## 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

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

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)

- (2) b^2 + (1 - b)^2 - 1/2

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.

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

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)