understandings document) and 10/9/15.
[1] Nadel, Ryvchin, Strichman: Efficient MUS extraction with resolution. FMCAD’13
[2] Belov, Marques-Silva: MUSer2: An efficient MUS extractor. JSAT

17. Can we reconstruct a proof of ? ?
BL are derived based on a meta-argument, using the fact that the previous formula was unsatisfiable.
Our strategy: use the BL only when it is worth it…
Hence… we do not have a deductive proof that uses less than all of the clauses.

18. Repeating the process We apply two delays to encourage proofs of :
Initial delay until BL are computed / used
Many instances are solved fast without them
If proved unsat with BL, continue for a bounded amount of time with the hope to find a proof of .
If all else fails…
do not use BL until the next proof of .
In between such proofs, use redundancy-removal (BL = :c)
The advantage of the second type of delay is that it helps SAT cases

19. Experiments We compare to Path Strengthening [1] –
a prefix of clauses without siblings in cone(c) c2 ? c c3 c1 ’
[1] Nadel, Ryvchin, Strichman: Efficient MUS extraction with resolution. FMCAD’13

20. Results SAT’11 comp. benchmarks: 6% improvement.
Sat02¯ unsat benchmarks: 10% improvement.
Sat02¯
Sat02¯

21. Why such a small improvement ?
Benchmarks
With a 15 min. timeout, we can only compute MUS for easy instances.
Typically each iteration solved < 1 sec.
The P-time cost vs. exp-time benefit does not play much of a role in such formulas.

22. Why such a small improvement ?
A diminishing value of extra assumptions:
Sat02¯

23. Summary
We showed a P-time algorithm to extract Backbone Literals in an incremental setting.
A new type of incremental data
Challenge: fix the negative-feedback loop problem !
Implemented in HaifaMUC-1.3
…. Looking for collaborations on this topic … .

24. Optimization #1 for min. unsat core
Suppose
last proof used assumptions set A
(hence it is a proof of :A)
Clause c’ was not used in proof, and
Clause c’ is not a root of a clause in an l-cut, for l2A.
Hence c’ is not necessary for the proof and can be removed.
Apply this for every clause c’ not used in the proof.
Problem: for each l 2 A, find an l vertex cut (the highest possible, so as to minimize roots).
Solution: Go bottom up with l, stop at c if it has a parent p such that l  p.LitSet (and assert that l 2 c.LitSet).
Wrong! Core + core(BL) is insufficient.
There is some confusion here with negation, since the assumptions are negation of literals in the BL set.
So it is really:
A \subset \neg l, l \in BL(c’)

25. Used to prove :l1 and :l2
Can be removed
Used in proof
of (l1 Ç l2) c
l1-cut
l2-cut
Wrong! Core + core(BL) is insufficient.

26. The case of min. unsat core (MUC)
In contrast to general incremental SAT, where clauses are removed between instances from outside,
… in MUC we try to remove as many clauses as possible.

27. Optimization for min. unsat core
Suppose
last proof used assumptions set A
hence it is a proof of :A
Clause c’ was not used in proof,
A µ BL(c’)
Hence c’ is not necessary for the proof and can be removed.
Apply this for every clause c’ not used in the proof.
 n c’ ) :A
 n c’ ) A
This is pf_unsatopt in the code.
There is some confusion here with negation, since the assumptions are negation of literals in the BL set.
So it is really:
A \subset \neg l, l \in BL(c’)