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A digital communication channel has an error rate of one bit per every thousand transmitted Errors are rare, but when they occur, they tend to occur in bursts that affect many consecutive bits If a single bit is transmitted, we might model the probability of an error as 1 1000 However, if the previous bit was in error, because of the bursts, we might believe that the probability that the next bit is in error is greater than 1 1000 In a thin lm manufacturing process, the proportion of parts that are not acceptable is 2% However, the process is sensitive to contamination problems that can increase the rate of parts that are not acceptable If we knew that during a particular shift there were problems with the lters used to control contamination, we would assess the probability of a part being unacceptable as higher than 2%.

In a manufacturing process, 10% of the parts contain visible surface aws and 25% of the parts with surface aws are (functionally) defective parts However, only 5% of parts without surface aws are defective parts The probability of a defective part depends on our knowledge of the presence or absence of a surface aw These examples illustrate that probabilities need to be reevaluated as additional information becomes available The notation and details are further illustrated for this example Let D denote the event that a part is defective and let F denote the event that a part has a surface aw Then, we denote the probability of D given, or assuming, that a part has a surface aw as P1D F2 .

5.4 11.105

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This notation is read as the conditional probability of D given F and it , is interpreted as the probability that a part is defective, given that the part has a surface aw Because 25% of the parts with surface aws are defective, our conclusion can be stated as P1D F2 025 Furthermore, because F denotes the event that a part does not have a surface aw and because 5% of the parts without surface aws are defective, we have that P1D F 2 005 These results are shown graphically in Fig 2-12..

5% defective P(D F ) = 0.05

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the notation used in the original description of MD5 [122], it is awkward for analysis of the algorithm. Strangely, Wang numbers the bits of 32-bit words from 1 to 32 (right-to-left). Some authors use 6 for the modular difference and A for the XOR difference, while we use A for the modular difference. The bottom line is that considerable care must be taken when attempting to analyze results culled from a variety of papers, since it is not a trivial task to translate the results into a consistent notation.

Table 2-3

10 30 40

EXAMPLE 2-16

Wang s MD5 attack is a differential attack. Recall that Dobbertin s MD4 attack [42] uses subtraction modulo 232 as the difference operator. Wang s attack uses this same modular difference for inputs. However, some parts of the MD5 attack require more detailed information than modular subtraction provides, so a kind of precise differential [157] is also employed. This differential combines modular subtractions with information on the precise location of the bit differences. In effect, this precise differential includes both a modular difference and an XOR difference, and also provides additional information beyond what these two standard differentials provide. To motivate this precise differential, consider the pair of bytes given by y = 00010101 and y = 00000101 and another pair of bytes z = 00100101 and z = 00010101. Then

Table 2-3 provides an example of 400 parts classi ed by surface aws and as (functionally) defective. For this table the conditional probabilities match those discussed previously in this section. For example, of the parts with surface aws (40 parts) the number defective is 10. Therefore, P1D F2 10 40 0.25

and of the parts without surface aws (360 parts) the number defective is 18. Therefore, P1D F 2 18 360 0.05

= z/ -

In Example 2-16 conditional probabilities were calculated directly. These probabilities can also be determined from the formal de nition of conditional probability.

De nition The conditional probability of an event B given an event A, denoted as P1B A2 , is P1B A2 for P1A2 0. P1A B2 P1A2 (2-5)

= O O O ~ O O O O= 24,

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