[Cryptography] Unicity distance of Playfair

mok-kong shen mok-kong.shen at t-online.de
Thu Mar 24 14:35:25 EDT 2016


Am 24.03.2016 um 16:20 schrieb Jerry Leichter:
>> One point I don't yet understand: You wrote: "Using Playfair we
>> either can't distinguish IJ, or we can't distinguish YZ." Clearly
>> you refer 'IJ' to input and 'YZ' to output. But 'IJ' has a (variable)
>> space of 25*25 (like 'I' has a space of 25), isn't it? So 'IJ'
>> would be an alphabet of 25*25 in my current view.
> Playfair uses a 5x5 table - 25 cells.  Each input character has to correspond to a unique cell, but there are 26 letters.  So traditionally we replace every I in the input with a J (or every X with a Y).  Now the input has only 25 different characters.
>
> The replacement means "we can't distinguish I [from] J (or X [from] Y)" after decoding - which has no significant effect on the understandability of English-language text.
>                                                          -- Jerry

I suppose there is misunderstanding between us.

My point is that since Playfair processes digrams (as single units),
one should accordingly take that fact into consideration while
determining its unicity-distance. Consider the analogy of a
hypothetical (because very impractical owing to its size) Vigenere
substitution table for digrams, i.e. one has entries of aa, ab, ac,
.... zz instead of a, b, c, .... z. Then the "alphabet" concerned in
that case is clearly of size 26**2 and not 26. Do you agree?

M. K. Shen




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