IN solving the group of twenty-five-letter United States military service "double transposition" cryptograms by the method given two weeks ago, a numerical series was obtained which could be used in translating other cryptograms.
Last week the above-mentioned series was used to find the original cipher key for use in deciphering similarly keyed cryptograms of any length. In this article the same key will be determined from another numerical series.
To obtain this other numerical series, here shown at (c), merely arrange the numbers of the consecutive series (b) in the order indicated by the previously determined series (a). Thus 1 (a) is above 9 (b), which accordingly becomes the first number of (c); and so on.
The length of the key will usually be shown by the fixed difference occurring between various sequences in the series. To facilitate the search for these sequences, it is convenient to add to the entire series such numbers as may be considered possible key lengths.
(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) 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 (c) 9 6 5 7 23 20 19 21 11 8 10 2 25 22 24 4 1 3 16 13 12 14 18 15 17 (6) 15 12 11 13 29 26 25 27 17 14 16 8 31 28 30 10 7 9 22 19 18 20 24 21 23 (7) 16 13 12 14 20 27 26 28 18 15 17 9 32 29 31 11 8 10 23 20 19 21 2S 22 24 (8) 17 14 13 15 31 28 27 29 19 16 18 10 33 30 32 12 9 11 24 21 20 22 26 23 25 (9) 18 15 14 16 32 29 28 30 20 17 19 11 34 31 33 13 10 12 25 22 21 23 27 24 26
Upon examination, the sequences underlined in the new series thus formed are found also in (c). There are two short sequences in the (9) line; but the length and number of them in the (7) line point more definitely to a key of that length.
A careful examination should make this easily clear. (This is the method of Cours de Cryptographie, a French work by Colonel M. Givierge.)"
Proceeding upon this assumption, the series 1—2—3—4—...—message order—and 9—6—5—7—23...—second transposition order—may now be written in lines of seven numbers each, as shown at (d) and (f). Given these two diagrams, the problem of reconstructing the first transposition diagram (e) presents no particular difficulty.
For the sequences found by the above method form columns or parts of columns of this diagram. And, from the method of encipherment, its horizontals are made up of the verticals of (d), the numbers of which increase by 7, the length of the key. The order in which the verticals of (d) are found transcribed in (e), of course, determines the key, 4—1—6—2—5—3—7—key word PARAPET.
P A R A P E T
4———1———6———2———5———3———7
(d) 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 . . .
(e) 2 9 16 23 4 11 18
25 6 13 20 1 8 15
22 5 12 19 3 10 17
24 7 14 21 . . .
(f) 9 6 5 7 23 20 19
21 11 8 10 2 25 22
24 4 1 3 16 13 12
14 18 15 17 . . .
The above method and that given last week afford an interesting illustration of how it is often possible to reach the same objective by more than one path. It is only fair to add, however, that the present cryptograms were very favorable for solution by either method.
Speaking of keys, the selection of a key for this cipher is an important matter. For certain key lengths under certain conditions will produce cryptograms that almost decipher themselves.
Cipher No. 115, herewith, represents such a combination of unfavorable circumstances. This example has been correctly enciphered. And both transpositions have been properly carried out.
Both cryptograms offered in No. 112, last week's problem in the above type of cipher, were enciphered with the key 8-1-9-3-7-11-5-10-2-4-6, derived from the key phrase HAND GRENADE. The first communication (a) was in double transposition cipher. In the second communication (b) one of the transpositions was, supposedly, inadvertently omitted in encipherment.
Here are the translations: (a) ENEMY CIPHER SYSTEMS AND MACHINES HAVE FALLEN INTO OUR POSSESSION. (IHET, nulls), (b) HOSTILE CAVALRY HAS OCCUPIED VILLAGE TWO MILES NORTH OF OUR PRESENT POSITION.
Last week's straight substitution Cipher No. 110 (L. Rinker and H. Kramer) used the order of letters on the standard typewriter keyboard as the cipher alphabet, thus: Q=a; W=b; E=c... M=z. The message: True cipher fans never abandon the chase until they have captured their "quarry."
The groups ZIT, ZITN, and ZITOK, should have been obvious as the, they, and their, respectively, and should have afforded an immediate clew as to the alphabetic structure. For another cipher of this type try No. 113, a fine example of the "Q. and A. "cryptogram.
Cipher No. 111 (M. Walker) conveyed the message: "Imagination is necessary to the poet, the artist, and not less to the cryptographer." Transposition was effected by numbering every third letter in rotation, skipping those already so numbered, and using these numbers to show the order of letters in the cryptogram.
Any desired count may be used in preparing the key to this cipher. But the length of the message is also a determining factor as to the order of transposition. Here is the beginning and end of the sixty-eight-number key used in this instance: 23-55-1-66-24-2-39 ... 68-22-48-38. A good cipher, this, and not too difficult to be practical.
Turning now to this week's ciphers, besides those already mentioned, we also have with us in No. 114 another interesting transposition cipher. This is a single transposition type in which the columns of a rectangle are transposed by means of a numerical key, and about which its author would offer you this bit of counsel: "I call this a quiz. There are plenty of clews—and Q's—The number of letters should help materially."
CIPHER No. 113 (Anthony Ranieri, Brooklyn, New York).
Question: DFV QJRP, "RH RQ J TVUPRHRVU HFJH TVUKYVUHQ AQ— UVHH J HFBVYG"? Answer: XYVNBY TIBNBIJUP.
CIPHER No. 114 (Arthur Bellamy, Boston, Massachusetts).
QEEEA YEEEU SNDSI DSELQ ACULS TIWQA UULOL QCAXS UFYUQ ITEAQ SBQRE URSEL WVUKS TASOU IUYUQ LALOE Z
CIPHER No. 115.
EICWS UNNTO VONIE LIFEN EFECM RERNO YIRAJ OMWAL TIORS TTERV PTOGL SCUNI RREOT ANPSE FDEGR XTNNI MFTEY ENNOE TFLNE MIOER