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Toward the beginning of Neal Stephenson's Cryptonomicon, Avi gives Randy two strings of sixteen supposedly random hexadecimal numbers as encryption keys:

AF  10  06  E9  99  BA  11  07  64  C1  89  E3  40  8C  72  55
67  81  A4  AE  FF  40  25  9B  43  0E  29  8D  56  60  E3  2F

If you change the hexadecimals to decimals and change the decimals to letters, modulo 25 without J, like Lawrence does with his one-time-pads later in the book, you get:

z   k   f   h   c   l   r   g   z   s   m   b   o   p   o   k
c   d   o   y   m   o   m   e   r   o   q   q   l   v   b   w

The first sequence contains two Os and the second contains three Os. The probability of randomly picking at least five Os out of twenty-five letters given thirty-two picks is less than 1%, which makes it seem like the numbers are not actually random.

Additionally, there are six prime numbers in each sequence:

E9  11  07  C1  89  E3
67  25  43  29  E3  2F

(all primes in base sixteen)

What does this mean?

Nippers
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1 Answers1

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There are 9 primes between 0 and 25, so having six primes when picking sixteen numbers at random in this interval perfectly makes sense. In addition, this is just an encoding of the letters: we could have associated the letters to numbers between 1 and 26, or 8 and 33... So checking for primality has no reason to be relevant here.

Moreover, the probability of having five O's might be low, but this is just one letter; have you looked at the probability that "some" letter appears five times? It should be quite reasonable. To me, you are seeing patterns where there is nothing to search for: given a small enough random sequence, one can always find many such patterns.

It took me three attempts on this random sequence of letters generator to get the sequence pmrvkpehkyonfrkhxrnpnznbmnfmluwh, which contains five n's.

Geoffroy Couteau
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