Page:The Mystery of the Sea.djvu/480

466 bols used; there would still be required five for each letter to be infolded. We have therefore to try another process of reduction, that affecting the variety of symbols without reference to the number of times, up to five, which each one is repeated.

Take therefore the Baconian Biliteral and place opposite to each item the number of symbols required. The first, (aaaaa) requires but one symbol "a," the second, (aaaab) two, "a" and "b;" the third (aaaba) three, "a" "b" and "a;" and so on. We shall thus find that the 11th (ababa) and the 22nd (babab) require five each, and that the 6th, 10th, 12th, 14th, 19th, 21st, 23rd and 27th require four each. If, therefore, we delete all these biliteral combinations which require four or five symbols each—ten in all—we have still left twenty-two combinations, necessitating at most not more than two changes of symbol in addition to the initial letter of each, requiring up to five quantities of the same symbol. Fit these to the alphabet; and the scheme of cipher is complete.

If, therefore, we can devise any means of expressing, in conjunction with each symbol, a certain number of repeats up to five; and if we can, for practical purposes, reduce our alphabet to twenty-two letters, we can at once reduce the biliteral cipher to three instead of five symbols.

The latter is easy enough, for certain letters are so infrequently used that they may well be grouped in twos. Take "X" and "Z" for instance. In modern printing in English where the letter "e" is employed seventy times, "x" is only used three times, and "z" twice. Again, "k" is only used six times, and "q" only three times. Therefore we may very well group together "k" and "q," and "x" and "z." The lessening of the Elizabethan alphabet thus effected would leave but twenty-two letters, the same number as the combinations of the biliteral remaining after the elision. And further, as "W"