Lightweight multimedia encryption: Algorithms and performance analysis.  Page 120 
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bit stream 0 1 0 … 1 0 1 0 1 1 1 0 … V (k  2) V (k  4) V (k) = V (k  2) +V(k  4) Figure A.2: The recursive relationship of V (k). i must fall into the interval S = {s, s + 1, . . . s + L}, s + L ≤ k containing L + 1 integers at a certain point. We have V (k) = Pi aiV (i), i ∈ S with appropriate coefficients ai. This completes the proof. A.2 Proof of Lemma 2 Let A = (a1a2 . . . aN) be a stream of N bits. Assume B is the maximum block size allowed in partitioning A. Notice that this restriction implies any rotation in partition cannot start with bit beyond aB. Thus, all R(N) possible rotation in partition RP(A, p, r) can be classified into the following B categories. 1) those starting with a1; 2) those starting with a2; ... B) finally, those starting with aB. We denote the total number of each category by R1(N), R2(N), . . . , RB(N). Note this classification is mutuallyexclusive and allinclusive, meaning that any possible resultant bit stream A′ = RP(A, p, r) must belong to one and only one of the above categories. Thus, we have R(N) = B Xi=1 Ri(N) (A.1) Now let us look at each of the above categories. In category 1), a1 is fixed and we are left with N − 1 bits after a1 which we can freely partition and rotate. Thus R1(N) = R(N − 1). In category 2), it must be true that the first 2 ≤ k ≤ B bits are chosen as a block and a 1bit left rotation is performed. This is the only way A′ can start with a2. If k = 2 (A′ starts with a2a1), we have N − 2 bits left over and total number of possible rotation in partition is clearly R(N − 2). k = 3 (A′ starts with a2a3a1) corresponds to R(N − 3). Finally for k = B we have the number R(N − B). Notice that this classification with different values of k is again mutuallyexclusive and allinclusive. Hence we end up with: R2(N) = N−2 X k=N−B R(k) 110
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Title  Lightweight multimedia encryption: Algorithms and performance analysis.  Page 120 
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Full text  bit stream 0 1 0 … 1 0 1 0 1 1 1 0 … V (k  2) V (k  4) V (k) = V (k  2) +V(k  4) Figure A.2: The recursive relationship of V (k). i must fall into the interval S = {s, s + 1, . . . s + L}, s + L ≤ k containing L + 1 integers at a certain point. We have V (k) = Pi aiV (i), i ∈ S with appropriate coefficients ai. This completes the proof. A.2 Proof of Lemma 2 Let A = (a1a2 . . . aN) be a stream of N bits. Assume B is the maximum block size allowed in partitioning A. Notice that this restriction implies any rotation in partition cannot start with bit beyond aB. Thus, all R(N) possible rotation in partition RP(A, p, r) can be classified into the following B categories. 1) those starting with a1; 2) those starting with a2; ... B) finally, those starting with aB. We denote the total number of each category by R1(N), R2(N), . . . , RB(N). Note this classification is mutuallyexclusive and allinclusive, meaning that any possible resultant bit stream A′ = RP(A, p, r) must belong to one and only one of the above categories. Thus, we have R(N) = B Xi=1 Ri(N) (A.1) Now let us look at each of the above categories. In category 1), a1 is fixed and we are left with N − 1 bits after a1 which we can freely partition and rotate. Thus R1(N) = R(N − 1). In category 2), it must be true that the first 2 ≤ k ≤ B bits are chosen as a block and a 1bit left rotation is performed. This is the only way A′ can start with a2. If k = 2 (A′ starts with a2a1), we have N − 2 bits left over and total number of possible rotation in partition is clearly R(N − 2). k = 3 (A′ starts with a2a3a1) corresponds to R(N − 3). Finally for k = B we have the number R(N − B). Notice that this classification with different values of k is again mutuallyexclusive and allinclusive. Hence we end up with: R2(N) = N−2 X k=N−B R(k) 110 