resolution into its components. The method is apparently not applicable
to relays because the relaying operator makes up a new tape and this will
vary in minor details, notably punctuation. But if Berlin sends the same
message to both Rome and Paris the same tape will be used on automatic
transmission and the two texts will be letter perfect. An occasional
garble, even if on the 5th impulse, will not invalidate the procedure
although an omitted letter of course will. The real risk is interruption
of the tape with stretches of hand transmission or "go-backs", that is,
repetitions of portions of what has already been transmitted. "Go-backs"
in the crib can be found and corrected for but if the crib happens to be
in a portion of the unknown cipher text which includes "hand" or "go-backs"
it will not be possible to lodate it. A possible aid to setting cribs is
provided by the fact that where a transmission includes several messages
the receipts, which are sent in clear, state the exact time at which each
message was concluded. Correlation of these times with time of interception
and with each other should make it possible to deteraine approximately the
portion or the message which contains the crib. However, this information
has not been used because the Germans make mistakes and it is therefore
not too reliable and because once the necessary preliminary work has been
done it is almost as easy to try all possible settings.
8. In the description which follows I will assume that the X2 + P5
limitation is employed. The method could not be used without any limitation
but is fully applicable if only the X2 limitation is used. To develop the
theory of this procedure we use the usual notation - a = proportion of
crosses in MT; b proportion of crosses in Delta Psi and assume that ab = 1/2
Let a' = proportion of crosses in MB. Then, with X2 + P5 limitation, which
is assumed to produce equal numbers of dots and crosses
1 - a = 1/2(1 - a') or a' = 2a - 1
If P5two back + X2one back = x, the chance that MT = . is the same as the
chance that MB =. , namely 1 - a'. Therefore the chance that DeltaPsi'2 = . is
1 - a' + a'(l - b) = 1 - a'b = 1 - 2ab + b = b
If P5two back + X2one back =. MT must be x and the chance that
DeltaPsi'2 = x is b. Consequently the chance is b that
DeltaPsi'2 + P5two back + X2one back = x
or that
DeltaPsi'2.+Delta X2 + P5two back = X2one back + DeltaX2 + x
or that
DeltaZ2 + DeltaP2 +P5two back = X2one back + DeltaX2 + x
The left hand side of the last equation becomes known when plain and cipher
texts are properly juxtaposed. X2one back+DeltaX2 (which is called X2hat) is
unknown but it must be periodic. Therefore, DeltaZ2 + DeltaP2 +P5two back
combined with itself at an interval of 31 should yield a high proportion of
coincidences (or