/* 

  Waiting for a truck in Picat.

  https://brainstellar.com/puzzles/probability/1005
  """
  On a given highway, trucks arrive at the station according to 
  a Poisson process with Lambda = 0.1/minute. This means that after 
  a truck is just passed, the time for the next truck to arrive is an 
  exponential random number with average arrival time of 10 minutes. 
  Your car just broke on this highway, and you are waiting for the next 
  truck for hitchhiking, what is your expected waiting time? On average 
  how many minutes ago the last truck left?

  Solution: Using memoryless property of exponential distribution, the 
            expected waiting time is 10 minutes. This also holds backwards, 
            hence the expected time last truck passed is also 10 minutes. 
            But this does not violate the total inter-arrival time of 10 
            minutes, because if your car breaks at a random time, you are 
            more likely to be in long interval than a short one.
  """


  Cf my Gamble model gamble_waiting_for_a_truck.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.

main => go.

/*

  var : t1
  Probabilities (truncated):
  9: 0.1290000000000000
  10: 0.1259000000000000
  11: 0.1145000000000000
  8: 0.1134000000000000
  .........
  1: 0.0004000000000000
  23: 0.0002000000000000
  25: 0.0001000000000000
  24: 0.0001000000000000
  mean = 10.0045

  var : t2
  Probabilities (truncated):
  19: 0.0900000000000000
  18: 0.0855000000000000
  20: 0.0848000000000000
  21: 0.0842000000000000
  .........
  6: 0.0003000000000000
  39: 0.0002000000000000
  38: 0.0002000000000000
  37: 0.0001000000000000
  mean = 20.0379

  var : t1,t2
  Probabilities (truncated):
  [9,19]: 0.0193000000000000
  [10,18]: 0.0163000000000000
  [9,17]: 0.0160000000000000
  [9,20]: 0.0156000000000000
  .........
  [2,8]: 0.0001000000000000
  [2,7]: 0.0001000000000000
  [1,11]: 0.0001000000000000
  [1,10]: 0.0001000000000000
  Mean (truncated)
  mean = [[9,19] = 0.0193,[10,18] = 0.0163,[9,17] = 0.016,[9,20] = 0.0156,[10,20] = 0.0153,[9,18] = 0.0151,[8,17] = 0.0146,[11,22] = 0.0143,[11,21] = 0.014,[10,19] = 0.014,[10,21] = 0.0133,[8,18] = 0.013,[11,20] = 0.0127,[8,19] = 0.0126,[12,21] = 0.0122,[12,22] = 0.0121,[7,16] = 0.0121,[8,16] = 0.0119,[11,23] = 0.0118,[9,21] = 0.0117]

  var : diff
  Probabilities (truncated):
  10: 0.1295000000000000
  9: 0.1208000000000000
  8: 0.1149000000000000
  11: 0.1146000000000000
  .........
  22: 0.0005000000000000
  1: 0.0004000000000000
  25: 0.0002000000000000
  24: 0.0002000000000000
  mean = 10.0334

*/
go ?=>
  reset_store,
  run_model(10_000,$model,[show_probs_trunc,mean]),
  nl,
  % show_store_lengths,nl,
  % fail,
  nl.
go => true.
  
  
model() =>
  Lambda = 1/0.1,
  T1     = poisson_dist(Lambda),
  T2     = T1 + poisson_dist(Lambda),

  Diff = abs(T1-T2),
  
  add("t1",T1),
  add("t2",T2),
  add("t1,t2",[T1,T2]),
  add("diff",Diff).