T HE opener on this week's bill is the promised explanation of cipher No., 50 (J. Levine) of three weeks ago. This clever cipher is of the double substitution variety; the first substitution employing a "three-unit" alphabet, and the second using words of one, two, and three syllables to represent these units.
The object of this cipher is to convey a secret message without appearing to do so. And this remarkable feat is accomplished by cloaking the communication in an "open letter," which may be about any subject whatever not liable to arouse suspicion.
Three-unit alphabets of various kinds are found in very early works on cryptography. Mr. Levine's is trinumeral, being formed of the various combinations of the figures 1, 2, and 3, as follows:
111 A 211 J 311 S 112 B 212 K 312 T 113 C 213 L 313 U 121 D 221 M 321 V 122 E 222 N 322 W 123 F 223 O 323 X 131 G 231 P 331 Y 132 H 232 Q 332 Z 133 I 233 R 333 —
To decipher the cryptogram, part of which is given at (a), first substitute the figures 1, 2, and 3, for words of 1, 2, and 3, syllables, respectively, as shown at (b). Regrouping these figures by threes (c), it only remains to substitute letters (d) for numbers from the cipher alphabet. The first two message letters are thus found to be "G-E." The whole cryptogram, similarly treated, yields the information: "GET AWAY QUICK."
(a) Come tomorrow at one thirty. Forgot ... (b) 1 3 1 1 2 2 ... (c) 131 122 ... (d) G E ...
The efficacy of this cipher depends on the ruse of syllable substitution escaping detection. Penetrate this subterfuge and "the jig is up." For the three-unit alphabet, once suspected, can readily be solved by the usual simple substitution methods.
It should be interesting and instructive to compare this cipher with others of a similar nature that have previously been published in these columns.
The Bacon biliteral cipher, for example— see issue of April 25, 1925—uses a five-unit alphabet with anything capable of two differences, such as bells, trumpets, written or printed characters, and so on.
The ancient Greek telegraph of Polybius—see issue of March 28, 1925—employs a two-unit alphabet with combinations of from one to five torches. Cipher No. 49, of three weeks ago, was similar to this, only using candies of five different flavors instead of torches.
William Blair devised a four-unit alphabet—see issues of July 25 and August 15, 1925—which could be used with dots, types, and so forth. Finally, in the Morse alphabet, letters are represented by symbols of 1, 2, 3, 4, and more, units; these being transmissible as electrical impulses, light flashes, flag signals, and so on. These few examples should afford a good idea of the possibilities in systems of this kind.
One way to solve last week's cryptogram No. 53 (M. Walker) would be to assume that the termination "HUR," occurring twice, stood for "ING." Substitute these letters in the fifth and sixth groups, "SEUR XHBSHU," and they become "x-KG -I-xIN," with the unknown letter x in common, pointing to "HANG WITHIN" as a probable translation.
Again substituting in the last group, "EPPNBHBN," we have" AyyzTlTz," which becomes "APPETITE" by calling y and z, P and E, respectively. Other groups follow in a similar manner, the full translation being: "Beautiful, luscious, blushing peaches hang within reach, tempting frugivorous bucolic youngster's appetite."
Mr. Walker used the frequency table of August 15, 1925, as his alphabet, taking the letters in the order of their descending frequencies: A=E; B=T; C = 0; and so on.
You probably observed that all the groups in No. 54 contained an even number of letters, suggesting the use of two-unit symbols. Proceeding upon this supposition, the second, fourth, and sixth groups easily identify themselves as THE, THAT, and THIS, respectively, after which the rest comes easy.
The message was: "MARRY THE GIRL THAT SOLVES THIS." Did you ask your best girl to help you with this cipher as we suggested? The alphabet here is interesting. By writing the last half reversed below the first half, letters represented by reversible pairs are brought together: G=OZ; T=ZO; H=FS; S=SF; and so on.
You will find some distinct novelties in this week's ciphers. No. 56 combines, for what is probably the first time, two popular fads that are sweeping the country, the cryptogram, and the "question and answer" craze. Two different alphabets have been used; one for the question, and another for the answer. The solution of either part, however, is bound to help with the other. Try this, and let us know how you like it.
In No. 57 you are meeting,a problem of more than average difficulty. To help you along, Mr. Newell has enciphered the same message, using two different keywords.
See that wall in No. 58? It is supposed to be a section of a wall surrounding the estate of an old miser who has hidden his gold somewhere, and apparently left no clew as to its whereabouts. Maybe the wall has something to say about this! At any rate, see how much you can make it tell you.
CIPHER No. 56.
Question: WZSV EJ VZP HKIVISEV HSIAP KI JHPSRETQ HKIVISEV JMJVPF? Answer: GXRA RA P AJAGOT NC AGPEWPQWRVREL ZOQANEPY KX- PQPKGOQRAGRKA AN GXPG GXOJ TPJ BO PKKSQPGOYJ WOAKQRBOW RE HNQWA.
CIPHER No. 57 (F. H. Newell, Tonkawa, Oklahoma).
(a) B FKHB JHF M& && FHFJ FE HADJ: R&N NKJ BA IJKFE. (b) G P&BG IJC KC IJ QNKI DB FBM&: GFI OMM &B &&&&&.
CIPHER No. 58.
The solutions to ciphers Nos. 56 and 57 will be published next week, and the full explanation of No. 58 in three weeks. Also look for the key to No. 44 (J. McA. T.) in next week's issue.
Fans are requested, as usual, to include answers and full explanations with ciphers submitted to this department for publication.