# Encrypt twice with DES using same key ? $\operatorname{DES}(\operatorname{DES}(x,k), k)$

Does encrypting with $$\operatorname{DES}$$ in the following way $$\operatorname{DES}(\operatorname{DES}(x,k), k)$$ make $$\operatorname{DES}$$ as strong as $$\operatorname{2DES}$$ with 2 keys?

Because we have to run $$\operatorname{DES}$$ twice now for each key in order to find it - first run $$\operatorname{DES}(x,k)$$ and the output is the input as text for the second round of encryption? So overall $$2*2^{56} = 2^{57}$$ just like meet in the middle complexity, right ?

Does encrypting with $$\operatorname{DES}$$ in the following way $$\operatorname{DES}(\operatorname{DES}(x,k), k)$$ make $$\operatorname{DES}$$ as strong as $$\operatorname{2DES}$$ with 2 keys?

No.

So overall $$2*2^{56} = 2^{57}$$ just like meet in the middle complexity, right ?

This time estimate ignores the memory requirement that the meet-in-the-middle attack on 2DES requires.

Memory is cheap, but it's not free. The estimate of $$2^{57}$$ operations to break 2DES ignores the overhead of both the memory and the interprocessor communication (it's unlikely that a single processor will be able to perform $$2^{57}$$ DES operations in an acceptable period of time). As these objections can be addressed, it's not a lie to ignore them, but they are nontrivial considerations.

In contrast, finding the key for $$DES(DES(x,k), k)$$ requires no memory, and can be trivially parallelized.

• ... and should be feasible using a couple of thousand USD on modern cloud platforms. – SEJPM Nov 15 '18 at 9:56