Fill list in SWI-Prolog

I am trying to fill a list of given length N with numbers 1,2,3,...,N.

I thought this could be done this way:

开发者_如何学JAVAcreate_list(N,L) :-
    length(L,N),
    forall(between(1,N,X), nth1(X,L,X)).

However, this does not seem to work. Can anyone say what I am doing wrong?

First things first: Use clpfd!

:- use_module(library(clpfd)).

In the following I present zs_between_and/3, which (in comparison to my previous answer) offers some more features.

For a start, let's define some auxiliary predicates first!

equidistant_stride([]    ,_).
equidistant_stride([Z|Zs],D) :-
   equidistant_prev_stride(Zs,Z,D).

equidistant_prev_stride([]     ,_ ,_).     % internal predicate
equidistant_prev_stride([Z1|Zs],Z0,D) :-
   Z1 #= Z0+D,
   equidistant_prev_stride(Zs,Z1,D).

Let's run a few queries to get a picture of equidistant_stride/2:

?- Zs = [_,_,_], equidistant_stride(Zs,D).
Zs = [_A,_B,_C], _A+D#=_B, _B+D#=_C.

?- Zs = [1,_,_], equidistant_stride(Zs,D).
Zs = [1,_B,_C], _B+D#=_C, 1+D#=_B.

?- Zs = [1,_,_], equidistant_stride(Zs,10).
Zs = [1,11,21].

So far, so good... moving on to the actual "fill list" predicate zs_between_and/3:

zs_between_and([Z0|Zs],Z0,Z1) :-
   Step in -1..1,
   Z0 #= Z1 #<==> Step #=  0,
   Z0 #< Z1 #<==> Step #=  1,
   Z0 #> Z1 #<==> Step #= -1,
   N #= abs(Z1-Z0),    
   (  fd_size(N,sup)
   -> true
   ;  labeling([enum,up],[N])
   ),
   length(Zs,N),
   labeling([enum,down],[Step]),
   equidistant_prev_stride(Zs,Z0,Step).

A bit baroque, I must confess...

Let's see what features were gained---in comparison to my previous answer!

?- zs_between_and(Zs,1,4).      % ascending  consecutive integers
Zs = [1,2,3,4].                 % (succeeds deterministically)

?- zs_between_and(Zs,3,1).      % descending consecutive integers (NEW)
Zs = [3,2,1].                   % (succeeds deterministically)

?- zs_between_and(Zs,L,10).     % enumerates fairly
  L = 10, Zs =       [10]       % both ascending and descenting (NEW)
; L =  9, Zs =     [9,10]
; L = 11, Zs =    [11,10]
; L =  8, Zs =   [8,9,10]
; L = 12, Zs = [12,11,10]
; L =  7, Zs = [7,8,9,10]
...

?- L in 1..3, zs_between_and(Zs,L,6).
  L = 3, Zs =     [3,4,5,6]
; L = 2, Zs =   [2,3,4,5,6]
; L = 1, Zs = [1,2,3,4,5,6].

Want some more? Here we go!

?- zs_between_and([1,2,3],From,To).
  From = 1, To = 3
; false.

?- zs_between_and([A,2,C],From,To).
  A = 1, From = 1, C = 3, To = 3    % ascending
; A = 3, From = 3, C = 1, To = 1.   % descending

I don't have a prolog interpreter available right now, but wouldn't something like...

  isListTo(N, L) :- reverse(R, L), isListFrom(N, R).
  isListFrom(0, []).
  isListFrom(N, [H|T]) :- M is N - 1, N is H, isListFrom(M, T).

reverse can be done by using e.g. http://www.webeks.net/prolog/prolog-reverse-list-function.html

So tracing isListTo(5, [1, 2, 3, 4, 5])...

  isListTo(5, [1, 2, 3, 4, 5])
  <=> isListFrom(5, [5, 4, 3, 2, 1])
  <=> 5 is 5 and isListFrom(4, [4, 3, 2, 1])
  <=> 4 is 4 and isListFrom(3, [3, 2, 1])
  <=> 3 is 3 and isListFrom(2, [2, 1])
  <=> 2 is 2 and isListFrom(1, [1])
  <=> 1 is 1 and isListFrom(0, [])
  QED

Since PROLOG will not only evaluate truth, but find satisfying solutions, this should work. I know this is a vastly different approach from the one you are trying, and apologize if your question is specifically about doing loops in PROLOG (if that is the case, perhaps re-tag the question?).

Here's a logically pure implementation of predicate zs_from_to/3 using clpfd:

:- use_module(library(clpfd)).

zs_from_to([],I0,I) :-
   I0 #> I.
zs_from_to([I0|Is],I0,I) :- 
   I0 #=< I,
   I1 #= I0 + 1,
   zs_from_to(Is,I1,I).

Let's use it! First, some ground queries:

?- zs_from_to([1,2,3],1,3).
true.

?- zs_from_to([1,2,3],1,4).
false.

Next, some more general queries:

?- zs_from_to(Zs,1,7).
  Zs = [1,2,3,4,5,6,7]
; false.

?- zs_from_to([1,2,3],From,To).
From = 1, To = 3.

Now, let's have some even more general queries:

?- zs_from_to(Zs,From,2).
  Zs =            [], From in 3..sup
; Zs =           [2], From =  2
; Zs =         [1,2], From =  1
; Zs =       [0,1,2], From =  0
; Zs =    [-1,0,1,2], From = -1
; Zs = [-2,-1,0,1,2], From = -2
...

?- zs_from_to(Zs,0,To).
  Zs = [],          To in inf.. -1
; Zs = [0],         To = 0
; Zs = [0,1],       To = 1
; Zs = [0,1,2],     To = 2
; Zs = [0,1,2,3],   To = 3
; Zs = [0,1,2,3,4], To = 4
...

What answers do we get for the most general query?

?- zs_from_to(Xs,I,J).
  Xs = [],        J#=<I+ -1
; Xs = [I],       I+1#=_A, J#>=I, J#=<_A+ -1
; Xs = [I,_A],    I+1#=_A, J#>=I, _A+1#=_B, J#>=_A, J#=<_B+ -1
; Xs = [I,_A,_B], I+1#=_A, J#>=I, _A+1#=_B, J#>=_A, _B+1#=_C, J#>=_B, J#=<_C+ -1
...

Edit 2015-06-07

To improve on above implementation of zs_from_to/3, let's do two things:

1. Try to improve determinism of the implementation.
2. Extract a more general higher-order idiom, and implement zs_from_to/3 on top of it.

Introducing the meta-predicates init0/3 and init1/3:

:- meta_predicate init0(2,?,?).
:- meta_predicate init1(2,?,?).
init0(P_2,Expr,Xs) :- N is Expr, length(Xs,N), init_aux(Xs,P_2,0).

init1(P_2,Expr,Xs) :- N is Expr, length(Xs,N), init_aux(Xs,P_2,1).

:- meta_predicate init_aux(?,2,+).  % internal auxiliary predicate
init_aux([]    , _ ,_ ).
init_aux([Z|Zs],P_2,I0) :-
   call(P_2,I0,Z),
   I1 is I0+1,
   init_aux(Zs,P_2,I1).

Let's see init0/3 and init1/3 in action!

?- init0(=,5,Zs).             % ?- numlist(0,4,Xs),maplist(=,Xs,Zs).
Zs = [0,1,2,3,4].

?- init1(=,5,Zs).             % ?- numlist(1,5,Xs),maplist(=,Xs,Zs).
Zs = [1,2,3,4,5].

Ok, where do we go from here? Consider the following query:

?- init0(plus(10),5,Zs).      % ?- numlist(0,4,Xs),maplist(plus(10),Xs,Zs).
Zs = [10,11,12,13,14].

Almost done! Putting it together, we define zs_from_to/2 like this:

z_z_sum(A,B,C) :- C #= A+B.

zs_from_to(Zs,I0,I) :-
   N #= I-I0+1,
   init0(z_z_sum(I0),N,Zs).

At last, let's see if determinism has improved!

?- zs_from_to(Zs,1,7).
Zs = [1,2,3,4,5,6,7].          % succeeds deterministically

If I understood correctly, the built-in predicate numlist/3 would do. http://www.swi-prolog.org/pldoc/man?predicate=numlist/3