Lightweight multimedia encryption: Algorithms and performance analysis.  Page 124 
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A.6 Proof of Lemma 6 A general binary sequence s can be treated as an output of an binary information source S, which emits 0 and 1 according to its statistical structure. Thus, the statistical property of s reflects the statistical structure of information source S. The nth order entropy for S is defined as H(n)(S) =Xd p(d) 1 log p(d) , (A.6) where the summation is over all possible nbit subsequence d. For a sufficiently long sequence s, the probability of a particular subsequence d can be calculated as the number of occurrence of d divided by the length of s, i.e., p(d) = # of occurrence of d in s length of s . (A.7) For instance, the first and the secondorder entropies are computed by H(1)(S) = p(0) log 1 p(0) + p(1) log 1 p(1) , and H(2)(S) = p(00) log 1 p(00) + p(01) log 1 p(01) + p(10) log 1 p(10) + p(11) log 1 p(11) . We examine the input binary sequence A and the output binary sequence C as a result of the RPB encryption. Since the RPB operation only reorders certain bit blocks, the total number of 0 and 1 remains the same after passing the RPB unit. It is clear that p(0) and p(1) of output C is equal to those of input A, respectively. Hence, by the definition of the firstorder entropy, we have H(1)(A) = H(1)(C). The situation becomes more complicated for the secondorder entropy. The number of digrams 00, 01, 10, 11 will vary under the RPB operation due to rotation of bit blocks. The change of digrams after a bit block rotation is illustrated in Fig. A.3. Before block rotation: b a After block rotation: b a x0 … xm y0 … yn y0 … yn x0 … xm Figure A.3: Change of digrams under the RPB operation. The bit block x0 . . . xm is rotated behind bit block y0 . . . yn. b and a are the bit before and after the rotation, either 0 or 1. By examining all occurrences of digrams, we can find the following changes: b x0 → b y0, xm y0 → yn x0, yn a → xm a. For example, if ( x0, xm, y0, yn) = ( 0, 0, 1, 1), then the changes are: b0 → b1, 01 → 10, 1a → 0a. (A.8) 114
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Title  Lightweight multimedia encryption: Algorithms and performance analysis.  Page 124 
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Full text  A.6 Proof of Lemma 6 A general binary sequence s can be treated as an output of an binary information source S, which emits 0 and 1 according to its statistical structure. Thus, the statistical property of s reflects the statistical structure of information source S. The nth order entropy for S is defined as H(n)(S) =Xd p(d) 1 log p(d) , (A.6) where the summation is over all possible nbit subsequence d. For a sufficiently long sequence s, the probability of a particular subsequence d can be calculated as the number of occurrence of d divided by the length of s, i.e., p(d) = # of occurrence of d in s length of s . (A.7) For instance, the first and the secondorder entropies are computed by H(1)(S) = p(0) log 1 p(0) + p(1) log 1 p(1) , and H(2)(S) = p(00) log 1 p(00) + p(01) log 1 p(01) + p(10) log 1 p(10) + p(11) log 1 p(11) . We examine the input binary sequence A and the output binary sequence C as a result of the RPB encryption. Since the RPB operation only reorders certain bit blocks, the total number of 0 and 1 remains the same after passing the RPB unit. It is clear that p(0) and p(1) of output C is equal to those of input A, respectively. Hence, by the definition of the firstorder entropy, we have H(1)(A) = H(1)(C). The situation becomes more complicated for the secondorder entropy. The number of digrams 00, 01, 10, 11 will vary under the RPB operation due to rotation of bit blocks. The change of digrams after a bit block rotation is illustrated in Fig. A.3. Before block rotation: b a After block rotation: b a x0 … xm y0 … yn y0 … yn x0 … xm Figure A.3: Change of digrams under the RPB operation. The bit block x0 . . . xm is rotated behind bit block y0 . . . yn. b and a are the bit before and after the rotation, either 0 or 1. By examining all occurrences of digrams, we can find the following changes: b x0 → b y0, xm y0 → yn x0, yn a → xm a. For example, if ( x0, xm, y0, yn) = ( 0, 0, 1, 1), then the changes are: b0 → b1, 01 → 10, 1a → 0a. (A.8) 114 