/* 

  Cute list in Picat.

  https://theweeklychallenge.org/blog/perl-weekly-challenge-191/#TASK2
  """
  Task 2: Cute List
  Submitted by: Mohammad S Anwar

  You are given an integer, 0 < $n <= 15.

  Write a script to find the number of orderings of numbers that form a
  cute list.

  With an input @list = (1, 2, 3, .. $n) for positive integer $n, an
  ordering of @list is cute if for every entry, indexed with a base of 1, either

  1) $list[$i] is evenly divisible by $i
  or
  2) $i is evenly divisible by $list[$i]

  Example

  Input: $n = 2
  Ouput: 2

  Since $n = 2, the list can be made up of two integers only i.e. 1 and 2.
  Therefore we can have two list i.e. (1,2) and (2,1).

  @list = (1,2) is cute since $list[1] = 1 is divisible by 1 and
  $list[2] = 2 is divisible by 2.
  """

  Solutions for N=2..15 (0.2s)

  2 = 2
  3 = 3
  4 = 8
  5 = 10
  6 = 36
  7 = 41
  8 = 132
  9 = 250
  10 = 700
  11 = 750
  12 = 4010
  13 = 4237
  14 = 10680
  15 = 24679


  This program was created by Hakan Kjellerstrand, hakank@gmail.com
  See also my Picat page: http://www.hakank.org/picat/

*/

import cp.

main => go.


go =>
  N::2..15,
  indomain(N),
  A = findall(L,cute_list(N,L)),
  Len = A.len,
  println(N=Len),
  fail,
  nl.

go2 :-
  N::15..15,
  indomain(N),
  time(A = findall(L,cute_list(N,L))),    
  % time(A = findall(L,cute_list2(N,L))),
  Len = A.len,
  println(N=Len),
  % fail,
  nl.


cute_list(N,L) :-
  L = new_list(N),
  L :: 1..N,
  all_different(L),
  foreach(I in 1..N)
    L[I]mod I#=0 #\/ I mod L[I] #= 0
  end,
  % solve($[ffc,split],L).
  solve($[constr],L).


% Alternative (Cf cute_list.pl SWI Prolog)
cute_list2(N,L) :-
        L = new_list(N),
        L :: 1..N,
        all_different(L),
        check2(1,L),
        % solve($[ffc,split],L).
        solve($[constr],L).        

check2(_Ix,[]).
check2(Ix,[H|T]) :-
        (H mod Ix #= 0 #\/ Ix mod H #= 0),
        Ix1 is Ix+1,
        check2(Ix1,T).