From: Joachim Schimpf <jschimpf_at_...311...>

Date: Fri, 1 Jul 2016 13:02:46 +0100

Date: Fri, 1 Jul 2016 13:02:46 +0100

On 30/06/16 15:01, Annick Fron wrote: > Hi, > > Your mail helped me a lot. > I rewrote the program , but got some problems to define that all columns must be different. > I wrote something, but which gives me sometimes the same columns. > What is wrong with it ? ... > > differentcols(M,Nb,N):- > (for(I,1,Nb),param(M,N,Nb) do > Col1 is M[I], > Iplus1 is I + 1, > (for(J,Iplus1,Nb),param(M,Col1,N) do > Col2 is M[J], > not(equalcol(Col1,Col2)))). > > equalcol(X,Y,N):-dim(X,[N]),dim(Y,[N]),X#::0..1, Y#::0..1, > ( for(J,1,N),param(X,Y) do A is X[J], B is Y[J], A#=B). Two problems: 1. you are confusing rows and columns (M[I] is the Ith row) 2. you cannot use Prolog negation (not/1 or \+/1) to negate constraints You could do the following: diff_cols(A) :- dim(A, [M,N]), ( for(J,1,N-1), param(A,M,N) do ( for(K,J+1,N), param(A,M,J) do % compare columns J and K ( for(I,1,M), foreach(B,Bs), param(A,J,K) do B #= (A[I,J] #\= A[I,K]) ), sum(Bs) #> 0 % at least one different element ) ). However, depending on your actual problem, it might be better to have a constraint that requires not only that the columns are different, but that they are lexicographically ordered. This will eliminate symmetric solutions, i.e. solutions that differ only in that the columns are permuted. Of course this is only correct if your problem is indeed symmetric with regard to the column order. To impose such an order (which subsumes the difference constraint), use: ordered_cols(A) :- dim(A, [M,N]), ( for(J,1,N-1), param(A,M) do lex_lt(A[1..M,J], A[1..M,J+1]) ). -- JoachimReceived on Fri Jul 01 2016 - 12:02:57 CEST

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