Encodings into SAT Combinatorial Problem Solving (CPS) Enric Rodr - PowerPoint PPT Presentation

Encodings into SAT Combinatorial Problem Solving (CPS) Enric Rodr´ ıguez-Carbonell May 29, 2020

What is an encoding? Language of SAT solvers: CNF propositional formulas ■ To solve combinatorial problems with SAT solvers, ■ constraints have to be represented in this language An encoding of a constraint C into SAT is ■ a CNF F that expresses C , so that there is a bijection solutions to C ⇐ ⇒ models of F 2 / 23

Examples: AMO constraints An AMO constraint is of the form x 0 + . . . + x n − 1 ≤ 1 ■ where each x i is 0-1 (At Most One of the variables can be true) Quadratic encoding. ■ Variables: the same x 0 , . . . , x n − 1 ◆ Clauses: for 0 ≤ i < j < n , x i ∨ x j ◆ � n = O ( n 2 ) clauses � Requires ◆ 2 Other encodings try to use fewer clauses, ■ at the cost of introducing new variables 3 / 23

Examples: AMO constraints Logarithmic encoding. Let m = ⌈ log 2 n ⌉ . Then: ■ Variables: the x i and new variables y 0 , y 1 , . . . , y m − 1 ◆ Clauses: for 0 ≤ i < n , 0 ≤ j < m ◆ x i ∨ y j if the j -th digit in binary of i is 1 ■ x i ∨ y j otherwise ■ Requires O (log n ) new variables, O ( n log n ) clauses ◆ Heule encoding. ■ If n ≤ 3 , the encoding is the quadratic encoding. ◆ If n ≥ 4 , introduce an auxiliary variable y and encode (recursively) ◆ x 0 + x 1 + y ≤ 1 and x 2 + · · · + x n − 1 + y ≤ 1 . Requires O ( n ) new variables, O ( n ) clauses ◆ Other encodings exist (see next) ■ 4 / 23

Consistency and Arc-Consistency Let us consider an encoding of a constraint C such that ■ there is a correspondence between maps of the variables of C to their domains, and partial assignments of the boolean variables of the encoding The encoding is consistent if ■ whenever M is not compatible with any solution to C , unit propagation on the boolean assignment of M leads to a conflict The encoding is arc-consistent if ■ it is consistent, and ◆ unit propagation discards arc-inconsistent values ◆ (i.e., values without a support) These are good properties for encodings: ■ SAT solvers are very good at unit propagation! 5 / 23

Consistency and Arc-Consistency In the case of an AMO constraint x 0 + . . . + x n − 1 ≤ 1 : ■ Consistency ≡ if there are two true vars x i in M or more, ■ then unit propagation should give a conflict Arc-consistency ≡ Consistency + if there is one true var x i in M , ■ then unit propagation should set all others x j to false The quadratic, logarithmic and Heule encodings ■ are all arc-consistent 6 / 23

Cardinality Constraints A cardinality constraint is of the form x 1 + . . . + x n ⊲ ⊳ k ■ where each x i is 0-1 and ⊲ ⊳ ∈ {≤ , <, ≥ , >, = } AMO are a particular case of card. constraints where k = 1 and ⊲ ⊳ is ≤ ■ Without loss of generality we may assume ⊲ ⊳ is < , i.e., ■ x 1 + . . . + x n < k Naive encoding. ■ Variables: the same x 1 , . . . , x n ◆ Clauses: for all 1 ≤ i 1 < i 2 < . . . < i k ≤ n , ◆ x i 1 ∨ x i 2 ∨ . . . ∨ x i k � n � This is clauses! ◆ k 7 / 23

Adders Again, other encodings try to use fewer clauses, ■ at the cost of introducing new variables Adder encoding. ■ Build an adder circuit by using bit-adders as building blocks: x s Full Adder y c z ↔ XOR( x, y, z ) s ↔ ( x ∧ y ) ∨ ( x ∧ z ) ∨ ( y ∧ z ) c where XOR( x, y, z ) is short for ( x ∧ y ∧ z ) ∨ ( x ∧ y ∧ z ) ∨ ( x ∧ y ∧ z ) ∨ ( x ∧ y ∧ z ) 8 / 23

Adders Encodings of this kind are not arc-consistent. ■ Consider x + y + z ≤ 0 . ■ Then unit propagation should propagate x, y, z . Let us encode the constraint with a full adder ■ The encoding is the Tseitin transformation of s, c and ■ ↔ XOR( x, y, z ) s ↔ ( x ∧ y ) ∨ ( x ∧ z ) ∨ ( y ∧ z ) c Note that ■ → ( x ∨ y ∨ z ) ∧ ( x ∨ y ∨ z ) ∧ ( x ∨ y ∨ z ) ∧ ( x ∨ y ∨ z ) s → ( x ∨ y ) ∧ ( x ∨ z ) ∧ ( y ∨ z ) c Unit propagation cannot propagate anything! ■ 9 / 23

Sorting Networks Sorting Network encoding. ■ Pass x 1 , . . . , x n as inputs to a circuit that sorts (say, decreasingly) n bits. Let y 1 , . . . , y n be the outputs of this circuit. Then if the constraint to be encoded is � n i =1 x i ≥ k , then add clause y k ◆ � n i =1 x i ≤ k , then add clause y k +1 ◆ � n i =1 x i = k , then add clauses y k , y k +1 ◆ 10 / 23

Sorting Networks How to build such a sorting circuit? ■ A possibility is to implement mergesort ■ In what follows: so-called odd-even sorting networks ■ The basic block of odd-even sorting networks are 2-comparators ■ 11 / 23

2-comparators A 2-comparator is a sorting network of size 2: ■ it has 2 input variables ( x 1 and x 2 ) ◆ it has 2 output variables ( y 1 and y 2 ) ◆ y 1 is true if and only if at least one of the input variables is true ◆ (i.e., it is the maximum or disjunction) y 2 is true if and only if both two input variables are true ◆ (i.e., it is the minimum or conjunction) 12 / 23

2-comparators Clauses: ■ ← ← x 1 ∨ x 2 ← x 1 y 2 , x 2 y 2 , y 1 , → → x 1 ∧ x 2 → x 1 y 1 , x 2 y 1 , y 2 Graphical representation: ■ y 1 x 1 y 2 x 2 Some simplifications are possible: ■ For ≥ constraints: top three clauses suffice ◆ For ≤ constraints: bottom three clauses suffice ◆ For = constraints: all clauses needed ◆ 13 / 23

2-comparators Clauses: ■ ← ← x 1 ∨ x 2 ← x 1 y 2 , x 2 y 2 , y 1 , ← ← x 1 ∨ x 2 ← x 1 y 1 , x 2 y 1 , y 2 Graphical representation: ■ y 1 x 1 y 2 x 2 Some simplifications are possible: ■ For ≥ constraints: top three clauses suffice ◆ For ≤ constraints: bottom three clauses suffice ◆ For = constraints: all clauses needed ◆ 13 / 23

Merge Networks From now on we assume that n is a power of two ■ (if not, pad with variables set to false) A merge network takes as input two ordered sets of variables of size n and ■ produces an ordered output of size 2 n . Let ( x 1 , . . . , x n ) and ( x ′ 1 , . . . , x ′ n ) be the inputs. ■ We recursively define a merge network as follows: If n = 1 , a merge network is a 2-comparator: ■ Merge( x 1 ; x ′ 1 ) := 2-Comp( x 1 , x ′ 1 ) . 14 / 23

Merge Networks For n > 1 : Let us define ■ ( z 1 , z 3 , . . . , z 2 n − 1 ) = Merge( x 1 , x 3 , . . . , x n − 1 ; x ′ 1 , x ′ 3 , . . . x ′ n − 1 ) , ( z 2 , z 4 , . . . , z 2 n ) = Merge( x 2 , x 4 , . . . , x n ; x ′ 2 , x ′ 4 , . . . , x ′ n ) , ( y 2 , y 3 ) = 2-Comp( z 2 , z 3 ) , ( y 4 , y 5 ) = 2-Comp( z 4 , z 5 ) , . . . ( y 2 n − 2 , y 2 n − 1 ) = 2-Comp( z 2 n − 2 , z 2 n − 1 ) Then, Merge( x 1 , x 2 , . . . , x n ; x ′ 1 , x ′ 2 , . . . , x ′ n ) := ( z 1 , y 2 , y 3 , . . . , y 2 n − 1 , z 2 n ) 15 / 23

Merge Networks z 1 x 1 z 1 z 2 x 2 y 2 z 3 x 3 y 3 z 4 x 4 y 4 Merge n =2 Merge n =2 y 5 x ′ z 5 1 y 6 x ′ z 6 2 y 7 x ′ z 7 3 z 8 x ′ z 8 4 16 / 23

Merge Networks Sketch of the proof of correctness of Merge : By IH: { x 1 , x 3 , . . . , x n − 1 , x ′ 1 , x ′ 3 , . . . , x ′ n − 1 } = { z 1 , z 3 , . . . , z 2 n − 1 } x ′ 2 , x ′ 4 , . . . , x ′ By IH: { x 2 , x 4 , . . . , x n , n } = { z 2 , z 4 , . . . , z 2 n } x ′ 1 , x ′ 2 , . . . , x ′ Hence { x 1 , x 2 , . . . , x n , n } = { z 1 , z 2 , . . . , z 2 n } And ( y 2 , y 3 ) = 2-Comp( z 2 , z 3 ) implies { y 2 , y 3 } = { z 2 , z 3 } ( y 4 , y 5 ) = 2-Comp( z 4 , z 5 ) implies { y 4 , y 5 } = { z 4 , z 5 } . . . ( y 2 n − 2 , y 2 n − 1 ) = 2-Comp( z 2 n − 2 , z 2 n − 1 ) implies { y 2 n − 2 , y 2 n − 1 } = { z 2 n − 2 , z 2 n − 1 } So { x 1 , x 2 , . . . , x n , x ′ 1 , x ′ 2 , . . . , x ′ n } = { z 1 , y 2 , y 3 , . . . , y 2 n − 2 , y 2 n − 1 , z 2 n } 17 / 23

Merge Networks Let us prove outputs are sorted decreasingly. For 1 ≤ i < n − 1 let us see: z 2 i ≥ z 2( i +1)+1 : ■ Let us see z 2( i +1)+1 = 1 implies z 2 i = 1 If z 2( i +1)+1 = z 2 i +3 = z 2( i +2) − 1 = 1 there are i + 2 1’s in odd x, x ′ Let p be the number of 1’s in odd x Let q the number of 1’s in odd x ′ Then p + q = i + 2 As x, x ′ is ordered decreasingly, there are p − 1 1’s in even x , q − 1 1’s in even x ′ So altogether there are ( p − 1) + ( q − 1) = p + q − 2 = i 1’s in even x, x ′ Hence z 2 i = 1 18 / 23

Merge Networks Let us prove outputs are sorted decreasingly. For 1 ≤ i < n − 1 let us see: z 2 i ≥ z 2( i +1)+1 : proved ■ 18 / 23

Merge Networks Let us prove outputs are sorted decreasingly. For 1 ≤ i < n − 1 let us see: z 2 i ≥ z 2( i +1)+1 : proved ■ z 2 i ≥ z 2( i +1) : by IH ■ 18 / 23

Merge Networks Let us prove outputs are sorted decreasingly. For 1 ≤ i < n − 1 let us see: z 2 i ≥ z 2( i +1)+1 : proved ■ z 2 i ≥ z 2( i +1) : by IH ■ z 2 i +1 ≥ z 2( i +1)+1 : by IH ■ 18 / 23