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  APPENDIX B

  ON THE REDUCTION OF THE NUMBER OF SYMBOLS IN BACON'S BILITERAL CIPHER

  When I examined the scripts together, both that of the numbers and thoseof the dots, I found distinct repetitions of groups of symbols; but nocombinations sufficiently recurrent to allow me to deal with them asentities. In the number cipher the class of repetitions seemed moremarked. This may have been, however, that as the symbols were simplerand of a kind with which I was more familiar, the traces or surmiseswere easier to follow. It gave me hope to find that there was somethingin common between the two methods. It might be, indeed, that bothwritings were but variants of the same system. Unconsciously I gave myattention to the simpler form--the numbers--and for a long weary timewent over them forward, backward, up and down, adding, subtracting,multiplying, dividing; but without any favorable result. The onlyencouragement which I got was that I got additions of eight and nine,each of these many times repeated. Try how I would, however, I could notscheme out of them any coherent result.

  When in desperation I returned to the dotted papers I found that thismethod was still more exasperating, for on a close study of them I couldnot fail to see that there was a cipher manifest; though what it was, orhow it could be read, seemed impossible to me. Most of the letters hadmarks in or about them; indeed there were very few which had not.Examining more closely still I found that the dots were disposed inthree different ways: (a) in the body of the letter itself: (b) abovethe letter: (c) below it. There was never more than one mark in thebody of the letter; but those above or below were sometimes single andsometimes double. Some letters had only the dot in the body; and others,whether marked on the body or not, had no dots either above or below.Thus there was every form and circumstance of marking within these threecategories. The only thing which my instinct seemed to impress upon mecontinually was that very few of the letters had marks both above andbelow. In such cases two were above and one below, or _vice versa_; butin no case were there marks in the body and above and below also. Atlast I came to the conclusion that I had better, for the time, abandonattempting to decipher; and try to construct a cipher on the lines ofBacon's Biliteral--one which would ultimately accord in some way withthe external conditions of either, or both, of those before me.

  But Bacon's Biliteral as set forth in the _Novum Organum_ had fivesymbols in every case. As there were here no repetitions of five, I setmyself to the task of reducing Bacon's system to a lower number ofsymbols--a task which in my original memorandum I had held capable ofaccomplishment.

  For hours I tried various means of reduction, each time getting a littlenearer to the ultimate simplicity; till at last I felt that I hadmastered the principle.

  Take the Baconian biliteral cipher as he himself gives it and knock outrepetitions of four or five aaaaa: aaaab: abbbb: baaaa: bbbba: andbbbbb. This would leave a complete alphabet with two extra symbols foruse as stops, repeats, capitals, etc. This method of deletion, however,would not allow of the reduction of the number of symbols used; therewould still be required five for each letter to be infolded. We havetherefore to try another process of reduction, that affecting thevariety 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 itemthe number of symbols required. The first, (aaaaa) requires but onesymbol "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 fivesymbols each--ten in all--we have still left twenty-two combinations,necessitating at most not more than two changes of symbol in addition tothe initial letter of each, requiring up to five quantities of the samesymbol. Fit these to the alphabet; and the scheme of cipher is complete.

  If, therefore, we can devise any means of expressing, in conjunctionwith 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 fivesymbols.

  The latter is easy enough, for certain letters are so infrequently usedthat they may well be grouped in twos. Take "X" and "Z" for instance.In modern printing in English where the letter "e" is employed seventytimes, "x" is only used three times, and "z" twice. Again, "k" is onlyused six times, and "q" only three times. Therefore we may very wellgroup together "k" and "q," and "x" and "z." The lessening of theElizabethan alphabet thus effected would leave but twenty-two letters,the same number as the combinations of the biliteral remaining afterthe elision. And further, as "W" is but "V" repeated, we could keep aspecial symbol to represent the repetition of this or any other letter,whether the same be in the body of a word, or if it be the last ofone word and the first of that which follows. Thus we give a greaterelasticity to the cipher and so minimise the chance of discovery.

  As to the expression of numerical values applied to each of thesymbols "a" and "b" of the biliteral cipher as above modified, suchis simplicity itself in a number cipher. As there are two symbolsto be represented and five values to each--four in addition to theinitial--take the numerals, one to ten--which latter, of course, couldbe represented by 0. Let the odd numbers according to their values standfor "a":

  a=1 aa=3 aaa=5 aaaa=7 aaaaa=9

  and the even numbers according to their values stand for "b":

  b=2 bb=4 bbb=6 bbbb=8 bbbbb=0

  and then? Eureka! We have a Biliteral Cipher in which each letter isrepresented by one, two, or three, numbers; and so the five symbols ofthe Baconian Biliteral is reduced to three at maximum.

  Variants of this scheme can of course, with a little ingenuity, beeasily reconstructed.