# Would XDES be more secure than 3DES, where X is a number > 3?

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

• – hardyrama Jan 21 '19 at 14:10

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).

Example:

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.

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