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Asked: 2026-05-26T23:56:00+00:00

So here it is, for my sudoku project I have a list, such as

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So here it is, for my sudoku project I have a list, such as :

L = [_G1-0:0:0,_G19-0:6:2,_G22-0:7:2,_G25-0:8:2].

and I want to filter this to get only the free variables :

[_G1, _G19, _G22 and _G25].

So wanting that, I wrote this predicate :

remove_position([], []) :- !.
remove_position([X-_|Xs], [X|Rs]) :- remove_position(Xs, Rs).

When calling

?- remove_position(L, Result).

SWI-Prolog answers

false.

So I edited the predicate a little :

remove_position([], []) :- !.
remove_position([X-_:_:_|Xs], [X|Rs]) :- remove_position(Xs, Rs).

and now it works as expected :

?- remove_position(L, Result).
Result = [_G1, _G19, _G22, _G25].

Still, I can’t see what’s wrong with the first one, for me, SWI-Prolog should match 0:0:0 with _ just as well as with _:_:_. Is there something I do not get ? Or is it a SWI-Prolog bug ?

Thanks for any enlightenment !

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1 Answer

  1. Editorial Team

    When you have a term _G1-2:3:4 without any parentheses, you have to consider the associativity and precedence of the operators in order to find out how the term is parsed.

    The term _G1-2:3:4 is the same as (_G1-2):(3:4), not _G1-(2:3:4), and so the unification of _G1-2:3:4 and X-_ fails.

    If you add parentheses to your terms, like _G1-(2:3:4), then your original remove_position/2 predicate will work.

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