book cover
From FLYNN's July 7, 1928


Edited by M. E. Ohaver
A New Solvers' List—Also a New Method for Determining
the Key Length of a Multiple Alphabet Cipher

EDITOR'S NOTE.— "Solving Cipher Secrets" will not appear again until the issue of August 4, when it will appear in this space again.

IT frequently happens that a problem in a multiple alphabet system, such as the Porta cipher with which we have been experimenting, will not provide the necessary repeated groups for determination of the period, or key length by the well known Kasiski method.

Even in such instances, however, it will often be possible to find the period mathematically, and without any need of guessing at the contents of the message, by recourse to another simple process which we shall now describe.

The method in question is best applied to short cryptograms, or to short sections of longer cryptograms, and depends on the fact that oftentimes a given letter will be enciphered at each occurrence in the same cipher alphabet, the key length being determined by factoring the intervals formed by that letter.

Last week's Porta Cipher No. 193 will be used in explaining the method. In this particular example, to be sure, the groups CSV—16 and 64 and FV—49 and 53—with intervals of 48 and 4, respectively, are sufficient to indicate a key of four letters by the Kasiski method. But it will also serve our present purpose.

    5    10      15     20

   25     30     35     40

   45     50     55     60


Perhaps the best procedure for this method is to prepare a table showing all the letters and their places in the cryptogram, similar to that herewith. Then, taking the least frequent letters first, try for the period by factoring their intervals. Here, for example, J, L, and W, each used twice, and at intervals of 4, 20, and 4, respectively, seem to point to a four-letter key:

A  51                    N  26-47
B  23-27-28-39-61-63     O  24-57
C  16-32-48-64           P 
D  20                    Q  11-12-45
E  15-33                 R  25-31
F  10-29-49-53           S  17-37-65
G  2-4-14-19-40-46       T  6-36-52
H  3                     U  43-44-56-59
I                        V  8-13-18-21-35-50-54-66
J  30-34                 W  5-9
K  1-58                  X  55
L  42-62                 Y  7-38-60
M  41                    Z  22

Glancing over the table, we now find two other letters, C and S, which seem to support the four-letter key supposition. For example, S, occurring as the seventeenth, thirty-seventh and sixty-fifth letters, provides the intervals 20, 48, and 28, with the predominating factor, 4, as before. And the intervals of C further bear out the hypothesis.

Once the period is known, solution may proceed by any desired method. The determination of a single letter in any Porta alphabet, of course, automatically fixes the whole alphabet.

Trying the most used symbols in the present four alphabets for E, T, A, O, and so on, would soon lead to the solution here. Or the solver could look for THE, which occurs twice in the message. This cryptogram used the key word CRAG, the message being: "You will find a horse at the old fortress. Ride down the coast and take a ship to sea." How did you make out with it?

Now for No. 191, Mrs. Fitzpatrick's "limerick crypt." The groups PXT—saw, TXP—was, and X—a, afforded a ready clew. With this lead other groups follow in rapid succession. An interesting crypt, though easy. Here is the translation:

Lindbergh  saw  a  very  queer  sight
In  Mexico, ’twas  a  bullfight.
  “Amazing!”  folks  cried.
  “Please  keep  the  bull  tied.”
But  our  hero  enjoyed  it  all  right.

No. 192, the transposition problem by L. Bekasi, conveyed the message: "This system is a simple one requiring no tables to be memorized, no kind of apparatus besides paper and pencil, and is fairly difficult of solution." The message was first transcribed by fives (a). Alternate letters of these groups were then interchanged (b ) the cipher being written as at (c).




This week's ciphers start out with an unusually interesting crypt by John R. Edwards. Note that every letter of the alphabet is used, and take a look at that "f. t."

Now turn to No. 195, about which its author writes: "I wonder how many fans can decipher this, and determine the key!" It is only fair to add that the cipher is of the fixed substitution class, each number standing for one letter and no other. Easy to solve if you catch on. But what has Mr. Hood used as key ?

No. 196 is an interesting variation of the Porta system. By using the numerical alphabet, results practically identical with those of the Porta cipher can be had without the special table. Thus, to encipher the message (a), "Come to me," using the key (b), PAL, first replace letters of both key and message with the proper numbers. This illustrates the method:

A  B  C  D  E  F  G  H  I  J  K  L  M
1  2  3  4  5  6  7  8  9 10 11 12  13
N  O  P  Q  R  S  T  U  V  W  X  Y  Z

(a) C  O  M  E  T  O   M  E
    3  2  13 5  7  2   13 5
(b) P  A  L  P  A  L   P  A
    3  1  12 3  1  12  3  1
          25       14 16
          13       13 13
(c) 6  3  12 8  8  1   3  6
(d) S  C  Y  U  H  A   P  S 

Then add the numbers thus brought together, subtracting 13 when the sum exceeds 13, and you have the series (c). Substituting for (c) you have the cipher (d), which can be grouped by fives, if desired, for transmission. Note that each substitute (d) is taken from the opposite line of the key from that of the letter it represents.

For example, C—top line—is represented by S—bottom line; O—bottom line—by C—top line; and so on. Problem No. 196 was prepared in the manner just described, but using another key word. There are plenty of clews, so we are going to expect plenty of answers. Try to solve it.

CIPHER No. 194, by John R. Edwards, Scranton, Pennsylvania.

Abcdcbef  ghijbcklfm  frxp  ibonp  jh  mkaa-
fefon  tfig, f. t.,  ribot  Ubckgp  npf  aktsern-
kqf  efuefgfonrnkbo  ba  Rstsgn  vrg  grem-
bohy,  vknp  Pfjefvg  xreofckro.  Akegn
ifron  xbowstrc  pruukofgg  vpkcf  crnnfe
dfun  rvrh  kokzsknbsg  ikgabensof.

CIPHER No. 195, by J. Lloyd Hood, Bastrop, Texas.


CIPHER No. 196.


A battle of no mean proportions is being waged these days among the fans for supremacy in our solvers' lists. Line-up for Ciphers Nos. 152 to 163, inclusive, is shown on the list herewith. Alfred N. Pray, fourth on the last

list, has now forged to first place with nine scalps to his credit. Lieutenant Commander E. H. Barber and J. Lloyd Hood, each with seven solutions, occupy second and third places. Other solvers are to be commended for their efforts, and congratulated for their success.

Keep your answers coming, fans. Another solvers' list will be published in a few weeks. Watch for it!