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*To*: rene@xxxxxxxxxx*Subject*: VMS words and Roman numerals*From*: Jorge Stolfi <stolfi@xxxxxxxxxxxxx>*Date*: Wed, 27 Dec 2000 23:42:17 -0200 (EDT)*Cc*: voynich@xxxxxxxx*Delivered-to*: reeds@research.att.com*In-reply-to*: <3A4A698E.4A60F050@voynich.nu>*References*: <3A4A0DBA.BC28F9CE@voynich.nu> <200012271855.eBRIt0J00982@coruja.dcc.unicamp.br> <3A4A698E.4A60F050@voynich.nu>*Reply-to*: stolfi@xxxxxxxxxxxxx*Sender*: jim@xxxxxxxxxxxxx

> [rene:] A base-60 system as used by the Babylonians and > understood (to the best of my knowledge) by later cultures is > one interesting possibility [to generate the factor `12'] that > comes to mind immediately. The base numbers could be: > 1,5,10,20,60,300,600,1200,3600 Hmmm, only 9, not 12. Hm, these could be the nine yes/no slots in the binomial part of the code. Indeed the Babylonians (and the Greek, Roman, Chinese...) used a digit-position code: with different sets of symbols for each position, omitting the zeros. Unfortunately, all but the Romans had several choices per slot; so the word length distribution for those numerals is not symmetrical. But your remark made me realize that the *Roman* system, unlike the others, is actually quite similar to the binary bit-position code, except that it allows multiple I/X/C letters. Here is the length distribution d_k for the Roman "digits" from 0 to 9, without the subtractive notation: k d_k words - --- ----------- 0 1 () 1 2 I V 2 2 II VI 3 2 III VII 4 2 IIII VIII 5 1 VIIII The length of a Roman numeral between 0 and 999 will be the sum of three variables, each with this distribution (one for each decimal position). With a couple of unix hacks, I computed the number R_k of distinct Roman numerals in 0-999 with each given length k: k R_k --- --- 0 1 (empty) 1 6 (I, V, X, L, C, D) 2 18 (II, VI, XI, XV, XX, LI,... DC) 3 38 4 66 5 99 6 128 7 144 8 144 9 128 10 99 11 66 12 38 13 18 14 6 15 1 (DCCCCLXXXXVIIII) This distribution is not quite a binomial distribution, but, thanks to the law of large numbers, it is not very far from one --- specifically, to binomial(15,k), except for a constant factor: k R_k binm ratio --- --- ---- ----- 0 1 1 1.000 1 6 15 0.400 2 18 105 0.171 3 38 455 0.084 4 66 1365 0.048 5 99 3003 0.033 6 128 5005 0.026 7 144 6435 0.022 8 144 6435 0.022 9 128 5005 0.026 10 99 3003 0.033 11 66 1365 0.048 12 38 455 0.084 13 18 105 0.171 14 6 15 0.400 15 1 1 1.000 The match between R_k and the binom(15,k) distribution is not as good as in the case of the VMS words (the ratio varies from 0.02 to 0.05 over the significative range), but it is close enough to be suggestive. So perhaps we do not need to assume nine independent X/empty slots in the VMS words. Perhaps there are only (say) three slots, each of which may be filled with a "digit" string of length between 0 and 3. Let d_k be the number of distinct "digits" of each length k, in a given slot. It is not necessary that these counts be in the ratio 1:3:3:1. As long as they are symmetrical (i.e. d_0=d_3 and d_1=d_2), the word length distribution will be symmetrical and approximately binomial. All the best, --stolfi

**Follow-Ups**:**Re: VMS words and Roman numerals***From:*Rene Zandbergen

**References**:**RE: On the word length distribution***From:*Rene Zandbergen

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