Question

Earlier in the class we discussed how a 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

Question:-

Earlier in the class we discussed how a 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:-

• 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 2^{112} and for double DES, intuitively double the key space is required which is nothing but 2 × 2^{112} = 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.

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