Lightweight multimedia encryption: Algorithms and performance analysis.  Page 121 
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Continuing the same line of reasoning we have the following equation: Ri(N) = N−i X k=N−B R(k) Finally, summing up Ri(N), we arrive at R(N) = B Xi=1 Ri(N) = R(N − 1) + B Xi=2 N−i X k=N−B R(k) = R(N − 1) + N−2 X k=N−B (N − 1 − k)R(k) (A.2) Rearranging the terms of the above equation, we obtain another recursive relationship of R(N): R(N) = 2R(N − 1) + N−3 X k=N−B R(k) (A.3) If we define the following recursive sequence S(N) = ½ 2S(N − 1) N > 0 1 N = 0 the solution is apparently S(N) = 2N. From the above definitions of R(N) and S(N), it is clear that if R(N0) > S(N0) for some N0, then R(N) > S(N) for all N > N0. Now, it is readily checked that R(6) = 65 > S(6) = 64. Thus, we come to the conclusion that R(N) > 2N for N ≥ 6. This completes the proof. A.3 Proof of Lemma 3 Let C(N) to denote the number of different ciphertexts by applying all different RPB oper ations to A. There are R(N) possible ways to do an RPB operation and the range size is C(N). Thus, by the pigeon hole principle, there must exist a ciphertext A′ such that at least ⌈R(N)/C(N)⌉ RPB operations transform A to A′. That is, we have Alias(A,A′) ≥ ⌈R(N)/C(N)⌉ Now, let us estimate the size of C(N). Note that an RPB operation only alters the order of bits in plaintext A as a result of random rotation. Since the total number of 0’s in ciphertext A′ remains the same after any RPB operation, any ciphertext C also contains Z 0’s. The total number of Nbit binary sequences containing Z 0’s is ¡N Z¢. However, we claim that C(N) must be strictly less than ¡N Z¢ due to the upper bound of partitioned block’s length pi < B. 111
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Title  Lightweight multimedia encryption: Algorithms and performance analysis.  Page 121 
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Full text  Continuing the same line of reasoning we have the following equation: Ri(N) = N−i X k=N−B R(k) Finally, summing up Ri(N), we arrive at R(N) = B Xi=1 Ri(N) = R(N − 1) + B Xi=2 N−i X k=N−B R(k) = R(N − 1) + N−2 X k=N−B (N − 1 − k)R(k) (A.2) Rearranging the terms of the above equation, we obtain another recursive relationship of R(N): R(N) = 2R(N − 1) + N−3 X k=N−B R(k) (A.3) If we define the following recursive sequence S(N) = ½ 2S(N − 1) N > 0 1 N = 0 the solution is apparently S(N) = 2N. From the above definitions of R(N) and S(N), it is clear that if R(N0) > S(N0) for some N0, then R(N) > S(N) for all N > N0. Now, it is readily checked that R(6) = 65 > S(6) = 64. Thus, we come to the conclusion that R(N) > 2N for N ≥ 6. This completes the proof. A.3 Proof of Lemma 3 Let C(N) to denote the number of different ciphertexts by applying all different RPB oper ations to A. There are R(N) possible ways to do an RPB operation and the range size is C(N). Thus, by the pigeon hole principle, there must exist a ciphertext A′ such that at least ⌈R(N)/C(N)⌉ RPB operations transform A to A′. That is, we have Alias(A,A′) ≥ ⌈R(N)/C(N)⌉ Now, let us estimate the size of C(N). Note that an RPB operation only alters the order of bits in plaintext A as a result of random rotation. Since the total number of 0’s in ciphertext A′ remains the same after any RPB operation, any ciphertext C also contains Z 0’s. The total number of Nbit binary sequences containing Z 0’s is ¡N Z¢. However, we claim that C(N) must be strictly less than ¡N Z¢ due to the upper bound of partitioned block’s length pi < B. 111 