Here, the cryptogram, or a substantial portion of it, would be written across a sheet of quadrille paper, and the probable word would be written at one side, where each of its letters will govern one row of decipherments. The first letter, S in the figure, has been used to decipher the whole row of cryptogram-letters, giving every possible key-letter which can produce S. The second letter, U, has been used to decipher them all again (except the very first letter; we do not expect a word UPPLIES). The third letter, P, has been used to decipher them all a third time; and soon. The resulting rows of decipherment include all key-letters which could have produced S, then U, then P, and so on. To read them consecutively, beginning at any cryptogram letter, start immediately below that letter, and read diagonally downward to the right. The first diagonal gives key CYK. . . , the second gives AFS. . . , and so on to the fifth diagonal, showing the key as T C O M E T C O. (If it is desired that these possible keys should come out standing in a horizontal position, then the decipherments may be made diagonally.) F. R. Carter, the originator of this scheme, does not necessarily make all of the decipherments which are included in the figure. He begins with the assumption that his key will be a recognizable word; having deciphered in full the first three rows, he abandons all of those diagonals which cannot develop into words. If, in the end, he is forced to conclude that his key was incoherent, no decipherments have been erased; he may still go back and develop the rest of his diagonals, in the hope that one will begin repeating.
The more difficult of our two cases, that in which we have no probable words other than the, and, which, that, have, but, etc., can follow exactly the routine
| Figure 90 Deciphering with the Probable Word SUPPLIES - Routine of F.R.CARTER Cryptogram fragment: .... U S Z H L W D B P B G G F S ......... Probable word: S C A H P T/ E L J X J O O N A ......... U Y F N R C/ J H V H M M L Y ......... P K S W H O/ M A M R R Q D ......... P S W H O M/ A M R R Q D ......... L A L S Q E/ Q V V U H ......... I O V T H T//Y Y X K ......... E Z X L X C//C B O ......... S O// (Key: COMET) |
outlined in Fig. 90; but in this case there must be two separate work-sheets. Here, it is usually better to forget words and start at once with the list of normally frequent trigrams, THE, AND, THA, ENT, ION, TIO, etc. The key-fragments which are deciphered by these will be very short, and very numerous; a great many of them will be very good usable sequences, and perhaps the correct key-sequence will not look quite so inviting as others which are incorrect. It becomes necessary, then, to have a second work-sheet on which we may take these fragments one by one and try them as keys. If any one of them is a fragment of the original key, it must bring out fragments of plaintext, and must bring them out at some regular interval. If the scheme of Fig. 90 is the one preferred, the second work-sheet may be prepared exactly like the first, and used in the same way. The only difference is as follows: On the first work-sheet, where the figure shows the word SUPPLIES, a supposed trigram (THE, AND, etc.) will have been used to bring out supposed key-fragments; on the second work-sheet, one of these supposed key-fragments will have been used. These new rows of decipherment may then be examined to find out whether any of the new diagonals contain apparent plaintext fragments, and, if so, whether these occur at a regular interval.
For this kind of work, however, Ohaver has offered us another routine which requires somewhat more preparation than Carter’s but which is well worth the extra trouble, especially if it be remembered that a trigram-search is never necessary except with the shortest of cryptograms. For the longer cryptograms, we have easier methods. Ohaver’s plan can be examined in Fig. 91.
The cryptogram, shown at the top of this figure, contains 26 letters; therefore, remembering that each letter, except the final two, may begin a cipher-trigram, it contains 24 trigrams. The preparation of the two work-sheets requires that these 24 cipher-trigrams be written out in full on both sheets. This work should be done in ink, or on the typewriter. Then, too, for a reason which will be explained in a moment, it is well that the first of these work-sheets be prepared with a great deal of space, say seven or eight lines, between its rows of trigrams. Now, considering the first work-sheet, shown at (a) of the figure: The upper row shows the 24 cipher-trigrams as originally written out. We have been working down the trigram list, using every normally frequent trigram as a trial key, and have failed to find THE, AND, THA, or ENT, which means that we have done quite a lot of tedious work. We have now reached the normally frequent trigram ION, and this we have
| Figure 91 L N F V E O L N V M R N G Q F H H R N H I R V F E B, (a) Trial Sheet No. 1ION LNF NFV FVE VEO EOL OLN LNV NVM VMR MRN RNG NGQ AZS FRI XHR NQB WAY GXA DZI FHZ NYE EDA JZT FSD GQF QFH FHH HHR HRN RNH NHI HIR IRV RVF VFE FEB YCS IRU XTU ZTE ZDA JZU FTV ZUE ADI JHS NRR XQO (b) Trial Sheet No. 2EDA LNF NFV FVE VEO EOL OLN LNV NVM VMR MRN RNG NGQ HKF JCV BSE RBO ALL KIN HKV JSM RJR ION NKG JDQ GQF QFH FHH HHR HRN RNH NHI HIR IRV RVF VFE FEB CNF MCH BEH DER DON NKH JEI DFR EOV NSF RCE BBB (c) Testing out the Period 5 D A E D A E D A E D A E D A E D L N F V E O L N V M R N G Q F H H R N H I R V F E B I N . . A L L . . I O N . . B E H . . D F R . . A Y (TION?) (FRIDAY?) |
applied as a trial key, assuming one by one that each of the 24 trigrams represents ION. We have, then, 24 decipherments on the second row, and any one of these 24 deciphered trigrams might be a fragment of the original key. However, it is natural to assume that a trigram FRI or WAY is more likely than one such as XHR or NQB, and those fragments which look like usable sequences have been underscored in the figure. These are to be tested first. At (b), we have the other work-sheet, the upper row, as before, showing the 24 possible cipher-trigrams. Here, we have already failed in our tests for key-fragments FRI, WAY, DZI, NYE, which means that we have done some more tedious work, and we have now arrived at the possible key-fragment EDA. If this sequence, EDA, is actually a portion of the original key, it must not only bring out fragments of a plaintext message, but must bring them out at some constant distance apart. The point at which we found this is the tenth trigram, and here it may be advisable to remind that this begins at the tenth cryptogram letter; that is, every trigram presents only one new letter, so that to find a completely different trigram in either direction, we must count backward or forward a distance of three trigrams.
Beginning, then, at the tenth trigram, and examining every third trigram in both directions, we find that our key-fragment has given us the following decipherments: HKF, RBO, HKV, ION, CNF, DER, JEI, NSF. These are largely incoherent; but, in addition, it must not be overlooked that on the continuously-written cryptogram, these would be consecutive, giving us a message H K F R B O. . . Applied at interval 3, then, our key-fragment EDA, will not decipher us a message; therefore, the period of this cryptogram, using this key, cannot be 3.
To examine for the possibility of a period 4, we start again with our tenth trigram, and examine every fourth decipherment in both directions; our series, this time, is JCV, KIN, ION, MCH, NKH, NSF. Most of these are usable, and the first one might be due to nulls, initials, and so on; but here again we have the reminder that with each trigram representing only one new letter, these are almost consecutive, starting at the second cryptogram letter, so that our message, with each fourth letter missing, will be as follows: * J C V * K I N * I O N. . . . Unless we can think of some letters which would fill these gaps and provide plaintext, our period is not 4.