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‎Supporting algorithm 1 (VB).bas‎

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Attribute VB_Name = "Module1"
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'##############################################################################################
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'# John Wiley & Sons, Inc. #
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'# #
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'# Book: Markov Chains: From Theory To Implementation And Experimentation #
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'# Author: Dr. Paul Gagniuc #
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'# Data: 01/09/2016 #
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'# #
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'# Description: #
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'# Supporting algorithm 1. A two states Markov Chain simulator based on the proportions of #
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'# letters. Two letter sequences with predetermined proportions of “W” and “B” letters are #
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'# used for the representation of two jars. The chance of a letter chosen at random from #
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'# one of the two sequences is directly dictated by the proportions of “W” and “B” letters. #
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'##############################################################################################
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Dim Jar(0 To 1) As String
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Private Sub main()
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draws = 17
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Jar(0) = "WWBBBBBBBB"
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Jar(1) = "WWWWWBBBBB"
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a = Draw(0) ' Draws start from jar "W"
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z = z & " Jar W[" & a & "],"
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For i = 1 To draws
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If a = "W" Then
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a = Draw(0)
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z = z & " Jar W[" & a & "],"
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Else
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a = Draw(1)
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z = z & " Jar B[" & a & "],"
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End If
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MsgBox z
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Next i
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End Sub
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Function Draw(ByVal S As Integer) As String
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Randomize
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randomly_choose = Int(Rnd * Len(Jar(S)))
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ball = Mid(Jar(S), randomly_choose + 1, 1)
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Draw = ball
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End Function

‎Supporting algorithm 10 (VB).bas‎

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Attribute VB_Name = "Module1"
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'##############################################################################################
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'# John Wiley & Sons, Inc. #
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'# #
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'# Book: Markov Chains: From Theory To Implementation And Experimentation #
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'# Author: Dr. Paul Gagniuc #
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'# Data: 01/09/2016 #
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'# #
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'# Description: #
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'# Supporting algorithm 10. Predictions based on sequences produced by n-state #
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'# Markov machines. This example also uses a DNA sequence as a model. However, #
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'# the algorithm allows for an unlimited number of letters (observations). #
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'# Previously, the vector – matrix multiplication cycle was declared manually #
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'# with a range of expressions. Here, the multiplication cycle is made #
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'# iteratively. For a prediction on more than 4 states, the matrix elements #
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'# and the number of vector components can be increased to cover a new #
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'# prediction requirement. Note that “ExtractProb” function is not shown. #
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'# However, when the above algorithm is used the “ExtractProb” function #
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'# must be present. #
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'##############################################################################################
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Dim M(1 To 4, 1 To 4) As String
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Private Sub main()
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Dim v(0 To 3, 0 To 1) As Variant
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Call ExtractProb("TACTTCGATTTAAGCGCGGCGGCCTATATTA")
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chain = 5
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v(0, 0) = 1
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v(1, 0) = 0
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v(2, 0) = 0
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v(3, 0) = 0
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v(0, 1) = 0
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v(1, 1) = 0
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v(2, 1) = 0
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v(3, 1) = 0
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For k = 1 To chain
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For i = 0 To 3
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For j = 0 To 3
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v(i, 1) = v(i, 1) + (v(j, 0) * M(j + 1, i + 1))
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Next j
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Next i
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For i = 0 To 3
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v(i, 0) = v(i, 1)
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v(i, 1) = 0
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Next i
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A = v(0, 0)
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T = v(1, 0)
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c = v(2, 0)
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G = v(3, 0)
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MsgBox "V(" & k & ")=[" & A & " | " & T & " | " & c & " | " & G & "]"
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Next k
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End Sub
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Function ExtractProb(ByVal s As String)
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Ea = "A"
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Et = "T"
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Eg = "G"
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Ec = "C"
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For i = 1 To 4
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For j = 1 To 4
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M(i, j) = 0
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Next j
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Next i
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Ta = 0
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Tt = 0
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Tg = 0
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Tc = 0
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For i = 2 To Len(s) - 1
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DI1 = Mid(s, i, 1)
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DI2 = Mid(s, i + 1, 1)
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If DI1 = Ea Then r = 1
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If DI1 = Et Then r = 2
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If DI1 = Eg Then r = 3
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If DI1 = Ec Then r = 4
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If DI2 = Ea Then c = 1
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If DI2 = Et Then c = 2
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If DI2 = Eg Then c = 3
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If DI2 = Ec Then c = 4
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M(r, c) = Val(M(r, c)) + 1
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If DI1 = Ea Then Ta = Ta + 1
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If DI1 = Et Then Tt = Tt + 1
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If DI1 = Eg Then Tg = Tg + 1
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If DI1 = Ec Then Tc = Tc + 1
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Next i
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For i = 1 To 4
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For j = 1 To 4
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If i = 1 Then M(i, j) = Val(M(i, j)) / Ta
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If i = 2 Then M(i, j) = Val(M(i, j)) / Tt
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If i = 3 Then M(i, j) = Val(M(i, j)) / Tg
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If i = 4 Then M(i, j) = Val(M(i, j)) / Tc
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Next j
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Next i
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End Function
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‎Supporting algorithm 11 (VB).bas‎

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Attribute VB_Name = "Module1"
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'##############################################################################################
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'# John Wiley & Sons, Inc. #
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'# #
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'# Book: Markov Chains: From Theory To Implementation And Experimentation #
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'# Author: Dr. Paul Gagniuc #
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'# Data: 01/09/2016 #
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'# #
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'# Description: #
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'# Supporting algorithm 11. The conversion of measurements to states. #
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'# A range of values is divided into 4 equal regions. Each region #
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'# corresponds to a state: “A”, “B”, “C”, and “D”. The numeric values #
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'# are associated with a representative letter based on their position #
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'# over the regions. Thus, the initial values are listed as letters (observations). #
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'##############################################################################################
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Private Sub main()
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Dim Inp() As String
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Dim R As String
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R = "159,82,187,194,179,115,197,102,105,104,95,126,74,143,143,127,98," & _
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"70,92,170,168,182,149,85,137,100,170,180,61,177,86,195,198,182,150," & _
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"197,103,103,186,100,96,196"
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Inp = Split(R, ",")
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Lu = 200
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Ld = 60
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n = 4
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Pr = (Lu - Ld) / n
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For i = 0 To UBound(Inp)
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s = (Inp(i) - Ld) / Pr
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s = Split(s, ".")(0)
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If s = 0 Then l = "A"
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If s = 1 Then l = "B"
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If s = 2 Then l = "C"
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If s = 3 Then l = "D"
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Obs = Obs & l
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Reg = Reg & s & ","
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Next i
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MsgBox "Reg=" & Reg & vbCrLf & "Obs=" & Obs & vbCrLf
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End Sub

‎Supporting algorithm 12 (VB).bas‎

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Attribute VB_Name = "Module1"
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'##############################################################################################
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'# John Wiley & Sons, Inc. #
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'# #
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'# Book: Markov Chains: From Theory To Implementation And Experimentation #
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'# Author: Dr. Paul Gagniuc #
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'# Data: 01/09/2016 #
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'# #
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'# Description: #
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'# Supporting algorithm 12. Prediction based on a 3x3 transition matrix. #
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'# Known transition probability values are directly used from a transition #
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'# matrix for highlighting the behavior of an absorbing Markov Chain. #
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'##############################################################################################
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Dim Jar(1 To 3, 1 To 3) As String
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Private Sub main()
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Dim v(0 To 2, 0 To 1) As Variant
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Jar(1, 1) = 0.33
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Jar(1, 2) = 0.33
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Jar(1, 3) = 0.33
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Jar(2, 1) = 0.5
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Jar(2, 2) = 0.5
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Jar(2, 3) = 0
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Jar(3, 1) = 0
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Jar(3, 2) = 0
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Jar(3, 3) = 1
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chain = 5
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v(0, 0) = 1
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v(1, 0) = 0
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v(2, 0) = 0
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v(0, 1) = 0
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v(1, 1) = 0
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v(2, 1) = 0
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For k = 1 To chain
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For i = 0 To 2
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For j = 0 To 2
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v(i, 1) = v(i, 1) + (v(j, 0) * Jar(j + 1, i + 1))
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Next j
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Next i
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For i = 0 To 2
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v(i, 0) = v(i, 1)
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v(i, 1) = 0
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Next i
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A = v(0, 0)
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B = v(1, 0)
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C = v(2, 0)
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MsgBox "Step(" & k & ")=[" & A & " | " & B & " | " & C & "]"
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Next k
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End Sub

‎Supporting algorithm 13 (VB).bas‎

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Attribute VB_Name = "Module1"
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'##############################################################################################
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'# John Wiley & Sons, Inc. #
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'# #
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'# Book: Markov Chains: From Theory To Implementation And Experimentation #
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'# Author: Dr. Paul Gagniuc #
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'# Data: 01/09/2016 #
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'# #
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'# Description: #
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'# Supporting algorithm 13. Prediction framework based on a 4x4 transition matrix. #
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'# Known transition probability values are directly used from a transition matrix #
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'# for highlighting the behavior of a Markov Chain. A variable number of states #
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'# and configurations can be tested. For instance, a diagram with three states can #
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'# be molded on the algorithm. However, since the framework is made for a total of #
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'# 4 states the following modifications are required: all Jar(n, 4) elements are #
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'# set to zero (Jar(1, 4)=0; Jar(2, 4)=0; Jar(3, 4)=0; and Jar(4, 4)=0). Furthermore, #
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'# the molding of a diagram with two states on this framework involves two #
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'# modifications: all Jar(n, 4) elements are set to zero and all Jar(n, 3) elements #
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'# are set to zero. Any type of diagram configuration can be tested by following two #
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'# rules: 1) the absence of an arrow is indicated by zero, and 2) any value greater #
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'# than zero and less than or equal to 1 indicates an arrow. #
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'##############################################################################################
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Dim Jar(1 To 4, 1 To 4) As String
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Private Sub main()
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Dim v(0 To 3, 0 To 1) As Variant
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Jar(1, 1) = 1
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Jar(1, 2) = 0
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Jar(1, 3) = 0
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Jar(1, 4) = 0
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Jar(2, 1) = 0.5
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Jar(2, 2) = 0
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Jar(2, 3) = 0.5
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Jar(2, 4) = 0
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Jar(3, 1) = 0
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Jar(3, 2) = 0.5
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Jar(3, 3) = 0
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Jar(3, 4) = 0.5
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Jar(4, 1) = 0
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Jar(4, 2) = 0
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Jar(4, 3) = 1
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Jar(4, 4) = 0
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chain = 5
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v(0, 0) = 0
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v(1, 0) = 0
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v(2, 0) = 0
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v(3, 0) = 1
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v(0, 1) = 0
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v(1, 1) = 0
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v(2, 1) = 0
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v(3, 1) = 0
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For k = 1 To chain
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For i = 0 To 3
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For j = 0 To 3
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v(i, 1) = v(i, 1) + (v(j, 0) * Jar(j + 1, i + 1))
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Next j
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Next i
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For i = 0 To 3
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v(i, 0) = v(i, 1)
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v(i, 1) = 0
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Next i
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A = v(0, 0)
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B = v(1, 0)
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C = v(2, 0)
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D = v(3, 0)
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MsgBox "Step(" & k & ")=[" & A & "|" & B & "|" & C & "|" & D & "]"
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Next k
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End Sub

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