/* 

  From 11 to 25 in Picat.

  The Bonus problem from 
  "Can you solve the giant cat army riddle? - Dan Finkel"
  https://www.youtube.com/watch?v=YeMVoJKn1Tg&feature=youtu.be
  @4:35
  """
  Get from 11 to 25 
  Operations available: 
    - x*2 
    - x-3
  """

  This is the shortest and unique solutions (length 8),
  [11,8,5,10,7,14,28,25]

  Cf two_buttons_problem.pi

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

*/

import util.
import cp.

main => go.

op1(X,Y) => Y = X*2.
op2(X,Y) => Y = X-3.

go ?=>
  member(Len,4..100),
  println(len=Len),
  L = new_list(Len),
  L[1] = 11,
  L[Len] = 25,
  foreach(I in 2..Len)
     (
     op1(L[I-1],L[I])
     ;
     op2(L[I-1],L[I])
     )
  end,
  
  println(L), 

  % fail,
  
  nl.

go => true.



go2 ?=>
  _ = random2(),
  StartVal = random() mod 100,
  EndVal = random() mod 100,  
  StartVal != EndVal,
  println([start=StartVal,end=EndVal]),
  OpList1 = [*,//],
  Op1 = OpList1[1+random() mod 2], 
  OpList2 = [+,-],
  Op2 = OpList2[1+random() mod 2],
  Val1 = 2+random() mod 10,
  Val2 = 2+random() mod 10,
  F1 =.. [Op1,Val1],
  F2 =.. [Op2,Val2],
  println([f1=F1,f2=F2]),
  % bp.assert($(f1(X,Y) :- Y #= F1(X))),
  % bp.assert($(f2(X,Y) :- Y #= F2)Y)),  
  
  println([op1=Op1,val1=Val1,op2=Op2,val2=Val2]),

  member(Len,3..20),
  println(len=Len),
  L = new_list(Len),
  L[1] #= StartVal,
  L[Len] #= EndVal,
  foreach(I in 2..Len)
     (
     /*
     L[I] = apply(Op1,Val1,L[I-1])
     ;
     L[I] = apply(Op2,Val2,L[I-1])     
     */
     L[I] = apply(Op1,L[I-1],Val1)
     ; 
     L[I] = apply(Op2,L[I-1],Val2)
     % f1(L[I-1],L[1])
     % ;
     % f2(L[I-1],L[1])
     )
  end,

  (println(L),true ; fail),

  % fail,
  
  nl.

go2 => true.