THE old adage, "Union is strength," may apply well enough to such cases as the bundle of sticks. But exactly the converse holds good with most ciphers. For example, take the United States military service double transposition system described last week. The larger the "bundle" of cryptograms of the same length and key in this cipher, the easier it can be broken.
Thus, the four similar twenty-five-letter cryptograms of this type published last week as problem No. 106 can readily be translated by the multiple anagram method employed two weeks ago in solving the improved Myszkowski single transposition cipher.
The various principles involved in this method have already been treated in sufficient detail. Briefly, to solve the present cryptograms, it is only required to prepare a series of paper strips, numbered from 1 to 25, inclusive, each of which bears the corresponding letters of the several cryptograms, and to rearrange these strips in such order as to develop all four translations, as shown herewith:
17 12 18 16 3 2 4 10 1 11 9 21 20 22 24 19 25 23 7 6 8 14 5 15 13 S E N D R E E N F O R C E M E N T S A T O N C E A W E A R E A D V A N C I N G A T N O O N T O D A Y O U R S U P P L I E S A R E E X H A U S T E D T R E N E M Y F O R C E S A R E R E T R E A T I N G E
Divided into words the four messages revealed by this arrangement of strips, with the "nulls" which were added in encipherment to fill out the last five-letter groups, stand as follows:
(a) SEND REENFORCEMENTS AT ONCE. A (b) WE ARE ADVANCING AT NOON TODAY. (c) OUR SUPPLIES ARE EXHAUSTED. TR (d) ENEMY FORCES ARE RETREATING. E
The use of nulls in this system, so that all cipher groups will consist of five letters, tends to reduce error in operation and transmission. And at the same time it renders analysis somewhat more difficult, since the decipherer does not know where the nulls are located.
But there will never be more than four nulls in any cryptogram. And all the nulls in any group of similar cryptograms, collated for analysis by the present method, will be located on the same strips. Here, for example, they are on strips 15 and 13. Further, the fact that cryptograms cannot occur in all lengths, but only in multiples of five letters, makes it about five times as easy to collect a group of cryptograms of the same length for solution.
To go on with the story, however, the numerical series, 17-12-18-16-3, et cetera, just determined, may now be employed to decipher any similar cryptogram of the same length and key by merely numbering its letters serially from 1 to 25, and arranging them in the order designated by this series.
But to decipher a similarly keyed cryptogram of a different length, the original numerical key used in performing the two separate transpositions must first be determined. Can you find this key for the above examples? If so you will be able to decipher No. 109 in this issue, which has been enciphered in the same key. See next week's cipher article for full details.
To better acquaint the uninitated with the straight substitution type of cipher, last week's No. 104 is appended with its full translation. In this variety of cipher a given message letter is invariably represented by the same cipher letter; and, conversely, a given cipher letter is always the symbol for the same message letter.
1 2 3 4 5 6 7 8 9 10
Cipher: PWC DCQC LRRQCWCSGNES EX L UEDNSA CONT WLG RHP DLSI
Message: THE MERE APPREHENSION OF A COMING EVIL HAS PUT MANY
11 12 13 14 15 16
NSPE L GNPHLPNES EX HPDEGP YLSACQ.
INTO A SITUATION OF UTMOST DANGER.
Such cryptograms can be solved in numerous ways. Many fans like to begin by spotting E, which is usually represented by a predominating symbol. Here we have the symbols C, E, L, P, and S, used seven times each, and N six times, any one of which might be E.
So mulling things over for another clew, we find the three-letter word (1) PWC, ending in one of these predominating symbols, and therefore likely material for THE, which frequently begins a sentence. Another valuable clew lies hidden in the termination (13) -PNES, the letters of which are also used in (11) NSPE. Taking P=T (from PWC=THE), these two groups can easily represent -TION and INTO, respectively.
If N=1, however, the single-letter word (5-12) L, must be A. Substituting assumed values in (13) GNPHLPNES, we have -IT-ATION, obviously SITUATION. Proceeding in a similar manner the entire message is soon unfolded.
We have been regularly featuring one cryptogram of this sort in each issue for some time, along with more difficult ciphers. Do you vote for a continuance of this plan? If so, how about voicing your approval by contributing an occasional cryptogram for publication? Alphabets may be arranged in any order whatever.
Last week's No. 105 conveyed the message: "Nothing is so hard but search will find it out." Normal word divisions were observed in the cryptogram, and each letter was enciphered by counting forward in the alphabet the number of places indicated by its exact position in the selected word, thus:
Key: 1234567 12 12 1234 ... Message: NOTHING IS SO HARD ... Cipher: OQWLNTN JU TQ ICUH ...
Besides other new ciphers already mentioned, this week's layout also contains an interesting problem by Irving Bloom. To solve this, find the word spacers and the number of digits used for each letter. Then find how these groups of digits can be manipulated to produce simple numerical substitutes.
CIPHER No. 107.
MVLWZB MPZHG QWJFSW FKGZZVKS CXD RZ GZZK. PJJN XQQZKQVIZPD XKY DJF HVPP UVKY QWZC.
CIPHER No. 108 (Irving Bloom, Brooklyn, New York).
95946 46375 73848 40086 84959 50021 61212 15151 61510 08695 42980 06131 63839 49800 64635 45472 32216 10095 94646 32161 00413 27661 94510
CIPHER No. 109 (Double transposition).
TNECO CRFSA TEEAS SRSNS EYNDA XPAET NINOI DAGVR MFPAE SDIIN RPAIY FINLE ERTER EEEEM OVEET
There you are, ladies and gentlemen! Now put on your thinking caps, figure these out, and send in your answers! Solutions to all three will be published next week in FLYNN'S WEEKLY DETECTIVE FICTION.