IN our issue of March 28 we offered a sample of the famous Nihilist code which was a challenge to our readers to send us messages that Mr. Ohaver could not translate. Many of the correspondents doubted his ability to do it without the keyword and hoped that he would reveal the secret of his method.
It is no secret. Skillful cryptographers the world over know it, and as Mr. Ohaver says, this method is about as safe as a leaky rowboat in the middle of the Atlantic. It's all in knowing how.
In the department this week Mr. Ohaver explains one deciphering method. He says it's easy. Maybe it is. Try it for yourselves.
Incidentally he offers some more ciphers from readers and gives the keywords of a lot of Nihilist messages we have received. Your own may be among them.
F the numerous correspondents who submitted Nihilist ciphers for solution in response to our invitation in FLYNN'S for March 28, some were absolutely certain that their messages could not possibly be read without the keyword.
Others, not so confident as these, thought we might be able to decipher their cryptograms, saying, however, that they were completely in the dark as to how it could be done.
But almost to an individual they wanted to know, if we succeeded in deciphering their communications, by what method this could be accomplished.
A few of the more experienced fans succeeded without the key in deciphering the Nihilist cipher in FLYNN'S for April 25. But for the benefit of the many who were unable to do this, we have decided in response to insistent requests to publish here for the first time a full exposure of the method used in deciphering this kind of cipher.
To begin with, the Nihilist cipher, while bearing certain other earmarks that assist in its identification by the initiated, is easily recognizable from the fact that its numbers are never lower than 22 nor higher than 110.
If this cipher be carefully examined with a view to discovering weak points in its structure, it will be found to consist in the use of a number of cipher alphabets in succession. Lik e the Gronsfeld cipher, it is of the polyalphabetical type, each of its several alphabets being formed from the original simple numerical alphabet by modifying the latter with one of the numbers of the secondary or variable key.
A Nihilist cipher written with a key word of ten letters thus would use a fixed series of ten cipher alphabets in a fixed order, and, in cipher parlance, would be said to have a period of 10.
The Nihilist cipher is, as far as safety is concerned, just about as valuable as a rowboat shot full of holes in the middle of the Atlantic Ocean. This cipher cannot hold a secret, for it is as leaky as a sieve.
Under favorable conditions it may be solved by the method of trial guessing explained in the article on the Gronsfeld cipher in FLYNN'S for June 6. Or it may be resolved by a general method for polyalphabetical ciphers worked out by a German major, F. W. Kasiski, and described in his book, " Die Geheimschriften und die Dechiffrirkunst," published in 1863.
Treatment by the Kasiski method consists first in the mathematical determination of the period, and then in the development of the several alphabets. All of these theories will be fully treated in later articles. But for the present we shall confine ourselves to a much more ready method, that will neither require as long a message as does Kasiski's, nor as much time in its application.
Now, to get down to brass tacks, this special method consists in the determination, first, of the number of letters in the" key (that is, of the period of the cipher); and, second, of the identity of each one of these key letters. This latter step is, of course, equivalent to the determination of each cipher alphabet.
By the Kasiski method a period is discovered by finding out the only one that it could possibly be. But by this method, as paradoxical as it may seem, it is determined by discovering what it is impossible for it to be.
This is what detectives would call identification by elimination. If it is absolutely certain, say, that one man out of a dozen has committed a murder, and eleven can prove their innocence, then the twelfth must be guilty, even though there may be no direct evidence against him. Jus t so, if we can prove that the period of a cipher is nothing other than four, for example, then four must be the guilty party.
By consulting the full description of the Nihilist cipher in FLYNN'S for March 28 you will find that the twenty-five numbers of the original alphabet, or primary key, are formed from the various combinations of the figures 1, 2, 3, 4, 5, and that the numbers of the final cipher result from additions of two of these figures.
This being the case, it is possible to determine when two numbers cannot have been enciphered by the same key number by the following simple rule: It is impossible for any two numbers whose units or whose tens differ by more than 4 to have been enciphered by the same key number.
In taking these differences a zero in the units place is counted as ten. And the only exception to the rule is when a number ends in zero. This always results from the addition of two fives, and causes the tens digit to be increased by 1. Consequently, in applying the rule, 1 must always be deducted from the tens figure of any number ending in 0.
To illustrate the full application of the method, nothing could be much more appropriate than to use it in solving one of the recently submitted Nihilist ciphers. And for this purpose the cipher of Thomas J. Sullivan, New York City, has been selected as especially fit in illustrating all of the necessary points.
For the purposes of this explanation, his cipher has been rewritten and numbered by fives, making it easy to tell at a glance the serial position of any particular cipher group:
(5) | (10) | (15) |
———— | ———— | ———— |
55-38-85-48-78- | 79-48-49-62-55- | 89-102-30-96-69- |
(20) | (25) | (30) |
———— | ———— | ———— |
79-47-42-84-67- | 85-48-93-50-78- | 57-46-64-48-74- |
(35) | (40) | (45) |
———— | ———— | ———— |
39-73-59-77-78- | 44-55-59-74-30- | 84-87-58-78-54- |
(50) | (55) | (80) |
———— | ———— | ———— |
B5-69-95-50-63- | 50-59-49-85-83- | 58-93-29-53-68- |
(65) | (70) | (75) |
———— | ———— | ———— |
48-58-74-76-33- | 94-39-86-79-57- | 58-74-4466-105- |
(80) | (85) | (90) |
———— | ———— | ———— |
50-86-80-56 70- | 42-74-43-84-58- | 87-70-69-68-73- |
(93) | (100) | (105) |
———— | ———— | ———— |
72-58-72-47-96- | 48-68-67-75-46- | 57-95-39-78-79- |
In the subjoined table you will find all of the data required to determine the period of this particular cipher.
For example, if the period is 1, this would mean that the key consisted of but a single letter, and that consequently every number in the cipher had been equally increased by the same number. The discovery of any two numbers not so enciphered would eliminate this period as a possibility. Two such numbers are given in the first line of the table.
Incidentally, a Nihilist cipher with only a single letter as the key would be equivalent to a simple substitution cipher, and could be solved by any method commonly used with such ciphers.
In the case of a period of 2, every second letter would have to be enciphered with the same key number. Th e discovery of any two numbers separated by an interval of 2, which, according to the rule, were not so enciphered, would eliminate this period. Two such numbers are (group 9) 62, and (group 11) 39, which have a unit's difference of 7.
All of the other periods should be similarly dealt with. The particular groups used in illustrating intervals 7 and 10 were selected to show the application of the rule to numbers ending in 0. Here you will see the tens differences tabulated as 5 and 6, respectively, where actually they are only 4 and 5.
INTERVAL | GROUPS FORMING | DIFFERENCES | ||
---|---|---|---|---|
THE INTERVAL | TENS | UNITS | ||
1 | (gr.2)—38: | (gr. 8)—85 | 5 | |
2 | (gr.9)—62: | (gr.11)—39 | 7 | |
3 | (gr.6)—79: | (gr. 9)—62 | 7 | |
4 | (gr.9)—62: | (gr.l3)—30 | 8 | |
5 | (gr.4)—48: | (gr. 9)—62 | 6 | |
6 | (gr.6)—79: | (gr.12)—102 | 7 | |
7 | (gr.6)—79: | (gr.l3)—30 | (5) | |
8 | (gr.3)—85: | (gr.ll)—39 | 5 | |
9 | —— —— | —— —— | —— | —— |
10 | (gr.3)—85: | (gr.13)—30 | (6) | 5 |
11 | (gr.1)—55: | (gr.l2)—102 | 5 | |
12 | (gr.1)—55: | (gr.l3)—30 | 5 | |
I3 | (gr.6)—79: | (gr.l9)—64 | 5 | |
14 | (gr.4)—48: | (gr.18)—42 | 6 | |
15 | (gr.8)—49: | (gr.23)—93 | 5 | 6 |
etc. | etc. | etc. | etc. | etc. |
It is now necessary to mention some facts that might escape your attention. Th e elimination of any number as a period also eliminates all factors of that number as possible periods. For instance, if it is found that the period cannot be 12, neither can it be 1, 2, 3, 4, or 6, all of which being evenly divisible into 12 are thus automatically disposed of.
But the elimination of a given number as a period does not eliminate the multiples of such a number. Thus, if it is found that 3 cannot be the period, it does not follow that 6, 9, 12, 15, et cetera, must on this ground also be rejected.
It is always good policy to test each elimination by two or three trials with other groups. Also it is best before going further to verify the interval not eliminated by thorough tests for positive results throughout the cipher. In the present case there will be found no instance where any groups separated by an interval of 9 provide a difference larger than 4, as per the rule.
Having rejected every period but 9, it may therefore be tentatively assumed that the period of the cipher is 9, because it is not found possible for it to be anything else.
Things now begin to look pretty black for Number Nine, don't they? So far, nothing but circumstantial evidence has been uncovered. But perhaps a little snooping may reveal something definite.
Proceeding upon the assumption that there are nine letters in the key, the next step is to learn just what these letters are. For this purpose the cipher is now divided into periods of nine groups each, and arranged so as to form nine columns. In this way all of the groups enciphered by any particular key number are brought together into the same column.
  | (1) | (2) | (3) | (4) | (5) | (6) | (7) | (8) | (9) |
---|---|---|---|---|---|---|---|---|---|
  | ——— | ——— | ——— | ———- | ——— | ——— | ——— | ——— | ——— |
  | 55- | 38- | 85- | 48- | 76- | 79- | 46- | 49- | 62- |
  | 55- | 30- | 102- | 30- | 96- | 69- | 79- | 47- | 42- |
  | 64- | 67- | 85- | 46- | 93- | 50- | 78- | 67- | 46- |
  | 64- | 48- | 74- | 39- | 73- | 59- | 77- | 78- | 44- |
  | 65- | 59- | 74- | 30- | 84- | 87- | 58- | 78- | 54- |
  | 83- | 69- | 95- | 59- | 65- | 50- | 69- | 49- | 85- |
  | 83- | 68- | 93- | 29- | 53- | 68- | 48- | 88- | 74- |
  | 76- | 36- | 94- | 39- | 86- | 79- | 57- | 58- | 74- |
  | 44- | 66- | 105- | 50- | 86- | 69- | 56- | 76- | 42- |
  | 74- | 48- | 84- | 68- | 87- | 70- | 60- | 68- | 73- |
  | 72- | 58- | 72- | 47- | 96- | 48- | 68- | 67- | 75- |
  | 46- | 57- | 95- | 39- | 76- | 79- |   |   |   |
  | ——— | ——— | ——— | ——— | ——— | ——— | ——— | ——— | ——— |
KEY: | 31-L | 14-D | 51-V | 15-E | 42-R | 35-P | 24-IJ | 24-IJ | 31-L |
  |   | 15-E |   |   |   | 25-K | 25-K |   | |
  |   | 24-IJ |   |   |   |   | 34-0 | 34-0 |   |
  |   | 25-K |   |   |   | 35-P | 35-P |   |
Now, since in the Nihilist cipher any figure is the result of the addition of two of the figures 1, 2, 3, 4, or 5, it is clear that the units or tens figures in any column can only be from 1 to 5 greater than the units or tens figure respectively of the key number for that column. Here the only exception is, as above, when a number ends in 0, its tens figure must be reckoned as 1 less than its actual value.
It may now be mentioned that in the Nihilist cipher 2 is always the result of the addition of 1 and 1; and 0 always represents the sum of 5 and 5. Similarly, 22, 30, 102, and 110 are always the doubles of 11 (A), 30 (E), 51 (V), and 55 (Z) respectively. The presence of any of these figures or numbers in a cipher, as in columns 3 (V) and 4 (E) of the foregoing cryptogram, will thus always materially simplify operations.
To illustrate the method of determining the figures of the key, take column 1. Here the units run from 2 to 6. The 2 alone is evidence enough that the unit figure of the key is 1; but the 6 is additional proof, in that it is not larger than 1 by more than five. The tens in this same column run from 4 to 8. Consequently the tens figure of this key number must be 3, since 4-5-6-7-8 are within the prescribed limits of that figure. The first key number is thus found to be 31 (L).
When the cryptogram is long enough, it is possible to determine all of the letters of the key in this manner. And, further, since the cipher is thus mathematically decipherable, it is unnecessary to know anything of the contents of the message in order to solve it.
The method would work just as well if the message chanced to be, say, in French, and the key word happened to be German. And the decipherer would not have to understand a single word of either language in order to find the key.
But in a shorter cipher, it is not always possible by this method alone to narrow each letter of the key down to but a single possibility. As a general rule, as the length of the message decreases, or that of the key increases, the greater becomes the number of mathematically possible numbers for each letter of the key.
Such a cipher is solved by trying all possible combinations of the various key numbers, in search of those combinations that provide logical sequences of letters both in the key word and in the message. Th e climax is reached when the key is as long as, or even longer than, the message itself. Even such a refractory case is, however, possible of solution by the method just mentioned.
In the present case more than one key number is mathematically possible in columns 2, 7, and 8. In column 2, for example, the first figure of the key might be either 1 or 2; and the second, either 4 or 5. This makes possible for this column four different key numbers: 14 (D), 15 (E), 24 (IJ), or 25 (K). Similarly, both columns 7 and 8 could be 24 (IJ), 25 (K), 34 (O), or 35 (P). That these two happen to come out alike is, of course, purely accidental.
The key, insofar as it has yet been determined, will stand "as shown in the above periodic arrangement of the cryptogram.
Poor old Number Nine, now just about completely enmeshed in the web of his own making, seems to be nearing the end of his rope.
We can almost hear the rapping of the gavel, and the stern voice of the judge as he thunders: "Number Nine, stand up, and receive the sentence of this court! "
But let's hang around a little longer. Maybe we'll see this suspect get it in the neck.
Of the several key combinations possible with the additional letters, that forming the word LIVERPOOL fairly shouts its presence. And by actual trial it is the only one that will provide plausible results in all parts of the cryptogram.
For example, the second letter of the key can be either E or I. But E can be rejected, however, since the key LEVERP-L, by trial with the first period of the cipher gives ICONOT--L as the translation, while LIVERP- L gives IDONOT--L ( I DO NOT--L, et cetera).
Again, the ninth period as deciphered by the key L-VERP--L comes out C-YPTO--A. Here the word intended obviously is CRYPTOGRA(m), thus determining with a single word all the doubtful letters of the key.
Here is Mr. Sullivan's message completely deciphered with the key LIVERPOOL: "I DO NOT BELIEVE YOU CAN SOLVE THE NIHILIST CIPHER WITHOUT THE KEY WORD, AND I SUBMIT THIS CRYPTOGRAM IN SUPPORT OF MY CONTENTION."
So now you have the method. Quite lengthy to describe, it is simplicity itself to use.
The periods may be represented by a simple row of numbers, each interval being crossed out as it is eliminated. The cipher may be then divided by pencil marks into periods of the determined length, and most of the work done mentally from the original cryptogram.
To discover the key word is often but the work of two or three minutes.
Here are some Nihilist ciphers upon which to try out your newly acquired method. They are selected from those sent in response to the Nihilist challenge.
And, by the way, fans, let's try now to break the record. About two hundred of you submitted solutions to the Nihilist cipher in FLYNN'S for March 28.
Armed with this exposure, more than this should succeed in solving one or more of the following ciphers without the key words!
Here goes!
CIPHER No. 1. This one came unsigned from Meriden, Connecticut. However, the sender had used his name as the key word.
104-50-46-73-76-85-36-86-49-89-64-50-58- 86-94-78-65-65-30-97-96-69-69-76-87-78- 43-96-49-99-66-50-28-66-77-77-37-84-39- 98-96-50-46-73-74-69-45-76-67-108-86-80- 29-66-56-78-66-53-48-68-76-87-39-73-73- 98-45-53-48-79-106-69-60-94-66-98-45-66- 49-108-64-59-46-73.
We're on, Bill ! You'll find what you are after in this article.
CIPHER No. 2. Neil C. Bierce, police headquarters, Portsmouth, New Hampshire, sends in this sample:
66-30-55-69-65-65-65-65-96-75-48-65-60- 82-26-86-88-94-57-30-47-60-74-46-59-47- 66-53-46-53-99-74-SS-78-76-66-57-58-34- 99-73-35-57-57-53-84-29-37-78-73-35-66- 57-86-53-46-65-79-75-34-59-77-86-76-57- 46-60-74-45-65-55-53-55-59-
This looks as though some one might have been watching the clock.
CIPHER No. 3. This one was submitted by James Veale, D. D., South Ozone Park, Long Island. Rev. Mr. Veale enjoys FLYNN'S and finds this department a storehouse of curious and interesting information.
34-55-89-56-48-85-37-67-67-77-36-53-64- 58-77-47-59-57-45-35-57-34-70-66-46-39- 89-57-60-84-26-48-70-46-30-84-24-66-79- 47-28-73-28-67-46-65-30-74-47-67-79-47- 48-86-28-66-50-66-59-66-46-46.
CIPHER No. 4. Here is a specimen from an enthusiastic fan, Fredolf A. Holmberg, of Augusta, Kansas:
59-26-74-68-57-77-30-75-59-78-59-57-58- 48-66-28-67-58-50-58-53-77-38-56-26-84- 59-69-48-57-96-47-57-28-56-68-47-30-73- 68-39-84-30-57-87-59-48-56-59-37-57-50- 56-59-77-26-54-75-39.
We would be interested to receive full details of the scheme mentioned in this cipher.
CIPHER No. 5. J. Fleming Jones, State Game and Fish Ranger, Okemah, Oklahoma, says:
97-79-80-47-53-78-78-57-74-76-79-38-57- 76-68-84-74-59-56-46-96-57-67-85-67-88- 59-48-53-37-65-53-86-88-47-69-66-68-77- 57-74-66.
And you can bet your last nickel that we will do as he suggests.
CIPHER No. 6. Henry Koester, Newburgh, New York, whose cipher follows, believes in more ways than one that his message is in desperate need of his key word.
63-55-26-86-38-47-26-54-47-60-84-58-48- 26-44-47-30-56.
The keys and solutions to all of the above ciphers will be found in next Solving Cipher Secrets.
The first of the Gronsfeld ciphers published in FLYNN'S for June 6 was enciphered by the same key used by Verne for the cryptogram in his Giant Raft; and the message was a translation from the original French of that same cipher. The second of the two Gronsfeld ciphers, using a longer key, in connection with a shorter message, was expected to be much more difficult of solution. Here are the keys and solutions of these two ciphers:
CIPHER No. 1. Key: 432513. Solution: "THE REAL AUTHOR OF THE ROBBERY OF THE DIAMONDS AND OF THE MURDER OF THE SOLDIERS WHO ESCORTED THE CONVOY, COMMITTED DURING THE NIGH TOF THE TWENTY-SECOND OF JANUARY, ONE THOUSAND EIGHT HUNDRED AND TWENTY-SIX, WAS NOT THUS JOAM DACOSTA, UNJUSTLY CONDEMNED TO DEATH; IT WAS I, THE WRETCHED SERVANT OF THE ADMINISTRATION OF THE DIAMOND DISTRICT. YES, I ALONE, WHO SIGN THIS WITH MY TRUE NAME, ORTEGA."
CIPHER No. 2. Key: 465812462. Solution: JULES VERNE WOULD NOT HAVE THOUGHT IT POSSIBLE FOR YOU TO HAVE DECIPHERED THIS WITHOUT A KEY.
Many polyalphabetical ciphers, such as the Nihilist and Gronsfeld ciphers, can be solved by a general method that does not require any information as to the particular system used.
When a cryptogram is believed, however, to be in some particular cipher, often a special method can be successfully applied. The method of trial guessing described in last Solving Cipher Secrets, while not as effective as some others, may often be used to advantage, either alone, or in combination with other methods.
And it is especially adaptable, as you have already, learned, to the Gronsfeld cipher. The solutions to the three Nihilist ciphers in the June 6 issue are intentionally withheld so as to give every one a chance to solve them by the method dealt with in this article.
The solutions to these ciphers will be published, however, in the next Solving Cipher Secrets, together with those of the Nihilist ciphers in this issue.
The solution to the cipher by Hyman Wacks, in FLYNN'S for May 16, is as follows:
YOUR LIFE IS IN DANGER. AVOID ME. POSTPONE ATTEMPT ON FIRST NATIONAL.
This is a simple substitution cipher in which the key is formed by using the fifth letter of the alphabet, E, as the substitute for A; the fifth letter after (namely, J ) for; the fifth after that, O, for C; and so on.
This cryptogram does not use all of the letters of the alphabet, but by determining the principle on which the key is based, the entire cipher alphabet, as given below, may be reconstructed:
Normal Alphabet: | A | B | C | D | E | F | G | H | I | J | K | L | M | N | O | P | Q | R | S | T | U | V | W | X | Y | Z |
Cipher Alphabet: | E | J | O | T | Y | D | I | N | S | X | C | H | M | R | W | B | G | L | Q | V | A | F | K | P | U | Z |
Here, fans, is a letter that doubly deserves your attention. For not only has it come a long distance, but it contains, as well, an ingenious cipher to test your skill.
DEAR SIR:
The following message is written in a secret code known to smugglers:
AH MING: YRT HSRKNVMG WFV HSRMBL NZIF. ED.
If this message is solved, or cannot be solved, please let me know the results. The readers of your magazine will have a hard time deciphering it.
S. HUTCHINSON.
Honolulu, Hawaiian Islands.
Here is an opportunity for you to benefit from your ability as a decipherer:
DEAR SIR:
I will give the money for one year's subscription to FLYNN' S to the first person solving the following cipher:
14 3 3 4 2 0 10 2 4 17 18 11 14 0 4 1 6 9 1 20 3 3 3 13 0 10 8 9 4 8 2 0 11 3 7 0 21 14 5 10 15 14 6 26 3 22 20 5 0 10 0 11 23 15 12 11 5 1 10 7 7 5 14 0 7 9 24 0 8 4 11 11 9 2 10 3 9 2 1 3 0 3 13 13 13 14 14 21 5 9 0 1 2 12 0 4 0 16 4 3 4 12 0 18 1 9 4 5 5 4 23 6 3 3 9 17 7 5 8 2 4 2 15 7 4 13 1 4 2 9
The only condition to this offer is that your solution must be received within one month from the date of the issue in which the cipher is published.
J. LEVINE, Long Beach, California.
Here is a cipher from Mr. L. W. Harke, Chicago, Illinois. It is not of extreme difficulty. Neither is it as easy as it looks:
O A B A E A E A A O A E B A B T O after E A E A H O B T 8 after D O A C A A D O B C after H B T 7 plus 1 0 7 plus 1 after 8 A I O 8 A A D after C A S O B T A B T B T A E after 4 then 4 O B after 4 O A H E A D B E O B B E XII after E A E A H O after 6 E B T A B C A B E.
RED.
The following interesting cryptogram is submitted by Mr. J. C. Bell, 16910 Endora Road, Cleveland, Ohio. Our correspondent is absolutely certain that his system is one which no living man can decipher. Here it is:
7-55-54-28-13-8-33-0-3-21-39-32-29-0-5-9-23- 13-0-9-52-20-2-X-35-0-6-23-53-32-14-42-17- X-0-8-19-26-23-25-53-5-0-4-49-21-40-16-0-2- 47-19-31-49-45-3-28-10-0-12-43-0-13-37-2- 31-XX-X-47-X-0-1-2-36-48-12-9-X-7-46-0- 17-14-46-9-0-14-15-41-0-11-1-28-40-34-12- 54-0-15-5-32-44-38-16-3-0-13-36-25-3-0-16- 39-27-0-20-8-17-2-26-12-X-31-0-18-44-15-0- 29-19-20-45-7-l3-30-55 11-30-0-17-0-15-22-23.
Mr. Bell says that until his cipher is solved he is going to stubbornly insist that it cannot be done. Try your hand, fans, and look for some more of these challenge ciphers at an early date.
Some more of the fans who submitted ciphers in response to the recent invitation to solve Nihilist ciphers will find their names and their key words listed here.
Another group will be published in the next Solving Cipher Secrets.