[Cryptography] Unicity distance of Playfair

[Ray Dillinger]

On 03/25/2016 02:21 AM, mok-kong shen wrote:
> Am 25.03.2016 um 01:38 schrieb Ray Dillinger:

> Your sentence "Every cipher that has the same key length will have
> the same unicity distance when used to encrypt English." is IMHO
> evidently a "consequence" of these "ciphers" which have been examined
> all have alphabet of size 26 (excepting Playfair, which we are arguing
> here). So your logic is apparently problematical here.

It's subject to a few assumptions.  As others have pointed
out, Playfair is a block cipher of two-letter blocks, so
calculating its unicity distance in terms of letters is
valid for an even number of letters.  The Unicity distance
will always be a multiple of the block size, so if the
calculation in letters had come out to, eg, 26.8 or something
you'd have to round up to 28 instead of 27.

Also the redundancy measure I used was a very specific figure
applicable to most hand ciphers:  It was the redundancy of
English text represented in letters, over the space of all
possible strings of letters the same length.

But if we're talking about the redundancy of English text as
represented in a 676-character alphabet (digraphs) over the
space of all possible digraph strings the same length, the
sizes of the units (and therefore the redundancy per unit)
are the only thing that's changed.

You have the same ratio of non-English strings to English
strings at a length of N digraphs, that you have at a length
of 2N letters.  So your redundancy per digraph is twice as much
as the redundancy per letter.

You get a unicity distance half as big as it was with letters
because you doubled the redundancy in the denominator without
changing the number of possible keys.  So you get unicity =
13 digraphs, which happens to be 26 letters.

> Anyway, your post of 24.03. employed the term "aphabet". In a post
> of mein to J. Leichter I used the hypothetical (because impractical)
> case of digraphic Vigenere. Do you agree at all that in that case
> the "alphabet" is 26**2 instead of 26?

Sure, you could calculate any cipher's unicity distance in
pairs instead of letters.  But ultimately all it will tell
you is that unicity measured in digraphs is half (rounding up)
the unicity distance in characters.  Because each digraph
is twice as much plaintext as each letter, that's the same
amount of plaintext either way.

Anyway, Viginere (26x26 or 676x676) can have any key length,
so a unicity measurement requires making an assumption about
the key length.  If the sequence across the table is not known
to your opponent (ie, if that's part of the key) then the
676x676 version would have a much larger practical unicity
distance just because there are MANY more keys.

				Bear

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