Now, considering Fig. 153: Our five recovered diagonals are imagined to be those of a well-mixed square, so that we have no discoverable sequences. It has
| Figure 153 Five Complete Diagonals Recovered From Any Alphabet No.1 Y C E I V S H B R N T K L F U A W M P G O D X Z Q D E S I N U C B T R F G H L A O P Q K M X Y Z V W |
been found that the letter e, in alphabets 1, 2, 3, and 4, has taken, successively, the substitutes B, H, Q, and Z, and that the cipher-letter E has served, successively, as the substitute for x, u, f, and o. The four letters B H Q Z, then, must all have stood in a single column in exactly the order named; and the four letters X U F O must have stood in another column, two positions to the left of the first, but with a minor difference in the order: some other letter (the one on the same line as E) must have intervened between X and U. The order in this column, then, must have been U F O X. Our first step toward combining the five diagonals is that of adjusting four of them so as to set up the column B H Q Z. This automatically sets up four letters of the other column U F * X (D) — in the figure, the X D is present, but has not been adjusted to the U F * — after which, the fifth diagonal can be added to the others by placing the O of the column U F O X (D). Now, since the letter E has taken, successively, all four of its substitutes and all four of its originals, it must, in alphabet 1, have been standing on the top row. Two parallel lines (if desired) can be ruled across the set-up to show the top and bottom of the square, and two others (placed anywhere, so long as they mark a width of five columns) may be ruled to show the two sides. The outside letters, Y, V, and D, may all be transferred to the opposite ends of their diagonals, after which the rhomboid is easily adjusted to the form of a square.
It will be seen from what precedes that group-by-group encipherment offers little, if anything, that is new, and no problems or theories which the student could not figure out with what he has learned of substitutions. Solving the actual cryptograms, of course, could present some difficulties, according to the individual case.
Of the practice cryptograms to follow, No. 140 follows the plan of the puzzlers, and should not prove very difficult in spite of its brevity. No. 141 is also a “Phillips,” while No. 142 has been enciphered with a mixed-alphabet slide applied to five-letter groups.
140. By NEMO. ("Phillips." Probable word: ASSOCIATION)
N D Q T F Q Z C N G B U Z H X N L U K Y F T E E W N R G U R M O X Q X
E Q Z L B G X H W F F N R P X P X V D D F I T G S E W R T I I T Z X E
R V W A R I S P E Y I G R Q C.
141. By PICCOLA. (This is the real McCoy - in 1938. But times change).
(a) K G E U H C K T S X P C K N C A D F X Q C B X T
(b) O U T U I U B F S B Q A P H N B Y Z X X L R U G
(c) O F U O S K H Q T K P W Q F E T B W W X P K B O K G H
(d) B L A M R P G X B W G C W K Q Z I A Q C U H Y R C
(e) G U C O S B B L P S B Q D K P G P K D S R C T B L I
(f) X R O S U I T T F G Y P C M C K F T F X O S R B H O A G M
(g) B O I B V B U K E E B D K B C O B Y W B T B M U H O O A B
(h) Y C U Y U T B I T F H S A N P H C W T.
142. By PICCOLA. (Direct examination - Snowball vs. Snowball).
(q) H F X L F M B L R N I N J W P Z Z G I S B B O Z X S F S H R
H T A T M R O F V ? (a) S X F U R R W X I Q S S.
(q) U F V H C N T I T T F O E J X O G N S G X U S O E H I V L X
E A T ? (a) U R Q Q T W E X U W I T O S.
(q) H C W R U Q U I T T F Y O Q I U D R S G X Z W H G F E T P C
J E M K Q F I N D O E E Z B L ? (a) U R Q Q T W.
(q) U R S U F U G J R E D V T O V E C Z X Z U Z G X S D S H C Q
K E X Z W I O V I R M H D W B D Z R R M ? (a) U R Q T D H T T H I
J S S.
(q) X C Q H R E M Z L T T O P A V H U L E B O Y O G N U F V X T
Z L E K S W F A V N ? (a) X R L L Q E W L T C O S P W V C Z L E B
D W Y I Z F I S R D T W C E Z T T A S G O L E.
143. By PICCOLA. (Can you guess what cipher? "Foregoing" refers to [No. 141]).
R N N G T R I O O H E I T T A F N D E N O G E L G E Y F I R D A I S E.
CHAPTER XX
Vigenère with Key-Progression
Before leaving the study of multiple-alphabet ciphers, we will consider briefly the process which, in its simplest form, would be that shown in Fig. 154. The initial key, in each of three examples, is A, and a long key has been formed by causing the initial one to progress in the normal alphabet according to an agreed index. In the first example, the progression index is 1, in the second, it is 2, and in the third, it is 25 (or minus 1). The resulting long key will govern a period, which is 26 or 13, according to whether the progression index is odd or even. This encipherment,