/* 

  The Case of Forgotten Phone number in Picat.

  From Logic Puzzles and Logic Programming 
  """
  At a party, someone suggested that they should give Frank a
  ring. Unfortunately, there was no one at present who knew his
  phone number. All they could collect was some little bits of
  information:

  (1) He had a six-figure number.
  (2) The second half of the number, that is, the number formed
      by the last three figures, was equal to four times the first
      one.
  (3) The two figures in the middle of the number were
      identica1.
  (4) The second figure was equal to twice the first.
  (5) And the third figure in the phone number was two times
     the second one or two plus the second one.

  What was Frank’s phone number?
  """

  Note: My cp version (go/0) first reported two solutions:

    [0,0,0,0,0,0]
    [1,2,4,4,9,6]

  I guees we can rule out the first solution as a not
  valid phone number, so I require that the first digit
  is not 0.

  The Prolog program (go2/0) (op.cit) has a unique solution:
    Frank’s phone number is: 124-496

  For fun: go3/0 (using gen/2) generates similar problems but with
  different constants in the description.

  The original problem statement is phrased like with this approach:
  """
  [hints = [4,3,4,2,2],x = [1,2,4,4,9,6]]
     Problem statement:
        (1) He had a six-figure number.
        (2) The second half of the number, that is, the number formed
            by the last three figures, was equal to 4 times the first one.
        (3) The two figures in position 3 and 4 were identica1.
        (4) The second figure was equal to 2 times the first.
        (5) And the third figure in the phone number was 2 times
            the second one or 2 plus the second one

        Frank's telephone number is [1,2,4,4,9,6]
        (This is the original problem.)
  """

  Another version is:
  """
  [hints = [6,3,5,2,3],x = [1,2,5,7,5,0]]
  Problem statement:
        (1) He had a six-figure number.
        (2) The second half of the number, that is, the number formed
            by the last three figures, was equal to 6 times the first one.
        (3) The two figures in position 3 and 5 were identica1.
        (4) The second figure was equal to 2 times the first.
        (5) And the third figure in the phone number was 3 times
            the second one or 3 plus the second one

        Frank's telephone number is [1,2,5,7,5,0]
  """
  
  There are 20 such problem statements.


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

*/

import v3_utils.
import cp.


main => go.

%
% CP version
%
go ?=>
  forgotten_phone_number(X),
  println(X),
  fail,  
  nl.
go => true.

% Prolog version
go2 ?=>
  start,
  fail,
  nl.
go2 => true.


%
% Generate some similar problems.
%
go3 ?=>
  All = findall([X,Hints],gen(X, Hints)),
  NumSols = new_map(),
  foreach(A in All)
    % println(A)
    NumSols.put(A[1],NumSols.get(A[1],0)+1)
  end,
  foreach(A in All)
    if NumSols.get(A[1]) == 1 then
      print_hints(A[2],A[1]),
    end
  end,
  nl.
go3 => true.


forgotten_phone_number(X) =>
  % (1) He had a six-figure number.
  N = 6,
  X = new_list(N),
  X :: 0..9,

  X[1] #!= 0, % remove the trivial solution [0,0,0,0,0,0]

  % (2) The second half of the number, that is, the number formed
  %     by the last three figures, was equal to four times the first
  %     one.
  % to_num(X[4..6],X46),
  % to_num(X[1..3],X13),
  % X13*4 #= X46,
  4*(100*X[1]+10*X[2]+X[3]) #= 100*X[4]+10*X[5]+X[6],

  % (3) The two figures in the middle of the number were
  %     identica1.
  X[3] #= X[4],
  
  % (4) The second figure was equal to twice the first.
  X[1]*2 #= X[2],
  
  % (5) And the third figure in the phone number was two times
  %    the second one or two plus the second one.
  X[3] #= 2*X[2] #\/ X[3] #= 2+X[2],
  
  solve(X).



to_num(List, Base, Num) =>
        Len = length(List),
        Num #= sum([List[I]*Base**(Len-I) : I in 1..Len]).

to_num(List, Num) =>
       to_num(List, 10, Num).

%
% Below is the Prolog program, op.cit. page 37
%
start :-
      assertz($phone(x,y)), % Dummy for phone/2
      retract($phone(x,y)), % Remove dummy     
      digitt(D1),
      D2 is 2*D1, digitt(D2),
      ( D3 is 2*D2 ; D3 is 2+D2 ), digitt(D3),
      half(D1,D2,D3, FIRST), FIRST != 0,
      SECOND is 4*FIRST, SECOND < 1000,
      half(D3,_D5,_D6, SECOND),
      remember(FIRST, SECOND),
      fail.
      
start :-
      number_of_results(N) ,
      out(N).
      
half(D1,D2,D3, N):-
      digitt(D1), digitt(D2), digitt(D3),
      Y1 is 100*D1, Y2 is 10*D2, N is Y1 + Y2 + D3.
half(D1,D2,D3, N) :-
      integer(N), N > 0,
      N < 1000,
      D1 is N div 100, Y is N mod 100,
      D2 is Y div 10, D3 is Y mod 10.

remember(F, S):-
      bp.phone(F, S),
      !, fail.
remember(F, S) :-
      assert($phone(F, S ) ).
      
number_of_results(many):-
      bp.phone(F, S),
      bp.phone(Fl, SI),
      ( F != Fl ; S != SI ), !.
number_of_results(1) :-
      bp.phone(_F, _S ), !.
number_of_results(0).

out(many):-
     nl, print("The phone number is not unique, "),
     print("\nthe folks have to make some trials."), nl, nl,
     print("The possible numbers are:"), nl,
     out.
out(1):-
     retract($phone(F, S)),
     nl, print("Frank’s phone number is: "),
     print(F), print("-"), print(S), nl, nl.
out(0):-
     nl,
     print("The pieces of information do not "),
     print("determine a phone number."),
     nl.

out :-
    retract($phone(F, S)),
    tab(1), print(F), print("-"), print(S), nl,
    out.
out :- nl.

digitt(0).
digitt(1).
digitt(2).
digitt(3).
digitt(4).
digitt(5).
digitt(6).
digitt(7).
digitt(8).
digitt(9).


%
% Generate a puzzle
%
gen(X,Hints) =>
  % (1) He had a six-figure number.
  N = 6,
  X = new_list(N),
  X :: 0..9,

  Hints = new_list(5),
  Hints :: 2..9, % Note all are #> 1

  % (2) The second half of the number, that is, the number formed
  %     by the last three figures, was equal to four times the first
  %     one.
  % to_num(X[4..6],X46),
  % to_num(X[1..3],X13),
  Hints[1]*(100*X[1]+10*X[2]+X[3]) #= 100*X[4]+10*X[5]+X[6],

  % (3) The two figures in the middle of the number were
  %     identica1.
  % Two figures Hints[2] and Hints[3] are identical
  % X[2] #= X[3],
  Hints[2] :: 1..N,
  Hints[3] :: 1..N,
  Hints[2] #!= Hints[3],
  element(Hints[2],X,Hints[3]),
  
  % (4) The second figure was equal to twice the first.
  % (4) The second figure was equal to Hints[4] times the first.
  X[1]*Hints[4] #= X[2],
  
  % (5) And the third figure in the phone number was two times
  %    the second one or two plus the second one.
  % X[3] #= 2*X[2] #\/ X[3] #= 2+X[2],
  X[3] #= Hints[5]*X[2] #\/ X[3] #= Hints[5]+X[2],

  Vars = X ++ Hints,
  solve(Vars).


print_hints(Hints,X) =>
  println([hints=Hints,x=X]),
  println("Problem statement:"),
  println("\t(1) He had a six-figure number."),
  println("\t(2) The second half of the number, that is, the number formed"),
  printf("\t    by the last three figures, was equal to %d times the first one.\n", Hints[1]),
  printf("\t(3) The two figures in position %d and %d were identica1.\n",Hints[2],Hints[3]),
  printf("\t(4) The second figure was equal to %d times the first.\n", Hints[4]), 
  printf("\t(5) And the third figure in the phone number was %d times\n",Hints[5]),
  printf("\t    the second one or %d plus the second one\n",Hints[5]),
  nl,
  printf("\tFrank's telephone number is %w\n",X),
  if X == [1,2,4,4,9,6] then
      println("\t(This is the original problem.)")
  end,

  nl,
  nl.