string manipulation

**kylefoley76** · 02-13-2014, 03:32 PM

I am trying to transform

(((p v q) > (r v (p v q))) > ((p & s) v (r v s))) > (p & r)

into

~(~(~(p v q) v (r v (p v q))) v ((p & s) v (r v s))) v (p & r)

So it seems that the algorithm should be:

1. start with the left most character, go right, look for >, when the first > is found, go left and count the open and closed parentheses, when the open and closed parentheses equal each other then replace the open parentheses that eventually equaled the amount of closed parentheses with ~(
2. then return to the first > found, then go right and look for another > and repeat the procedure until all > have been transformed.

I used one of Rick's macros from another post to try to do it myself but I had problems with getting Excel to process ">" so I had to change the symbol to "t"

I did eventually get it to work but then I realized that the required algorithm is more complicated since it can only handle one t per sentence.

Code:

Function matim(ByVal S As String) As String
  Dim X As Long, t As Long, Total As Long
  t = InStr(1, S, "t", vbTextCompare)
  
  For X = InStrRev(S, ")", t) To 1 Step -1
    If Mid(S, X, 1) = ")" Then
      Total = Total + 1
    ElseIf Mid(S, X, 1) = "(" Then
      Total = Total - 1
    End If
    If Total = 0 Then
      S = Application.Replace(S, X, 0, "~")
      Exit For
    End If
  Next
  S = Replace(S, "t", "v", , , vbTextCompare)
  matim = S
End Function

That code doesn't work.

My final bit of string manipulation will then be to transform v into &

This is done as follows:

(p v q) = ~(~p & ~q)

So the algorithm would go:

1. Start at letter 1, go right, find the first v.
2. Go right and look for either ( or a letter. Whichever one you find place a ~ to the left of it.
3. Find either the first letter to the right or the first ( and put a ~ to the left of it.
4. Then go left and find either the first letter or the first ).
5. If one finds a letter place a ~ to the left of it.
5a. Then continue left and place transform the next ( to ~(
6. If one finds a ) then count it and continue left. As soon as the number of ) equals the number of (, then replace ( with ~(
6a. Then continue left and it is always the case that the next character one finds is either ~ or (. If ~ do nothing and continue left and replace the first ( with ~(
7. Return to the beginning of the newly transformed string and repeat steps one through 6.

Input
(~(~(~(p v q) v (r v (p v q))) v ((p & s) v (r v s))) v (p & r))

1.
(~(~(~~(~p & ~q) v (r v (p v q))) v ((p & s) v (r v s))) v (p & r))
2.
(~(~~(~~~(~p & ~q) & ~(r v (p v q))) v ((p & s) v (r v s))) v (p & r))
3.
(~(~~(~~~(~p & ~q) & ~~(~r & ~(p v q))) v ((p & s) v (r v s))) v (p & r))
4.
(~(~~(~~~(~p & ~q) & ~~(~r & ~~(~p & ~q))) v ((p & s) v (r v s))) v (p & r))
5.
(~~(~~~(~~~(~p & ~q) & ~~(~r & ~~(~p & ~q))) & ~((p & s) v (r v s))) v (p & r))
6.
(~~(~~~(~~~(~p & ~q) & ~~(~r & ~~(~p & ~q))) & ~~(~(p & s) & ~(r v s))) v (p & r))
7.
(~~(~~~(~~~(~p & ~q) & ~~(~r & ~~(~p & ~q))) & ~~(~(p & s) & ~~(~r & ~s))) v (p & r))
8.
~(~~~(~~~(~~~(~p & ~q) & ~~(~r & ~~(~p & ~q))) & ~~(~(p & s) & ~~(~r & ~s))) & ~(p & r))

https://www.youtube.com/channel/UCnxwq2aGJRbjOo_MO54oaHA
https://www.youtube.com/watch?v=2oT4qrHmDMY&lc=UgwviLabd7r_3KpP6wh4AaABAg. 9h5lFRmix1R9h78GftO_iE
https://www.youtube.com/watch?v=2oT4qrHmDMY&lc=UgyG714V_k7odQMrTz14AaABAg. 9h740K6COOA9h77HSGDH4A
https://www.youtube.com/watch?v=2oT4qrHmDMY&lc=UgyG714V_k7odQMrTz14AaABAg. 9h740K6COOA9h76fafzcEJ
https://www.youtube.com/watch?v=2oT4qrHmDMY&lc=UgyG714V_k7odQMrTz14AaABAg. 9h740K6COOA9h759YIjlaG
https://www.youtube.com/watch?v=2oT4qrHmDMY&lc=UgyG714V_k7odQMrTz14AaABAg. 9h740K6COOA9h74pjGcbEq
https://www.youtube.com/watch?v=2oT4qrHmDMY&lc=UgyG714V_k7odQMrTz14AaABAg
https://www.youtube.com/watch?v=2oT4qrHmDMY&lc=UgzJJUDVv2Mb6YGkPYh4AaABAg. 9h5uPRbWIZl9h7165DZdjg
https://www.youtube.com/channel/UCnxwq2aGJRbjOo_MO54oaHA

https://www.youtube.com/channel/UCnxwq2aGJRbjOo_MO54oaHA

**p45cal** · 02-15-2014, 12:42 PM

Originally Posted by kylefoley76

but then I realized that the required algorithm is more complicated since it can only handle one t per sentence.

Check out the Split and Join. I haven't looked in detail at what you want to do with the portions of the string, but this snippet will give you an idea of how you might go about it:

Code:

Sub blah()
zzz = "(((p v q) > (r v (p v q))) > ((p & s) v (r v s))) > (p & r)"
yyy = Split(zzz, ">")
For i = LBound(yyy) To UBound(yyy)
  yyy(i) = aFunctionToProcessTheSubString(yyy(i))
Next i
result = Join(yyy, " glug ")
MsgBox result
End Sub

**snb** · 02-15-2014, 04:14 PM

**kylefoley76** · 02-15-2014, 09:54 PM

Originally Posted by snb

I'd prefer you to tell us what's the aim of all this.
I do not exclude the possibility there's a much simpler approach to it.

It's all about building software to solve logical problems. So in logic implication (p > q) is equal to (~p v q) and disjunction is equal to conjunction (p v q) = ~(~p & ~q). Even more difficult is equivalence: (p <-> q) = (p > q) & (q > p)

**kylefoley76** · 02-19-2014, 11:44 PM

Originally Posted by p45cal

Check out the Split and Join. I haven't looked in detail at what you want to do with the portions of the string, but this snippet will give you an idea of how you might go about it:

Code:

Sub blah()
zzz = "(((p v q) > (r v (p v q))) > ((p & s) v (r v s))) > (p & r)"
yyy = Split(zzz, ">")
For i = LBound(yyy) To UBound(yyy)
  yyy(i) = aFunctionToProcessTheSubString(yyy(i))
Next i
result = Join(yyy, " glug ")
MsgBox result
End Sub

It says there is a function not defined at this line of code:

aFunctionToProcessTheSubString(yyy(i))