我正在尝试使用Prolog来实现乘以两个多项式的多项式乘法。下面是SML中给出的代码,但我需要在Prolog中使用它。 SML的测试解决方案是poly_mult([1.0,5.0,1.0],[3.0,~10.0,15.0]);并将返回val it = [3.0,5.0,~32.0,65.0,15.0]:真实列表我试图在Prolog中编写代码,但我所拥有的是不正确的。有人可以帮忙吗?谢谢!
IN SML
fun poly_add (M,nil) = M
| poly_add (nil,N) = N
| poly_add ((m:real)::mr, n::nr) = (m+n)::poly_add(mr,nr);
fun scalar_mult (nil,m) = nil
| scalar_mult((m:real)::mr,n) = (m*n)::scalar_mult(mr,n);
fun poly_mult (M,nil) = nil
| poly_mult (M,n::nr) = poly_add (scalar_mult(M,n), 0.0::poly_mult(M,nr));
我在PROLOG中有什么
poly_add(Constant,[],Constant) :- !.
poly_add([],Constant2,Constant) :- !.
poly_add([Head1 | Head2], [Tail | Tail2], [HeadSum |TailSum]) :-
HeadSum is Head + Tail,
poly_add(Head2, Tail2, TailSum).
scal_mult([],Head,[]) :- !.
scal_mult([Head | Head2], [Tail], [HeadMult | TailMult]) :-
HeadMult is Head * Tail,
scal_mult(Head2, Tail, TailMult).
poly_mult(Constant,[],[]) :- !.
poly_mult([Constant], [Tail | Tail2]) :-
poly_add(scal_mult([Constant, Tail]), [0 | poly_mult] ([Constant, Tail2]).
答案 0 :(得分:3)
poly_sum(Ms,[],Ms) :- Ms = [_|_].
poly_sum([],Ns,Ns).
poly_sum([M|Ms], [N|Ns], [S|Ss]) :-
S is M+N,
poly_sum(Ms, Ns, Ss).
scal_prod([],_Sc,[]).
scal_prod([M|Ms], Sc, [P|Ps]) :-
P is M*Sc,
scal_prod(Ms, Sc, Ps).
poly_prod(_,[],[]).
poly_prod(Ms,[N|Ns], Xs2) :-
poly_prod(Ms,Ns, Xs1),
scal_prod(Ms, N, Ps),
poly_sum(Ps, [0.0|Xs1], Xs2).
?- poly_prod([1,2,3,4],[5,6,7],Xs).
Xs = [5.0, 16.0, 34.0, 52, 45, 28] ;
false.