| Figure 102 A common happening in all PERIODIC ciphers: Vigenère key: C O M E T C O M E T C O M E T C O M Plaintext: T H E R E I S A N O T H E R Q U E S REPEATED SEQUENCE: V V Q V . . . . . . V V Q V . . . . |
which is twice the key-length, 5). This does not mean that all repeated sequences found in Vigenère cryptograms are periodic. Often, they are purely accidental; oftener still, they will be due to the repetition of alphabets in the key itself, especialiy if it is such a word as CORCORAN or DESDEMONA. But a distinct majority of them, according to Kasiski, are caused by periodicity; and if all of the repeated sequences found in a given cryptogram be examined to find what the separating interval is in each case, and if all of these intervals be factored, the factor which predominates will betray the period of the cryptogram.
In order to have a look at the Kasiski method, we will consider the cryptogram shown in Fig. 103; and, to approximate a more troublesome case, we will assume that no repeated sequences can be found except those few which have been underscored in the figure. With the cryptogram-letters serially numbered, in the manner shown, the distance apart of any two of them is readily learned by subtraction. The digram CH is found beginning at the 1st letter and again at the 46th letter; 46 minus 1 equals 45, their distance apart. Thus, if CH is one of the periodic repetitions, the period could be 15, 9, 5, or even 3. The trigram UBF, 8th and 63d letters, shows an interval 55; here, the period could be 11 or 5. Notice that a period 5 has been indicated by both.
Now let us look at Fig. 104, where the method of presentation is once more a debt to M. E. Ohaver. In this figure, the repeated sequences have been listed, and each one is accompanied by the two serial numbers of its two first letters, together with the interval which was obtained by subtraction. Ohaver’s process provides a column for each possible factor, beginning with 2 and carried as far as desired. Opposite each interval, its various possible factors may then be noted in their correct columns. In the average case, the correct period will be pointed out by the column showing the largest number of entries. But in this connection, it must be taken
| Figure 103 5 10 15 20 25 30 C H G S L F A U B F X U P H S J D A G Y X M N Z U W W J P D 35 40 45 50 55 60 J S U P L G C G F K R N I M F C H K O A Q A V X O N N U I L 65 70 75 80 85 90 N S U B F N D V P K A I P L S N M Q O H M E U I L B L K Q W 95 100 105 110 115 120 N D V I Y X U I I A Q E U U Y J W C O K O E N M P W W J J J 125 130 135 140 145 150 Q I U O V C M W D O X F C O L F S K U L V B W U N R V G T B 155 160 165 170 B S Q N L U E P H A Q T Q X V A K Q O E |
into consideration that small factors 2 and 3, and even factors 4 and 5, are usually present in considerable numbers, partly as accidentals, but also because they are factors of the period itself; that is, if the period is 6, there will surely be factors 2 and 3 for every factor 6, and there will usually be a few extra appearances due to accidental repetition.
Now, considering our tabulation, and ignoring the fact that short periods like 2 and 3 are seldom encountered, we find that factors 3 and 5 are present in equal numbers. Often, we are faced with exactly this problem. Here are two factors which have appeared in approximately equal numbers. Which one of these actually represents the period? Ohaver’s recommendations include these: Where two factors seem almost equally prominent, select the larger if it is a multiple of the smaller. If one factor is not a multiple of the other, try to select a period which is a multiple of both (as 15 here, includes both 3 and 5). He points out also that the factor which is the correct key-length will usually be accompanied, in the tabulation, by quite a number of its own multiples, growing gradually fewer and fewer as their size increases. In this respect, our factors 3 and 5 are both disappointing. If we consider factor 5, we find factors 10 and 15, but not growing fewer; instead the number increases. We find no factor 20, but we do find a factor 25; another increase. Or, if we consider factor 3, we find factors 6 and 9, but no factor 12, and then a sudden increase in the number of factors 15. This is a case in which the decryptor would play safe by selecting the period 15.
The student who cares to examine this matter more closely may do so by
| Figure 104 Tabulation for Finding Period M.E.OHAVER Repeated Positions - POSSIBLE FACTORS of INTERVALS Sequence Intervals C H 46 - 1 = 45 3 5 9 15 U B F 63 - 8 = 55 5 11 U P 33 - 12 = 21 3 7 21 S U 62 - 32 = 30 2 3 5 6 10 15 P L 73 - 34 = 39 3 13 W W J 116 - 26 = 90 2 3 5 6 9 10 15 18 N D V 91 - 66 = 25 5 25 The factors found in the largest number of DIFFERENT intervals are 3 and 5. |