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So far, this chapter has described the use of Monte Carlo methods to evaluate integrals. The other application is to study the behavior of random processes. These are problems in which the input is stochastic and the results uncertain. The Monte Carlo method is particularly effective when the process is strongly nonlinear or involves many uncertain inputs, which may be distributed differently. To perform such a study, the analyst generates a random value for each uncertain variable and performs the calculations necessary to yield a solution for that set of values. This gives one sample of the process. The trials are repeated many times, giving many samples of the process. Once a large number of runs have been completed, it is possible to study the output statistically and to obtain values of means, variances, probabilities of various percentiles, and other statistical parameters. Two important points should be noted: 1. Regardless of the number of stochastic variables, each run gives one sample of the process. Hence, increasing the number of stochastic input variables does not increase the number of runs for the same level of accuracy. 2. The technique is essentially a repeated sampling of the stochastic process, and the methods of Part 2 and particularly 11 can be used to examine the accuracy and precision of the results.

quick check shows that 70597243 is the hexadecimal value for the characters CrYp, and 39586554 for the characters TeX9. Cryptex is simply verifying the header and printing an error message if there is a mismatch. The following code sequence is the one you re after (because it decides whether the function returns 1 or prints out the bad password message). This sequence compares two 16-byte sequences in memory and prints the error message if there is a mismatch. The first sequence starts at 00405038 and is another global variable whose contents are unknown at this point. The second data sequence starts at 00406070, which is a part of the header global variable you looked at before, that starts at 00406058. This is apparent because earlier ReadFile was reading 0x28 bytes into this address 00406070 is only 0x18 bytes past the beginning, so there are still 0x10 (or 16 in decimal) bytes left in this buffer. The actual comparison is performed using the REPE CMPS instruction, which repeatedly compares a pair of DWORDs, one pointed at by EDI and the other by ESI, and increments both index registers after each iteration. The number of iterations depends on the value of ECX, and in this case is set to 4, which means that the instruction will compare four DWORDs (16 bytes) and will jump to 00401234 if the buffers are identical. If the buffers are not identical execution will flow into 0040121E, which is where we wound up. The obvious question at this point is what are those buffers that Cryptex is comparing Is it the actual passwords A quick look in OllyDbg reveals the contents of both buffers. The following is the contents of the global variable at 00405038 with whom we are comparing the archive s header buffer:

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The buffer that originated in the archive s header contains the following:

The unbiased estimate of the mean of a random process is simply the mean of the values returned by the process n 1 xi (17.28) E[ X ] = n i=1 and the standard deviation of this estimate is related to the standard deviation of all the values by X X = (17.29) n If the process is Normally distributed, the variance is 2 distributed. It can be shown that the upper con dence limit at the (1 ) level is ( 2 )1 = (n 1)s 2 2 ,n 1 (17.30)

The two are obviously different, and are also clearly not the plaintext passwords. It looks like Cryptex is storing some kind of altered version of the password inside the file and is comparing that with what must be an altered version of the currently typed password (which must have been altered with the exact same algorithm in order for this to work). The interesting questions are how are passwords transformed, and is that transformation secure would it be somehow possible to reconstruct the password using only that altered version If so, you could extract the password from the archive header.

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