D ID you ever arrange a set of dominoes on end in the form of a circle, so that a touch of a finger to any one of them would upset all the others in the ring?
Something very much like this is going to happen here to the autokey cipher described last week. For the peculiar weakness of this system is that the discovery of but a single number will unlock the rest, the "fate" of any number deciding that of its immediate neighbors.
In the first place, the number of possible keys is limited. Thus the initial key number in the present illustrative example must run somewhere from 1 to 20, since it must be less than the first number, 21, of the cryptogram, in line (d).
Now it should be clear that if all these trial keys be successively applied, one of the results must be identical with (b), in which form the cipher can be translated by the usual and quite simple substitution processes.
The number of trial keys can be further narrowed down by considering the smallest numbers in a given cryptogram, and also by rejecting all keys which would result in values less than 1, or more than 26. Negative values have been avoided here, although they would in nowise complicate the solution.
To start things going, take any number from 1 to 20, say 12, as an initial key number, and reduce the cryptogram to the form given in line (f).
(a) W A T C H T H E M A N I N T H E W H I T E M A S K
(b) 11 2 16 12 3 16 3 10 17 2 21 8 21 16 3 10 11 3 8 16 10 17 2 26 13
(c) 10 11 2 16 12 3 16 3 10 17 2 21 8 21 16 3 10 11 3 8 16 10 17 2 26
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(d) 21-13-18-28-15-19-19-13-27-19-23-29-29-37-19-13-21-14-11-24-26-27-19-28-39
(e) 12 9 4 14 14 1 18 1 12 15 4 19 10 19 18 1 12 9 5 6 18 8 19 0 28
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(f) 9 4 14 14 1 18 1 12 15 4 19 10 19 18 1 12 9 5 6 18 8 19 0 28 11
Now an examination of (f) will show that the numbers occupying the odd and even places are larger or smaller throughout by a fixed amount than the corresponding numbers in line (b). Here this constant difference is 2; the odd series, 9.14.1 ..., being 2 units smaller, and the even series, 4.14.18 ..., 2 units larger, than the numbers above them in line (b).
Consequently the effect of this simple twist was to reduce the cipher (d) from one using 26 alphabets in mixed order, to one using only 2 alphabets alternately, in line (f).
The first domino is beginning to tip, folks. Now watch them all take a sympathetic tumble!
These two alphabets can now be reduced to a single alphabet by the method given in FLYNN'S WEEKLY for December 18, 1926. But if the message happens to contain any repeated groups of letters at odd intervals, a more direct and more quickly applicable method may apply.
Recurrent groups at even intervals cannot be used by this method, since the similar letters of such groups fall in the same alphabets. Thus 19-13, line (d), becomes 1-12 in both cases in line (f). In the case of odd intervals, however, such letters fall in both alphabets, calling both substitutes into play.
For example, 27-19, line (d), assumes the two forms, 15-4 and 19-0, in line (f). Assuming that these two sequences represent the same letters, the common difference 4 (19-15=4, and 4-0=4), indicates that one alphabet—that containing the above 19 and 4—is larger than the original cipher alphabet by two units; while the other alphabet—containing the 15 and 0— is smaller by two units.
By alternately adding and subtracting this 2, line (f) is made identical with (b), where the cipher can be solved, as already stated, by straight substitution methods. The order of discovery in this case would probably be somewhat as follows:
— — T — H T H E — — — — — T H E — H — T K — — — — — — T — H T H E — — N I N T H E — H I T E — — — — W — T — H T H E — — N I N T H E W H I T E — — — — W A T C H T H E M A N I N T H E W H I T E M A S K
When these repeated groups result—as they often do—from different letter sequences, they are of no use by this method. Such groups may sometimes be recognized, however, by the fact that their differences, as above, may be odd as well as even. While natural repeated groups always give an even difference, as 4 in this case under discussion.
In cipher No. 19 (James Olden), published last week, the repeated group 19-23— with the odd interval 55—is the target for attack, allowing a speedy reduction to the form 19-20-15-12-5..., where analysis reveals a straight A=1 ... Z=26 alphabet. Mr. Olden's initial key, 4, was hidden as the sixth from last number in the cryptogram itself. Here is a sample of his encipherment, and the message in full:
(a) S T O L E N S I L K ...
(b) 19 20 15 12 5 14 19 9 12 11 ...
(c) 4 19 20 15 12 5 14 19 9 12 ...
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(d) 23-39-35-27-17-19-33-28-21-23 ...
Message: Stolen silk shipment located in old warehouse southwest corner of Canal Street, facing river. Gang prepared to move same at midnight. Plan your raid that none escape capture. (Signed) RIORDAN,
Acting Chief.
Last week's No. 20 (Geo. Abbott) had this to say in typewriter cipher: "Never stumble twice over the same stone is good advice for the solver of ciphers." The fans no doubt spotted 5*3 in this at once as substitute for THE, after which the rest would be easy.
The first "crypt" received in response to our recent invitation to readers to submit some ciphers of that type is offered for your solution herewith. You should find it plain sailing.
CIPHER No. 21 (A. MacDonald, Jr., Aylraer, Quebec. Canada).
IQY SDMFI LE N RVDK VR SLFQYD YKFXVMLBU N KYIQVO VR ELKFXY XLIYDNX EHPEILIHILVB.
The next cipher is an autokey, like those described above. The message has something to say about a skeleton. Can you find out what?
CIPHER No. 22.
33-26-41-31-31-40-26-34-34-24-38-37-37-31- 19-25-31-31-35-27-27-31-11-21-31-18-14-30- 48-39-36-45-27-27-31-11-23-47-40-32-33-17- 14-33-30-31-30-19-33-46-31-33-51-43-33-33- 25-9-21-45-34-26-25-15-29-47-40-32-33-17- 22-39-31-20-15.
Speaking of "crypts," we have not yet heard from "Primrose"—John Q. Boyer—how our readers made out with the National Puzzlers' League crypts in the March 19 issue of this magazine; but this information should soon be forthcoming.
At this writing only two answers—from "Arty Ess," editor of the Enigma, and "Primrose "—have been received to our own crypt. No. 5 in the above issue. Incidentally, no solutions were submitted to Mr. Davidson's cipher, No. 1 in the issue of March 5.
The weekly department idea seems to have struck a responsive chord in the hearts and minds of our readers. Almost every mail brings letters boosting the weekly feature. Have you voiced your own opinion? We should be glad indeed to hear from you.
The answers to ciphers Nos. 21 and 22 will be published next week. You will also find in the next article a full explanation of Fletcher Pratt's challenge cipher No. 12, and the solutions to the code messages in the April 23 issue.
All this, and more, in next week's department!