How to resolve the algorithm Pseudo-random numbers/Combined recursive generator MRG32k3a step by step in the 11l programming language

Published on 12 May 2024 09:40 PM
#11l

How to resolve the algorithm Pseudo-random numbers/Combined recursive generator MRG32k3a step by step in the 11l programming language

Table of Contents

Problem Statement

numbers as shown above. are as shown above
repetitions of Is as follows:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Pseudo-random numbers/Combined recursive generator MRG32k3a step by step in the 11l programming language

Source code in the 11l programming language

V a1 = [Int64(0), 1403580, -810728]
V m1 = Int64(2) ^ 32 - 209
V a2 = [Int64(527612), 0, -1370589]
V m2 = Int64(2) ^ 32 - 22853
V d = m1 + 1

T MRG32k3a
   [Int64] x1, x2

   F (seed_state = 123)
      .seed(seed_state)

   F seed(Int64 seed_state)
      assert(seed_state C Int64(0) <.< :d, ‘Out of Range 0 x < #.’.format(:d))
      .x1 = [Int64(seed_state), 0, 0]
      .x2 = [Int64(seed_state), 0, 0]

   F next_int()
      ‘return random int in range 0..d’
      V x1i = (sum(zip(:a1, .x1).map((aa, xx) -> aa * xx)) % :m1 + :m1) % :m1
      V x2i = (sum(zip(:a2, .x2).map((aa, xx) -> aa * xx)) % :m2 + :m2) % :m2
      .x1 = [x1i] [+] .x1[0.<2]
      .x2 = [x2i] [+] .x2[0.<2]
      V z = ((x1i - x2i) % :m1 + :m1) % :m1
      R z + 1

   F next_float()
      ‘return random float between 0 and 1’
      R Float(.next_int()) / :d

V random_gen = MRG32k3a()
random_gen.seed(1234567)
L 5
   print(random_gen.next_int())

random_gen.seed(987654321)
V hist = Dict(0.<5, i -> (i, 0))
L 100'000
   hist[Int(random_gen.next_float() * 5)]++
print(hist)

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