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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 $\operatorname{DES}$?

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    $\begingroup$ 2-DES is only 57-bit secure because of meet-in-the-middle attack, so k3 would decrease the security by 1 bit. $\endgroup$
    – DannyNiu
    Mar 26 '19 at 1:12
  • $\begingroup$ why only 1 bit will be affected? $\endgroup$
    – nanopro
    Mar 26 '19 at 2:28
  • $\begingroup$ @DannyNiu That would decrease the security of 2DES relative to DES by 1 bit. But the question is about the security of DES itself. $\endgroup$ Mar 26 '19 at 23:33
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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.

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  • $\begingroup$ I hope you dont mind, fixed some words in your nice answer. $\endgroup$
    – kodlu
    Mar 27 '19 at 1:56
  • $\begingroup$ @kodlu nope. Thanks. $\endgroup$
    – kelalaka
    Mar 27 '19 at 5:28

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