/* 

  Triangular distribution in Picat.

  From Mathematica TriangularDistribution
  """
  An executive is given an account of historical seasonal demands for a product in 
  millions of units. The minimum, maximum, and most likely demands are 1, 1.4, 
  and 1.25, respectively. Find the expected demand [...] using TriangularDistribution:

  td = TriangularDistribution[{1, 1.4}, 1.25];
  Mean[td]
  -> 1.21667
  """

  This version - with the third argument mode - is called triangular2_dist in Picat PPL.

  Picat> X = triangular2_dist_mean(1,1.4,1.25)
  X = 1.216666666666667

  For more on this distribution, see ppl_distributions.pi and ppl_distributions_test.pi


  Cf my Gamble model gamble_.rkt

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

*/

import ppl_distributions, ppl_utils.
import util.
% import ordset.

main => go.

/*
var : d
Probabilities (truncated):
1.39798860913652: 0.0001000000000000
1.397641478831406: 0.0001000000000000
1.397629226077137: 0.0001000000000000
1.39686844589101: 0.0001000000000000
.........
1.008381621027417: 0.0001000000000000
1.007177553370896: 0.0001000000000000
1.006692846511102: 0.0001000000000000
1.006692394252017: 0.0001000000000000
mean = 1.21669



*/
go ?=>
  reset_store,
  run_model(10_000,$model,[show_probs_trunc,mean % ,
                           % show_percentiles,
                           % show_hpd_intervals,hpd_intervals=[0.94],
                           % show_histogram,
                           % min_accepted_samples=1000,show_accepted_samples=true
                          ]),
  nl,
  % show_store_lengths,nl,
  % fail,
  nl.
go => true.
  
  
model() =>
  MinVal = 1,
  MaxVal = 1.4,
  Mode = 1.25,

  D = triangular2_dist(MinVal,MaxVal,Mode),
  add("d",D).