Programming challenge: wildcard exclusion in cartesian products

Dinko Tenev dinko.tenev at gmail.com
Fri Mar 17 11:05:45 CET 2006


Heh, here's a Prolog version:

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gen( _, 0, [] ) :- !.
gen( S, N, [H | T] ) :- member( H, S ), M is N - 1, gen( S, M, T ).

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Yep, that's it :)))

Here's how to test it:

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1 ?- gen([a, b, c], 3, X), print(X), nl, fail.
[a, a, a]
[a, a, b]
[a, a, c]
[a, b, a]
[a, b, b]
[a, b, c]
[a, c, a]
[a, c, b]
[a, c, c]
[b, a, a]
[b, a, b]
[b, a, c]
[b, b, a]
[b, b, b]
[b, b, c]
[b, c, a]
[b, c, b]
[b, c, c]
[c, a, a]
[c, a, b]
[c, a, c]
[c, b, a]
[c, b, b]
[c, b, c]
[c, c, a]
[c, c, b]
[c, c, c]

No
2 ?- gen([a, b, c], 3, X), not(member(X, [[a, _, _], [_, b, _], [_, _,
c]])), print(X), nl, fail.
[b, a, a]
[b, a, b]
[b, c, a]
[b, c, b]
[c, a, a]
[c, a, b]
[c, c, a]
[c, c, b]

No

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