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Regex – Identify special characters in Excel by column names

I found a code that uses RegEx to determine if a column contains special characters. I've just learned about RegEx and tried to edit that code, but I've found 3 issues: 1) It uses ActiveSheet, but I've read that it could cause problems. I just want to review the worksheet named "Consolidated". I have tried it with worksheets ("Consolidated"). Area in the line "For each", but it is not accepted. 2) This applies to the entire worksheet. I'd like to know if it's possible to apply it to a column based on its name (for example, first name) and not on its scope because the columns in the consolidated worksheet are not set. 3) Also mark the first line where the column names are located (which should not happen). I understand the logic or workflow for that, but I'm new to VBA, so I really want to ask you for advice.

Dim strPattern As String: strPattern = "[^a-z0-9-]"
Dimming RegEx as an object
Darken cell as area

Set regEx = CreateObject ("VBScript.RegExp")
regEx.Global = True
regEx.IgnoreCase = True
regEx.Pattern = strPattern

For each cell in ActiveSheet.Range ("A: Z") & # 39; Define your own scope here
If strPattern <> "" Then & # 39; If the cell is not empty
If regEx.Test (Cell.Value) Then & # 39; Check if there is a match
Cell.Interior.ColorIndex = 6 & # 39; If yes, change the background color
End If
End If
Next

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co.combinatorics – The chromatic list number of some special toroidal grid graphs

  • A list assignment $ L $ to the cornerstones of $ G $ is the assignment of a list
    to adjust $ L (v) $ from colors to each vertex $ v $ from $ G $; and a $ k $list mapping is a list mapping like $ | L (v) | geq k $for every corner point $ v $, If $ L $ is a list mapping too $ G $, then one $ L $Staining of $ G $ is a coloring (not necessarily correct) in which each vertex receives a color from its own list;

  • The graph $ G $ is $ k $-list-colourable or $ k $-Selectableif it is right $ L $-Colourable for everyone $ k $Lists assignment $ L $ to $ G $, The chromatic number $ chi (G) $ from $ G $ is the smallest number $ k $ so that $ G $ is
    $ k $-Farbbar. The List chromatic number $ chi_l (G) $is the smallest number $ k $ so that $ G $ is $ k $-list-colourable or $ k $-Selectable.

  • It has been proved that $ chi_l (G) geq chi (G) $, since when $ k < chi (G) $ then $ G $ is not $ L $-Farbbar
    if every vertex $ v $ from $ G $ is given the same list $ L (v) $ from $ k $ To dye.

Look at the graphic $ S_n $ what has as
Vertex set the $ n ^ 2 $ Cells of ours $ n times n $ Array with two adjacent cells if they are in the same row or column.
The graph $ S_n $ There no $ n $ Cells in a row are adjacent in pairs, we need at least $ n $ To dye.
In addition, any staining with $ n $ Colors corresponds to a latin square,
wherein the cells numbered with the same number form a color class. Since
Latin squares exist, as we have seen, we conclude $ chi (Sn) = n $and the Dinitz
Problem can be stated as $$ chi_l (S_n) = n? $$

By solving the Dinitz problem we know that the list is chromatic number of $ C_3 Box C_3 $ is 3, that is $ chi_l (C_3 Box C_3) = 3 $,

The method of attacking the Dinitz problem is we have to
Find an orientation of the graph $ S_n $ with outdegrees $ d _ + (v) ≤ n – 1 $ for all $ v $
and that ensures the existence of a kernel for all induced subgraphs.
I want to know that $ chi_l (C_3 Box C_5) $ and $ chi_l (C_5 Box C_5) $,
If anyone can give suggestions or comments, I will appreciate it. Many Thanks.

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General topology – Special multi-valued holomorphic function on open Riemann surfaces

To let $ M $ be an open Riemann surface and $ f $ a multi-valued holomorphic function $ M $ to $ mathbb {H} $, from where $ mathbb {H} $ is the upper half-plane. Suppose the monodroma of $ f $ exists in $ A = left { begin {pmatrix} a & b \ 0 & frac {1} {a} end {pmatrix}: a> 0, b in mathbb {R} right }. $ I suspect that $ M $ must be a hyperbolic Riemann surface (The surface that is not compact and carries a negative non-constant subharmonic function). But I can not prove it. This problem has been on my mind for a long time, and I will be grateful for any answers or suggestions.

Special case: when the monodroma of $ f $ exists in $ B = left { begin {pmatrix} 1 & b \ 0 & 1 end {pmatrix}: b in mathbb {R} right } $, then $ Im f $ is a positive harmonic function. We know that on a Riemann parabolic surface there is a non-constant positive harmonic function (the surface that is not compact and does not carry a negative non-constant subharmonic function). Since $ f $ is not a constant $ Im f $ is not a constant, $ M $ is a hyperbolic Riemann surface.

If the monodroma of $ f $ exists in $ C = left { begin {pmatrix} a & 0 \ 0 & a ^ {- 1} end {pmatrix}: a> 0 right } $I can prove it by looking at the maximum abelian coverage $ M $,