5.docx - 1 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

# 5.docx - 1 We have a symmetric encryption algorithm EK(M)=C...

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1. 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? Note that Google has around 2 million servers. Also, check the Internet and see what the

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Unformatted text preview: expected life time of the Sun is. Can you crack the key before that? According to 192bits key length the keyspace would be 2 192 = 6.277* 1057 possible keys Which will give us half of this amount for 50% of the keys : 50% of the keys = 0.5 * 6.277* 1057 = 3.1385 * 1057 The keys that can be calculated with the given machines in one year would be : 1,000,000*5,000,000 C1 per second Therefore, in one year = 5*1012*60*60*24*365 = 1.5768 * 1020 years The required time for decryption= 3.1385 * 1057 / 1.5768 * 1020 = 1.99 * 1037 Suns life is 5*109 years; therefore this time is not enough and much more time is needed even with the google servers! ...
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