Multiple values of a variable inbetween 0 and a number prolog

问题

So I've been trying to teach myself prolog and I think I'm coming along nicely. However, I'm sort of stuck at this one method I'm trying to make.

toN(N,A) A is equal to the integer values between 0 and N-1, generated in ascending order.

so toN(5,A) would be

A = 0;
A = 1;
A = 2;
A = 3;
A = 4.

I'm still new to prolog so I'm not exactly sure how to do this with multiple values. I had something like this:

toN(N,A) :- 0 < N, Nx is N-1, toN(Nx,A).
toN(N,A) :- 0 =< N, Nx is N-1, A = Nx.

However this just returns false. Nothing else. It seems perfectly fine to me

回答1:

Check if the Prolog implementation that you are using supports clpfd!

:- use_module(library(clpfd)).

The implementation of toN/2 gets declarative and super-concise:

toN(N,A) :- 
   A #>= 0, 
   A #<  N, 
   labeling([up],[A]).

You'll find more labeling options in the clpfd manual: SWI-Prolog clpfd, SICStus Prolog clpfd.

回答2:

Something like this should generate the sequence of integers between any two arbitrary endpoints:

sequence(X,Y,X) :- % to generate the integers between X and Y,
  integer(X) ,     % - the starting point must be bound
  integer(Y) ,     % - the endpoint must be bound
  range(X,Y,Z)     % - then we just invoke the worker
  .                %

range(X,X,X) .  % hand back the last item in the sequence if X and Y have converged.
range(X,Y,X) :- % otherwise, return an item
  X =\= Y .     % - if X and Y haven't converged.
range(X,Y,Z) :- % otherwise, 
  X < Y ,       % - if X < Y ,
  X1 is X+1 ,   % - increment X
  range(X1,Y,Z) % - and recurse down.
  .             %
range(X,Y,Z) :- % otherwise
  X > Y ,       % - if X > Y
  X1 is X-1 ,   % - decrement X
  range(X1,Y,Z) % - and recurse down
  .             %

With that general-purpose tool, you can simply say:

to_n(N,A) :- sequence(0,N,A).

回答3:

Your implementation does not fail: by backtracking it yields numbers from -1 to N-1

?- toN(5,A).
A = -1 ? ;
A = 0 ? ;
A = 1 ? ;
A = 2 ? ;
A = 3 ? ;
A = 4 ? ;
no

To eliminate the -1 you should just replace =< by < in your second clause as @false commented above.

An alternative implementation, maybe more readable, would be

Edit: inserted condition N>=0 in answer to @false comment below.

toN(N,A) :-
  N >= 0,
  toN(0,N,A).

toN(K,N,K).
toN(K,N,A) :-
  K < N-1,
  Kn is K+1,
  toN(Kn,N,A).

来源：https://stackoverflow.com/questions/30113105/multiple-values-of-a-variable-inbetween-0-and-a-number-prolog

标签

algorithm

variables

recursion

prolog

logic