# Captain Walter Fried's Fish Notes March 1944 to January 1945

### Page 19

Tony Sale's
Codes and Ciphers

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TOP SECRET CX/MSS Page 4

Report #F 46 (continued)

dots) when the crib is correctly placed. The run is designated
(DeltaZ2 + DeltaP2 +P5two back(Delta31 . The proportion of dots expected is
b ^2 + (1 - b)^2. For a crib of 800 letters with b = .65 this will give a score
about 2.5 sigma above random. This is not enough but we can also make
independent runs at intervals 62, 93, etc.
9. It is not difficult to determine the minimum length of crib
required. If we use as intervals all possible multiples of 31 the number
of values available for comparison (that is, the aggregate or dots and
crosses which can be counted and scored) with a crib or length N is
(1) 31c(c -1)1/2 + cd where N is expressed as
31c + d (d<31)
The "bulge" is
(2) b^2 + (1 - b)^2 - 1/2

so that the expected excess of dots at the right placement is the product
of (1) and (2). sigma is one half the square root of (1) so that if we
determine what score we will require in terms of sdigma it is easy to
determine the necessary length for a given value of b (which is unknown and
must be estimated from past experience). A fairly good approximation for (1)
is N^2 /64. If we want a score of 5sigma and assume that b = 2/3 we get
N = 360 (using the approximation).
10. However, there are both theoretical and practical limitations.
The theoretical liniltation arises from the fact that if, at a particular
position we get a high random proportion of dots, say 1/2 +lambda, on a run
at interval 31, then at other intervals we will not get random scores but
instead proportions of (1/2 + lambda)^2 + (1/2 - lambda)^2 = 1/2 +
2(lambda)^2 . This factor makes it necessary to require slightly higher
relative scores for runs made at several intervals than for runs at interval 31
only. However, it is almost negligible. The practical limitation which is much
more important arises from the dirninishing utility of the runs at higher
multiples of 31and the labor of making tapes.
11. For the ideal run we would make one tape contafining
(DeltaP2 + P5twoback)Delta3l, then (DeltaP2 + P5twoback)Delta62, etc, Up
to the maiimum available interval. A space would be left after the Delta31
portion equal to the difference in length between the cipher text and the crib
and the same amount of space after the Delta62, Delta93, and succeeding
portions. One additional space is left after the final block. The other tape
contains (DeltaZ2)Delta31, (DeltaZ2)Delta62, etc.with no spaces at all
between the blocks. When these two tapes are run against each other single
scores are produced using all available data. The extra space.at the end of
the crib tape causes the tapes to step against each other and thus tries all
juxtapositions. Symbols must be

(AES Note: this shows the real difficulties in producing tapes for Colossus I)