Lightweight multimedia encryption: Algorithms and performance analysis.  Page 73 
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of alias keys, A(N), plays a key role in RPB encryption’s ability to withstand the known plaintext attack. Due to the large amount of alias keys, a cryptanalyst cannot determine the correct key given only few plaintext/ciphertext pairs. However, it is interesting to ask, when a sufficient number of plaintext/ciphertext pairs are available for analysis, whether it is possible to exclude wrong alias keys and determine the correct key uniquely. Thus, we study this problem and estimate the number of plaintext/ciphertext pairs needed to launch a knownplaintext attack. The total computational cost of this attack is analyzed. Given a sufficient number of plaintext/ciphertext pairs, a cryptanalyst can proceed as follows. At first, he/she can select a given pair and calculate A(N) alias keys for this pair using a certain algorithm. Then, he can check all these A(N) keys against each available pair (Ai,Ci). If RPB(Ai, k) = Ci, then k is counted as one possible key. Otherwise, k must be a wrong alias key, which can be discarded. As more and more pairs are examined, the number of possible keys decreases. This process is repeated until only one key is left, which must be the correct key. For a general Nbit plaintext, let P(N) denote the expected number of plaintext/ciphertext pairs needed to uniquely determine the correct key. The following lemma offers an estimate for P(N). Lemma 5 With R(N), A(N) and P(N) defined in the above description, we have the fol lowing relationship: P(N) = R(N) R(N) − A(N) lnR(N). (3.5) Then, a rough estimate of P(N) is given by P(N) ≈ N ln 2. (3.6) 63
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Title  Lightweight multimedia encryption: Algorithms and performance analysis.  Page 73 
Repository email  cisadmin@lib.usc.edu 
Full text  of alias keys, A(N), plays a key role in RPB encryption’s ability to withstand the known plaintext attack. Due to the large amount of alias keys, a cryptanalyst cannot determine the correct key given only few plaintext/ciphertext pairs. However, it is interesting to ask, when a sufficient number of plaintext/ciphertext pairs are available for analysis, whether it is possible to exclude wrong alias keys and determine the correct key uniquely. Thus, we study this problem and estimate the number of plaintext/ciphertext pairs needed to launch a knownplaintext attack. The total computational cost of this attack is analyzed. Given a sufficient number of plaintext/ciphertext pairs, a cryptanalyst can proceed as follows. At first, he/she can select a given pair and calculate A(N) alias keys for this pair using a certain algorithm. Then, he can check all these A(N) keys against each available pair (Ai,Ci). If RPB(Ai, k) = Ci, then k is counted as one possible key. Otherwise, k must be a wrong alias key, which can be discarded. As more and more pairs are examined, the number of possible keys decreases. This process is repeated until only one key is left, which must be the correct key. For a general Nbit plaintext, let P(N) denote the expected number of plaintext/ciphertext pairs needed to uniquely determine the correct key. The following lemma offers an estimate for P(N). Lemma 5 With R(N), A(N) and P(N) defined in the above description, we have the fol lowing relationship: P(N) = R(N) R(N) − A(N) lnR(N). (3.5) Then, a rough estimate of P(N) is given by P(N) ≈ N ln 2. (3.6) 63 