Lamb’s method, applied to an actual cipher alphabet, can be seen in Fig. 146. The first half-chain, if started at A, will include the letters B and C, so that the second half-chain would probably have been started at D. But AD will not be a correctly adjusted digram. It is necessary to look for one which forms an alphabetical sequence, as FG; and when the two halves are adjusted so that F and G are together, other alphabetical, or nearly alphabetical, sequences are also found to be adjusted, as MN, VW, BD, KL, making it likely that we have found some digrams belonging to the original mixed alphabet. With this adjustment reached, it is found that pairs can be taken at the constant interval 9, AT-BD-FG-HJ. . . . . , thus bringing back the original cipher alphabet.
Presuming that the original alphabet did not contain these alphabetical sequences, then there are thirteen possible adjustments for the two half-chains, and any one of these could be the original alphabet (or its equivalent). But it must not be forgotten that in an actual case our key-frame always contains portions of other cipher alphabets; and if the foregoing principle has been well understood, it may be readily seen how we could make use of these in order to determine which of the thirteen possible adjustments is correct. Even the recovery of the 14th cipher alphabet (which results in thirteen 2-letter chains), would not be useless with this other information present always in every key-frame. The Type III slide, in fact, can often be reconstructed without possessing any fully recovered cipher alphabet. This cipher is very popular with members of the American Cryptogram Association, and is usually known, for no very good reason, as “the Quagmires cipher.”
In the case of the Type IV slide, we do not begin reconstruction with one complete cipher alphabet already in our possession. It becomes necessary that we recover two, the perfect case being that in which one is an odd-numbered alphabet and the
| Figure 147 The Alphabets from a TYPE IV Slide: Plaintext letters: 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 (1) 4th Alphabet: E W Y P V T S Q O N M K R J I H G F Z D X B A C U L (2) 7th Alphabet: Y S V Z Q O N M K J I H X G F D B A W C T U L P E R A CHAIN Started with ab ab ys vd qx nt jp gl bi sf du xm tz pw lr io fk uh me zc wa ry ov kq hn ej cg (ab) EW UZ BP GC JD NH SK WO ZT PX CR DL HA EF OI TM XQ RV LY AE FU IB MG QJ VN YS REARRANGEMENT of this CHAIN: ab bi io ov vd du uh hn nt tz zc cg gl lr ry ys sf fk kq qx xm me ej jp pw wa EW WO OI IB BP PX XQ QJ JD DL LY YS SK KF FU UZ ZT TM MG GC CR RV VN NH HA AE A Reconstructed EQUIVALENT Slide: Plaintext: a b i o v d u h n t z c g l r y s f k q x m e j p w CIPHER: E W O I B P X Q J D L Y S K F U Z T M G C R V N H A ORIGINAL Slide, Found by Taking Letters at the Interval 5: b u c s m a d z y x w v t ....... W X Y Z R E P L U C A B D ....... |
other an even-numbered one. We will follow this case in Fig. 147, where the two recovered alphabets are Nos. 4 and 7. This tableau, like the preceding one, has columns which are duplicates of its rows, and to see our preceding case again (with its one modification), let us begin by looking only at the three alphabets immediately below the heading. One of these, the plaintext alphabet, shown in lower-case letters, we will disregard for a moment, giving our attention only to the two cipher alphabets.
These two alphabets, like the two from the Type III slide, are running parallel in the imaginary tableau, so that we have, as before, a series of 26 vertical distances, EY, WS, YV, and so on, all known to be equal in the columnar direction and therefore known to be equal distances on any row. A chain may be started, exactly as in the other case, EY, YV, VQ, QM, MI. . . . . , resulting in a series of equally-spaced letters E Y V Q M I F A L R. . . . . . which is either the original cipher alphabet, or the original one with letters taken at some odd interval other than 13. It is, however, only the cipher alphabet; the mixed plaintext alphabet must still be found. This may be done, as in the case of the Type I plaintext alphabet, by using either of the two cipher alphabets which were first recovered and setting originals above their substitutes. If this is done with our cipher alphabet in the order E Y V Q M I F A L R. . . . . . , then the plaintext alphabet comes out in the order a c e h k o r w z m. . . . . . , and we have an equivalent slide. If we first rearrange the sliding alphabet (each 9th letter of the series E Y V Q. . . .), we obtain the plaintext alphabet also rearranged.
Continuing, now, with the rest of our figure: The method we have just seen was based on a tableau, and our equal intervals were all vertical. In the figure, we are dealing purely with horizontal distances, and our method is based, not on a tableau, but on a slide (as it was with the Type II). Our 4th and 7th (secondary) cipher alphabets, after all, are merely two different positions of the same slide. If we select any two letters, as a and b of the plaintext alphabet, and find that their substitutes are, respectively, E and W, in alphabet 4, then the lineal distance ab in the stationary alphabet must be exactly equal to the lineal distance EW in the sliding
| Figure 148 Some EXERCISES in the RECONSTRUCTION OF ALPHABETS Plaintext ....... 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 Exercise 1: Q * Z A X B O C N * E R F P V G * Y M U I * W * T L Exercise 2: U V D W S X K Y H Z C F R J Q L I N G P T O M E A B Exercise 3: H J G K F P E Q O R S T D M B U V W X A Y Z C L I N Exercise 4: (1) V N U X J Y Z D Q E M P O W C K R I A T L S B F G H (2) H S G J R K L N F P Q B U I V A W C X Y T Z D E M O Exercise 5: (1) G X Y Z M H A F T R L K E V Q U O J W I P N S B C D (2) E * G J I K * L B * * U T C V W * Q D X S * * * * * The keywords, all selected by Mr. A. F. SEMPER, do not contain any repeated letters. The TYPES of slide, respectively, are: I, III, III, IV, and III. In the 5th exercise, the completely recovered alphabet is a "number 14." (See also [practice-cryptogram No. 46]). |
one; if this were not true, the letters could not have coincided as they do. Then, if we find the same substitutes, E and W, in alphabet 7, and note that, in this position of the slide, they have coincided, respectively, with plaintext letters y and s, then the distance EW in the sliding alphabet must be exactly equal to the distance ys in the stationary one. It follows from this that ys and ab are equal in the stationary alphabet. If we begin again with the lineal interval ys, we find that this is equal to UZ of alphabet 4, and that UZ, found again in alphabet 7, is equal to vd. Here, then, is another interval, vd, which is equal to both ab and ys. And so we may continue, forming a chain made up of these known equal intervals, ab, ys, vd, qx, etc., for the plaintext alphabet, and EW, UZ, BP, GC, for the cipher alphabet. Sometimes we return to ab (EW) without having included all 26 letters, and in that case (unless the number of letters included is a divisor of 26), it becomes necessary to abandon ab, and try starting with some other interval, as ac.
Here, as in the case of the Type III slide, we sometimes obtain two 13-letter chains; but in the fortunate case of having recovered in full both an odd-numbered and an even-numbered cipher alphabet, we end the chain with every letter represented twice, both in the stationary alphabet and in the sliding one. Pairs of letters (representing horizontal intervals) can then be rearranged as in the other case, the second letter of one becoming the first letter of the next (in both series), and the dropping out of duplicated letters gives us an equivalent slide. In the figure, starting with ab (EW), we find a plaintext alphabet a b i o v d u. . . . . and a cipher alphabet E W O I B P X. . . . . . This, as mentioned, is one of eleven possible equivalent slides, of which one is the original. Here, the original can be found by taking letters at interval 21. In the figure, letters were taken at interval 5, a result of observing the sequence W X Y Z standing at that interval in the lower alphabet, and the slide comes out in reverse order. This is still an equivalent slide, and the decryptor may or may not care to decide which alphabet runs backward.