Time it takes to break DES via bruteforce

[ecy5maa]

Q. Given a pair of plaintext P and ciphertext C encrypted under a DES key, in the worst case, how long does an attacker have to try for finding a key K such that C = DES(K, P) which represents the DES encryption of P under key K?

 

Assume that the attacker can perform 10^13 encryptions per second. Give your answer to the nearest hours.

 

 

 

Allright so the book(Stallings) says that if the key is 56 bits long, then the total number of Keys is 7.2x10^16 keys. So I assume that this means that the time taken is simply (no. of keys)/(10^13) which means it will take 2 hours to find the key K.

 

Is this correct? If not can you suggest an alternative means to answer this?

[NetSplitter]

Wouldn't the key space be 2^56 instead of 7.2x10^16 (not that it matters because the answer is still the same hourly wise) - I've just never seen it written out like that.

 

2^56 / 7.2x10^16 = 7205 seconds = 2 hours.

 

I would believe that to be correct.

Useless side note: Although it would be VERY unlikely it would take two hours. On average brute force attacks will end up finding the correct key after half the keyspace has been attacked(central limit theorem)

 

Just curious, is this for a class, certification, or just self-study?

Edited December 11, 2011 by NetSplitter

[ecy5maa]

Its an exam question, worth 4 marks.

 

Thanks for your reply!!