Lightweight multimedia encryption: Algorithms and performance analysis.  Page 69 
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The answer to the first question is most likely positive since one is allowed to arbitrarily partition the bit stream provided that block size < B and rotate freely in each block. From the above 8bit plaintext example, it seems not so hard to obtain two alias keys by observing the bit stream pattern and do several trials. The second problem, i.e., to compute the accurate number of alias keys, is however not an easy one. Since alias keys are ciphertext dependent, there seems no quick formula to compute the number of alias keys for a given plaintext/ciphertext pair. Nonetheless, if we take into account all possible ciphertexts C for a plaintext A, we have the following conclusion regarding alias keys. Lemma 3 Let A = (a1a2 . . . aN) be a bit stream of length N containing Z 0’s and Alias(A,C) denote the number of alias keys for the plaintext/ciphertext pair (A,C). Then, there exists a ciphertext A′ such that Alias(A,A′) > »2N.µN Z¶¼ (3.3) In a statistically average sense, a random plaintext A contains half 0’s (Z = N/2). When the plaintext length N is large enough, we have Alias(A,A′) > p¼N/2 (3.4) The above lemma establishes the existence of alias keys. Refer to Appendix A.3 for a complete proof. The quantity p¼N/2 is however a conservative estimate of number of alias keys. Further analysis of the average number of alias keys will be given in Sec. 3.5.5 below. 3.5.3 CiphertextOnly Attack To evaluate the security strength of the proposed RPB encryption scheme, we consider ciphertextonly and known/chosenplaintext attacks in this and the next subsections. 59
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Title  Lightweight multimedia encryption: Algorithms and performance analysis.  Page 69 
Repository email  cisadmin@lib.usc.edu 
Full text  The answer to the first question is most likely positive since one is allowed to arbitrarily partition the bit stream provided that block size < B and rotate freely in each block. From the above 8bit plaintext example, it seems not so hard to obtain two alias keys by observing the bit stream pattern and do several trials. The second problem, i.e., to compute the accurate number of alias keys, is however not an easy one. Since alias keys are ciphertext dependent, there seems no quick formula to compute the number of alias keys for a given plaintext/ciphertext pair. Nonetheless, if we take into account all possible ciphertexts C for a plaintext A, we have the following conclusion regarding alias keys. Lemma 3 Let A = (a1a2 . . . aN) be a bit stream of length N containing Z 0’s and Alias(A,C) denote the number of alias keys for the plaintext/ciphertext pair (A,C). Then, there exists a ciphertext A′ such that Alias(A,A′) > »2N.µN Z¶¼ (3.3) In a statistically average sense, a random plaintext A contains half 0’s (Z = N/2). When the plaintext length N is large enough, we have Alias(A,A′) > p¼N/2 (3.4) The above lemma establishes the existence of alias keys. Refer to Appendix A.3 for a complete proof. The quantity p¼N/2 is however a conservative estimate of number of alias keys. Further analysis of the average number of alias keys will be given in Sec. 3.5.5 below. 3.5.3 CiphertextOnly Attack To evaluate the security strength of the proposed RPB encryption scheme, we consider ciphertextonly and known/chosenplaintext attacks in this and the next subsections. 59 