Prolog - Latin Square solution - prolog

Question

I am trying to write a program in Prolog to find a Latin Square of size N.
I have this right now:
delete(X, [X|T], T).
delete(X, [H|T], [H|S]) :-
delete(X, T, S).
permutation([], []).
permutation([H|T], R) :-
permutation(T, X),
delete(H, R, X).
latinSqaure([_]).
latinSquare([A,B|T], N) :-
permutation(A,B),
isSafe(A,B),
latinSquare([B|T]).
isSafe([], []).
isSafe([H1|T1], [H2|T2]) :-
H1 =\= H2,
isSafe(T1, T2).

Answer 1

using SWI-Prolog library:
:- module(latin_square, [latin_square/2]).
:- use_module(library(clpfd), [transpose/2]).
latin_square(N, S) :-
numlist(1, N, Row),
length(Rows, N),
maplist(copy_term(Row), Rows),
maplist(permutation, Rows, S),
transpose(S, T),
maplist(valid, T).
valid([X|T]) :-
memberchk(X, T), !, fail.
valid([_|T]) :- valid(T).
valid([_]).
test:
?- aggregate(count,S^latin_square(4,S),C).
C = 576.
edit your code, once corrected removing typos, it's a verifier, not a generator, but (as noted by ssBarBee in a deleted comment), it's flawed by missing test on not adjacent rows.
Here the corrected code
delete(X, [X|T], T).
delete(X, [H|T], [H|S]) :-
delete(X, T, S).
permutation([], []).
permutation([H|T], R):-
permutation(T, X),
delete(H, R, X).
latinSquare([_]).
latinSquare([A,B|T]) :-
permutation(A,B),
isSafe(A,B),
latinSquare([B|T]).
isSafe([], []).
isSafe([H1|T1], [H2|T2]) :-
H1 =\= H2,
isSafe(T1, T2).
and some test
?- latinSquare([[1,2,3],[2,3,1],[3,2,1]]).
false.
?- latinSquare([[1,2,3],[2,3,1],[3,1,2]]).
true .
?- latinSquare([[1,2,3],[2,3,1],[1,2,3]]).
true .
note the last test it's wrong, should give false instead.

Answer 2

Like #CapelliC, I recommend using CLP(FD) constraints for this, which are available in all serious Prolog systems.
In fact, consider using constraints more pervasively, to benefit from constraint propagation.
For example:
:- use_module(library(clpfd)).
latin_square(N, Rows, Vs) :-
length(Rows, N),
maplist(same_length(Rows), Rows),
maplist(all_distinct, Rows),
transpose(Rows, Cols),
maplist(all_distinct, Cols),
append(Rows, Vs),
Vs ins 1..N.
Example, counting all solutions for N = 4:
?- findall(., (latin_square(4,_,Vs),labeling([ff],Vs)), Ls), length(Ls, L).
L = 576,
Ls = [...].
The CLP(FD) version is much faster than the other version.
Notice that it is good practice to separate the core relation from the actual search with labeling/2. This lets you quickly see that the core relation terminates also for larger N:
?- latin_square(20, _, _), false.
false.
Thus, we directly see that this terminates, hence this plus any subsequent search with labeling/2 is guaranteed to find all solutions.

Answer 3

I have better solution, #CapelliC code takes very long time for squares with N length higher than 5.
:- use_module(library(clpfd)).
make_square(0,_,[]) :- !.
make_square(I,N,[Row|Rest]) :-
length(Row,N),
I1 is I - 1,
make_square(I1,N,Rest).
all_different_in_row([]) :- !.
all_different_in_row([Row|Rest]) :-
all_different(Row),
all_different_in_row(Rest).
all_different_in_column(Square) :-
transpose(Square,TSquare),
all_different_in_row(TSquare).
all_different_in_column1([[]|_]) :- !.
all_different_in_column1(Square) :-
maplist(column,Square,Column,Rest),
all_different(Column),
all_different_in_column1(Rest).
latin_square(N,Square) :-
make_square(N,N,Square),
append(Square,AllVars),
AllVars ins 1..N,
all_different_in_row(Square),
all_different_in_column(Square),
labeling([ff],AllVars).