Question

Meet in the Middle attack made double-DES (encrypting something twice in a row with two separate keys of DES) not nearly as secure as we might think. This came at the cost of storing many possible decryptions in a table. For this question I want you to calculate how much storage space it would take to store the information necessary to do a meet in the middle attack on double-AES (of 128 bits). State any assumptions you have to make. This storage size should be in gigabytes. This might be a large number.

Answer 1

Answer: The DES key given is 128 bits, in which every eighth bit is parity check, so overall 16 bits is for parity check remaining 112 bit key is for DES. The size of the key space which is nothing but storage space is
119
and for double DES, intuitively double the key space is required which is nothing but 2 ×
119
= $2^{113}$ bits. This we have to convert into gigabyte.

Approximately, 1 gigabyte = 8000000000 bits.

To convert $2^{113}$ bits to gigabyte we need to divide $2^{113}$ /8000000000. This will give you the answer in gigabyte which is ofcourse a large number.