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If there always existed a $k_3$ such that $\operatorname{DES}(k_2, \operatorname{DES}(k_1, M)) = \operatorname{DES}(k_3, M)$, how would that affect the security of $\operatorname{DES}$?

nanopro
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1 Answers1

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This question was tested years ago by Kaliski at. al Is the Data Encryption Standard a group? (Results of cycling experiments on DES as;

  • Is DES closed under functional composition?

They applied the cycling test and concluded that DES is not a group. Therefore, we don't expect that $$\operatorname{DES}(k_2, \operatorname{DES}(k_1, M)) = \operatorname{DES}(k_3, M)$$

As noted in the article, if there was such functional composition then a known-plaintext attack with $2^{28}$ would be able to break DES, on average.

kodlu
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kelalaka
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