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} else if (weather == "thunder" || weather == "lightning") { stayInside(); }

case "thunder": case "lightning": stayInside(); break;

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four known plaintexts Then one additional known plaintext will enable us to solve for the rotation, for a total requirement of five known plaintext blocks Suppose we have the required five known plaintexts Then we can solve for the Ki and L in (423) as follows For each of the 64 possible choices for L , we solve for one bit at a time, beginning from the low-order bit position It is necessary to keep track of carry bits and at some steps we must save multiple possible solutions (ie, some branching occurs) Also, it is not possible to uniquely recover all of the subkey words using this approach We can always recover K O , K2, K 1 3 ~ + 5 ;and K13R+8, but we only determine K1 CB K13R+6 and K2 @ K13~+7,or K1 @ K13~+7and K2 @ K13R+6, depending on the rotation L .





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The nal method that you can use to branch depending on a Boolean expression is the conditional operator, sometimes called the ternary operator because it is the only operator to take three operands ( :). The conditional operator behaves similarly to the basic if statement.

The conditional operator takes three expressions as operands. The rst is a logical expression to evaluate for truth, much like the expression passed into an if statement. This can be a comparison, a variable, a function call, or any expression that can evaluate to a Boolean value. It may be required to couch the expression in parentheses to ensure it s interpreted in the correct order; you may choose as a matter of style to always enclose the expression in parentheses. The logical expression is followed by a question mark. Immediately after the question mark is the expression that will be evaluated and returned if the logical statement is found to be true. This is followed with a colon, and nally, the expression to be returned if the rst value was found to be false. Here s a practical example. This time, you check for hail:

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See Problem 20 for a slightly simplified version of this subkey recovery problem Once we have recovered the subkey words and the shift, we are then in a situation similar to (421) At this point, if we are given any ciphertext, we use the recovered subkey words to obtain the XOR of words of the corresponding plaintext This is somewhat analogous to a one-time pad cipher where the key has been used more than once At a minimum this leaks information about the plaintext, and it may be possible to recover the plaintext directly from the ciphertext, provided that we have sufficient information about the plaintext In [82] it is claimed that if we simply know the plaintext is English, or random ASCII text, then it is possible to recover the plaintext.

(weather == "hail") bringMotorcycleHelmet() : bringStrawHat();

This is basically the same as writing:

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The recovery of the plaintext from the ciphertext is considered in Problem 21, where a slightly simplified version of the problem is given In practice, a similar approach could be used on actual Akelarre ciphertext..

if (weather == "hail") { bringMotorcycleHelmet(); } else { bringStrawHat(); }

The thing that makes the conditional operator unique is that the entire thing evaluates to the result of the correct branch, rather than simply executing it like an if..then..else statement. So if it s hailing, not only will bringMotorcycleHelmet() run, but the entire conditional expression will be replaced with whatever value bringMotorcycleHelmet() returns. So you can use conditional expressions within other expressions. The following:

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