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The approaches described previously in this chapter generate sets of random elds to be used in Monte Carlo simulation. This requires a large number of simulations if accurate results are desired. It can often be dif cult to perform sensitivity analyses on the parameters in a Monte Carlo analysis. Therefore, a direct calculation of the effects of correlation becomes attractive. Unfortunately, the computational dif culties have stood in the way of many such efforts. Baecher and Ingra (1981) describe a direct evaluation of correlation effects in the nite element analysis of settlements. Starting from Equation (19.1), they observe that a rst-order estimate of the expected displacements is E[u] = (E 1 [K])p (19.15)

3 EDX is then added to itself, twice This is equivalent of edx = edx 4, which means that so far you ve essentially calculated edx = edx 4,092 4 Finally, EDX is shifted back right by 10 bits, which is the equivalent of dividing by 1,024 The final formula is edx = edx 4092 1024 You might be wondering why Cryptex didn t just use the MUL instruction to multiply EDX by 4,092 and then apply the DIV instruction to divide the result by 1,024 The answer is that such code would run far more slowly than the one we ve just analyzed MUL and DIV are both relatively slow instructions, whereas ADD, SUB, and the shifting instructions are much faster.

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It is important to note that this sequence reveals an interesting fact about Cryptex: It was most likely compiled using some kind of an optimize-for-fast-code switch, rather than with an optimize-for-small-code switch That s because using the direct arithmetic instructions for division and multiplication would have produced smaller, yet slower, code The compiler was clearly aiming at the generation of high-performance code, even at the cost of a somewhat larger executable The result of this little arithmetic sequence goes right into the printf call that prints the current file entry This is quite illuminating because it tells you exactly what Cryptex was trying to calculate in the preceding arithmetic sequence: the file size In fact, it is quite obvious that because the file size is printed in kilobytes, the final division by 1,024 simply converts the file size from bytes to kilobytes.

where (E 1 [K]) is the inverse of the matrix of expected stiffnesses. The rst-order estimation of the variances and covariances requires partial differentiation of the displacement vector with respect to the element moduli: K u M +

The question now is, what was that original number and why was Cryptex multiplying it by 4,092 Well, it would seem that the file size is maintained by using some kind of fixed block size, which is probably somehow related to the cluster you saw earlier while decrypting the buffer The problem is that the cluster you were dealing with earlier was 4,104 bytes long, and here you re dealing with clusters that are only 4,092 bytes long The difference is not clear at this point The original number of clusters that you multiplied was taken from offset +8 in the current file entry structure, so you know that offset +8 contains the file size in clusters This raises the question of where does Cryptex store the actual file size It would not be possible to accurately recover encrypted files without creating them with the exact size they had originally.

Therefore Cryptex must also maintain the accurate file size somewhere in the archive file Other than the file size, the printf call also takes the file name, which is easily obtained by taking the address of offset +14 from ESI Keep in mind that ESI was incremented by 8 earlier, so this is actually offset +1C from the original data structure, which matches what you saw in our data dump, where the string started at offset +1C After printing the file name and size, the program loops back to print the next entry To reach the next item, Cryptex increments ESI by 0x98 bytes (152 in decimal), which is clearly the length of each entry This indicates that there is a fixed number of characters reserved for each file name Since you know.

4 Roll the boat on top of you. Helm and crew move across the boat together, sheeting in as they reach the new side of the boat. Turn the boat onto a close reach on the new tack

(19.16)

6

in which M i is the average modulus within element i. If the loads are independent of the element properties, u K K = u (19.17) M i M i and u M = K 1

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