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


1 Answer 1


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, 2019 at 17:33
  • $\begingroup$ Meet-in-the-middle attack requires the attacker to know the plaintext, how can the attacker already know the plaintext without decrypting the ciphertext? $\endgroup$
    – Flan1335
    Apr 6, 2023 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.