/* 

  Nearest sum of cubes of a number in Picat.

  http://stackoverflow.com/questions/25262578/find-the-nearest-small-number-x-where-x-can-be-represented-by-sum-of-cubes
  """
  I recently came across a problem where a number x is given and i have to find y and 
  below are the conditions

    - y < x > 1
    - y can be expressed as a^3 + b^3 (more than one combination)
    - example if 4105 is x then 4104 is the answer, where 4104 can be expressed as 
      16^3 + 2^3 and also 15^3 + 9^3
  """

  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.


go =>

  X = 4105,  
  check(X,Y1),
  println(y1=Y1),
  check2(X,Y2),
  println(y2=Y2),

  nl.

go2 =>
  L = [4105, 99998, 4104,1023,999,216,3],
  foreach(X in L)
    nl,
    println(x=X),
    (time3($check(X,Y1),T1),println(y1=Y1) ; true),
    (time3($check2(X,Y2),T2), println(y2=Y2) ; true),
    println(time=[T1,T2]),
    nl
  end,
  nl.

go3 => 
  Total1 = 0,
  Total2 = 0,
  Diffs = [],
  foreach(_ in 1..20)
     X = 1+random2() mod 1000000,
     println(x=X),
     ( time3($check(X,Y1),T1), println(y1=Y1) ; true),
     ( time3($check2(X,Y2),T2), println(y2=Y2) ; true),
     println(time=[T1,T2]),
     Total1 := Total1 + T1,
     Total2 := Total2 + T2,
     Diffs := Diffs ++ [Total1-Total2],
     nl
  end,
  println([total1=Total1,total2=Total2]),
  println("total1-total2"=Total1-Total2),
  println(diffs=Diffs),
  nl.

go4 => 
  foreach(X in 2..10000)
    (check2(X,Y), println([x=X,y=Y]) ; true)
    % (check2(X,Y) ; true)
    % (check(X,Y) ; true)
  end,
  nl.

go5 =>
  L = [4105, 99998, 4104,1023,999,216,3],
  foreach(X in L)
    check2(X,Y),
    println(y=Y)
  end,
  nl.

go6 => 
  X = 545459, % 943579, % 609835,
  time2(check(X,Y1)),
  println(y1=Y1),

  time(check2(X,Y2)),
  println(y2=Y2),

  time(check3(X,Y3)),
  println(y3=Y3),

  nl.

% check check2 on larger instances
go7 => 
  Times = [],
  foreach(_ in 1..10)
     X = 1+random2() mod 10000000,
     println(x=X),
     ( time3($check2(X,Y2),T2), println([y2=Y2,time2=T2]) ; true),
     ( time3($check3(X,Y2),T3), println([y2=Y2,time3=T3]) ; true),
     Times := Times ++ [[T2,T3,T2-T3]]
  end,
  println(times=Times),
  Diffs = [Diff : [_,_,Diff] in Times],
  println(diffs=Diffs),
  println(diffsSum=sum(Diffs)),
  nl.


% special time which just returns T
time3(Goal, T) =>
   statistics(runtime,_),
   (call(Goal); true),
   statistics(runtime, [_,End]),
   T = End / 1000.0.


%
% CP approach: This is in general slower than check2/2.
%
check(X,Y) =>

  [Y,A,B] :: 1..X-1,
  Y #= A**3 + B**3,
  A #=< B,
  A #< Y,
  B #< Y,

  % solve($[max(Y),degree,updown],[Y,A,B]).
  % solve($[max(Y),backward],[Y,A,B]).
  solve($[max(Y)],[B,A,Y]).

  % println(x=X),
  % Diff = X -Y,
  % printf("%d**3 + %d**3 = %d (< %d  diff: %d)\n",A,B,Y,X,Diff),
  % nl.


%
% Brute force + maxof: in general faster than check/2.
%
check2(X,Y) =>
  maxof(check2b(X,Y),Y).
  % maxof(check2b(X,Y),Y,$printf("\tytmp=%d\n", Y)).
check2b(X,Y) =>
  Cubes = [Cube  : I in 1..X, Cube = I**3, Cube <= X].reverse(),
  member(I,Cubes), 
  member(J,[K : K in Cubes, K >= I]),
  I <= J,
  Y = I + J,
  Y < X.

% maxof_inc/2 was introduced in Picat v0.6
check3(X,Y) =>
  maxof_inc(check2b(X,Y),Y).