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Frequently Asked Questions (FAQS);faqs.457 My impression from the black and white plates of the Voynich MS I've seen, are that the illustrations are very weird when compared to other 'illuminated' manuscripts of this time. Particularly I would say that there is emphasis on the female nude that is unusual for the art of this period. I can't say that I myself believe the images to have ANYTHING to do with the text. My own conjecture is that the manuscript is a one-way encipherment. A cipher so clever that the inventor didn't even think of how it could be deciphered. Sorta like an /etc/passwd file. Bibliography ------------ 1. William R. Newbold. _The Cipher of Roger Bacon_Roland G Kent, ed. University of Pennsylvania Press, 1928. 2. Joseph Martin Feely. _Roger Bacon's Cipher: The Right Key Found_ Rochester N.Y.:Joseph Martin Feely, pub., 1943. 3. _The Most Mysterious Manuscript_ Robert S. Brumbaugh, ed. Southern Illinois Press, 1978 Unix filters are so wonderful. Massaging the machine readable file, we find: 4182 "words", of which 1284 are used more than once, 308 used 8+ times, 184 used 15+ times, 23 used 100+ times. Does this tell us anything about the language (if any) the text is written in? For those who may be interested, here are the 23 words used 100+ times: 121 2 115 4OFAE 114 4OFAM 155 4OFAN 195 4OFC89 162 4OFCC89 101 4OFCC9 189 89 111 8AE 492 8AM 134 8AN 156 8AR 248 OE 148 OR 111 S9 251 SC89 142 SC9 238 SOE 150 SOR 244 ZC89 116 ZC9 116 ZOE Could someone email the Voynich Ms. ref list that appeared here not very long ago? Thanks in advance... Also... I came across the following ref that is fun(?): The Voynich manuscript: an elegant enigma / M. E. D'Imperio Fort George E. Mead, Md. : National Security Agency(!) Central Security Service(?), 1978. ix, 140 p. : ill. ; 27 cm. The (?!) are mine... Sorry if this was already on the list, but the mention of the NSA (and what's the CSS?) made it jump out at me... -- Ron Carter | rcarter@nyx.cs.du.edu rcarter GEnie 70707.3047 CIS Director | Center for the Study of Creative Intelligence Denver, CO | Knowledge is power. Knowledge to the people. Just say know. Distribution: na Organization: Wetware Diversions, San Francisco Keywords: From sci.archaeology: >From: jamie@cs.sfu.ca (Jamie Andrews) >Date: 16 Nov 91 00:49:08 GMT > > It seems like the person who would be most likely to solve >this Voynich manuscript cipher would have >(a) knowledge of the modern techniques for solving more complex > ciphers such as Playfairs and Vigineres; and >(b) knowledge of the possible contemporary and archaic languages > in which the plaintext could have been written. An extended discussion of the Voynich Manuscript may be found in the tape of the same name by Terence McKenna. I'm not sure who is currently publishing this particular McKenna tape but probably one of: Dolphin Tapes, POB 71, Big Sur, CA 93920 Sounds True, 1825 Pearl St., Boulder, CO 80302 Sound Photosynthesis, POB 2111, Mill Valley, CA 94942 The Spring 1988 issue of Gnosis magazine contained an article by McKenna giving some background of the Voynich Manuscipt and attempts to decipher it, and reviewing Leo Levitov's "Solution of the Voynich Manuscript" (published in 1987 by Aegean Park Press, POB 2837 Laguna Hills, CA 92654). Levitov's thesis is that the manuscript is the only surviving primary document of the Cathar faith (exterminated on the orders of the Pope in the Albigensian Crusade in the 1230s) and that it is in fact not encrypted material but rather is a highly polyglot form of Medieval Flemish with a large number of Old French and Old High German loan words, written in a special script. As far as I know Levitov's there has been no challenge to Levitov's claims so far. Michael Barlow, who had reviewed Levitov's book in Cryptologia, had sent me photocopies of the pages where much of the language was described (pp.21-31). I have just found them, and am looking at them now as I am typing this. Incidentally, I do not believe this has anything to do with cryptology proper, but the decipherment of texts in unknown languages. So if you are into cryptography proper, skip this. Looking at the "Voynich alphabet" pp.25-27, I made a list of the letters of the Voynich language as Levitov interprets them, and I added phonetic descriptions of the sounds I *think* Levitov meant to describe. Here it is: Letter# Phonetic Phonetic descriptions (IPA) in linguists' jargon: in plain English: 1 a low open, central unrounded a as in father e mid close, front, unrounded ay as in May O mid open, back, rounded aw as in law or o as in got (British pronunciation) 2 s unvoiced dental fricative s as in so 3 d voiced dental stop d 4 E mid, front, unrounded e as in wet 5 f unvoiced labiodental fricative f 6 i short, high open, front, i as in dim unrounded 7 i: long, high, front, unrounded ea as in weak 8 i:E (?) I can't make head nor tail of Levitov's explanations. Probably like "ei" in "weird" dragging along the "e": "weeeird"! (British pronunciation, with a silent "r") 9 C unvoiced palatal fricative ch in German ich 10 k unvoived velar stop k 11 l lateral, can't be more precise from description, probably like l in "loony" 12 m voiced bilabial nasal m 13 n voiced dental nasal n 14 r (?) cannot tell precisely from Scottish r? description Dutch r? 15 t no description; dental stop? t 16 t another form for #15 t 17 T (?) no description th as in this? th as in thick? 18 TE (?) again, no description or ET (?) 19 v voiced labiodental fricative v as in rave 20 v ditto, same as #19 ditto (By now, you will have guessed what my conclusion about Levitov's decipherment was) In the column headed "Phonetic (IPA)" I have used capital letters for lack of the special international phonetic symbols: E for the Greek letter "epsilon" O for the letter that looks like a mirror-image of "c" C for c-cedilla T for the Greek letter "theta" The colon (:) means that the sound represented by the preceding letter is long, e.g. "i:" is a long "i". The rest, #21 to 25, are not "letters" proper, but represent groups of two or more letters, just like #18 does. They are: 21 av 22a Ev 22b vE 23 CET 24 kET 25 sET That gives us a language with 6 vowels: a (#1), e (#1 again), O (#1 again), E (#4), i (#6), and i: (#7). Letter #8 is not a vowel, but a combination of two vowels: i: (#7) and probably E (#4). Levitov writes that the language is derived from Dutch. If so, it has lost the "oo" sound (English spelling; "oe" in Dutch spelling), and the three front rounded vowels of Dutch: u as in U ("you", polite), eu as in deur ("door"), u as in vlug ("quick"). Note that out of six vowels, three are confused under the same letter (#1), even though they sound very different from one another: a, e, O. Just imagine that you had no way of distinguishing between "last", "lest" and "lost" when writing in English, and you'll have a fair idea of the consequences. Let us look at the consonants now. I will put them in a matrix, with the points of articulation in one dimension, and the manner of articulation in the other (it's all standard procedure when analyzing a language). Brackets around a letter will mean that I could not tell where to place it exactly, and just took a guess. labial dental palatal velar nasal m n voiced stop d unvoiced stop t k voiced fricative v (T) unvoiced fricative f s C lateral l trill (?) (r) Note that there are only twelve consonant sounds. That is unheard of for a European language. No European language has so few consonant sounds. Spanish, which has very few sounds (only five vowels), has seventeen distinct consonants sounds, plus two semi-consonants. Dutch has from 18 to 20 consonants (depending on speakers, and how you analyze the sounds. Warning: I just counted them on the back of an envelope; I might have missed one or two). What is also extraordinary in Levitov's language is that it lacks a "g", and *BOTH* "b" and "p". I cannot think of one single language in the world that lacks both "b" and "p". Levitov also says that "m" occurs only word-finally, never at the beginning, nor in the middle of a word. That's true: the letter he says is an "m" is always word-final in the reproductions I have seen of the Voynich MS. But no language I know of behaves like that. All have an "m" (except one American Indian language, which is very famous for that, and the name of which escapes me right now), but, if there is a position where "m" never appears in some languages, that position is word-finally. Exactly the reverse of Levitov's language. What does Levitov say about the origin of the language? "The language was very much standardized. It was an application of a polyglot oral tongue into a literary language which would be understandable to people who did not understand Latin and to whom this language could be read." At first reading, I would dismiss it all as nonsense: "polyglot oral tongue" means nothing in linguistics terms. But Levitov is a medical doctor, so allowances must be made. The best meaning I can read into "polyglot oral tongue" is "a language that had never been written before and which had taken words from many different languages". That is perfectly reasonable: English for one, has done that. Half its vocabulary is Norman French, and some of the commonest words have non-Anglo-Saxon origins. "Sky", for instance, is a Danish word. So far, so good. Levitov continues: "The Voynich is actually a simple language because it follows set rules and has a very limited vocabulary.... There is a deliberate duality and plurality of words in the Voynich and much use of apostrophism". By "duality and plurality of words" Levitov means that the words are highly ambiguous, most words having two or more different meanings. I can only guess at what he means by apostrophism: running words together, leaving bits out, as we do in English: can not --> cannot --> can't, is not --> ain't. Time for a tutorial in the Voynich language as I could piece it together from Levitov's description. Because, according to Levitov, letter #1 represent 3 vowels sounds, I will represent it by just "a", but remember: it can be pronounced a, e, or o. But I will distinguish, as does Levitov, between the two letters which he says were both pronounced "v", using "v" for letter #20 and "w" for letter #21. Some vocabulary now. Some verbs first, which Levitov gives in the infinitive. In the Voynich language the infinitive of verbs ends in -en, just like in Dutch and in German. I have removed that grammatical ending in the list which follows, and given probable etymologies in parentheses (Levitov gives doesn't give any): ad = to aid, help ("aid") ak = to ache, pain ("ache") al = to ail ("ail") and = to undergo the "Endura" rite ("End[ura]", probably) d = to die ("d[ie]") fad = to be for help (from f= for and ad=aid) fal = to fail ("fail") fil = to be for illness (from: f=for and il=ill) il = to be ill ("ill") k = to understand ("ken", Dutch and German "kennen" meaning "to know") l = to lie deathly ill, in extremis ("lie", "lay") s = to see ("see", Dutch "zien") t = to do, treat (German "tun" = to do) v = to will ("will" or Latin "volo" perhaps) vid = to be with death (from vi=with and d=die) vil = to want, wish, desire (German "willen") vis = to know ("wit", German "wissen", Dutch "weten") vit = to know (ditto) viT = to use (no idea, Latin "uti" perhaps?) vi = to be the way (Latin "via") eC = to be each ("each") ai:a = to eye, look at ("eye", "oog" in Dutch) en = to do (no idea) Example given by Levitov: enden "to do to death" made up of "en" (to do), "d" (to die) and "en" (infinitive ending). Well, to me, that's doing it the hard way. What's wrong with just "enden" = to end (German "enden", too!) More vocabulary: em = he or they (masculine) ("him") er = her or they (feminine) ("her") eT = it or they ("it" or perhaps "they" or Dutch "het") an = one ("one", Dutch "een") "There are no declensions of nouns or conjugation of verbs. Only the present tense is used" says Levitov. Examples: den = to die (infinitive) (d = die, -en = infinitive) deT = it/they die (d = die, eT = it/they) diteT = it does die (d = die, t = do, eT = it/they, with an "i" added to make it easier to pronounce, which is quite common and natural in languages) But Levitov contradicts himself immediately, giving another tense (known as present progressive in English grammar): dieT = it is dying But I may be unfair there, perhaps it is a compound: d = die, i = is ...-ing, eT = it/they. Plurals are formed by suffixing "s" in one part of the MS, "eT" in another: "ans" or "aneT" = ones. More: wians = we ones (wi = we, wie in Dutch, an = one, s = plural) vian = one way (vi = way, an = one) wia = one who (wi = who, a = one) va = one will (v = will, a = one) wa = who wi = who wieT = who, it (wi = who, eT = it) witeT = who does it (wi = who, t = do, eT = it/they) weT = who it is (wi = who, eT = it, then loss of "i", giving "weT") ker = she understands (k = understand, er =she) At this stage I would like to comment that we are here in the presence of a Germanic language which behaves very, very strangely in the way of the meanings of its compound words. For instance, "viden" (to be with death) is made up of the words for "with", "die" and the infinitive suffix. I am sure that Levitov here was thinking of a construction like German "mitkommen" which means "to come along" (to "withcome"). I suppose I could say "Bitte, sterben Sie mit" on the same model as "Bitte, kommen Sie mit" ("Come with me/us, please), thereby making up a verb "mitsterben", but that would mean "to die together with someone else", not "to be with death". Let us see how Levitov translates a whole sentence. Since he does not explain how he breaks up those compound words I have tried to do it using the vocabulary and grammar he provides in those pages. My tentative explanations are in parenthesis. TanvieT faditeT wan aTviteT anTviteT atwiteT aneT TanvieT = the one way (T = the (?), an = one, vi =way, eT = it) faditeT = doing for help (f = for, ad = aid, i = -ing, t = do, eT = it) wan = person (wi/wa = who, an = one) aTviteT = one that one knows (a = one, T = that, vit = know, eT = it. Here, Levitov adds one extra letter which is not in the text, getting "aTaviteT", which provide the second "one" of his translation) anTviteT = one that knows (an =one, T = that, vit = know, eT = it) atwiteT = one treats one who does it (a = one, t = do, wi = who, t = do, eT = it. Literally: "one does [one] who does it". The first "do" is translated as "treat", the second "one" is added in by Levitov: he added one letter, which gives him "atawiteT") aneT = ones (an = one, -eT = the plural ending) Levitov's translation of the above in better English: "the one way for helping a person who needs it, is to know one of the ones who do treat one". Need I say more? Does anyone still believe that Levitov's translations are worth anything? As an exercise, here is the last sentence on p.31, with its word-for-word translation by Levitov. I leave you to work it out, and to figure out what it might possibly mean. Good luck! tvieT nwn anvit fadan van aleC tvieT = do the ways nwn = not who does (but Levitov adds a letter to make it "nwen") anvit = one knows fadan = one for help van = one will aleC = each ail ==> cryptology/swiss.colony.p <== What are the 1987 Swiss Colony ciphers? ==> cryptology/swiss.colony.s <== Did anyone solve the 1987 'Crypto-gift' contest that was run by Swiss Colony? My friend and I worked on it for 4 months, but didn't get anywhere. My friend solved the 1986 puzzle in about a week and won $1000. I fear that we missed some clue that makes it incredibly easy to solve. I'm including the code, clues and a few notes for those of you so inclined to give it a shot. 197,333,318,511,824, 864,864,457,197,333, 824,769,372,769,864, 865,457,153,824,511,223,845,318, 489,953,234,769,703,489,845,703, 372,216,457,509,333,153,845,333, 511,864,621,611,769,707,153,333, 703,197,845,769,372,621,223,333, 197,845,489,953,223,769,216,223, 769,769,457,153,824,511,372,223, 769,824,824,216,865,845,153,769, 333,704,511,457,153,333,824,333, 953,372,621,234,953,234,865,703, 318,223,333,489,944,153,824,769, 318,457,234,845,318,223,372,769, 216,894,153,333,511,611, 769,704,511,153,372,621, 197,894,894,153,333,953, 234,845,318,223 CHRIS IS BACK WITH GOLD FOR YOU HIS RHYMES CONTAIN THE SECRET. YOU SCOUTS WHO'VE EARNED YOUR MERIT BADGE WILL QUICKLY LEARN TO READ IT. SO WHEN YOUR CHRISTMAS HAM'S ALL GONE AND YOU'RE READY FOR THE TUSSLE, BALL UP YOUR HAND INTO A FIST AND SHOW OUR MOUSE YOUR MUSCLE. PLEASE READ THESE CLUES WE LEAVE TO YOU BOTH FINE ONES AND THE COARSE; IF CARE IS USED TO HEED THEM ALL YOU'LL SUFFER NO REMORSE. Notes: The puzzle comes as a jigsaw that when assembled has the list of numbers. They are arranged as indicated on the puzzle, with commas. The lower right corner has a drawing of 'Secret Agent Chris Mouse'. He holds a box under his arm which looks like the box the puzzle comes in. The upper left corner has the words 'NEW 1987 $50,000 Puzzle'. The lower left corner is empty. The clues are printed on the entry form in upper case, with the punctuation as shown. Ed Rupp ...!ut-sally!oakhill!ed Motorola, Inc., Austin Tx. ==> decision/allais.p <== The Allais Paradox involves the choice between two alternatives: A. 89% chance of an unknown amount 10% chance of $1 million 1% chance of $1 million B. 89% chance of an unknown amount (the same amount as in A) 10% chance of $2.5 million 1% chance of nothing What is the rational choice? Does this choice remain the same if the unknown amount is $1 million? If it is nothing? ==> decision/allais.s <== This is "Allais' Paradox". Which choice is rational depends upon the subjective value of money. Many people are risk averse, and prefer the better chance of $1 million of option A. This choice is firm when the unknown amount is $1 million, but seems to waver as the amount falls to nothing. In the latter case, the risk averse person favors B because there is not much difference between 10% and 11%, but there is a big difference between $1 million and $2.5 million. Thus the choice between A and B depends upon the unknown amount, even though it is the same unknown amount independent of the choice. This violates the "independence axiom" that rational choice between two alternatives should depend only upon how those two alternatives differ. However, if the amounts involved in the problem are reduced to tens of dollars instead of millions of dollars, people's behavior tends to fall back in line with the axioms of rational choice. People tend to choose option B regardless of the unknown amount. Perhaps when presented with such huge numbers, people begin to calculate qualitatively. For example, if the unknown amount is $1 million the options are: A. a fortune, guaranteed B. a fortune, almost guaranteed a tiny chance of nothing Then the choice of A is rational. However, if the unknown amount is nothing, the options are: A. small chance of a fortune ($1 million) large chance of nothing B. small chance of a larger fortune ($2.5 million) large chance of nothing In this case, the choice of B is rational. The Allais Paradox then results from the limited ability to rationally calculate with such unusual quantities. The brain is not a calculator and rational calculations may rely on things like training, experience, and analogy, none of which would be help in this case. This hypothesis could be tested by studying the correlation between paradoxical behavior and "unusualness" of the amounts involved. If this explanation is correct, then the Paradox amounts to little more than the observation that the brain is an imperfect rational engine. ==> decision/division.p <== N-Person Fair Division If two people want to divide a pie but do not trust each other, they can still ensure that each gets a fair share by using the technique that one person cuts and the other person chooses. Generalize this technique to more than two people. Take care to ensure that no one can be cheated by a coalition of the others. ==> decision/division.s <== N-Person Fair Division Number the people from 1 to N. Person 1 cuts off a piece of the pie. Person 2 can either diminish the size of the cut off piece or pass. The same for persons 3 through N. The last person to touch the piece must take it and is removed from the process. Repeat this procedure with the remaining N - 1 people, until everyone has a piece. (cf. Luce and Raiffa, "Games and Decisions", Wiley, 1957, p. 366) There is a cute result in combinatorics called the Marriage Theorem. A village has n men and n women, such that for all 0 < k <= n and for any set of k men there are at least k women, each of whom is in love with at least one of the k men. All of the men are in love with all of the women :-}. The theorem asserts that there is a way to arrange the village into n monogamous couplings. The Marriage Theorem can be applied to the Fair Pie-Cutting Problem. One player cuts the pie into n pieces. Each of the players labels some non-null subset of the pieces as acceptable to him. For reasons given below he should "accept" each piece of size > 1/n, not just the best piece(s). The pie-cutter is required to "accept" all of the pieces. Given a set S of players let S' denote the set of pie-pieces acceptable to at least one player in S. Let t be the size of the largest set (T) of players satisfying |T| > |T'|. If there is no such set, the Marriage Theorem can be applied directly. Since the pie-cutter accepts every piece we know that t < n. Choose |T| - |T'| pieces at random from outside T', glue them together with the pieces in T' and let the players in T repeat the game with this smaller (t/n)-size pie. This is fair since they all rejected the other n-t pieces, so they believe this pie is larger than t/n. The remaining n-t players can each be assigned one of the remaining n-t pie-pieces without further ado due to the Marriage Theorem. (Otherwise the set T above was not maximal.) ==> decision/dowry.p <== Sultan's Dowry A sultan has granted a commoner a chance to marry one of his hundred daughters. The commoner will be presented the daughters one at a time. When a daughter is presented, the commoner will be told the daughter's dowry. The commoner has only one chance to accept or reject each daughter; he cannot return to a previously rejected daughter. The sultan's catch is that the commoner may only marry the daughter with the highest dowry. What is the commoner's best strategy assuming he knows nothing about the distribution of dowries? ==> decision/dowry.s <== Solution Since the commoner knows nothing about the distribution of the dowries, the best strategy is to wait until a certain number of daughters have been presented then pick the highest dowry thereafter. The exact number to skip is determined by the condition that the odds that the highest dowry has already been seen is just greater than the odds that it remains to be seen AND THAT IF IT IS SEEN IT WILL BE PICKED. This amounts to finding the smallest x such that: x/n > x/n * (1/(x+1) + ... + 1/(n-1)). Working out the math for n=100 and calculating the probability gives: The commoner should wait until he has seen 37 of the daughters, then pick the first daughter with a dowry that is bigger than any preceding dowry. With this strategy, his odds of choosing the daughter with the highest dowry are surprisingly high: about 37%. (cf. F. Mosteller, "Fifty Challenging Problems in Probability with Solutions", Addison-Wesley, 1965, #47; "Mathematical Plums", edited by Ross Honsberger, pp. 104-110) ==> decision/envelope.p <== Someone has prepared two envelopes containing money. One contains twice as much money as the other. You have decided to pick one envelope, but then the following argument occurs to you: Suppose my chosen envelope contains $X, then the other envelope either contains $X/2 or $2X. Both cases are equally likely, so my expectation if I take the other envelope is .5 * $X/2 + .5 * $2X = $1.25X, which is higher than my current $X, so I should change my mind and take the other envelope. But then I can apply the argument all over again. Something is wrong here! Where did I go wrong? In a variant of this problem, you are allowed to peek into the envelope you chose before finally settling on it. Suppose that when you peek you see $100. Should you switch now? ==> decision/envelope.s <== Let's follow the argument carefully, substituting real numbers for variables, to see where we went wrong. In the following, we will assume the envelopes contain $100 and $200. We will consider the two equally likely cases separately, then average the results. First, take the case that X=$100. "I have $100 in my hand. If I exchange I get $200. The value of the exchange is $200. The value from not exchanging is $100. Therefore, I gain $100 by exchanging." Second, take the case that X=$200. "I have $200 in my hand. If I exchange I get $100. The value of the exchange is $100. The value from not exchanging is $200. Therefore, I lose $100 by exchanging." Now, averaging the two cases, I see that the expected gain is zero. So where is the slip up? In one case, switching gets X/2 ($100), in the other case, switching gets 2X ($200), but X is different in the two cases, and I can't simply average the two different X's to get 1.25X. I can average the two numbers ($100 and $200) to get $150, the expected value of switching, which is also the expected value of not switching, but I cannot under any circumstances average X/2 and 2X. This is a classic case of confusing variables with constants. OK, so let's consider the case in which I looked into the envelope and found that it contained $100. This pins down what X is: a constant. Now the argument is that the odds of $50 is .5 and the odds of $200 is .5, so the expected value of switching is $125, so we should switch. However, the only way the odds of $50 could be .5 and the odds of $200 could be .5 is if all integer values are equally likely. But any probability distribution that is finite and equal for all integers would sum to infinity, not one as it must to be a probability distribution. Thus, the assumption of equal likelihood for all integer values is self-contradictory, and leads to the invalid proof that you should always switch. This is reminiscent of the plethora of proofs that 0=1; they always involve some illegitimate assumption, such as the validity of division by zero. Limiting the maximum value in the envelopes removes the self-contradiction and the argument for switching. Let's see how this works. Suppose all amounts up to $1 trillion were equally likely to be found in the first envelope, and all amounts beyond that would never appear. Then for small amounts one should indeed switch, but not for amounts above $500 billion. The strategy of always switching would pay off for most reasonable amounts but would lead to disastrous losses for large amounts, and the two would balance each other out.