HAS it ever been your lot to invent a cipher which yon believed insoluble, only later to devise a method of solution for it yourself?
Such was the experience of Charles Winsor with his double—transposition system, No. 127, in the January 28 issue, the keys to which have been withheld all this time pending receipt of methods of solution from our readers. In fact, Mr. Winsor's answer was the very first submitted.
Before delving into the solution, however, let us first briefly run through the method of encipherment, using the numbers 1 to 29 to represent the twenty-nine letters of the message, MOVE YOUR DIVISION AT ONCE. Should the reader so desire, he may, of course, repeat the process with the message itself.
In effecting the first transposition, the columns of (b) are taken out in the order indicated by the key (a), and the resultant series is transcribed by successive horizontals into (d). For the second and final transposition (e) take the columns of (d) in the order indicated by the second key (c).
L I N C O L N
(a) 3——2——5——1——7——4——6
(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 26 27 28
29 . . . . . .
G E T T Y S B U R G
(c) 3——2——7——8—10——6——1——9——5——4
(d) 4 11 18 25 2 9 16 23 1 8
15 22 29 6 13 20 27 3 10 17
24 7 14 21 28 5 12 19 26 .
(e) 16-27-12-11-22-7-4-15-24-8-17-1-10-26-9-
20-5-18-29-14-25-6-21-23-3-19-2-13-28.
The problem facing the decipherer here is, given the series (b) and (c), to determine the intermediate series (d), and the two numerical keys (a) and (c).
At first sight this might seem like a tough proposition. But Mr. Winsor found that results could be had by the same methods used for the United States Army double transposition cipher, as described in the issues of December 24 and December 31, 1927, the common difference between groups indicating the length of the first key, the number of such groups the length of the second key, and the groups themselves forming sections of the intermediate stage (d). Our limited space will not permit a full solution. But by referring to the above articles the reader should have no difficulty in performing it himself.
Our correspondent also found that the system could be solved by trial and error, disregarding the length of the second key, and trying one length after another for the first key until the right length was found.
For example, trying seven columns for the first transposition, and knowing that 16 must come in the first horizontal of (d), we have the series 2-9-16-23 from (b), and 16-27-12... from (e) as a framework about which to group the remaining numbers, thus:
. 2 9 16 23 . . . . 27 . . . . . 12 . .
So after all, this system turns out to be no more difficult to solve than the army cipher after which it was modeled. Had it proved to be all that our correspondent expected of it, we should have had a simple, practical system in which single messages would have been safe even though several messages in the same key, but of a different length, had already been combined for solution by other methods.
As matters stand Mr. Winsor deserves credit both for his effort to evolve such a cipher and his method of solving it. Other solutions submitted to this cipher will be published in our next solvers' list.
Here is the answer to John Q. Boyer's No. 152, of last week: "SNOWY CRACK, ARCTIC CLEFT, ICY GAP, FROZEN CHASM, ALBEIT RISKY, OFT TEMPT SKIING ACROBAT." A, pretty word picture, this, but not an easy crypt! Note that wicked double I in SKIING. Did you get it ?
No. 153, by Raymond Wallace, used the following alphabet of coupled pairs in which A=Y, Y=A; P=B, B=P; and so on. H and W acted as their own symbols. Phonetic equivalents were used for QU (KW ) and X (KS and GZ). The message:"This cipher only requires a short table which can be memorized in a few minutes."
A E I P T C K F S L M Y U O B D J G V Z R N
To decipher No. 154, by M. L. Harris, transcribe the cryptogram by descending verticals, left to right, to form the subjoined nine by eleven rectangle. Then read by successive horizontals, left to right, with due regard for phonetic values, and omitting the italicized letters, which are only nulls, and you will get the message: TU HYMN HU IN ZI... which, in ordinary spelling, becomes:"To him who in the love of Nature holds communion with her visible forms, she speaks a various language."
T U H Y M M H U I N A Z I L U V U V N A C B H U R R H O L D Z K O O M M U N Y U N W Y B Z H U R R V I Z I B F U L L F O U R M M Z V S H I S P I K S A V H A R I U S L A N G W O I J R A Z O B H F V Z
Every cipher on this week's list is of exceptional interest and merit. The first is a cleverly constructed O. and A. cryptogram, with the two parts in different simple substitution alphabets. Mrs. Fowler's message is very unusual.
In No. 156, C. E. Roe is offering readers of this department their first "limerick crypt." A straight substitution alphabet has been used.
A single rule, simple and easy to remember, is all you need to decipher Dr. Farrell's No. 157. The system is simplicity itself, but without the rule you may have a tough time of it.
CIPHER No. 155 (Mrs. M. L. Fowler, Kansas City, Kansas).
Question: WHAT YGHANLIZ NEB GAIZTMZ NMXMTYQ-YUH SPAN YGETYQ-HIM MRALPN UGLY ? Answer: BUYFO-RNINE LQESYNS FLYNN.
CIPHER No. 156 (C. E. Roe, Hudson, Massachusetts).
Baestus, fly paxdu at Straigs, Edbsejdu ry lde omygod, "Flsr s mai kyg sed!" Stu ld osau, "Y bk zgddt, Ao ar bsttdeo kyg bdst, Ye uy kyg ednde ry bk naiged?"
CIPHER No. 157 (Dr. G. A. Farrell, Montgomery, Alabama).
11—13.29.1.6—20.14.24.20.18.12.2—21.3.16.6—
28.12.10.28—12.18.33.9.6—2.14—24.7.17.27.
17.6—2.29—21.16.2.19.17.24.8.21—17.41.1—
32.14.28.12.6.36.40—0.13.3—3.39.17.7—18—
20.7.23.2.38—18.8.0.38.6.8.18—28.6.36.16.8—
3.7—4.38.11.29.9—1.15-3.3.15.18.24.8.32.