Floyd Algorithm, need help with path Matrix..

[Kb24]

Hi i'm new here.. Got a little coding problem..
I'm creating an excel macro to determine shortest route between a number of nodes using Floyd's algorithm..
I had no problem calculating and displaying the distance,
but I can't get the path Matrix to give correct result..

From the network pic below, the correct path for T1 to T2 is supposed to be :
T7-T8-T3-T4
But my program gives me T8 - T4..

So i need help fixing the path algorithm in Xls file here :

http://ifile.it/j8dc1a9/floyd_no_cycle.xls

I hope anyone familiar with this algorithm could help me, I'm stuck at this problem for weeks.. :(

Thanks in advance.

[Zachm]

Sub Button1_Click()
Dim i, j, k, n, x
ActiveWorkbook.Sheets("Input").Activate
n = ActiveWorkbook.Sheets("Input").Cells(1, 8).Value


'Clear output sheet
For i = 1 To 50
For j = 1 To 50
Sheets("Output").Cells(3 + i, 1 + j).Value = Null
Sheets("Routes").Cells(3 + i, 1 + j).Value = Null
Next j
Next i

'Copy weights to output sheet
For i = 1 To n
For j = 1 To n
y = y + 1
Sheets("Output").Cells(3 + i, 1 + j).Value = Sheets("Input").Cells(3 + i, 1 + j)
If Sheets("Input").Cells(3 + i, 1 + j).Value = "" Then
Sheets("Output").Cells(3 + i, 1 + j).Value = 10 ^ 15
Else
Sheets("Routes").Cells(3 + i, 1 + j).Value = Sheets("Routes").Cells(3 + i, 1)
End If
Next j
Sheets("Output").Cells(3 + i, 1 + i).Value = 0
Next i

'Run algorithm
Sheets("Output").Activate

For k = 1 To n
For i = 1 To n
For j = 1 To n
If Sheets("Output").Cells(3 + i, 1 + j) > (Sheets("Output").Cells(3 + i, 1 + k) + Sheets("Output").Cells(3 + k, 1 + j)) Then
Sheets("Output").Cells(3 + i, 1 + j) = Sheets("Output").Cells(3 + i, 1 + k) + Sheets("Output").Cells(3 + k, 1 + j)
Sheets("Routes").Cells(3 + i, 1 + j) = Sheets("Routes").Cells(3 + k, 1 + j)
End If
Next j
Next i

'If x = 8 Then
'k = n
' End If
Next k
End Sub


I marked the lines where I changed things in red.
You initialized the path matrix wrongly and you updated it with the wrong nodes.

Can you understand your errors ?

Regards,
Zachm