So that if the digram VI is present at all, the key-length must be 8, 9, 11, or something longer. Since the key-length 11 is possible in both cases, this is the one which tempts; when it fails, the remaining two can be tried. The student may decide for himself whether a trigram IVI is possible, considering the distance apart of the two I’s. It will be readily understood how this method, in combination with the one first explained, could be used, say, in a cryptogram where the suspected word is CIPHER, with the low-frequency letter P occurring only once.
Totally aside from analysis, there are many ways in which the key-length can become known, or suspected. If the correspondence is a military one, it may have been learned by espionage, perhaps through careless talk on the part of an enlisted man; or, because of careless habits on the part of the authority providing the keys, in having confined himself always to certain lengths. Knowing the key-length is two-thirds of the battle. It enables us, as in our former case, to mark off the cryptogram into its approximate column-lengths, making it easier to know the approximate whereabouts of any several letters supposed to form a sequence. It even enables us to prepare a block, which, cut apart to form paper strips, will effect a mechanical solution almost as easily as in the case of the completed unit.
Such a block, for our foregoing cryptogram (Fig. 30), can be studied in Fig. 38, and is explained as follows: An 8-unit key, used on a 75-letter text, calls definitely for three 10-letter columns and five 9-letter columns, and these columns have become eight segments in the cryptogram. If all three of the long columns were taken off first, then the arrangement shown at (a) has every letter in its proper column. And if all three of these were taken off last, then the arrangement shown at (b) has every letter in its proper column. With blocks shown for the two extreme cases, it can be seen that the block at (c) is a combination-block, in which one of
Figure 38
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the two extremes has been superimposed upon the other, so that every column in block (c) shows every letter which it could possibly have contained. By concealing the letters of the “cap,” we have a duplicate of block (a); and by changing the alignment, so as to bring all of the topmost letters into the same row, we have block (b), with a “cap” attached at the bottom.
Comparing block (c) with the two above it: If the first column of (a) was actually a short one, then its last letter, X, belongs at the top of the second column. The making of this transfer would cause the second column to have eleven letters, so that it would become necessary also to transfer the last letter of the second column to the top of the third; this third column would then have too many letters, and its last letter would have to be transferred to the top of the fourth, which at present has only nine and may have another. But if the second column was also short, then there are two of its letters which belong at the top of column 3. And if this column, too, was a short one, it has three transferable letters at the bottom.
To prepare such a block, first write the cryptogram as at (a), and mark off its transferable (uncertain) letters by the following rule: One for the first column, two for the second, and so on, until the number is equal to the number of long columns, which is the maximum number possible. But if the final row is more than half filled, the maximum will not be reached, and a check may be made by marking off letters from right to left: zero for the last column, one for the next-to-last, two for the third-to-last, and so backward to the number which equals the number of long columns. Having marked off the transferable letters, form the “bonnet” by copying these, in each case, at the top of the following column, preferably making some clear distinction to show the duplication. For this latter purpose, many solvers use red ink. In this kind of work, as we saw in a previous case, the spacing must be accurate both laterally and vertically, since many of the letters belonging to the same sequence are not found on the same row. A few of the strips cut from block (c) have been matched at (d), where the beginning was made from the common suffix -ABLE. The duplicated letters A H Y have shown up plainly, partly by the style in which the letters are written, and partly, too, by the fact of consecutive column-numbers, 3 and 4. This same thing is true of the letters X T N, column-numbers 4 and 5. These numbers, it must not be forgotten, are also the serial numbers of the cryptogram segments, and thus are the key-numbers. With the eight strips correctly matched, and any misplaced columns transferred to their own side of the block, the strip-numbers as they stand across the top will reproduce the numerical key.
The matching of strips is generally a purely mechanical process, in which impossibilities are not considered. However, having before us a block (a) or (b), it is possible to apply the principle used with our former digram VI, and find out in advance whether certain letters found on two strips can possibly have stood in sequence. Nor is the cutting apart of the strips really necessary; it is merely a convenient method for dispensing with mental effort.
Now suppose we consider this same cryptogram on the theory that its key-length cannot be determined, or restricted to certain possibilities. Our first step is to select, somewhere in the cryptogram, a segment which is to be set up vertically on a sheet of paper to act as a trial column. If we select it from the body of the cryptogram, we shall have to make it a rather long segment, since we are uncertain as to whether it represents one column or parts of two. We should do this, however, if the body of the cryptogram shows Q, or any other letter or series of letters likely to be vulnerable. Otherwise, we know definitely that one of the columns begins with the first letter of the cryptogram, and that another column ends with the final letter of the cryptogram, and one or the other of these two segments is usually chosen, preferably the one containing the largest number of vulnerable letters. If we have a probable word, and find that its letter P, or M, or G, is the only one in the cryptogram, we select the segment which contains this P, or M, or G.
| Figure 39 Tests, in Attempting to Judge Column-Length E G 20 ( 15) N G 75 ( 83) T E 94 ( 74) H O 46 ( 42) V I 19 ( 14) C C 12 ( 6) C R 7 ( 12) O L 17 ( 36) T B 14 ( 8) X X - ( -) R C 14 ( 7) E S 145 (115) M U 13 ( 7) U T 45 ( 35) S L 6 ( 9) (First column of figures is taken from Meaker's chart; the second is from Ohaver's). |