Supporting algorithm 15 (VB).bas

Attribute VB_Name = "Module1"
'##############################################################################################
'# John Wiley & Sons, Inc.                                                                    #
'#                                                                                            #
'# Book:   Markov Chains: From Theory To Implementation And Experimentation                   #
'# Author: Dr. Paul Gagniuc                                                                   #
'# Data:   01/09/2016                                                                         #
'#                                                                                            #
'# Description:                                                                               #
'# Supporting algorithm 15. A Markov Chain framework for simulation.                          #
'# The probability values present inside a 4x4 transition matrix (P)                          #
'# are directly used for an automatic generation of the letter                                #
'# combination that make up the representation of four jars. Thus,                            #
'# the four letter sequences have a calculated proportion of “A”,                             #
'# “B”, “C” and “D” letters. The chance of a letter chosen at random                          #
'# from one of the four sequences is directly dictated by the                                 #
'# proportions of “A”, “B”, “C” and “D” letters.                                              #
'##############################################################################################

Dim P(0 To 4, 0 To 3) As Variant
Dim Jar(1 To 4) As Variant

Private Sub main()

P(0, 0) = "A"
P(0, 1) = "B"
P(0, 2) = "C"
P(0, 3) = "D"

P(1, 0) = 0
P(1, 1) = 1
P(1, 2) = 0
P(1, 3) = 0

P(2, 0) = 0.33
P(2, 1) = 0
P(2, 2) = 0.33
P(2, 3) = 0.33

P(3, 0) = 0
P(3, 1) = 1
P(3, 2) = 0
P(3, 3) = 0

P(4, 0) = 0
P(4, 1) = 0
P(4, 2) = 1
P(4, 3) = 0

For j = 1 To 4
    Jar(j) = Fill_Jar(j)
Next j

draws = 100
a = Draw(1)

For i = 1 To draws
    For j = 0 To 3
        If a = P(0, j) Then
            a = Draw(j + 1)
            z = z & P(0, j)
            GoTo 1
        End If
    Next j
1:
Next i

MsgBox z

End Sub

Function Fill_Jar(ByVal S As Variant) As Variant

Ltot = 100

For i = 0 To 3
    a = Int(Ltot * P(S, i))
        For j = 1 To a
            b = b & P(0, i)
        Next j
Next i

Fill_Jar = b
End Function

Function Draw(ByVal S As Variant) As Variant
    Randomize
    randomly_choose = Int(Rnd * Len(Jar(S)))
    ball = Mid(Jar(S), randomly_choose + 1, 1)
    Draw = ball
End Function