Would XDES be more secure than 3DES, where X is a number > 3? Like 10DES... If so why aren't we using it?


Would XDES be more secure than 3DES, where $X > 3$ ?

Yes. This answer about if 4DES / 5DES is possible illustrates it quite well.

It would however make more sense, if you really want to make it more secure, to have $n$-DES, where $n = 2k-1 \space | \space k \in \mathbb{N}$, or in simple terms: $n$-DES, where $n$ is an odd number.

This comes from the Meet-in-the-middle attack (MITM attack).


DES has a keylength of $2^{56}$. Double DES only has a keylength of $2^{57} (= 2 \times 2^{56})$. Only by using 3DES you start to really improve the security from single-DES: $2^{56} \times 2^{56} + 2^{56} \approx 2^{112}$.

Why aren't we using XDES with large X?

Because of AES. It's not feasible to have MITM-attack on AES.

Using XDES instead of AES would have more drawbacks than benefits, especially from a computational view.

  • 1
    $\begingroup$ You can also add the small block-size; sweet32 $\endgroup$ – kelalaka Jan 21 '19 at 17:33

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