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Figure 24.6 Chaotic masking scheme. (a) A digital message; (b) the transmitted signal encoded; (c) the recovered signal; (d) the recovered signal filtered; (e) the original and filtered messages. then filtered (Figure 24.6(d)). A simple visual comparison of the original and the filtered messages (Figure 24.6(e)) shows the excellent performance of the method. Another interesting application of this methodology is the encoded transmission of digital images. As an example, let us consider the image given in Figure 24.7, which consists of a 100 100 matrix of gray-level intensities ranging from 0 to 255. In this case, we can use the couple of drive-response drivers to encode and decode the image. In the chaotic masking method, the chaotic signal from the transmitter is used to encode the list of gray-level intensities after a transient of 20 iteration steps. Afterwards, the scheme displayed in Figure 24.5 is applied to recover the original image by using the synchronization process to clean the chaotic signal out. Figure 24.8 shows the transmitted and received images. A simple visual comparison of Figures 24.7 and 24.8(b) shows the excellent performance of the method. Note that no information about the appearance of the initial image can be obtained from the noisy transmitted signal, shown in Figure 24.8(a).



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If you need to test the condition of something against several possible outcomes, as in the last example when you were testing for rain, snow, and sun, it might be best to use a switch statement. A switch is just like a giant if..else if chain where each if tests the same expression (in this example, the weather). You can check for any number of different cases or possible outcomes for the expression you re evaluating and even de ne a default behavior in case none of the cases match. Notice that break statements are used after each case. The break statement tells the conditional statement or loop to cancel the rest of its operation without going any further. Unlike the if statement, which executes code in a block enclosed with curly braces ({}), when working with switches, you must include break after each case to prevent the code from continuing to the next case.





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In this form, we have the XOR of output bits written in terms of the XOR of input bits, with a n unknown rotation L . If Akelarre did not employ its input and output transformations, then in (4.21), A would be plaintext and C would be the corresponding ciphertext, and the ciphertext would immediately provide some information about the plaintext. Of course, we must take the input and output transformations into account. As in Figure 4.12, we denote the plaintext block as X and the corresponding ciphertext block as Y . As above, A is the input t o the first round and C is t h e output of the last round. Let D be the result after C passes through the keyed rotation in the output transformation, but before the key is added and XORed. Let lo be the amount of the rotation in the output transformation. Then from (4.21) we have

switch (weather) { case "rain": bringUmbrella(); break; case "snow": bringCoat(); break; case "sun: bringSunscreen(); break; default: stayInside(); }

3 ) =

This is equivalent to writing:

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if (weather == "rain") { bringUmbrella(); } else if (weather == "snow") { bringCoat(); } else if (weather == "sun") { bringSunscreen(); } else { stayInside(); }

Chaotic masking is not totally efficient, because only a part of the transmitted power carries information, most being used to mask information. Chaotic switching (also known as chaotic shift keying) allows a higher level of efficiency and security [13].

(4.22)

It s not required, but you can use a default block to de ne a case that executes if none of the other cases in the switch statement match.

You can intentionally omit the break statement after a case. This allows the subsequent cases to run in addition to the current one, if they also match the condition. For example, you could replace

which relates the plaintext X t o the ciphertext Y , modulo the unknown rotation L arid the unknown subkey words K O K1, Kz, Ks, K13~+5, I ~ R + ~ , , K K 1 3 R f 7 , and KI:WM. Given sufficient known plaintext, we can solve for the unknown shift and subkey words in (4.23). Since each known plaintext yields one 64-bit equation, in principle, we can solve for t h e eight unknown 32-bit subkey words using

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