prolog - How to create a list of fixed length and sums up to a certain number while choosing from its elements from other list -

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assuming have list l=[1,2,3] want create list o length of o n , total sum of elements s has created using elements l if contain duplicates.

examples n=5 , s=7:

o=[2,2,1,1,1] o=[2,1,2,1,1] o=[3,1,1,1,1] o=[1,3,1,1,1]

here's variant similar @wouterbeek's solution doesn't use separate accumulator:

summate_tokens(types, sum, n, [x|t]) :-     n > 0,     member(x, types),     sum1 sum - x,       % subtract running sum     n1 n - 1,     summate_tokens(types, sum1, n1, t). summate_tokens(_, 0, 0, []).

it implicitly handles length of solution counter, n. work kind of numbers in types argument.

with results:

?- summate_tokens([1,2,3], 7, 5, tokens). tokens = [1, 1, 1, 1, 3] ; tokens = [1, 1, 1, 2, 2] ; tokens = [1, 1, 1, 3, 1] ; tokens = [1, 1, 2, 1, 2] ; tokens = [1, 1, 2, 2, 1] ; tokens = [1, 1, 3, 1, 1] ; tokens = [1, 2, 1, 1, 2] ; tokens = [1, 2, 1, 2, 1] ; tokens = [1, 2, 2, 1, 1] ; tokens = [1, 3, 1, 1, 1] ; tokens = [2, 1, 1, 1, 2] ; tokens = [2, 1, 1, 2, 1] ; tokens = [2, 1, 2, 1, 1] ; tokens = [2, 2, 1, 1, 1] ; tokens = [3, 1, 1, 1, 1] ; false.  ?-

if types , sum limited non-negative numbers, including sum1 >= 0 after sum1 sum - x improve performance.


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