Lightweight multimedia encryption: Algorithms and performance analysis.  Page 122 
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Suppose the first 0 of A occurs at bit position z1, and the ith 0 at bit position zi. Because pi < B, the first 0 cannot appear at a bit position after z1 + B in ciphertext A′. Thus, beginning from bit position z1 + B to the end, A′ can contain at most Z − 1 0’s. Likewise, A′ can contain at most Z −i 0’s after bit position zi +B. This limitation prohibits many bit patterns from being produced as a result of an RPB operation. An accurate analysis of C(N) becomes a quite complicated combinatorial problem since it requires complete knowledge of all zi’s of the particular A. In this study, we take C(N) < ¡N Z¢ as an upper bound estimate. From Lemma 2, we know that R(N) > 2N. Hence, we conclude Alias(A,A′) ≥ ⌈R(N)/C(N)⌉ > »2N.µN Z¶¼ which is Eq. (3.3). A typical random bit stream A contains half 0’s and half 1’s on the average. By substitut ing N = 2Z into above equations and using Sterling’s formula for factorial n! ≈ √2¼n(n/e)n, we obtain C(N) < µ2Z Z ¶ = 2Z! Z!Z! = √4¼Z(2Z/e)2Z 2¼Z(Z/e)2Z = 22Z √¼Z = 2N p¼N/2 Finally, we end up with Alias(A,A′) > 2N. 2N p¼N/2 = p¼N/2 This completes the proof. A.4 Proof of Lemma 4 First, we arbitrarily pick two random Nbit plaintexts, A1 and A2, and two Nbit ciphertexts, A′ 1 and A′ 2. By definition of A(N), there are roughly A(N) keys k1 ∈ K1 that encrypt A1 to A′ 1 and A(N) keys k2 ∈ K2 that encrypt A2 to A′ 2. Now, consider the plaintext concatenation A1A2 and ciphertext concatenation A′ 1A′ 2. It is obvious that a concatenation key k = k1k2 of any k1 ∈ K1 and any k2 ∈ K2 would encipher the plaintext A1A2 to ciphertext A′ 1A′ 2. In other words, all such k’s are alias keys for the 2Nbit plaintext A1A2, while the average number of a general 2Nbit plaintext is by definition A(2N). Hence, we get A(2N) ≥ # of such k = k1 k2 = A2(N). By repeatedly applying the above reasoning to m plaintexts, we have A(mN) ≥ Am(N), m is an integer. (A.4) It is a well known result that a function satisfying the above equation must be of the form cN. In addition, we have already demonstrated in Lemma 3 that A(N) > p¼N/2 for at least one ciphertext. Therefore, c > 1 when N is sufficiently large (otherwise, A(N) → 0). This completes the proof. 112
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Title  Lightweight multimedia encryption: Algorithms and performance analysis.  Page 122 
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Full text  Suppose the first 0 of A occurs at bit position z1, and the ith 0 at bit position zi. Because pi < B, the first 0 cannot appear at a bit position after z1 + B in ciphertext A′. Thus, beginning from bit position z1 + B to the end, A′ can contain at most Z − 1 0’s. Likewise, A′ can contain at most Z −i 0’s after bit position zi +B. This limitation prohibits many bit patterns from being produced as a result of an RPB operation. An accurate analysis of C(N) becomes a quite complicated combinatorial problem since it requires complete knowledge of all zi’s of the particular A. In this study, we take C(N) < ¡N Z¢ as an upper bound estimate. From Lemma 2, we know that R(N) > 2N. Hence, we conclude Alias(A,A′) ≥ ⌈R(N)/C(N)⌉ > »2N.µN Z¶¼ which is Eq. (3.3). A typical random bit stream A contains half 0’s and half 1’s on the average. By substitut ing N = 2Z into above equations and using Sterling’s formula for factorial n! ≈ √2¼n(n/e)n, we obtain C(N) < µ2Z Z ¶ = 2Z! Z!Z! = √4¼Z(2Z/e)2Z 2¼Z(Z/e)2Z = 22Z √¼Z = 2N p¼N/2 Finally, we end up with Alias(A,A′) > 2N. 2N p¼N/2 = p¼N/2 This completes the proof. A.4 Proof of Lemma 4 First, we arbitrarily pick two random Nbit plaintexts, A1 and A2, and two Nbit ciphertexts, A′ 1 and A′ 2. By definition of A(N), there are roughly A(N) keys k1 ∈ K1 that encrypt A1 to A′ 1 and A(N) keys k2 ∈ K2 that encrypt A2 to A′ 2. Now, consider the plaintext concatenation A1A2 and ciphertext concatenation A′ 1A′ 2. It is obvious that a concatenation key k = k1k2 of any k1 ∈ K1 and any k2 ∈ K2 would encipher the plaintext A1A2 to ciphertext A′ 1A′ 2. In other words, all such k’s are alias keys for the 2Nbit plaintext A1A2, while the average number of a general 2Nbit plaintext is by definition A(2N). Hence, we get A(2N) ≥ # of such k = k1 k2 = A2(N). By repeatedly applying the above reasoning to m plaintexts, we have A(mN) ≥ Am(N), m is an integer. (A.4) It is a well known result that a function satisfying the above equation must be of the form cN. In addition, we have already demonstrated in Lemma 3 that A(N) > p¼N/2 for at least one ciphertext. Therefore, c > 1 when N is sufficiently large (otherwise, A(N) → 0). This completes the proof. 112 