Lightweight multimedia encryption: Algorithms and performance analysis.  Page 42 
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an equal probability. Thus, the algorithm has to know KHS zk+1 to determine the correct ek+1 and in turn decrypt pk+1. In summary, all decryption steps are independent of each other. To compute the correct plaintext, the algorithm has to know the corresponding KHS sequence. Hence, if an algorithm exists to decrypt an RECencrypted ciphertex, such an algorithm can also be used to give a particular pseudorandom sequence produced by the underlying KHS generation algorithm. In this sense, we say that breaking the RECbased scheme is at least as difficult as breaking the underlying KHS generation algorithm. Next, let us examine the complexity of a bruteforce attack. There are two approaches to attack the KHS of a given ciphertext. The first method is to try all possible KHS to see whether the decrypted plaintext is semantically meaningful. The second strategy is a brute force search for the secret seed s since the underlying KHS generation algorithm is available to the public. A random seed value is picked to calculate a KHS, which is used to decrypt the ciphertext until a meaningful result is encountered. For the first method, the length of the plaintext is L so that it takes L decryption steps to recover the plaintext. As analyzed earlier, all these steps are independent of each other. In each step, there are E possible choices that are equally likely. Therefore, the entire search space is O(EL). For the second method, since the seed is a random rbit binary number, all rbit numbers occur with an equal likelihood. In other words, the seed may be produced as a result of tossing an unbiased coin r times, recording 0 for the head and 1 for the tail. Thus, the cryptanalyst has no better strategy other than searching all 2r possible rbit binary numbers. The entire search space in this case is O(2r). Thus, it is concluded that the computational complexity of a bruteforce attack to RECbased schemes is min(2r,EL). This completes the proof. 32
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Title  Lightweight multimedia encryption: Algorithms and performance analysis.  Page 42 
Repository email  cisadmin@lib.usc.edu 
Full text  an equal probability. Thus, the algorithm has to know KHS zk+1 to determine the correct ek+1 and in turn decrypt pk+1. In summary, all decryption steps are independent of each other. To compute the correct plaintext, the algorithm has to know the corresponding KHS sequence. Hence, if an algorithm exists to decrypt an RECencrypted ciphertex, such an algorithm can also be used to give a particular pseudorandom sequence produced by the underlying KHS generation algorithm. In this sense, we say that breaking the RECbased scheme is at least as difficult as breaking the underlying KHS generation algorithm. Next, let us examine the complexity of a bruteforce attack. There are two approaches to attack the KHS of a given ciphertext. The first method is to try all possible KHS to see whether the decrypted plaintext is semantically meaningful. The second strategy is a brute force search for the secret seed s since the underlying KHS generation algorithm is available to the public. A random seed value is picked to calculate a KHS, which is used to decrypt the ciphertext until a meaningful result is encountered. For the first method, the length of the plaintext is L so that it takes L decryption steps to recover the plaintext. As analyzed earlier, all these steps are independent of each other. In each step, there are E possible choices that are equally likely. Therefore, the entire search space is O(EL). For the second method, since the seed is a random rbit binary number, all rbit numbers occur with an equal likelihood. In other words, the seed may be produced as a result of tossing an unbiased coin r times, recording 0 for the head and 1 for the tail. Thus, the cryptanalyst has no better strategy other than searching all 2r possible rbit binary numbers. The entire search space in this case is O(2r). Thus, it is concluded that the computational complexity of a bruteforce attack to RECbased schemes is min(2r,EL). This completes the proof. 32 