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Ki = key[i (mod N ) ]
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next i j=O f o r i = 0 t o 255 j = ( j Si Ki) (mod 256) swap(S2, S j ) next i
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The RC4 keystream is generated one byte at a time. An index is determined based on the current contents of and the indexed byte is selected as the keystream byte. Similar to the initialization routine, at each step the permutation S is modified so that S always contains a permutation of {0,1,2,. . . ,255). The keystream generation algorithm appears in Table 3.10.
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In 2000, Fluhrer, Mantin, and Shamir [51]published a practical attack on RC4 encryption as it is used in the Wired Equivalent Privacy (WEP) protocol. In WEP, a non-secret 24-bit initialization vector, denoted as IV, is prepended to a long-term key and the result is used as the RC4 key. Note that the role of the IV in WEP encryption is analogous to the role that the message indicator (MI) plays in the World War I1 cipher machines discussed in the
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The attack does highlight a shortcoming in the RC4 initialization process-a ing that can be fixed without modifying the underlying RC4 algorithm. shortcom-
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Table 3.10: RC4 Keystream Generator RC3 Keystreani
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previous chapter. As with the 511 in the WWII cipher machines. the W E P IV is necessary t o prevent messages from being sent in depth. Recall that two ciphertext messages are in depth if they were encrypted using the same key. Alessages in depth are a serious threat t o a stream cipher. In WEP. Trudy, the crypt analyst, knows many ciphertext messages (packets) and their corresponding IVs, and she would like t o recoaw the long-term key. The Fluher-Mantin-Shamir attack provides a clever. efficient, and elegant way to do just that. This attack has been successfully used t o break real WEP traffic [145]. Suppose that for a particular message, the three-byte initialization vector is of the form (3.8) where V can be any byte \-due. Then thcse three IV bytes become K O ,Kl and Kz in the RC-2 initialization algorithm of Table 3.9, while K3 is the first byte of the unknown long-term key. T h a t is. the message key is
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here V is known to Trudy. but K 3 , K4, K j . are unknown. To understand the attack. we need t o carefull\ consider what happens to the table S during the RC4 initialization phase when K is of the foiin in (3.9). In the RC4 initialization algorithm in Table 3.9 we fir5t set S t o the identity permutation, so that we have
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Suppose that K is of the form in (3.9). Then at the i = 0 initialization step. we compute the index j = 0 + SO+ K O= 3 and elements i and j are swapped. resulting in the table
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At the next step, i = 1 and j = 3 S1 t K1 = 3 1 255 = 3, since the addition is modulo 256. Elements i and j are again swapped, giving
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3.4 RC4
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assuming that, after reduction modulo 256, we have 6 V K3 > 5 V . If this is not the case, then 6 V K3 will appear to the left of 5 V ,but this has no effect on the success of the attack. Now suppose for a moment that the RC4 initialization algorithm were to stop after the i = 3 step. Then if we generate the first byte of the keystream according to the algorithm in Table 3.10, we find i = 1 and j = Si = S = 0, 1 so that t = S SO= 0 3 = 3. Then the first keystream byte would be 1