Prolog: DRY way of creating lists declaratively

I'm doing some experiments with Prolog a开发者_如何学Gond having difficulties with the following rule:

row(Row, Matrix, [R1,R2,R3,R4]) :-
  cell(1, Row, Matrix, R1),
  cell(2, Row, Matrix, R2),
  cell(3, Row, Matrix, R3),
  cell(4, Row, Matrix, R4).

This rule extracts one row from a matrix, given its row number. For example,

row(2, [1,2,3,4,5,6,7,8], X)
 X = [5,6,7,8]

What nags me is that there is lots of repetition in that code. After finishing with 4x4 matrices, I will have to deal with 9x9 ones. And the code can get very non DRY.

Is there a way to extract that repetition out?

Thanks.

Edit: The complete code giving me trouble is here: https://github.com/kikito/7-languages-in-7-weeks/blob/master/3-prolog/day-3/sudoku-refactor.pl

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After writing my first answer, I realized that you can also simplify your program using findall

 row(Row, Matrix, L) :- findall(X,cell(_,Row,Matrix,X),L).

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I would think about changing the representation to a list of lists rather than a flat list, then selecting a row becomes very easy. You can just use the built-in nth1/3:

 :- use_module(library(lists)). % I use Sicstus Prolog
 row(N,M,X) :- nth1(N,M,X).
 cell(R,C,M,X) :- nth1(R,M,Y), nth1(C,Y,X).
 column(N,M,X) :- findall(Y,(nth1(_,M,Z), nth1(N,Z,Y)),X).

 m([[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]).
 example(Row,Cell,Column) :- m(M), row(2,M,Row), cell(2,3,M,Cell), column(2,M,Column).

 %| ?- example(A,B,C).
 %A = [5,6,7,8],
 %B = 7,
 %C = [2,6,10,14] ? ;
 %no