在prolog中将两个列表相乘

Question

I am currently working with prolog and want to multiply two lists together but in a certian way. For example:

[1,2,3] and [4,5,6] are my two lists.

I want to preform the following actions:

(1*4)+(2*5)+(3*6) = 32

Such that the first element of each list is multiplied to each other then added with the second elements multiplied together etc.

Is this possible to go in Prolog?

I know in other languages you can do a recursive function with takes the head of the list and the tail (the rest of the entries). This allows for a simple multiplication but I do not think that is possible in prolog?

Answer 1

Using built-ins:

mult(X, Y, Z) :- Z is X * Y.

sum_prod(A, B, SumProd) :-
    maplist(mult, A, B, Prods),
    sumlist(Prods, SumProd).   % In GNU Prolog this is sum_list

Using simple recursion:

sum_prod([A|As], [B|Bs], SumProd) :-
    sum_prod(As, Bs, SP),
    SumProd is SP + A*B.
sum_prod([], [], 0).

Using tail recursion:

sum_prod(A, B, SumProd) :-
    sum_prod(A, B, 0, SumProd).
sum_prod([A|As], [B|Bs], Acc, SumProd) :-
    Acc1 is Acc + A*B,
    sum_prod(As, Bs, Acc1, SumProd).
sum_prod([], [], Acc, Acc).

Answer 2

If all items of your lists are integers and your Prolog implementation offers clpfd , you can simply use the clpfd built-in predicate scalar_product/4 , like this:

?- scalar_product([1,2,3],[4,5,6],#=,Product).
Product = 32.

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Edit: You may also be interested in the related question " Prolog: Multiplying 2 lists with 1 of them not instantiated? ", particularly in this answer .

Answer 3

as an alternative to 'hand coded' loops, using library( aggregate ) and nth1 /3:

sum_prod(A,B,S) :-
    aggregate(sum(M), I^X^Y^(nth1(I,A,X), nth1(I,B,Y), M is X*Y), S).