| Figure 150 1. A F S X O S G Y F O N P Y O A K O A D G F Z K S Z O Y Z Y L A W A C F. 2. H O U A L H L E D H D L Y G A V D W A K. 3. K O N B K A X U O N H I Q P L B A Z F F S Y F D R R L Y F. 4. T B A A M A F Q E Z U M A I X G F S K B. 5. D K O A C Y B Y E N I M O W D L E G A D O H C Y U U R G. 6. T B B X T O M M D A S I A A Y D Z. 7. O U S U B U L O I Y G A K X M A K W E L. 8. A K R U W A N A L O N N F M S K A X E U. General Frequency Count: 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 25 8 3 10 6 11 7 5 5 . 11 10 7 7 15 2 2 4 8 3 10 1 5 6 12 6 Frequency Count on Alphabet 1, Only: 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 6 5 1 1 . 1 . 1 . . 4 1 1 1 5 . . 1 2 3 4 . 1 2 . . |
not possibly be its own substitute, can be assumed, by frequency, as a, rather than i.
These first correct substitutions are all shown on the left side of Fig. 151, on the lines marked (a). Surely the next identification would be the m of seems, and probably, too, the l of settl. . . With the vowel a already identified, the repeated OU would be tried as an, and the repeated KO as ha, using the digram list. These are the substitutions marked (b), and from this it is but a step to the assumptions marked (c). On the right side of this figure, we are proceeding into alphabet 2. A frequency count here has shown that the leading letters of this alphabet are A, O, Y, two of which, A and O, were also leaders in alphabet 1. It is one peculiarity of the “Phillips” cipher that a change in alphabets means a change in only fifteen of the substitutes, the remaining ten continuing to represent the same originals as in the preceding alphabet. Concerning A, we can see, from the third and fourth messages, that it has not continued to represent t; but O, in the sixth message, has rather suggested the word all and even the expression all right, which would carry us on into the third alphabet. From this point onward, then, we are in the same fortunate position as the decryptor who intercepts his message partly in cipher and partly in plaintext. With the context as a guide, we need not worry as to what happens at the ninth group.
Presumably, during all of this time, we have been recording substitutes in a
| Figure 151 First Alphabet Second Alphabet 1. A F S X O S G Y F O ... (a) t . . . a . . . n a (b) t . . m a . . o n a (c) t r y m a . . o n a (Try ma...) 2. H O U A L H L E D H ... (a) . a . t . . . . . . (b) . a n t . . s u r . (c) c a n t . e s u r e (Can't be sure...) 3. K O N B K A X U O N ... (a) . a . e . i . . a . (b) h a . e h i m . a t (c) h a v e h i m . a t (Have him ...) 4. T B A A M A F Q E Z ... (a) s e t t . i . g . a (b) s e t t l i n g u . (c) s e t t l i n g u p (Settling up ...) 5. D K O A C Y B Y E N ... (a) . . a t . . . . . . (b) . h a t . o y o u t (c) w h a t . o y o u t (What do you t...) 6. T B B X T O M M D A ... (a) s e e . s a l l . i (b) s e e m s a l l r i (c) s e e m s a l l r i (Seems all ri...) 7. O U S U B U L O I Y ... (a) a . . . e . . a . . (b) a n . n e . s a . o (c) a n y n e w s a . o (Any news abo...) 8. A K R U W A N A L O ... (a) t . . . . i . i . a (b) t h . n . i t i s a (c) t h i n k i t i s a (Think it is a...) Frequency Count on Alphabet 2, Only: 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 5 1 . 2 3 2 1 2 1 . . 3 2 3 5 . 1 . 1 . 2 . . 1 4 1 |
key-frame. We have recovered fifteen of these in alphabet 1, and also a number in alphabet 2. Those recovered from alphabet 1 are shown at the top of Fig. 152. O is the substitute for a, A is the substitute for t, T is the substitute for s, and S is the substitute for y. Thus, if the cipher is “Phillips,” then, in the original key-square, the two letters A O were consecutive on one of the diagonals, the two letters T A were consecutive, the two letters S T, and the two letters Y S, so that the complete diagonal must have been Y S T A O, and even though the letter o was not used at all in alphabet 1, we know that its substitute must have been Y, since these diagonals may be considered to begin with any one of the five letters. By beginning at c-H, and following out another such chain, we may find another complete diagonal, C H K W D; and, in addition, we may find parts of diagonals. All of these are shown at (b).
Whether or not we can go further than this, without consulting other cipher alphabets, depends upon whether or not the original key-square contained some of those alphabetical, or nearly alphabetical, sequences which so often betray the poorly-mixed alphabet; usually, these are most easily found toward the X Y Z end
| Figure 152 An Alphabet No. 1 - Taken from a Key-Frame: Plaintext: a b c d e f g h i . k l m n o p q r s t u v w x y z SUBSTITUTES: O . H . B . . K R . W M X U . . . F T A . N D . S . (a) A T S Y Y O A T S S T A (b) O *Y C E I L *V S H B R M N T K F *X U A *W O D (c) (d) (e) N D .│S│ S . N D . S I N D E . N D . S I N D E S U C│.│T . T . U C B T R U C T . U C . T R U C B .│H│L A H L A . . H L A F . L A . . H L A F . H │.│K M O . K M O . . K M O . K M O . . K M O . . │ │V W X Y . V W X Y . V W X Y V W X Y . V W X Y . |
of the normal alphabet. In the given case (b), we are able to find the letters V, W, X, and Y, each standing on a separate diagonal; thus, by readjusting the beginning-points of their diagonals so as to place these letters at the bottom, we are able to set together four of these diagonals in the order shown at (c), leaving only the part-diagonals E B and I R F still to be added. Their length will show that I R F belongs to the missing diagonal, but E B, by its length, could belong to any one of three diagonals. Further developments can be carried out with the rhomboid adjustment of (c), or, if this is confusing, the conversion to a square can be made immediately. The student may decide for himself which he prefers of the two developments marked (d) and (e). Notice that this restoration of the key-square can take place not only from a single alphabet, but with only 15 substitutes known in that alphabet. But without the aid of alphabetical sequences, we must, in the first place, have 20 substitutes, four for each diagonal, in order to recover the full diagonals, after which, each one is entirely independent of the other four, so that they cannot be adjusted and combined without consulting one or more of the other alphabets. Here, the method varies a little, according to just what we can recover, though a hasty glance at the perfect case will serve to show the general path for all. To see this as rapidly as possible, we will assume that we have recovered from alphabet 1 the full five diagonals, and that, in alphabets 2, 3, and 4, we have discovered the substitutes for e, and also the originals for which E has been the substitute.
A careful consideration of the cipher itself will show that no letter can have more than four different substitutes: the four letters in the next column to its right which are not on the same line with itself. Also, that no letter may act for more than four different originals: the four letters in the next column to its left which are not on the same line with itself. Any letter, in order to take all four substitutes and act for all four originals in four successive alphabets, must have started on the top line, which is the moving one.