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Added DetailsI am studying DES for my Cryptoexam .As far as I understood , the initial plaintext is permuted and then the left half is called L0 and Right half is called R0.

After that for each round the process is as follows

L(i) = R (i-1)

R(i) = L(i-1) XOR f(R(i-1), K(i))

This relation holds till the last round

i..e L(16) = R(15)

and R(16) = L(15) XOR f(R(15),K(16))

After the last round the text is interchanged into R(16)L(16) and the inverse permutation is applied.This is what Stinson Book says.

But When I read the same topic from Forouzan , he says that in the last round there is no swapper i..e L(16) != R(15) and R(16) = R(15) . I searched on net regarding this but didn't find anything useful. Is my understanding wrong? Excerpt here

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  • $\begingroup$ You can incorporate the last swap into the final permutation. $\endgroup$ – Paŭlo Ebermann Sep 13 '14 at 20:49
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According to the following link (Slide 5) and to what I studied last semester,

http://www.ee.ic.ac.uk/pcheung/teaching/ee4_network_security/L02DESIDESAES.pdf

During the final round (Round 16) before the inverse permutation, the left and right halves of the bits will be swapped then the inverse permutation will be applied.

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