Challenge "Monoalphabetic substitution with camouflage — Part 3"  

  By: admin on Dec. 7, 2011, 3:13 p.m.

In this challenge a variant has been used that allows to benefit from the camouflage effect and get a ciphertext not longer than the plaintext.

 Last edited by: admin on Oct. 31, 2021, 2:54 a.m., edited 1 time in total.

Re: Challenge  

  By: Veselovský on Aug. 4, 2016, 11:05 a.m.

Hello… a small remark.

You can read in the PDF file:

The plaintext is split into two parts at any random position (but perhaps it would be wise to choose the difference between the lengths of the two parts not too large in order to maximize the camouflage effect)."

Some of you interpret it as the lengths of the two parts are almost the same. But what was meant is that if you have plaintext of length, say, 500 it would not be wise to split it into 20 and 480 as it would be as easy to solve as an ordinary substitution cipher.
The difference between the lengths should be considered relative to the length of the larger part.
In both examples above the difference is 20 (260-240=20, 22-2=20), but for 500 this difference is small and for 24 the same difference is too large.
To be mathematically precise we should consider expression (L2-L1)/L2 where L2>=L1.
(L2-L1)/L2=1-L1/L2. Or simple consider only L1/L2 or L2/L1, as this expression depends only on this ratio.
It is up to the solver to decide what values of L2/L1 are more and what are less secure. For this challenge this ratio is unknown to the solver.

Best regards

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