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Challenge "M-138 — Part 2"

Challenge "M-138 — Part 2"  

  By: Javex on Dec. 24, 2014, 12:39 a.m.

This is part 2 in a series about the M-138. Part 2 is a partly-known plaintext challenge (48/100).
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 Last edited by: admin on Oct. 31, 2021, 2:55 a.m., edited 1 time in total.

Re: Challenge "M-138 — Part 2"  

  By: Bart13 on Dec. 24, 2014, 2:11 a.m.

The plaintext is:
"TWO THINGS ARE INFINITE THE UNIVERSE AND HUMAN STUPIDITY"

But that's 48 letters (56 including the spaces, but they should be ignored).
So are there 2 letters missing, or has the plaintext 48 letters?

I might not be counting correctly, but even in that case the plaintext seems very appropriate … [HTML_REMOVED]

Re: Challenge "M-138 — Part 2"  

  By: be on Dec. 24, 2014, 3:41 p.m.

The plaintext is:
"TWO THINGS ARE INFINITE THE UNIVERSE AND HUMAN STUPIDITY"

But that's 48 letters (56 including the spaces, but they should be ignored).
So are there 2 letters missing, or has the plaintext 48 letters?

I might not be counting correctly, but even in that case the plaintext seems very appropriate … [HTML_REMOVED]

Hello Bart13,
you are completely right:

  • We don't count the white spaces (blanks), and
  • The given plaintext has a length of 48 characters, and
  • The sought-after ciphertext has a length of 52 characters.
    We'll gonna fix this in the challenge.

Best regards and merry Xmas, BE

Re: Challenge  

  By: Gerben on Dec. 20, 2019, 8:46 p.m.

Hi,

I have the solution of this nice challenge! However this site says that this solution is a wrong answer… What can be wrong?
I have 52 letters in capital, no spaces and a nice text…

Re: Challenge  

  By: tryone144 on Dec. 24, 2019, 3:31 a.m.

Hi,

I have the solution of this nice challenge! However this site says that this solution is a wrong answer… What can be wrong?
I have 52 letters in capital, no spaces and a nice text…

Hello Gerben,

The format is right, but your solution has not been 100% correct.
Just two characters are wrong, but they should be easy to fix if you know the english word. (Try to google each word if you are not sure.)

Best Regards,
Bernd

Re: Challenge  

  By: itnomad on Feb. 17, 2021, 3:02 p.m.

One question: The challenge description says:

The plaintext of part 2 consists of 100 letters which means that each strip in the frame is used four times.

However, I have a total of 100 strips. Does that implicitly mean that only 25 of the set of 100 strips are in use, reducing the possible keyspace?

Thanks!
Alex.

Re: Challenge  

  By: Theofanidis on Feb. 17, 2021, 6:32 p.m.

One question: The challenge description says:

The plaintext of part 2 consists of 100 letters which means that each strip in the frame is used four times.

However, I have a total of 100 strips. Does that implicitly mean that only 25 of the set of 100 strips are in use, reducing the possible keyspace?

Thanks!
Alex.

Dear Alex (itnomad)
That is correct
Split both the ciphertext and known plaintext in groups of 25 letters
Then compare first column of ciphertext with first column of known plaintext
Then the second, the third, etc
With a few examinations you can find the offset and some of the stripes
Then continue until you find readable text
Kind Regards
George Theofanidis


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