On Weakening Strategies for PB Solvers SAT Conference – July 6th, 2020 Daniel Le Berre 1 , Pierre Marquis 1 , 2 , Romain Wallon 1 1 CRIL, Univ Artois & CNRS 2 Institut Universitaire de France
Pseudo-Boolean (PB) Constraints PB solvers generalize SAT solvers to consider in which • the degree d is a non-negative integer 1/14 • normalized PB constraints ∑ n i = 1 a i l i ≥ d • cardinality constraints ∑ n i = 1 l i ≥ d • clauses ∑ n i = 1 l i ≥ 1 ≡ ∨ n i = 1 l i • the coeffjcients a i are non-negative integers • l i are literals, i.e., a variable v or its negation ¯ v = 1 − v
to learn new constraints Generalized Resolution The generalized resolution proof system [Hooker, 1988] is used in PB These two rules are used during confmict analysis (saturation) 2/14 solvers as the counterpart of the resolution proof system: b ¯ al + ∑ n l + ∑ n i = 1 a i l i ≥ d 1 i = 1 b i l i ≥ d 2 (cancellation) ∑ n i = 1 ( ba i + ab i ) l i ≥ bd 1 + ad 2 − ab ∑ n i = 1 a i l i ≥ d ∑ n i = 1 min( a i , d ) l i ≥ d
to learn new constraints Generalized Resolution The generalized resolution proof system [Hooker, 1988] is used in PB These two rules are used during confmict analysis (saturation) 2/14 solvers as the counterpart of the resolution proof system: b ¯ al + ∑ n l + ∑ n i = 1 a i l i ≥ d 1 i = 1 b i l i ≥ d 2 (cancellation) ∑ n i = 1 ( ba i + ab i ) l i ≥ bd 1 + ad 2 − ab ∑ n i = 1 a i l i ≥ d ∑ n i = 1 min( a i , d ) l i ≥ d
to learn new constraints Generalized Resolution The generalized resolution proof system [Hooker, 1988] is used in PB These two rules are used during confmict analysis (saturation) 2/14 solvers as the counterpart of the resolution proof system: b ¯ al + ∑ n l + ∑ n i = 1 a i l i ≥ d 1 i = 1 b i l i ≥ d 2 (cancellation) ∑ n i = 1 ( ba i + ab i ) l i ≥ bd 1 + ad 2 − ab ∑ n i = 1 a i l i ≥ d ∑ n i = 1 min( a i , d ) l i ≥ d
to learn new constraints Generalized Resolution The generalized resolution proof system [Hooker, 1988] is used in PB These two rules are used during confmict analysis (saturation) 2/14 solvers as the counterpart of the resolution proof system: b ¯ al + ∑ n l + ∑ n i = 1 a i l i ≥ d 1 i = 1 b i l i ≥ d 2 (cancellation) ∑ n i = 1 ( ba i + ab i ) l i ≥ bd 1 + ad 2 − ab ∑ n i = 1 a i l i ≥ d ∑ n i = 1 min( a i , d ) l i ≥ d
to learn new constraints Generalized Resolution The generalized resolution proof system [Hooker, 1988] is used in PB These two rules are used during confmict analysis (saturation) 2/14 solvers as the counterpart of the resolution proof system: b ¯ al + ∑ n l + ∑ n i = 1 a i l i ≥ d 1 i = 1 b i l i ≥ d 2 (cancellation) ∑ n i = 1 ( ba i + ab i ) l i ≥ bd 1 + ad 2 − ab ∑ n i = 1 a i l i ≥ d ∑ n i = 1 min( a i , d ) l i ≥ d
Generalized Resolution The generalized resolution proof system [Hooker, 1988] is used in PB These two rules are used during confmict analysis (saturation) 2/14 solvers as the counterpart of the resolution proof system: b ¯ al + ∑ n l + ∑ n i = 1 a i l i ≥ d 1 i = 1 b i l i ≥ d 2 (cancellation) ∑ n i = 1 ( ba i + ab i ) l i ≥ bd 1 + ad 2 − ab ∑ n i = 1 a i l i ≥ d ∑ n i = 1 min( a i , d ) l i ≥ d to learn new constraints
Analyzing Confmicts 4 b The constraint we obtain here is no longer confmicting! 20 2 h 2 g 2 f 3 d 8 e 15 c 15 a 6 d c 5 a Suppose that we have the following constraints: 7 h g f 4 e 6 c 6 b This confmict is analyzed by applying the cancellation rule as follows: (confmict) (reason for b) 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6
Analyzing Confmicts 4 b The constraint we obtain here is no longer confmicting! 20 2 h 2 g 2 f 3 d 8 e 15 c 15 a 6 d c 5 a Suppose that we have the following constraints: 7 h g f 4 e 6 c 6 b This confmict is analyzed by applying the cancellation rule as follows: (confmict) (reason for b) 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6
Analyzing Confmicts 4 b The constraint we obtain here is no longer confmicting! 20 2 h 2 g 2 f 3 d 8 e 15 c 15 a 6 d c 5 a Suppose that we have the following constraints: 7 h g f 4 e 6 c 6 b This confmict is analyzed by applying the cancellation rule as follows: (confmict) (reason for b) 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6
Analyzing Confmicts 4 b The constraint we obtain here is no longer confmicting! 20 2 h 2 g 2 f 3 d 8 e 15 c 15 a 6 d c 5 a Suppose that we have the following constraints: 7 h g f 4 e 6 c 6 b This confmict is analyzed by applying the cancellation rule as follows: (confmict) (reason for b) 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6
Analyzing Confmicts 5 a The constraint we obtain here is no longer confmicting! 20 2 h 2 g 2 f 3 d 8 e 15 c 15 a 6 d c 4 b 7 Suppose that we have the following constraints: h g f 4 e 6 c 6 b This confmict is analyzed by applying the cancellation rule as follows: (confmict) b) 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6 (reason for ¯
Analyzing Confmicts 5 a The constraint we obtain here is no longer confmicting! 20 2 h 2 g 2 f 3 d 8 e 15 c 15 a 6 d c 4 b 7 Suppose that we have the following constraints: h g f 4 e 6 c 6 b This confmict is analyzed by applying the cancellation rule as follows: (confmict) b) 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6 (reason for ¯
Analyzing Confmicts Suppose that we have the following constraints: b) (confmict) This confmict is analyzed by applying the cancellation rule as follows: The constraint we obtain here is no longer confmicting! 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6 (reason for ¯ 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6 15 a + 15 c + 8 e + 3 d + 2 f + 2 g + 2 h ≥ 20
Analyzing Confmicts Suppose that we have the following constraints: b) (confmict) This confmict is analyzed by applying the cancellation rule as follows: The constraint we obtain here is no longer confmicting! 3/14 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6 (reason for ¯ 6 ¯ b + 6 c + 4 e + f + g + h ≥ 7 5 a + 4 b + c + d ≥ 6 15 a + 15 c + 8 e + 3 d + 2 f + 2 g + 2 h ≥ 20
Weakening To preserve the confmict, the weakening rule must be used: (weakening) (partial weakening) Weakening can be applied in many different ways! 4/14 al + ∑ n i = 1 a i l i ≥ d ∑ n i = 1 a i l i ≥ d − a al + ∑ n k ∈ N i = 1 a i l i ≥ d 0 < k ≤ a ( a − k ) l + ∑ n i = 1 a i l i ≥ d − k
Weakening To preserve the confmict, the weakening rule must be used: (weakening) (partial weakening) Weakening can be applied in many different ways! 4/14 al + ∑ n i = 1 a i l i ≥ d ∑ n i = 1 a i l i ≥ d − a al + ∑ n k ∈ N i = 1 a i l i ≥ d 0 < k ≤ a ( a − k ) l + ∑ n i = 1 a i l i ≥ d − k
Weakening To preserve the confmict, the weakening rule must be used: (weakening) (partial weakening) Weakening can be applied in many different ways! 4/14 al + ∑ n i = 1 a i l i ≥ d ∑ n i = 1 a i l i ≥ d − a al + ∑ n k ∈ N i = 1 a i l i ≥ d 0 < k ≤ a ( a − k ) l + ∑ n i = 1 a i l i ≥ d − k
Weakening To preserve the confmict, the weakening rule must be used: (weakening) (partial weakening) Weakening can be applied in many different ways! 4/14 al + ∑ n i = 1 a i l i ≥ d ∑ n i = 1 a i l i ≥ d − a al + ∑ n k ∈ N i = 1 a i l i ≥ d 0 < k ≤ a ( a − k ) l + ∑ n i = 1 a i l i ≥ d − k
Weakening To preserve the confmict, the weakening rule must be used: (weakening) (partial weakening) Weakening can be applied in many different ways! 4/14 al + ∑ n i = 1 a i l i ≥ d ∑ n i = 1 a i l i ≥ d − a al + ∑ n k ∈ N i = 1 a i l i ≥ d 0 < k ≤ a ( a − k ) l + ∑ n i = 1 a i l i ≥ d − k
Weakening To preserve the confmict, the weakening rule must be used: (weakening) (partial weakening) Weakening can be applied in many different ways! 4/14 al + ∑ n i = 1 a i l i ≥ d ∑ n i = 1 a i l i ≥ d − a al + ∑ n k ∈ N i = 1 a i l i ≥ d 0 < k ≤ a ( a − k ) l + ∑ n i = 1 a i l i ≥ d − k
least one of the constraints, the derived constraint is Different Weakening Strategies The original approach [Dixon, 2002; Chai & Kuehlmann, 2003] successively weakens away literals from the reason, until the saturation rule guarantees to derive a confmicting constraint As the operation must be repeated multiple times, its cost is not negligible Another solution is to take advantage of the following property: As soon as the coeffjcient of the literal to cancel is equal to 1 in at guaranteed to be confmicting [Dixon, 2004] 5/14
least one of the constraints, the derived constraint is Different Weakening Strategies The original approach [Dixon, 2002; Chai & Kuehlmann, 2003] successively weakens away literals from the reason, until the saturation rule guarantees to derive a confmicting constraint As the operation must be repeated multiple times, its cost is not negligible Another solution is to take advantage of the following property: As soon as the coeffjcient of the literal to cancel is equal to 1 in at guaranteed to be confmicting [Dixon, 2004] 5/14
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