Sun 01 Dec 2019

## Tips & Tricks 173 - Color Code for Official Color of Excel

By |Sunday, December 01st, 2019|Categories: Tips and Tricks|Tags: , , , , , , , |0 Comments

Everybody sees the dark green color of Excel.

The color code for this color is (more…)

Sat 04 Mar 2017

## Solution - Challenge 59 - Clean the Problem Workbook Data

By |Saturday, March 04th, 2017|Categories: Solutions|Tags: , , , , , |0 Comments

Below is a possible solution to the problem Challenge 59 - Clean the Problem Workbook Data

Formula to convert would be which you need to drag down would be

=--SUBSTITUTE(SUBSTITUTE(A2,UNICHAR(8237),""),UNICHAR(8236),"")

Data has UNICHAR(8237) and UNICHAR(8236) prefixed and suffixed which need to be replaced by above formula.

Sat 04 Feb 2017

## Challenge 59 - Clean the Problem Workbook Data

By |Saturday, February 04th, 2017|Categories: Challenges|Tags: , , , , , |1 Comment

Download the the workbook from the below link. The Challenge before you is to write a formula to clean the data. If you perform any mathematical operation on the data, it will return #VALUE error. You need to clean the data through a formula.

Challenge 59 - Problem Workbook

The solution to the above problem will be posted after a month i.e. on 4-Mar-17.

Mon 19 Dec 2016

## Solution - Challenge 56 – Cryptography Challenge 5 – Fully Functional Caesar’s Shift Cipher Decrypter

By |Monday, December 19th, 2016|Categories: Solutions, VBA||0 Comments

Below is a possible solution to Challenge 56 – Cryptography Challenge 5 – Fully Functional Caesar’s Shift Cipher Decrypter

Sat 19 Nov 2016

## Challenge 56 – Cryptography Challenge 5 – Fully Functional Caesar’s Shift Cipher Decrypter

By |Saturday, November 19th, 2016|Categories: Challenges, VBA||1 Comment

We made ourselves familiarized with Caesar's Shift in Challenge 39 - Cryptography Challenge 1 - Caesar's Shift Cipher and made its decrypter in Challenge 53 - Cryptography Challenge 4 - Caesar's Shift Cipher Decrypter. We also made a fully functional encrypter in Challenge 40 – Cryptography Challenge 2 – Fully Functional Caesar’s Shift Cipher based on VBA solution as formula based solution has its own limitation.

The challenge before you is to write a VBA function to create a fully functional Caesar's Shift Cipher Decrypter. This should take cell or text as first argument and amount of shift as second argument to generate the Decrypted Text.

Hence, to generate answer in A3, we have called the function as =DecryptCS(A2,C2)

Note - The shift is only for the English Alphabets and case sensitive. Any other character like comma, space etc. should appear as they are.

The Excel file related to this challenge can be downloaded from Challenge 56 – Cryptography Challenge 5 – Decryption - Fully Functional Caesar’s Shift Cipher

The answer to this challenge would be published after a month i.e. on 19-Dec-16.

Tue 08 Nov 2016

## Solution - Challenge 53 - Cryptography Challenge 4 - Caesar's Shift Cipher Decrypter

By |Tuesday, November 08th, 2016|Categories: Solutions||0 Comments

Below is a possible solution to the Challenge 53 - Cryptography Challenge 4 - Caesar's Shift Cipher Decrypter.

Put following formula and drag down

=IFERROR(CHAR(IF(CODE(LOWER(G2))-\$B\$3<97,CODE(G2)-\$B\$3+26,CODE(G2)-\$B\$3)),"")

The Excel file containing the solution can be downloaded from Solution - Challenge 53 - Decryption - Caesar's Shift Cipher

Sat 08 Oct 2016

## Challenge 53 - Cryptography Challenge 4 - Caesar's Shift Cipher Decrypter

By |Saturday, October 08th, 2016|Categories: Challenges||2 Comments

We published a cryptography challenge Challenge 39 - Cryptography Challenge 1 - Caesar's Shift Cipher. The purpose of the challenge was to create an Excel based Encrypter.

This time, we need to create the de-crypter for the same. The Excel file related to this challenge can be downloaded from Challenge 53 - Decryption - Caesar's Shift Cipher

Tue 27 Sep 2016

## Solution - Challenge 50 - Cryptography Challenge 3 - Generate Tabula Recta

By |Tuesday, September 27th, 2016|Categories: Solutions||0 Comments

Below is a possible solution to the Challenge 50 - Cryptography Challenge 3 - Generate Tabula Recta

Put following formula and drag down and right

=CHAR(MOD(ROWS(\$1:1)+COLUMNS(\$A:A)-2,26)+65)

A workbook containing the above solution can be downloaded from Solution - Challenge 50 - Cryptography Challenge 3 - Generate Tabula Recta

Sat 27 Aug 2016

## Challenge 50 - Cryptography Challenge 3 - Generate Tabula Recta

By |Saturday, August 27th, 2016|Categories: Challenges||1 Comment

Tabula Recta is supposed to be one of the foundation stones of cryptography. This was one of the earliest forms of encryption. The more about this can be read at https://en.wikipedia.org/wiki/Tabula_recta

One look at below and you will understand what it is. In upcoming challenges in future, Tabula Recta will be used. The first row is from A to Z and first column is from A to Z (Yellow row and column). You need to populate all other values by a single formula. (Yellow row and column is something which you can put as constant. But white area is something which you need to populate)

The solution to the above challenge will be published after a month i.e. on 27-Sep-16.

Mon 09 May 2016

## Solution - Challenge 40 – Cryptography Challenge 2 – Fully Functional Caesar’s Shift Cipher

By |Monday, May 09th, 2016|Categories: Solutions, VBA||1 Comment

Below is a possible solution to the challenge - Challenge 40 – Cryptography Challenge 2 – Fully Functional Caesar’s Shift Cipher

1. Make a backup of your workbook
2. Open your workbook and ALT+F11
3. Locate your Workbook name in Project Explorer Window
4. Right click on your workbook name > Insert > Module
5. Copy paste the Macro code given below
6. Save your file as .xlsm
7. Call your function as =EncryptCS(A1,\$B\$2)

```Function EncryptCS(PText, Shift) As String
Dim Ws As Worksheet
Dim i As Long
Dim CText As String
Dim PArr, CArr
Application.Volatile
PArr = Split(StrConv(PText, vbUnicode), vbNullChar)
ReDim CArr(UBound(PArr) - 1)
For i = LBound(PArr) To UBound(PArr) - 1
If Abs(77.5 - Asc(UCase(PArr(i)))) < 13 Then
If Asc(LCase(PArr(i))) + Shift > 122 Then
CArr(i) = Chr(Asc(PArr(i)) + Shift - 26)
Else
CArr(i) = Chr(Asc(PArr(i)) + Shift)
End If
Else
CArr(i) = PArr(i)
End If
Next i
EncryptCS = Join(CArr, "")
End Function```

A workbook containing the above solution can be downloaded from Solution - Challenge 40 – Cryptography Challenge 2 – Fully Functional Caesar’s Shift Cipher.