Propia and CHRs Transforming Constraint Specification into Constraint Behaviour

2 Motivation for Propia and CHRs  Encode constraints in readable form  Set up model without leaving choice points  Support constraint propagation NB: Never introduce choice points in constraint setup Why not? repeatedly setting up constraints on different branches partially blind searching

3 The noclash Example noclash(S,T)  (T  S+5)  (S  T) Specification: Encoding as ECLiPSe clauses: noclash(S,T) :- ic:(T >= S+5). noclash(S,T) :- ic:(S >= T). Model Setup: setup(S) :- S::1..10, constraint(noclash(S,6)).

4 The noclash constraint in ECLiPSe Prolog constraint(Goal) :- call(Goal). This program leaves a choice point ?- setup(S). yes S = 1 noclash(S,T) :- ic:(T >= S+5). noclash(S,T) :- ic:(S >= T). setup(S) :- S::1..10, constraint(noclash(S,6)).

5 noclash with Propia constraint(Goal) :- Goal infers most. S{[1, 6..10]} This program does not create any choice points ?- setup(S). yes noclash(S,T) :- ic:(T >= S+5). noclash(S,T) :- ic:(S >= T). setup(S) :- S::1..10, constraint(noclash(S,6)).

6 noclash with CHRs constraint(noclash(S,T)) ic:(T = T). constraint(noclash(S,T)) ic:(S = S+5). This program does not create any choice points ?- setup(S). yes S{1..10} noclash(S,T) :- ic:(T >= S+5). noclash(S,T) :- ic:(S >= T). setup(S) :- S::1..10, constraint(noclash(S,6)).

7 Choice Points and Efficiency Experiment sum(Products,Raw1,Raw2,Profit) :- ( foreach(Item,Products), foreach(R1,R1List), foreach(R2,R2List), foreach(P,PList) do product(Item,R1,R2,P) ), Raw1 #= sum(R1List), Raw2 #= sum(R2List), Profit #= sum(PList). product( 101,1,19,1). product( 102,2,17,2). product( 103,3,15,3). product( 104,4,13,4). product( 105,10,8,5). product( 106,16,4,4). product( 107,17,3,3). product( 108,18,2,2). product( 109,19,1,1). £1 1pin 19nuts £2 2pins 17nuts £P R1pins R2nuts Raw1 = R1 Raw2 = R2 Profit= P Products:

8 Choice Points and Efficiency product_plan(Products) :- length(Products,9), Raw1 #=< 95, Raw2 #=< 95, Profit #>= 40, sum(Products,Raw1,Raw2,Profit), labeling(Products). sum(Products,Raw1,Raw2,Profit) :- ( foreach(Item,Products), foreach(R1,R1List), foreach(R2,R2List), foreach(P,PList) do product(Item,R1,R2,P) ), Raw1 #= sum(R1List), Raw2 #= sum(R2List), Profit #= sum(PList). infers most 33 secs Time for all solutions: Time to first solution: 1 hour 95>= R1+R1+R1+R1+R1+R1+R1+R1+R1 95>= R2+R2+R2+R2+R2+R2+R2+R2+R2 40=< P +P +P +P +P +P +P +P +P

9 Propia  Annotations  Most Specific Generalisation  Algorithm

10 Propia’s infers Annotation Propia annotation says what you want to infer:  Goal infers consistent Fail as soon as inconsistency can be proven  Goal infers unique Instantiate as soon as unique solution exists  Goal infers ac Maintain finite domains covering all solutions  Goal infers most Use strongest available representation covering all solutions

11 Annotation Results p(1,1). p(1,3). p(1,4). ?- p(X,Y) infers consistent. No information extracted ?- p(X,3) infers unique. X=1 ?- p(X,Y) infers most. X=1 (assuming ic is not loaded) ?- p(X,Y) infers most. X=1,Y{[1,3,4]} (assuming ic is loaded) ?- p(X,Y) infers ac. X=1,Y{[1,3,4]} (ic must be loaded)

12 More “global” Consistency pp(X,Y,Z) :- p(X,Y), p(Y,Z). ?- pp(X,Y,Z) infers most. X=1, Y=1, Z{[1,3,4]} C1 C5 C4 C2 C3 C6 C7 C8 Problem Constraints (C1,C4,C5) infers most

13 Most Specific Generalisation The Opposite of Unification! [1,A] [1,1] [1,_] [1,1] [2,2] [X,X] “most specific generalisations”

14 Most Specific Generalisation over Finite Domains 1 2 X::{[1,2]} f(1,2) f(2,4) f(X::{1,2},Y::{2,4}) “most specific generalisations”

15 Naïve Propia Algorithm Goal infers most Find all solutions to Goal, and put them in a set Find the most specific generalisation of all the terms in the set member(X,[1,2,3]) infers most Find all solutions to member(X,[1,2,3]): {1,2,3} Find the most specific generalisation of {1,2,3}: X::{1,2,3}

16 Propia Algorithm Goal infers most Find one solution S to Goal The current most specific generalisation MSG = S Repeat Find a solution NewS to Goal which is NOT an instance of MSG Find the most specific generalisation NewMSG of MSG and NewS MSG := NewMSG until no such solution remains

17 Example - without finite domains p(1,2). p(1,3). p(2,3). p(2,4). p(X,Y) infers most MSG: 1st Iteration: p(1,2) 2nd Iteration: p(1,_) 3rd Iteration: p(_,_) No more iterations

18 Propia Algorithm for Arc Consistency p(1,2). p(1,3). p(2,3). p(2,4). This algorithm only works for predicates defined by ground facts p(X,Y) infers ac is implemented as: element(I,[1,1,2,2],X), element(I,[2,3,3,4],Y).

19 CHR - Constraints Handling Rules  CHR -- set of rewrite rules for constraints  Can be mixed with normal Prolog code H 1...,H n |.  CHR -- set of rewrite rules for constraints  Can be mixed with normal Prolog code H 1...,H n |. Constraint Store add, “tell”“ask”remove

20 Clauses, Constraints, Goals prec(X1,3,X2) X1::4..6 X2=6 Constraint Store ?- prec(X,D,X) Goal D::-inf..0 prec(X,N,Y) X=Y | ic:(N=<0) Clause HeadBody Guard prec(X,D,X)

21 Clauses, Constraints, Goals le(A,B) Constraint Store ?- le(B,C) Goal le(A,C) le(X,Y1), le(Y2,Z) ==> Y1=Y2 | le(X,Z) Clause HeadBodyGuard le(B,C)

22 Constraint Store Simplification Rule le(A,B) ?- le(B,A) Goal le(B,A) le(A,B) le(X,Y), le(Y,X) X=Y Clause HeadBody A=B

23 Weak Guard Entailment Test nc(X,Y) ==> Y<X | Y=0 ?- Y<X, nc(X,Y). X = X Y = Y

24 Definition of “less than or equal” (le) :- constraints le/2. le(X,Y), le(Y,Z) ==> le(X,Z). le(X,Y), le(Y,X) X=Y. le(X,Y) ==> X=<Y. le(X,Y) X=<Y | true. Constraint Store ?- le(1,2) Goal le(1,2) true

25 Entailment Testing Constraint Store le(A,B) ?- min(A,B,Min) Goal min(A,B,Min) le(X,Y) \ min(X,Y,Z) ==> Z=X Clause HeadBody Min=A

26 Entailment Testing – without CHR’s min1(X,Y,Z) :- ic: (Z=<X), ic: (Z=<Y), ge1(X,Y,Z). ?- ic: (X=<Y), ic: (X,Y,B). Yes B::0..1 ge1(X,Y,Z) :- ‘=<‘(X,Y,B), ge2(B,X,Y,Z). delay ge2(B,_,_,_) if var(B). ge2(0,_,Y,Z) :- Z=Y. ge2(1,X,_,Z) :- Z=X.

27 CHR-Defined Precedence prec(Time1,Duration,Time2) means that Time1 precedes Time2 by at least Duration. Note that: Time1 and Time2 may be variables or numbers Duration is a number Duration may be negative ( prec(S1,-5,S2) means that S2 precedes S1 by no more than 5)

28 Syntax and semantics for CHRs  Simplification rule: H 1...,H n |.  Propagation rule: H 1...,H n ==> |.  Simpagation rule: H 1..\..H n |.  Heads are match only - no binding of external vars.  Goals are any Prolog goals or constraints.  Constraint added to store if no rule applicable.  Simplification rule: H 1...,H n |.  Propagation rule: H 1...,H n ==> |.  Simpagation rule: H 1..\..H n |.  Heads are match only - no binding of external vars.  Goals are any Prolog goals or constraints.  Constraint added to store if no rule applicable.

29 CHRs for prec/3 %A timepoint precedes itself by a maximum duration of 0 prec(S,D,S) D=<0. ( implicit guard in the head ) %Two timepoints preceding each other by (at least) 0 are the same prec(S1,0,S2), prec(S2,0,S1) S1=S2. ( rule with two atoms in the head ) %Transitivity prec(S1,D1,S2),prec(S2,D2,S3) ==> D3 is D1+D2, prec(S1,D3,S3). ( propagation rule ) % prec(S1,D1,S2) is redundant prec(S1,D1,S2) \ prec(S1,D2,S2) D2>=D1 | true. ( simpagation rule )

30 Complete Precedence Entailment % Assume D is a number noclash(S,T)\ prec(T,D,S) D >= -5 | prec(T,0,S). noclash(S,T)\ prec(S,D,T) D >= 0 | prec(S,5,T). prec(S,D,T) ==> ic:(T>=S+D).

31 Global Consistency  Reasoning on Combinations of Constraints Propia and CHRs can apply consistency techniques to combinations of constraints  Propia for a priori combinations Propia can be applied to program-defined predicates These predicates combine sets of constraints selected in advance of program execution  CHRs for combining newly posted constraints CHR multi-headed rules can match combinations of constraints CHR multi-headed rules can match constraints newly posted during search

32 CHR Exercise Using CHR, axiomatise constraints “less than” and “minimum” Ensure logical completeness.

33 Propia and CHR Exercise Implement three constraints, 'and', 'or' and 'xor' in Propia in CHR (if you have time) The constraints are specified as follows: All boolean variables have domain [0,1]: 0 for 'false' 1 for 'true and(X,Y,Z) =def (X&Y) = Z or(X,Y,Z) =def (X or Y) = Z xor(X,Y,Z) =def ((X & -Y) or (-X & Y)) = Z

34 Testing Your Solution Suppose your constraints are called cons_and, cons_or and cons_xor Now write enter the following procedure: full_adder(I1,I2,I3,O1,O2) :- cons_xor(I1,I2,X1), cons_and(I1,I2,Y1), cons_xor(X1,I3,O1), cons_and(I3,X1,Y2), cons_or(Y1,Y2,O2). The test is : ?- full_adder(I1,I2,0,O1,1).