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I am doing some research on differential cryptanalysis. Most people who are familiar with that cryptanalysis technique do know the tutorial from Heys (Heys Tutorial). I came currently to the question, why is the in the final round no permutation, as in the other rounds? Heys says in the tutorial, that there would no purpose for that. I don't understand why there is no purpose, as in the other rounds. Can somebody clarify that to me?

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  • $\begingroup$ Does the final permutation add security? AFAIK, Heys explained that, too. $\endgroup$ – kelalaka Dec 2 '19 at 20:59
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    $\begingroup$ No, I know that already. But this is not clear to me. That is the reason me asking here that question. $\endgroup$ – chris000r Dec 2 '19 at 21:03
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Because a fixed and published permutation is known, linear and ever so reversible. It's very much like a bait & switch move, but without the switch(substitution). Security is built piecemeal from units of permutation & substitution(rounds). We can undo a singular publically published permutation, and it doesn't take any notice of a key(as it's fixed), thus it adds nothing security wise. Nobody really gets confused and the diffusion effects can easily be reversed.

As a kiddie, slightly similar example, think about ROT13 so called encryption.

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