We have a symmetric encryption algorithm E K (M)=C. Here K is the secret key, M

Question

We have a symmetric encryption algorithm EK(M)=C. Here K is the secret key, M is the plaintext, and C is the ciphertext. We (and the attacker) know that the key length is 192 bits. The attacker eavesdrops on the communication line and gets a copy of the ciphertext C1. Now the attacker decides to conduct the brute force attack and try every possible key to get the plaintext M1. Let us assume that there is only one possible M1 and if the attacker sees it, he will know that this is the correct one. The attacker has 1,000,000 machines, with each machine having the capabilities to try 5,000,000 decryption of C1 with different keys per second. If one machine finds the right key, it will automatically notify the attacker.

Now please answer, how many years (roughly) does the attacker need to try 50% of the keys?

Explanation / Answer

given, number of machines = 106 number of decryptions each machine can do = 5*106/sec number of decryptions that can be performed per sec = 5*106*106 = 5*1012 Given, we have 192 bits in the symmetric key. So, number of combinations = 10192 (Assuming the key is in Decimal Number System) we can decrypt 5*1012 per sec time required to decrypt half of 10192 keys = 10192/5*1012*2 = 10192/1013 sec = 10179 sec 10179 sec = 10179/60*60*24*365 years 3.18*e171 years

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