IN last week's continuation of the analysis of cipher No. 73, an interval of thirty-one letters was found to exist between two series of letters which presumably were located in adjacent columns in the original "set-up." In the present article, as advised last week, this interval will be used in the further elimination of key lengths.
To use one or more such intervals by the present method it is necessary to consider the column lengths in the various setups for the key lengths in question. This data would vary, of course, for each different length of cryptogram. The accompanying table has been prepared for the present cryptogram of seventy-two letters:
Key lengths Column lengths ×
5* 15 × 2; 14 × 3.
7 11 × 2; 10 × 5.
10* 8 × 2; 7 × 8.
11 7 × 6; 6 × 5.
13 6 × 7, 5 × 6.
14 6 × 2; 5 × 12.
15 5 × 12; 4 × 3.
16 5 × 8; 4 × 8.
17 5 × 4; 4 × 13.
19 4 × 15; 3 × 4.
20 4 × 12; 3 × 8.
et cetera et cetera
Key lengths not already eliminated by factoring—see issue of two weeks ago—are given in the first column. And each key length is followed by the number and length of columns in its particular set-up. Thus, with a five-letter key, there would be two columns of fifteen letters each, and three columns of fourteen letters each.
In the present type of transposition cipher, an interval of the above kind is always equivalent to the length of one or more columns in the set-up. Consequently, whether or not a given interval is possible in any case may easily be determined by taking the columns in various combinations to form that interval.
In computing these intervals the following general rules should be observed:
Application of this method will eliminate keys of five, ten, and fifteen letters in this case. For example, the key could not be one of five letters, for there is no combination of fourteens and fifteens which will add up to make thirty-one.
With a ten-letter key, the interval thirty-one could be formed by adding three columns of eight letters and one column of seven letters; but this set-up only contains two columns of eight, one less than the required number. Similarly, with a fifteen-letter key, three columns of five letters and four columns of four letters would make thirty-one; but there are only three columns of four letters in this set-up.
The other key lengths listed in the table cannot be eliminated by this method. Thus, with a seven-letter key, 11 × 2 plus 10 × 1 equals 31; with an eleven-letter key, 7 × 1 plus 6 × 4 equals 31; and so on.
Some of these remaining key lengths might be disposed of, it is true, by finding other intervals. And discovery of such intervals need not depend solely upon the method of matching certain letters, as described last week. For example, they may be found, but perhaps more laboriously, by trying any series of letters exhaustively with every other possible series in the cryptogram, in order to find the probable combinations.
But in this case there is still another card up the sleeve. In other words, besides factoring and computing intervals, there is yet another method which can be applied here to definitely indicate one of the remaining key lengths, and eliminate all the rest. Have you any idea how this can be done? If not, and you can't figure it out in the meantime, see next week's cipher article for full details.
Remember last week's No. 77 (J. A. Dockham), the "crypt" with the funny message? Here is the translation:
An antagonized and angered ant announced arrival of an extra pair of antique pants and an allowance of tainted candy cans to the village dump heap.
The humorous message is a capital way of making the simple substitution cipher more interesting. By using vowels as substitutes for vowels, and consonants for consonants, Mr. Dockham succeeded also in making his cryptogram pronounceable, "UG UGYUZIGATOS UGS, et cetera."
In this connection we have lately received a letter from J. R. Midford, Portland, Maine, suggesting a sort of contest on "pronounceable" or "speaking" ciphers. A good idea! Who has the best "speaking" cipher? They can be of any type whatever. Cryptograms of about the length you see in this department are best for publication.
Last week's transposition cipher No. 78, OTENS LINRT HIENU, et cetera, conveyed the message: Cowards taste death many times, the valiant die but once. The key word was CAESAR, and the set-up as follows:
C A E S A R 3 1 4 6 2 5 C O W A R D S T A S T E D E A T H M A N Y T I M E S T H E V A L I A N T D I E B U T O N C E
This cipher, which is of the same type as No. 73 being analyzed in this issue, offers a good chance to match series of letters by the method given last week. Columns 4-6-2, on account of the TH and THE, would probably provide the entering wedge. No. 82, below, is another transposition cipher of this type. Try it!
We also have with us this week No. 80, a very interesting simple substitution cipher. And finally, Mr. Walker is offering you a rare bit of sport in No. 81. "Dear Uncle," whoever he is, can no doubt read this cryptogram at sight.
CIPHER No. 80 (Ernest Brewster, Stockton, California).
DYVPUI WAAQTYAYUM OGZWUH YCG- XDYI, VPUZWVM GLGWMD IWDVOG- THY LWMOWU LGTIVU'D PBBWVY. XCGUD BQTMOYT DSQCCIOHHYTE QXPU TYCYGDY. MQTUSYE CPW- MYTD UYGTNE.
CIPHER No. 81 (M. Walker, Akron, Ohio).
DEAR UNCLE:
We have come back to town. We went
over to Canada for a week. You see, this
was our vacation. Almost every evening we
were out to a dance or a show. You know
we only get one week, while most people
get two. Guess you have two weeks in Au-
gust, don't you? We went one afternoon
out to the lake fishing. Used a quart of bait,
but had no luck. So we left our tackle with
a small boy and went boating. Good-by,
ALEX AND JOHN.
CIPHER No. 82.
DRBES EIRNT DSNAA HANSO GSNAC NORFE PCIWH HOSRT GEOTN EAEIE LUBMT SONNT IAFET OSATR RTFES YYEFR OOOVO OHPEC HTT
Solutions to ciphers Nos. 80 and 81 in this issue, and No. 76 of the October 1 issue, will be given next week. Explanation of No. 82 will follow in the magazine in three weeks.