LAST week's article outlined a method of determining the order of transposition in a number of United States military service "double transposition" ciphers of the same length and key, and faced the reader with the problem of finding the original key with which to decipher similarly keyed communications of any length.
To determine the key length, find the interval existing between certain numbers of the series, and at the same time between other similar numbers larger than these by one unit. This can be done either by inspection, or, much easier, by trying various lateral adjustments of two or more paper strips, each of which bears, evenly spaced, the above mentioned numerical series.
Thus, in the accompanying illustration, the numbers 16—9—23—1 3 on the upper "strip," and 17—10—24—14 on the lower occur at the interval 7 in both cases. This same position also provides 1—19—5 and 2—20—6 at this interval.
* * * *
17 12 18 16 3 2 4 10 1 11 9 21 20 22 24 19 25 23 7 6 8 14 5 15 13
17 12 18 16 3 2 4 10 1 11 9 21 20 22 24 19 25 23 7 6 8 14 5 15 13
* * *
Proceeding in the same manner until all the numbers have been tried or accounted for, the following sequences, formed by numbers at this same interval, are obtained from the present series: 16—9—23—13; 17—10—24—14; 18—11—25—15; 1—19—5; 2—20—6; 3—21—7; 4—22—8.
Taking 7 as the key length, the whole series 17—12—18, et cetera—here representing the message order—and 1—2—3 — 4, et cetera—second transposition, or cipher, order—may now be transcribed in lines of seven numbers each, as shown at (a) and (c), respectively, in the figure herewith. It will now be seen that the sequences found above form columns or parts of columns of (a).
Given (a) and (c), however, it is a comparatively easy matter to find the first transposition order (b), the lines of which are made up of the (a) columns in such a manner that the numbers, downward by columns, are in serial order.
P A R A P E T
4———1———6———2———5———3———7
(a) 17 12 18 16 3 2 4
10 1 11 9 21 20 22
24 19 25 23 7 6 8
14 5 15 13 . . .
(b) 12 1 10 5 16 9 23
13 2 20 6 17 10 24
14 3 21 7 18 11 25
15 4 22 8 . . .
(c) 1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25
In this case, for example, the columns of (a) may be arranged—as lines—in the following two groups, which may be combined to form (b) by merely placing 13—14—15 under 12.
12 1 19 5 16 9 23 13 2 20 6 17 10 24 14 3 21 7 18 11 25 15 4 22 8
The numerical order in any line of (b) will determine the original numerical key, 4-1-6-2-5-3-7—derived from the key word PARAPET—which may now be used to unlock any other communication in the same key, regardless of its length. Thus, last week's No. 109, so deciphered, yields the message: INTERCEPTED ENEMY MESSAGES INDICATE EXTENSIVE PREPARATIONS FOR AN EARLY OFFENSIVE.
The above method, though easy to apply, may seem quite complex in principle. But in reality it is very simple, depending upon the fact that the lines of (b) form the columns of (a); and that, consequently, such sequences in (b) as 1—19—5, 2—20—6, 3—21—7, et cetera, will be distributed at intervals of the key length in (a). Another method of finding the key to this cipher will be given next week.
It is very essential to perform both transpositions when using this cipher. Should one of the transpositions be omitted, the resultant single transposition cryptogram is liable to decipherment by methods such as have been given in recent articles. And the key, thus discovered, may be used in translating other properly enciphered double transposition cryptograms in the same key.
The two cryptograms offered in Cipher No. 112, herewith, illustrate this point. One of these—-we do not say which one—is a double transposition, and the other a single transposition; but the two are in the same key. Try both of these cryptograms, if necessary, by single transposition methods of analysis. Then use the key so found to translate the double transposition cipher.
Last week's straight substitution Cipher No. 107 conveyed the message: Cipher clews though unseeing may be seen; look attentively and you will find them. The groups QWJFSW—though—and QWZC—them—should, have led to an easy solution, For example, substitution in FKGZZVKS of the values thus found would give u--ee--g—unseeing. After which the rest should be a matter of minutes. For a new cryptogram of this type turn to No. 110 in this article.
Irving Bloom's No. 108 was a double substitution cipher conveying the message: "This is easy if you have the key," and using a straight five by five alphabetic square—combining U-V—in which 11=A, 12=B... 55=Z. Encipherment was accomplished by first replacing each letter of the message (a) by its two-figure substitute from the above alphabet, using 00 as a word spacer, as shown at (b).
These figures were then used as differences in obtaining the pairs of figures which replace them in turn at the second substitution (c). In this way four figures act as the symbol for each letter, and any letter has a large number of different symbols. Grouping by fives (d) is for convenience in transmission.
(a) T H I S (b) 4 5 2 3 2 4 4 4 (c) 95 94 64 63 75 73 84 84 00 (d) 95946 46375 73848 400
This week's ciphers, besides those already mentioned, include a highly interesting transposition type by M. Walker, based on a simple counting out process. Full explanations to all of these will be published next week.
CIPHER No. 110 (L. Rinker and H. Kramer, Muncie, Indiana).
ZKXT EOHITK YQFL FTCTK QVVQFRGF ZIT EIQLT XFZOS ZITN IQCT EQHZ- XKTR ZITOK "JXQKKN."
CIPHER No. 111 (M. Walker, Akron, Ohio).
ANIIE SROEE HRSNO ETHRT RHIIO NSTPT TATSE PARAS AHETL TOETE OISYM CTNGN ACYOT GDP
CIPHER No. 112.
(a) YTSNS MAVOF HALIS LNTOO
UNHEM SPSIY NSROE TPSNM
IEECC ISEHE NDEAA IEREH
(b) GRECO TACGO EOTHV IOOVU
ERSNE OLSPT APTTE IAILU
SHLIW HNSYD MFPCC ANRIL
SLERI
Keep your answers coming, fans! Get as many as you can. Also, don't forget to send full explanations along with your new ciphers.