AUTOKEY cipher No. 19, in the issue of May 14, stirred up more than the customary quota of enthusiasm. As you may recall, readers were left entirely to their own devices in handling this cipher, no suggestions for its solution being offered.
And, as it turned out, no suggestions were necessary. The fans located the weak spots in this cipher in double-quick time. Solutions were submitted by the following readers, in the order listed, those marked with an asterisk also.sending in their methods:
Most of the solvers turned the trick by beginning with a small number, trying all possible keys, and eliminating those which resulted at any point in numbers larger than twenty-six, or in zero or negative numbers.
The symbol 4 occurring in the fifth line of the cipher, for example, as shown below, could only have been keyed with 1, 2, or 3. But the first two of these possibilities are eliminated upon reaching symbol 7, where they result, respectively, inó1 and 0. By continuing the remaining series in both directions, reduction to a simple substitution cipher is quickly effected.
... 4—15—15—13—31—39—38—23—10—25—26——7 ... (1) 3 12 3 10 21 18 20 3 7 18 8 -1* (2) 2 13 2 11 20 19 19 4 6 10 7 0* (3) 1 14 1 12 19 20 18 2 5 20 6 1
The above problem met with such a rousing reception that we are offering herewith another cipher on the same basis. This new cipher uses a twenty-five-letter block alphabet—combining I-J—with a numerical key, both being derived from a common key word.
To prepare the alphabet first write down the key word, omitting all repeated letters after their first occurrence. Next write the remaining letters of the alphabet, in regular order, in lines of the same length as, and beneath, the key word. Finally, take the resulting arrangement by verticals, left to right, transcribing the letters into a five by five square, as shown on next page. To obtain the key numbers, write down nine letters of the key word—repeating if necessary —and number them from 1 to 9, according to their alphabetical places.
The first four key figures are now used with the last four horizontals in the block alphabet, the remaining five figures being employed to designate the columns. The letters in the top row use the single figures below the square as their substitutes. Any other letter is represented by the figure in the same line followed bv that in its column. Here, for example, using MYSTERY as the key word, M=i; r =78; and so on.
To encipher the illustrative message (a) merely substitute the proper numbers for the letters, (b), and then regroup by fives (c) for transmission. The last group may be be filled out with O's if desired.
M Y S T E R Y ————————————— M Y S T E R - A B C D F G H I K L N O P Q U V W X Y Z M Y S T E R Y M Y 2 7 S 6 1 4 8 3 9 M A I U Y 2 B K V S C 7 L W T D N 5 X E F O Z 6 K G P H Q 1 8 8 3 9 (a) F I N D — T H E— K E Y (b) 58 8 79 73 78 63 54 24 54 9 (c) 58879 73786 35424 54900
In deciphering with the key, the above process is merely reversed. The receiver knowing just which figures are used singly and which together, experiences no difficulty in properly dividing the numbers.
To determine these divisions without the key word, however, is a different proposition. No. 47 is the test problem. You know the system. All you lack is the key word. Can you solve it?
The best ideas among those offered will be presented in an early installment of the department. Try to have your answers in within two weeks. The translation and key word to this cipher will be published in the same article with our readers' methods of solution, but not before.
Now, to last week's answers. The two cryptograms in No. 42 used the key phrases
(a) "The golden key opens every door,"
(b) "There is no wisdom like silence."
By combining the two cryptograms, the pairs of letters VE-EM-YS-YL-OE ..., can easily be deciphered as symbols in a simple substitution alphabet. The message: Solve this and you kill, two birds with one stone. Had this message contained all, or nearly all, the letters of the alphabet, solution could also have been reached by multiple anagramming the two key phrases.
The key-phrase cipher, as mentioned in the January 22 issue, is; sometimes capable of other than the intended interpretation. Accordingly, it is representative of that class of ciphers to which standard No. 9, in FLYNN'S WEEKLY for May 28, has reference. Ciphers like No. 43, below, are not included in this class. Neither are multiple message systems which encipher two or more messages simultaneously in one cryptogram.
The "catch" in No. 43 was that the unsuspecting fan would probably take it for a "null" affair, in which every third letter was significant. This would give the apparent solution: "Watch your-step..." To reach the true solution, however, regroup the cipher by threes. Then, using an A=1... Z=26 alphabet, take the sums or differences of the first and third letters as indicated by the middle letters; M signifying addition, and E subtraction.
For example, WEC equals 23 minus 3, or 20, the equivalent of T, which thus becomes the first letter of the message. The real message in full: "The cipher should not prove too difficult, but it may fool some of the fans." Explanation of No. 44, as advised in the preceding issue, will be published in two weeks.
All of which being said and done, let's turn to this week's ciphers. Crypt fans will find something to their taste in No. 45. No. 46 is more difficult; but the veterans should be able to get it. No. 47 has already received full mention.
CIPHER No. 45.
QGK HZURI KUNQKIJ IYHI SHR EXIY LKQQVUD KXZYI IYGSO XJ KUJNQRJXOTU PQK KULURI YQTDGNJ EHILY QGI PQK YXS
CIPHER No. 46 (Herbert Lancaster, Bozeman, Montana).
ICQERVZ OCYF AGAYGDUNF AV C FBEG PU VKF VQBUVDJWU ZLPG NNP AV CLE RP AOGLS LCWAWUF
CIPHER No. 47.
13856 86767 79767 93138 14689 39397 93824 14682 32397 67638 39717 97679 71386 71481 59346 81337 23639 32368 13952 36397 25839 76793 82423 63971 47139 50000
The solutions to Nos. 45 and 46, also to the key-phrase cipher No. 41 of July 9, will be published next week.