Elementary cryptanalysis

by Helen Fouché Gaines

the alphabet we have been using, the first alphabet in the imaginary tableau is position 1 of the slide: C U L P E R Z Y X W V T. . . . . . , the second alphabet is position 2 of the slide: U L P E R Z. . . . . . . , the third alphabet is position 3 of the slide: L P E R Z Y. . . . . , and so on to the 26th alphabet, which is the final position of the slide: A C U L. . . . . But over and above this, it must be remembered that the columns of this imaginary tableau are duplicates of the rows, just as they are in the Vigenère tableau. We do not, of course, recover any of these alphabets in the order mentioned, since our plaintext alphabet of the key-frame must necessarily be arranged in its a-b-c order. For instance, the fourth alphabet, which, in the imaginary tableau, begins with its key-letter, P, and runs in the order P E R Z Y X W V T. . . . . , comes out in one of our examples (Type III slide) as L U P C Y A B D. . . . . , and in the other (Type IV slide) as E W Y P V T S. . . .

The Type III slide is, in many respects, the most interesting member of its family. With every alphabet taking exactly the same order (that is, the plaintext alphabet, the key-alphabet, and all cipher alphabets in the imaginary tableau), it parallels the Vigenère in every particular except the order of the 26 letters. It has a corresponding Beaufort form, and a corresponding variant in which complementary keys are based on the order of the mixed alphabet. Its 14th alphabet, like the N-alphabet of the Vigenère, is reciprocal throughout. And its first alphabet, like the A-alphabet of the Vigenère, is a duplicate of the plaintext alphabet. This was pointed out in connection with the slide of Fig. 138, where key-letter and index-letter were both C. There are two ciphers, then (the Type III slide and the Delastelle tableau), in which we are sometimes able to find, among a number of mixed

Figure 145 The Alphabets from a TYPE III Slide: Behavior of an EVEN-NUMBERED Alphabet 1st Alphabet (Always normal): 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 4th Alphabet: L U P C Y A B D F G H R I J K Z M X N O E Q S T V W For an EQUIVALENT SLIDE, follow the chain AL-LR-RX-XT-TO...... A L R X T O K H D C P Z W S N J G B U E Y V Q M I F L R X T O K H D C P Z W S N J G B U E Y V Q M I F A (1)........... ═ ═ ═ (2)................. ═══ ═══ ═══ ═══ ═══ To find the ORIGINAL SLIDE from the EQUIVALENT one: (1) Either take letters constantly at interval 9, which is the interval V-W-X: R Z Y X W V T S Q O N M K J I H G F D B A C U L P E (R) X W V T S Q O N M K J I H G F D B A C U L P E R Z Y (X) (2) Or: Spread the letters apart so that the alphabetical sequences K(JI)H, Z(YX)W, etc. are standing at the right interval, always maintaining the alphabet-length, 26, and intertwine. Both alphabets are the same in this slide: (The interbals A . . L . . R . . X . . T . . O . . K . . H . . D . are always odd, . C . . P . . Z . . W . . S . . N . . J . . G . . B 3, 5, 7, etc.) . . U . . E . . Y . . V . . Q . . M . . I . . F . . Behavior of an ODD-NUMBERED Alphabet 1st 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 3d Alphabet: U C L A Z B D F G H I E J K M R N Y O Q P S T V W X 1st HALF-CHAIN: 2d HALF-CHAIN: A U P R Y W T Q N K I G D (A) B C L E Z X V S O M J H F (B) Spread the letters of each half, trying interval 2, interval 4, interval 6, and so on, treating both halves alike, until the intertwining of the two will set up some alphabetical sequences: . A . U . P . R . Y . W . T . Q . N . K . I . G . D B . C . L . E . Z . X . V . S . O . M . J . H . F .

frequency counts, a single one which follows perfectly the graph of the Vigenère A-alphabet. Concerning the 14th alphabet, however, we are dealing altogether, here, with a 26-letter alphabet; and some of what follows is being explained on the theory that the number 26 contains no factors other than 2 and 13. If the student will give his careful attention to reasons, as well as to methods, he will be able to adjust these methods to alphabets of other lengths, as, for instance, the very common 25-letter alphabet met with in foreign texts. The first alphabet, of course, duplicates the plaintext alphabet regardless of what alphabet-length is being considered, and thus, whenever a Type III slide has been used, we are always in full possession of one of the cipher alphabets.

Now, granting that we have completed the decryptment of a message, we have before us a key-frame in which several cipher alphabets are at least partially recovered. With one alphabet fully known in advance, the recovery of another full alphabet usually enables us very quickly to restore the original slide, or an equivalent slide. The ideal case is that in which we recover one of the even-numbered alphabets (except No. 14). The recovery of an odd-numbered alphabet will, at times, leave us with thirteen possibilities; while the recovery of the 14th alphabet could be useless, provided we have no other information. The method of reconstruction can be followed in Fig. 145.

First, we have the perfect case, one in which an even-numbered alphabet (the 4th) has been recovered in full. We begin by writing this recovered cipher alphabet letter for letter below (or above) the one which is always known to us; thus the two substitutes for a are in a same column, the two substitutes for b are in a same column, the two substitutes for c are in a same column, and so on. The columns themselves are not in their original order, but the two alphabets, throughout, are running parallel, just as they would in the imaginary tableau, and thus the columnar distance is uniform which separates each pair of substitutes; that is, the vertical distances AL, BU, CP, DC, EY, etc., are all equal in the imaginary tableau. If these be rearranged in such a way that the last letter of each pair is the beginning letter of the next, we have a chain AL-LR-RX-XT-TO-OK. . . . . made up entirely of equal vertical intervals, from which the repeated letters may be dropped: A L R X T. . . . . , leaving us a series of 26 letters known to be equally spaced in the columns of the tableau. Then, remembering that the columns of this tableau are duplicates of its rows, we have also a series of 26 letters known to be equally spaced on the rows. That this series of letters, A L R X T. . . . , sliding against itself, produces exactly the results of the original series, the student may ascertain for himself; also that a number of other equivalent slides can be formed by taking letters of this series at any constant interval which is not divisible by 2 or 13. The total number possible is eleven, of which one was the original. An equivalent slide, of course, is all that we actually need for enciphering and deciphering cryptograms. But where alphabetical sequences existed in the original alphabet, two methods are shown for obtaining it without writing out the entire eleven possibilities: (1) Find, at some constant interval, the letters of an alphabetical, or nearly alphabetical, sequence, as the (reversed) V W X of the figure, standing at interval 9; the taking of all letters at this interval brings back the original order. (2) Find pairs of consecutive letters, as the (reversed) HK, WZ, GJ, which, if all spread apart to the same extent (some odd interval, as 3 of the figure), would then be standing at their normal alphabetical intervals, or nearly so. Lay out the 26 positions, and spread the entire alphabet, maintaining the common interval even after the 26th position is reached. The figure shows this on several rows; in practice, there is only one.

If the recovered alphabet is an odd-numbered one, the same plan is followed, but results in a chain of only 13 letters; it is necessary to begin with some other letter, not included among the first 13, and form another 13-letter chain. Having absolutely no additional information, we cannot combine these two halves with certainty unless the original alphabet contained some alphabetical, or nearly alphabetical sequences. Presuming that it did, the method ordinarily described for combining the two halves is that of the figure. Spread the letters of the two halves (plan 2 of the preceding case), treating both halves exactly alike, until a point is found at which the two halves can be intercombined to show alphabetical sequences.

For this case, however, George C. Lamb, the author of [Chapter X], suggests another plan which would seem to be more direct and less troublesome than the standard one. Lamb, incidentally, is to be congratulated here for his entirely new observation: If the two half-chains can be properly adjusted with reference to each other, each pair of letters, regardless of the order, will be a digram belonging to the original mixed alphabet. The reason for this division into halves, of the odd-numbered alphabets, is probably self-evident: One half contains only odd letters (1-3-5-7-9. . . . . . . . .) and the other contains only even letters (2-4-6-8-10. . . . . . .). If both halves were recovered in this order, and if one half were written directly below the other with letters 1-2 standing together, then the other pairs would also be standing together: 3-4, 5-6, 7-8, and so on. We seldom recover them in straight order; but whatever rearrangement has taken place in one of the halves has taken place, also, in the other half; should one be recovered with letters in the order 1-7-13-19-25-5-11. . . . (each third letter in a series 1-3-5-7. . . . .), then the other will be recovered with letters in the order 2-8-14-20. . . . . . (each third letter in a

Figure 146 Another Method for Combining the two Half-Chains of an Odd-Numbered Alphabet (Originated by GEORGE C. LAMB) With a Type III slide, based on the key-word EXCORIATE, the 7th alphabet, as recovered from a cryptogram would come out as shown: H K B L A M N P........ 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 H K B L A M N P G Q S U V W D Y Z F E J X C O T R I The second half-chain, started with D, must be re-adjusted so as to plece in correspondence the alphabetical sequences PQ, YZ, FG, MN, etc. 1st HALF-CHAIN: A H P Y R F M V C B K S E 2d HALF-CHAIN: (d l u x) T J Q Z I G N W O D L U X Each corresponding pair of letters was a digram in the original cipher alphabet. Taking some two letters, as FG, which form an alphabetical sequence, look for another pair, such as HJ, which may be its continuation. HJ having been found at the interval 9, try taking pairs at the interval 9: FG HJ KL MN PQ SU VW YZ EX CO RI AT BD (FG)

series 2-4-6. . . . . .), though neither half necessarily begins with the first letter of its series. If, then, we are able to place together letters 1-2, the other pairs will also be adjusted, perhaps in the order 1-2, 7-8, 13-14, 19-20, and so on. These pairs may then be taken at some regular interval and will bring back the original order 1-2, 3-4, 5-6, and so on.