Talk:Advantage (cryptography)

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Wait... [Either something's wrong or more explanation's needed][edit]

"a permutation selected at random from the (2⁶⁴)! possible permutations on 64-bit blocks." ... "Note from Stirling's approximation that (264)! is around 10^{3.47×10²⁰}, so even specifying which permutation is selected requires writing down a number too large to represent exactly in any real computer. "

There are 64! (~2²⁹⁶ or ~10⁸⁹) permutations of 64 items (bits or anything else,) NOT (2⁶⁴)!. The fact that the permutation may be specified by a 64-bit key rather than a 56-bit key doesn't change that, though a 56-bit key might further reduce the size of the utilized subset compared to specification by a 64-bit key...reducing the work function by 2^(64-56) = 2⁸ = 256. Specifying a particular 64P requires only 64 6-bit values, or 384-bits, not the given figure, so at most, the ratio is 2^(384-56) = 2³²⁸ = 5.53×10⁹⁸, which, though large, is far below what one would get using the given figure.

Either the quote from the article is flat-out wrong or there's missing information that would make clear how it isn't. — Preceding unsigned comment added by Ultracrypt (talk • contribs) 03:59, 28 March 2015 (UTC)[reply]

The article looks correct to me. DES takes a 64-bit input block and produces a 64-bit output block. Hence, for each key we have a permutation of a set with 2⁶⁴ elements (each possible input). 2A02:120B:2C4F:7900:55C8:3A06:7922:B43 (talk) 06:03, 1 April 2015 (UTC)[reply]

Then why not 64! × 2⁶⁴ ≈ 2³⁶⁰...i.e. every permutation of each value of a block? This remains far less than 10^{3.47×10²⁰}. Ultracrypt (talk) 06:59, 6 April 2015 (UTC)[reply]