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From FLYNN's January 7, 1928

SOLVING CIPHER SECRETS

Edited by M. E. Ohaver
IN WHICH IS APPLIED A METHOD EXPOSING THE VULNERABLE
SPOT IN LAST WEEK'S DOUBLE TRANSPOSITION CIPHER No 115

THE vulnerable spot in last week's double transposition Cipher No. 115 is due to the fact that the message assumed the form of a square when set up for encipherment, and that, consequently, it became subject to the easy method of resolution described in some detail later in this article.

It is possible for squares to occur in the system under discussion only through the use of keys of improper length, or through failure to add the necessary nulls to messages which are not multiples of five letters in length. A message of thirty-six letters with a key of six letters, for example, would make a six by six square if the four nulls were omitted. This can easily be seen by any one.

When communications are in multiples of five letters, on the other hand, squares are only possible with keys of 5, 10, 15, 20, or any other multiple of five letters. Thus, Cipher No. 115 used a key of ten letters with a message of one hundred letters, resulting naturally in a ten by ten square.

Application of the present method will be illustrated with a five by five cryptogram, formed by using the key, TOWER—4-2-5-1-3—as shown herewith in enciphering the message, SEND REENFORCEMENTS AT ONCE—A. The final A of this message is the null required to fill out the multiple of five letters, The illustration follows:

     T O W E R
     4-2-5-1-3

(a)  S E N D R
     E E N F O
     R C E M E
     N T N A T
     O N C E A

(b)  D F M A E
     E E C T N
     R O E T A
     S E R N O
     N N E S C

(c)  A T T N S
     F E O E N
     E N A O C
     D E R S N
     M C E R E

A glance at this encipherment will show that the whole lines of the message (a) form whole columns of the first transposition (b), the letters in all of these columns being similarly transposed. Further, the columns of (b) are preserved intact in the lines of the second and final transposition (c), only being changed in their order.

Hence, to decipher an example of this kind it is only necessary to transcribe the cryptogram by horizontals, left to right, into a square of the proper dimensions, and then to rearrange the columns thus formed so as to develop message sequences, words, and so forth, in all of the lines.

If the columns of the square are numbered in serial order, this rearrangement will also recover the numerical key, as shown at (d), where the key 4-2-5-1-3 is revealed. The whole message may then be read by taking the several lines in the order indicated by this key, and dividing into words.

    4-2-5-1-3
(d) N T S A T  1
    E E N F O  2
    O N C E A  3
    S E N D R  4
    R C E M E  5

As in many types of transposition ciphers, paper strips may here also be used to advantage. For example, there would be five strips in this case, each bearing a number and one of the above columns of five letters.

Last week's No. 115, in the same way, would require ten strips of ten letters each; and so on. This method is not at all difficult. In fact, it is great sport to try various combinations of columns or strips, and watch the message develop before your very eyes.

If you don't believe it, try No. 118, herewith, where the square results from the supposedly inadvertent omission of the nulls required to fill out the multiple of five letters. Also, if you have last week's issue handy, and you haven't already solved No. 115, try to get the solution by this method. The answer to No. 115, by the way, was promised this week; but on second thought, and for evident reasons, it will be better held over until next week.

Here is the solution to last week's "Q. and A." Cryptogram No. 113 (Anthony Ranieri), Question: WHO SAID, "IT IS A CONDITION THAT CONFRONTS US—NOT A THEORY"? Answer: GROVER CLEVELAND. This one should have quickly yielded to solution by inspection.

Thus, the single letter J, occurring twice, was probably A. Substituting this in HFJH, the first and last letters of which are alike, would give THAT. After which RH would become IT, and RQ, IS. Substituting known values in QJRP would give SAI-, obviously SAID. And so on. For another brain teaser of this sort try Cipher No. 116, herewith.

Last week's transposition Cipher No. 114 (Arthur Bellamy) expressed the message: "WE QUICKLY QUEST A SQUARE. NEXT, A 'Q ' USED AS A CLUE, AS 'Q' USUALLY FOLLOWED BY 'U,' SOLVES QUITE QUEER QUIZ."

Encipherment was performed by writing the message by horizontals, left to right, into a nine by nine square, as shown herewith, and then by taking the columns downward in the order indicated by the numerical key 4-7-1-2-6-5-8-3-9, derived from the key phrase QUEER QUIZ.

Q U E E R Q U I Z
4-7-1-2-6-5-8-3-9
W E Q U I C K L Y
Q U E S T A S Q U
A R E N E X T A Q
U S E D A S A C L
U E A S Q U S U A
L L Y I S F O L L
O W E D B Y U S O
L V E S Q U I T E
Q U E E R Q U I Z

In many transposition ciphers a QU combination is a ready-made clew. But this one contains so many Q's and U's that it is hard to tell which are in combination. The nine by nine square, however, quickly settles the matter. For columns of nine letters bring most of these Q's and U's together.

Before exposing you to the devastating fire of this week's ciphers, there is yet another of them. No. 117, that requires some comment. Mr. Napier has based this on the well known Vigenère table of twenty-six alphabets, using numbers instead of letters. More we cannot tell you without spoiling the problem.

CIPHER No. 116.

Question: OQFY BE F UFGMXBCYQ?

Answer: G  OZGLQ  SHZZ  ES  CDFJC-
LGFQ  RCDPCDVA  RWCLW  JQDPQJ
CF  PCSSCLHZF  FE  SCDP  FWQ  RGI
KQFRQQD  FWQ  QDFJGDLQ  GDP  LODFJGZ
LEBOGJFBQDF.

CIPHER No. 117 (P. A. Napier, Louisville, Kentucky).

14-19-19-10-10  16-16-20-20-3  3-25-25-22-22
21-21-2-2-2  2-20-20-25-25  7-7-26-26-23  23-
10-10-5-5  15-15-26-26-7  7-10-10-13-13  15-
15-26-26-8  8-15-15-13-13  8-8-26-26-15  15-
18-18-7.

CIPHER No. 118.

SREAN  KDSPA  ESTTY  METNA
USEDM  HNTEE  EERZE  IRENT
ATRSO  OTGTU  RRTSF  AFORC
EVRO