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Challenge "Monoalphabetic substitution with camouflage — Part 2"  

  By: admin on Dec. 7, 2011, 3:07 p.m.

In this challenge it is to test whether a monoalphabetic substitution with camouflage provides more security even though spaces were not removed prior to the encryption.
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 Last edited by: admin on Oct. 31, 2021, 2:54 a.m., edited 1 time in total.

Re: Challenge "Monoalp. substitution with camoufl. - Part 2"  

  By: bdcht on May 10, 2012, 11:50 a.m.

I have the entire fairy plaintext but the validation is refused;
I seems that '5th sentence counting from the end' is not clear enough. Define better 'counting' and 'sentence' please:

  • do you start counting from 0 or 1 ? (1st sentence from end is last sentence ?)
  • do you include the terminating dot in the sentence ?
  • do you include the initial space (encoded with a star) in the sentence ?

do you confirm that stars are separating words in the validator with all camouflage chars removed ?

I have assumed that counting starts from 1, and tried some combinations…

thx

Re: Challenge "Monoalp. substitution with camoufl. - Part 2"  

  By: Veselovský on May 10, 2012, 9:32 p.m.

I seems that '5th sentence counting from the end' is not clear enough. Define better 'counting' and 'sentence' please

Count the same way as you would count from beginning. It means that last sentence is first counting from the end.

do you include the terminating dot in the sentence ?

You can read in pdf: 'The last character of the solution is a period "."'

do you include the initial space (encoded with a star) in the sentence ?

No, it is not a part of a sentence.

do you confirm that stars are separating words in the validator with all camouflage chars removed ?

You can read in PDF: 'Keep in mind that "" in the plaintext represents " ", so after deciphering simply replace every asterisk by a space.' …so it means that in solution there are no asterisks as they all have been replaced by spaces'*

Re: Challenge "Monoalp. substitution with camoufl. - Part 2"  

  By: Veselovský on May 10, 2012, 9:36 p.m.

If this does not help, you can send me PM with your solution.

Re: Challenge  

  By: D3d4lu5 on July 29, 2017, 6:28 p.m.

The plaintext alphabet consists of the following 28 characters:
ABCDEFGHIJKLMNOPQRSTUVWXYZ*.

Find the 5th sentence counting from the end and provide it written in characters of the plaintext alphabet as a solution.

Translation of the german version:

The solution consists only of characters from the plaintext alphabet.

In my opinion this is misleading as space is not a character of the plaintext alphabet according to the first statement.

Re: Challenge  

  By: Veselovský on July 29, 2017, 10:24 p.m.

But at the end of the PDF is written:
"Keep in mind that "*" in the plaintext represents " ", so after
deciphering simply replace every asterisk by a space…"

Using "*" instead of " " was for practical reasons, as space is invisible.

  1. ABCDEFGHIJKLMNOPQRSTUVWXYZ
  2. ABCDEFGHIJKLMNOPQRSTUVWXYZ

Do you see any difference between the first and the second alphabet? The first contains space the second not, which is not visually obvious.

Re: Challenge  

  By: D3d4lu5 on July 31, 2017, 6:58 p.m.

"…replace every asterisk by a space and enjoy a comfortable reading." (not for entering the solution)

As I said, in my opinion the description is misleading and this challenge was one of the very few where my input was not accepted in the first attempt and I had to look into this thread.

So I decided to post a comment.

But I think most should be able to solve this in the second attempt (even though I needed 3), especially after looking into this thread. [HTML_REMOVED]


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