N devising a new cipher, one of the main objectives should be to elude known methods of analysis.
Because an inventor may not be altogether familiar with such methods, however, or that some apparently trivial but really important points may be overlooked, all new ciphers cannot be included in this category without further explanation.
In many instances complexities are introduced that are such, as Poe would say, "only in shadow." Oftentimes a device that is meant to throw the unauthorized decipherer into an inextricable maze can be torn off like a mask.
Just this can be done with the Radio Cipher, which was No. 3 in the October 2 issue of this magazine, and which will be dealt with in this article. It was to the first solver of this cipher, as you may well remember, that Mr. C. A. Castle, the originator of the cipher, offered the Superiorflex Radio outfit.
This system, as it happens, involves a process that can be made almost automatically to divest itself from the framework of the cipher, leaving the message in a simple alphabet, readable to the initiated almost at first sight.
How this can be accomplished should make an interesting story. But before we tell it we ought to mention that our readers owe their opportunity to win the radio not only to Mr. Castle's generosity, and interest in cryptography, but also, in a measure, to his grit. There is hardly another word to cover it.
For when Mr. Castle first submitted his proposition it was in the belief that his cipher was absolutely undecipherable. Yet when we advised him, as we promptly did, that his cipher could be solved, and that a number of solutions would probably be received, he, nevertheless, chose bravely to stick to his guns, rather than withdraw the offer.
As he explained it, what interested him most, if his cipher really could be solved, was how it could be done, which, we believe, is likewise of interest to all who read this.
To begin, that the cryptogram is one of numbers decides that it is of the substitution class. Of course, the principle of transposition may also be involved; or perhaps the numbers may be interspersed with nulls. But it is best to start with the supposition that all the numbers are significant, and that the message is expressed in its original or untransposed order. Starting with that idea, we proceed.
A glance at the following frequency table of the two hundred and sixteen numbers of this cryptogram—printed in groups of seven a little farther on—reveals that they range from 2 to 58, most of the numbers between these limits being used. A single glance will show what is meant.
2- 2 14- 9 26- 7 38-3 50-1 3- 8 15- 4 27- 5 30-2 51- 4- 5 16- 9 28- 2 40-2 52-1 5- 1 17- 6 29- 2 41- 53-2 6- 18- 3 30- 6 42-3 54- 7- 19- 1 31- 2 43-1 55- 8- 20- 4 32- 3 44-4 56-4 9- 21- 7 33- 6 45-6 57-1 10- 22- 6 34-14 46-1 58-1 11-10 23-14 35- 4 47-2 ———— 12-10 24- 2 36- 4 48-2 216 13- 7 25-15 37- 1 49-1
The first momentous question to be decided is whether these numbers represent the use of one or of several cipher alphabets. And, fortunately, by following the advice of our good friend, Kasiski—see FLYNN'S WEEKLY for August 7, 1926—this point can be settled in a few minutes by scanning the cryptogram for repeated groups.
Here is a list of them—enough evidence to hang a man—prepared to save you the trouble of looking them up yourself. This table, which ordinarily would be made from the original cryptogram numbered by fives, can be checked if desired with the accompanying transcription of it in groups of seven:
Groups Places 2-56 16; 142. 3-34 9; 30; 51; 139; 212. 3-34-30 0; 139. 4-23 125; 209. 11-48 23; 79. 11-44 77; 163. 12-25-32-3 27; 69. 12-21 5 5; 167. 12-S3-11 147; 161. 14-34 93; 201; 215. 14-12 104; 146. 15-13 6; 195. 15-12 54; 68. 16-34 184; 191. 17-23 74; 116; 130; 207. 20-21 42; 84. 21-33 81; 85. 23-13 117; 187. 25- 4 12; 173. 25-42 111; 123. 28-14 19; 143. 30-25 11; 106. 26-25 38; 136; 190. 34-26-25 37; 135. 16-45 88; 95; 165. 34-16-45 87; 94. 44-16 164; 103. 45-13 95; 99.
In a multiple alphabet cipher the probability of any recurrent group being periodic increases with its length. Hence, if the present cipher is of this type, the several longer groups in the table should be ample to determine the period:
Groups Places Intervals Factors 3-34-30 139- 9=130; 2-5-10-13-26-65-130. 12-25-32-3 69- 27= 42; 2-3-6-7-14-21-42. 12-53-11 161-147= 14; 2-7-14. 34-26-25 135- 37= 98; 2-7-14-49-98. 34-16-45 94- 87= 7; 7.
The largest predominating factor in the table is 7, so it is clear that seven alphabets have been used. The factor 14 occurs three times, it is true. But the one group, 34-16-45, with its interval of 7, decides the matter.
That the longer groups are not always periodic is here illustrated by 3-34-30. This repeated group, as will later be evident, is purely accidental, representing the sums of the following two different series of key and text numbers:
A S T E — C
Key: 1 24 15 2 10 20
E — T A S —
Message: 2 10 15 1 24 10
————————— —————————
Cipher: —3—34—30— —3—34—30—
(P) (139)
By the Kasiski method, which depends on repeated groups, it is clear that this cipher has a period of 7. However, since we are now outlining a general campaign against this type of cipher, and not just solving Mr. Castle's example, we shall offer still another method of finding the period, depending this time on repeated single characters.
This recurrent character principle has its own peculiar advantages, and can be used with many different kinds of ciphers. Sometimes it can be successfully applied to cryptograms that do not happen to contain any recurrent groups. In a cipher such as this turns out to be, it can best be applied to the highest and lowest numbers.
For it will presently be seen that such numbers can result from fewer combinations of key and text letters than can the middle numbers. Hence, the majority of any particular "outside" number will usually be found in a single cipher alphabet. The "middle" numbers, on the other hand, are, as a rule, more or less evenly distributed among several cipher alphabets.
In the following table this method is illustrated with the number 4, which occurs in the cryptogram at the five places given in the first column. The remainders are obtained by dividing the numbers—representing periods—in the top line into the numbers—representing places—in the first column:
Number Periods 4 2 3 4 5 6 7 8 9 10 ... —————————————————————————————————————————————————————— 13 1 1 1 3 1 6 5 4 3 ... 65 1 2 1 0 5 2 1 2 5 ... 125 1 2 1 0 5 6 5 8 5 ... 174 0 0 2 4 0 6 6 3 4 ... 209 1 2 1 4 5 6 1 2 9 ... Places Remainders
By this method the period will usually be the largest divisor in whose column any remainder occurs the most times. For example, the remainder 6 here occurs four times in column seven, thus indicating a period of 7. The numbers in column seven also show that the second 4 occurs in the second alphabet; and that all the other 4's occur in the sixth alphabet. In this way the table automatically indicates which ones of a given repeated character are in the same alphabet.
If the number 3 be tried in the same manner, six 2's and two 6's will fall in column seven. Not only does 3 thus indicate the same period as 4, but the similar remainders show that both these numbers result from the same small key numbers in similar alphabets.
The large number, 56, which gives four 3's in column seven, is additional evidence of a period of this length.
If desired, recurrent characters can be figured by taking their intervals, as in the Kasiski method. Or recurrent groups can be figured by the method just described. The group -17-23-, for example, will by this method give four 4's in column seven.
Now that the number of alphabets is decided, the cryptogram can be transcribed in lines of a length equal to, or a multiple of, the supposed period. This will cause the numbers in each alphabet to fall in the same column.
Here is the present cryptogram written off in groups of seven numbers each. Observe how the underlined recurrent groups, and also the recurrent 3's, 4's, and 56's, already mentioned, line up in their respective columns. (The Castle cipher in this form follows on the next page.)
The next step is to make a frequency table of the numbers in each of the several alphabets; that for the first alphabet being made from all the numbers in the columns numbered 1; that for the second alphabet from columns 2; and so on. (See table on page 636.)
Note that while the numbers of the original cryptogram covered a range of 57— from 2 to 58—in the table of alphabets there is only a difference of 33 between the highest and lowest numbers in alphabets 2-3-4-6, and of 32 in alphabets 1-5-7. Further, as shown in parenthesis at the foot of each column, there are only from thirteen to sixteen different numbers in each alphabet. This somewhat uniform distribution of the numbers can be taken as additional evidence that 7 is the period.
In fact, when other methods have failed, the period of a cipher can often be detected in just this way. Write the cryptogram in lines of various lengths, as has been done here in lines of seven numbers each. The arrangement providing the most even distribution of characters in the supposed alphabets will usually indicate the period.
The several alphabets which we thus seem to have found in this cipher may consist of anything from a nonrelated series of mixed alphabets, to a related series of methodized alphabets. And there are general methods, applicable in any case, that we expect to consider in subsequent articles. One such method is given on pages 101-108 of André Langie's Cryptography, as translated into English by J. C. H. Macbeth.
However, that the alphabets here have about equal ranges and similar distributions of frequencies and voids, or gaps, is evidence that, in this instance, they are probably all modifications by a series of key numbers of a single primary alphabet. Consequently a special method can be brought to bear.
Thus, the most used numbers in alphabets 2, 3, 4, 5, and 6, are 11, 34, 25, 23, and 12, respectively. Similarly in alphabets 1, 2, 3, and 7, the numbers 34, 15, 16, and 24, respectively, have not been used, there most being in each case four used numbers next above them, and one below.
1 2 3 4 5 6 7 1 2 3 4 5 6 7 ———————————————————————————————————————————— 7 14 23-11-26-19-23-15-13- 42- 3-34-30-25- 4-20- 21 28 24- 2-56-25-28-14-11- 35-11-48-16-38-12-25- 35 42 32- 3-34-29-16-22-25- 26-34-26-25-36-32-20- 49 56 21-11-46-18-46- 3-40 50- 3-34-27-15-12-21 63 70 23-12-66-30-23-31-12- 33- 4-26-49-15-12-25- 77 84 32- 3-57-17-23-27-11- 44-11-48-21-33-14-20- 91 98 21-33-34-16-45- 5-14- 21-14-34-16-45-13-20- 105 112 46-13-26-47-23-14-12- 30-25-26-40-23-25-42- 119 126 22-11-36-17-23-13-13- 31-33-39-25-42- 4-23- 133 140 23- 3-58-17-23-17-22- 22-34-26-25-38- 3-34- 147 154 30- 2-56-38-28-14-12- 53-11-27-26-35-34-33- 161 168 45-11-47-27-14-36-12- 53-11-44-16-46-12-21- 175 182 22-21-27-29-25- 4-43- 30-16-36-18-37-12-39- 189 196 26-16-34-18-23-13-33- 52-16-34-44-15-15-18- 203 210 22-36-26-25-14-34-17- 30-24-56-17-23- 4-23- 44- 3-34-35-14-34.
Again, the intervals between used and unused numbers can be considered. For example, there is an interval of 8 in every case between 22-30, 3-11, 26-34, and 12-20, the used pairs of characters in alphabets 1, 2, 3, and 7, respectively.
Such examples might be multiplied indefinitely. But enough has been shown to suggest the alignment given in the table, where the numbers in any given line represent, supposedly, the same text character.
Observe in the combined frequencies given in the last column that the total number of different characters provided by this arrangement is only 23, a favorable indication of a simple substitution alphabet.
1 2 3 4 5 6 7 Total ——————————————————————————————————————————————————————————————— (-19) (-0) (-23) (-14) (-12) (-1) (-9) 21-3 2-2 25-2 16-4 14-3 3-2 11-8 18 22-4 3-6 26-5 17-4 15-3 4-4 12-4 30 23-3 4-1 27-2 18-3 16-1 5-1 13-3 14 24-1 6- 88- 19-1 17- 6- 14-1 3 25- 6- 29- 20- 18- 7- 15- 26-2 7- 30- 21-1 19- 8- 16- 3 27- 8- 31- 22- 20- 9- 17-1 1 28- 9- 32- 23- 21- 10- 18- 29- 10- 33- 24- 22- 11- 19- 30-4 11-8 34-8 25-6 23-9 12-5 20-4 44 31-1 12-1 85- 26- 24- 13-3 21-2 7 32-2 13-1 36-2 27-2 25-2 14-4 28-1 14 33-1 14-1 37- 88- 26- 15-2 23-2 6 34- 15- 38- 29-2 27- 16- 24- 2 35-1 16-8 89-1 30-2 28-2 17-1 25-3 13 36- 17- 40- 31- 29- 18- 26- 37- 18- 41- 32- 80- 19- 27- 38- 19- 42- 33- 31- 20- 28- 39- 20- 43- 34- 32- 21- 29- 40- 21-1 44-1 35-1 33-1 28-1 30- 5 41- 28- 46-1 36- 34- 23- 31- 1 42-1 23- 46- 37- 35-1 24- 32- 2 43- 24-1 47-1 38-1 36-1 25-1 33-2 7 44-2 25-1 48-2 39- 37-1 26- 34-1 7 45-2 26- 49- 40-1 38-2 27-1 35- 6 46- 27- 50- 41- 39- 28- 36- 47- 28- 61- 42- 40- 29- 37- 48- 29- 52- 43- 41- 30- 38- 49- 30- 53- 44-1 42-1 31-1 39-1 4 50-1 31- 54- 45- 43- 32-1 40-1 3 51- 32- 55- 46- 44- 33- 41- 52-1 33-2 56-4 47-1 46-3 34-3 42-1 15 53-2 34-2 57-1 48- 46-1 35- 43-1 7 35-1 68-1 49-1 36-1 4 ——————————————————————————————————————————————————————————————— (16) (14) (13) (15) (14) (15) (16) (23)
Due to Mr. Castle having assigned one of the most used letters, A, to the smallest number in his primary alphabet, the present seven derived alphabets automatically align themselves by their first numbers. In other cases this upper margin of the adjusted alphabets might be more or less irregular, as is the lower margin here,
To proceed, if our supposition as to the origin of these alphabets is correct, any alphabet in the cipher can now be expressed in terms of any other alphabet by merely adding or subtracting to all its terms the common difference between any two numbers in the same line of such alphabets.
Suppose that we decide to reduce all the present alphabets to the size of No. 2, The differences between the several alphabets and No. 2 are shown in parentheses at the top of the table. Accordingly this series of numbers becomes a trial key, by which the original cryptogram can be reduced to the terms of only one of its alphabets. The resultant cipher in this instance proves to be of the simple substitution type, which is readable almost at sight. Here are the first seven numbers of Mr. Castle's cryptogram so treated:
Cipher: 23-11-25-19-23-15-13-
Trial Key: 19 0 23 14 12 1 9
—————————————————————
Reduction: 4-11- 2- 5-11-14- 4-
By modifying every group of seven in the same manner, the cryptogram takes on the following form:
4-11- 2- 5-11-14- 4- 23- 3-11-16-13- 3-11- 5- 2-33-11-16-13- 2- 16-11-25- 2-26-11-16- 13- 3-11-15- 4-21-16- 7-34- 3-11-24-31-11- 2-11-22- 4-34- 2-31- 31- 3-11-13- 3-11-12- 4-12-33-16-11-30- 3- 14- 4- 3-35- 3-11-16- 13- 3-34- 3-11-26- 2- 25-11-25- 7-21-13-11- 2-33-11- 2-33- 4- 5- 2-14-11- 2-33-12-11- 26-13- 3-33-11-13- 3- 11-25- 2-26-11-24-33- 3-11-13- 3-11-12- 4- 12-33-16-11-30- 3-14- 4- 3-35- 3-11-16-13- 3-34- 3-11-26- 2-25- 11- 2-33-24-16-13- 3- 34-11- 4-11-23-33-24- 26-11-24-13- 2-35- 3- 34-11-21- 2-33-11-12- 3-21- 4-15-13- 3-34- 11-16-13- 4-25-11-30- 7-16-11- 4-11-12-24- 33-16-11-30- 3-14- 4- 3-35- 3-11- 2-33- 8- 11-24-33- 3-11- 3-14- 25- 3-11-21- 2-33
The reader has no doubt observed that, up to this point, all the work on this cipher has been accomplished mathematically. It has not been required to assume the identity of a single number. As far as the analysis is concerned, the cryptogram might convey a message in some unknown language.
Since it is reasonably certain here, however, that the message is in English, it is time to speculate as to the significance of some of the numbers in the simplified form of the cryptogram.
Here is a frequency table of these numbers, showing also the values subsequently assigned to them:
2 A 18 11 — 44 20 29 3 E 30 12 D 7 21 C 5 30 B 4 4 I 13 H 14 22 G 1 31 F 3 5 M 14 L 6 23 K 2 32 6 15 P 2 24 O 7 33 N 15 7 U 16 T 13 25 S 7 34 R 7 8 Y 17 26 W 6 35 V 4 9 18 27 ——— 10 19 28 216
The number 11, occurring forty-four times, or 20.3 % of the two hundred and sixteen numbers of the cryptogram, at once engages the attention. This is probably a word space, since E, the most used letter, comprises only about 12.5% of average English text. This discovery further simplifies the cipher, for it now becomes a simple cipher with normal word divisions as will readily be seen.
The short groups as defined by the word space (11), and which supposedly represent whole words, can now be compared and analyzed. Here are just a few of them with their obvious meanings appended:
2 A
2-33 A-N
2-33-12 A-N-D
2-33-8 A-N-Y
13-3 H-E
16-13-3 T-H-E
16-13-3-34-3 T-H-E-R-E
16-13-2-16 T-H-A-T
16-13-4-25 T-H-I-S
4 I
12-4-12-33-16 D-I-D-N-T
By substituting these values in the longer words, additional values can be discovered, and Mr. Castle's message can gradually be developed in full, as follows:
I am like the man that saw the picture of a giraffe; he didn't believe there was such an animal. And when he saw one, he didn't believe there was another. I know Ohaver can decipher this, but I don't believe any one else can.
Mr. Castle's opinion of our prowess is appreciated. Nevertheless, we are glad to announce that he is in error. For a number of solutions to his cipher have already arrived. And others may be forthcoming.
The list of solvers, together with some of the answers, are scheduled for publication with the next cipher article.
One of the solutions was received from Dr. G. A. Ferrell, Bessemer, Alabama, who writes:
"I resent Mr. Castle comparing you to a giraffe, but am vain enough to think that he will put me in your class, with this cipher, anyway."
Never fear, doctor! Something tells us there will be a regular army of two-legged giraffes scattered over the country before this is over with.
Again to proceed, however, with our exposition, though the message has been translated, the cipher system itself does not yet stand fully revealed. To solve the final secrets of this cipher, the true alphabet must be discovered, and, if possible, reconstructed; and the method of using the key must be determined.
It was, of course, fairly obvious here that the primary alphabet is one unit smaller throughout than cipher alphabet No. 2. But in other cases, as in the test ciphers herewith, the trial alphabet might easily be more or less than, or equal to, the primary alphabet.
However, if the numbers of the modifying key are regulated by a key word used with the primary alphabet, the error in the trial alphabet can readily be settled by examining the equivalents of the trial key when increased and decreased by multiples of two, as follows:
—2 17 . 21 12 10 . 7
. . C D . . U
Trial 19 0 23 14 12 1 9
Key . . K L D . .
+2 21 2 25 16 14 3 11
C A S T L E —
+4 23 4 27 18 16 5 13
K I . . T M H
+6 25 6 29 20 18 7 15
S . . . . U P
The key word CASTLE — (note the dash) comes out clearly in the plus 2 line, indicating that the trial alphabet is "plus 1," or too large throughout by one unit. For the error, if any, in the trial alphabet will by this method always be one-half that in the trial key. When key letters are missing, they can often be supplied by context, providing additional values for the alphabet.
Only a moment's consideration will make this clear to the cipher enthusiast. The advantages to be derived from this particular cipher should be sufficiently clear, too.
Mr. Castle devised his alphabet, which also has substitutes for the ten figures, so that it could be reconstructed from memory. Beginning with A=1, the numbers were assigned to each fourth character in advance in the series, as is evident in the complete original alphabet herewith.
A 1 I 3 Q 5 Y 7 7 9 B 29 J 31 R 33 Z 35 8 37 C 20 K 22 S 24 1 26 9 28 D 11 L 13 T 15 2 17 0 19 E 2 M 4 U 6 3 8 — 10 F 30 N 32 V 34 4 36 G 21 O 23 W 25 5 27 H 12 P 14 X 16 6 18
The key word is, of course, changeable at will. Subjoined are a few letters of the present message, enciphered with the present key word, to illustrate the system:
C A S T L E — C A S T L E — 20 1 24 15 13 2 10 20 1 24 15 13 2 10 I — A M — L I K E — T H E — 3 10 1 4 10 13 3 22 2 10 15 12 2 10 23—11—25—19—23—15—13— 42—3—34—30—25—4—20
This system is thus similar to the Nihilist cipher given in FLYNN'S WEEKLY for March 28, 1925, which can also be solved by the present general method for ciphers of this type with unknown alphabets.
Should the primary alphabet of this kind of cipher be at hand, the key numbers can easily be found by identifying any of the numbers in the secondary alphabets. Or possibly advantage can be taken of peculiarities to prepare a special method, as that in the June 27, 1925, issue of FLYNN'S WEEKLY for the Nihilist cipher.
Whoever has followed through this explanation can now understand, did he not already, what cipher complexities are that exist only in shadow.
For here, apparently, is a very intricate system. Through the multiplicity of substitutes, and the fact that most of them have various meanings, all connection with a simple alphabet is apparently severed. But as has been pointed out in this department before it isn't always the most intricate cryptogram that proves the most successful. Here was a cipher that, to the layman, would appear the most invulnerable conveyor of secrets imaginable.
And yet this carefully reared edifice can be made to crash in ruin almost at a touch!
Here are some examples of this type of structure.
Can you loose the first stones?
CIPHER No. 1.
53-70-33-73-63-48-31-52-64-73-51-57-53-72-42- 73-62-53-48-60-65-70-32-63-59-72-50-65-55-48- 32-47-53-48-51-63-57-62-51-65-55-73-38-71-44- 64-51-47-65-74-32-63-59-72-55-60-59-62-51-52- 39-72-29-73-47-66-49-73-63-66-33-69-63-74-38- 73-39-72-29-53-53-72-37-72-39-66-51-52-55-72- 35-65-61-48-53-73-44-64-47-71-67-61-37-61-63- 53-27-71-50-74-41-71-53-66-51-70-57-55-53
CIPHER No. 2.
58-77-67-45-76-27-49-66-39-45-65-39-67-54-57- 45-66-40-65-57-48-86-56-49-77-44-39-48-77-38- 76-48-30-66-44-69-59-48-30-69-44-58-49-46-57 - 49-77-39-55-77-67-68-67-36-78-56-30-66-44-57- 49-47-30-45-47
Every one of our readers' ciphers, this time, presents some exceptionally interesting features.
Mr. VanderPyl's, for example, can be solved by several methods given in previous issues. The recurrent tg in the third and eighth groups is a dew, ever so slender, but still a clew, nevertheless, for the Kasiski principle to work on. Attack this tg properly, and a speedy disintegration of the cipher is sure to follow. But begin by reading Mr. VanderPyl's letter.
DEAR SIR:
The old Milwaukee Brewer in The. Prince of Pilsen used to say, " In Milwaukee everybody is chust as goot as somebody else and sometimes better." Which is just another way of saying that no matter how smart you may think you are there is always somebody just a little smarter.
Nevertheless, it seems to me that it should be not so difficult to devise a cipher that no one can figure out. Suppose, for instance, especially in a short one, some system is devised so that no letter shall appear in a sentence twice in the same character. Would not that be difficult to decipher, or has somebody tried that already?
I am no good at working anything out except a simple alphabetical cipher because I haven't the time to devote to the study of the thing.
Below is a cipher that I have used which is very simple—possibly some smart geezer can work it out. I don't know whether or not it is customary to inclose the solution, but in the event it is desired you will find it inclosed separately herewith.
R. A. VANDERPYL,
Executive Secretary, Conopus Club, Detroit, Michigan.
CIPHER No. 3 (R. A. VanderPyl).
iuukszlclq xzbfarri tgcfjbeowlxd oekxj oodqpttsd ykeispnxfqe bgmdzrcww mktg- gpyvncg vsqvdbf ocovngeimhr fpxaoprx gecmka
Ciphers having several substitutes for each letter, which we believe represent the type Mr. VanderPyl refers to in his second paragraph, have repeatedly been solved. And some of the methods, together with historical examples, will probably be presented in subsequent articles. However, a relatively short message in such a cipher, unless the alphabet is based on some systematic arrangement which is discovered, would probably escape solution.
Mr. Spence's cipher, No. 4, uses a simple numerical alphabet, the cryptogram being interspersed with nulls, or numbers of no value. The alphabet and the location of significant numbers are determined by the key, changeable at pleasure, which accompanies the cryptogram.
CIPHER No. 4 (T. P. Spence, Miami, Florida).
Key: B-23-1-5-6. 1-3. Message: 22-29-24-20 11-18 3-10 8-6-3-14 7-7-7-8 22-21-24-19 13-19 11-19-18 18-16 23-19-13-21 20-21-24-19 26-19 24-19 17-19-13-20 25-6-3-9 5-19-13 19-18-13.
In No. 5 we have a fine example of the concealment type of cipher. And we'll bet a plugged nickel that the receiver of this message won't keep the sender waiting. You'll agree when you read it.
CIPHER No. 5 (M. Walker, Akron, Ohio).
HAMAN DVILE EATSO WLSOA TSFRU ITINI TLACK SALAS DREAM SSMIL INGAT ATHRE ESPAR KLEBL AZEDA YMYAU NTKIT TYWAS EENLE AVEDE ASTOT HERIN ERTST RAYOL
Mr. Mihlhauser's No. 6, which he believes undecipherable, is an example of the very kind he asks for in his message; nor will we have to look very far for a method of solving it.
CIPHER No. 6 (W. N. Mihlhauser, Oklahoma City, Okla.).
74-93-51-11-43-51-00-74-42-21-93-91-43-81-00- 43-83-84-51-00-51-11-43-64-00-31-91-74-81-51- 73-43-00-54-91-44-81-00-91-82-43-44-73-42-31- 44-91-83-82-43-00-81-83-54-00-44-83-00-54-83- 73-94-00-44-81-51-84
In No. 7, where the autokey principle has been applied to the transposition cipher, Mr. Bellamy is offering our readers an ingenious and original construction. Since the order of transposition here is determined by the letters of the message itself, several messages of the same length would evade the method of multiple anagramming given in FLYNN'S WEEKLY for July 3, 1926.
CIPHER No. 7 (Arthur Bellamy, Boston, Mass.).
(a) This message—though in English—is intended for a German submarine captain.
LCROE EESRS DCOTS NTWLB MEKYM UOICA THIRI HTRTI A
(b) Here is one for a Northern general in 1865:
RTCYK HTAPT MUCRO DANRA EEISD OTAHR P
(c) This one is about ciphers:
SIRYT SIIEA SPVTO EUOST KWLNH HCNIO OHGSI TENYE
(d) And this last is on a military subject:
TKRKR BGRQF SAOEO MXCVD UEDYA DHURN OUOJI WIITZ HPRLI EETSE
Complete explanations of all the ciphers in this article will be printed in next Solving Cipher Secrets. Let us know, in the mean time, how you made out with them.
Although two different key words were used with the first six of the November 13 ciphers, we feel sure that our readers had no trouble by the method given in finding out which ones were in the same key. Here are the translations of all six, with the proper key word in each instance.
Cipher No. 7 was a code, based on a prearranged assignment of values, as follows:
Stockings (metal), size 9 (9 grains); Dress or Coat (hydroquinone), size 38 bust (38 grains); Shoes (sodium sulphite, dry), size 6½ (605 grains).
No values were assigned the last two items (sodium carbonate, dry; and water), but their inclusion was necessary that the cryptogram might pass the inspection of any one acquainted with such formulas without arousing suspicion.
In No. 8, Mr. Duree's "bootlegger's alphabet" used the even numbers in descending order, thus; 52=A; 50=B; 48=C ... 2=Z . His message: "I certainly do agree with Poe, for the human mind is surely ingenious."
Mr. Goldman's No. 9 used the following simple reciprocal substitution alphabet, in which A=B, B=A, C=D, D=C, and so on:
A C E G I K M O Q S U W Y B D F H J L N P R T V X Z
The vowels E, I, and U, were inserted, as required, to make the cipher groups pronounceable.
In lieu of the September 4 list of solvers, which will be combined with that of October 2 in the next cipher article, we are glad to be able to refer our readers to an interesting article on ciphers in the October, 1926, issue of Popular Mechanics. It is well worth reading.
As a parting word, don't forget that the names of those who solved the Radio Cipher, and at least some of the solutions, will be published in the next installment of this department.