Hi,
You wrote into the Qilang News group with your problem.
This is a solution in 17 lines of Qi for any n-product >= 2.
It falls short of your complete requirement since it uses
generate and then test, rather than interleaving the
two.
(define challenge
Patterns N X -> (filter (/. Y (member Y Patterns)) (n-product N X)))
(define n-product
2 X -> (cartesian-product l X X)
N X -> (cartesian-product c X (n-product (- N 1) X)))
(define cartesian-product
_ [ ] _ -> [ ]
c [X | Y] Z -> (append (map (/. W [X | W]) Z) (cartesian-product c Y
Z))
l [X | Y] Z -> (append (map (/. W [X W]) Z) (cartesian-product l Y
Z)))
(define filter
_ [] -> []
F [X | Y] -> (filter F Y) where (F X)
F [X | Y] -> [X | (filter F Y)])
(define member
_ [] -> false
X [Pattern | _] -> true where (query-prolog [[= Pattern X]])
X [_ | Patterns] -> (member X Patterns))
Notes:
Pattern filtering is done by a unification test within the member
function. You
can do this most easily by calling Qi Prolog using query-prolog.
Here's a test.
(42 -) (n-product 3 [a b c])
[[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]]
OK, remove all lists beginning [a a ....].
(43-) (challenge [[a a | X]] 3 [a b 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]]
Remove all lists beginning with a or b.
(51-) (challenge [[a | X] [b | X]] 3 [a b 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]]
Mark
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