Quote:Morten St George:
Per my investigations, the VMS was decoded between late 1585 and 1589 somewhere in Protestant-controlled Europe, possibly in Nérac, France, under the protection of King Henry of Navarre.
That's the time-period during which it may have been in the hands of those who attended the pharmacy and botanical garden in Krumlov where Jakub Horcicky was taken in to clean the Jesuit kitchen. That's where he was educated before going to Prague.
Maybe it originated in France, maybe in Switzerland, maybe in Provençal, maybe in Italy/Lombardy, but maybe also in southern Bohemia. France is only one of many possibilities.
------
If someone had decoded it (assuming there is language underneath it), don't you think they would have written at least SOMETHING somewhere in the manuscript to indicate at least one word of what it might say? Adding glosses to manuscripts was very common in those days.
(30-01-2018, 02:10 AM)-JKP- Wrote: You are not allowed to view links. Register or Login to view.Quote:Speculations. The VMS glyphs might represent numbers similar to how the Hebrew glyphs do, and perhaps even the same numbers. Thus, each glyph would correspond to a number. Imagine a sequence of glyphs (numbers) as follows: 2 + 60 + 5 + 20 + 10 + 3. These numbers add up to 100, and 1 + 0 + 0 add up to 1, Aleph, or be it, the Latin letter "a". And that's how a large number of VMS glyphs can reduce into a smaller number of Latin letters. I'm not making this up. It was standard cabala procedure incorporated into their Gematria.
The gematria angle has been explored countless of times by many different researchers (including me).
I was aware of kabbalah before I knew about the VMS so, of course, it was one of the first things I tried.
Reducing "a large number of VMS glyphs... into a smaller number of Latin letters" is not going to result in anything very intelligible because the VMS glyph-count is already somewhat restrained. Read some of the threads on entropy.
I haven't excluded the possibility of some kind of numeric system, I think it's possible, but I don't think it's based on gematria specifically. Not only is gematria not an "encryption" system (it's intended to find correspondences between "special" words, not for enciphering blocks of text), but it's impractical to read back if it is applied to every word (it leans dangerously toward one-way cipher territory).
I posted a Rosetta Stone of sorts earlier on in this thread. It is based on the assumption that a mini red star on top of a red star indicates the title of a poem that precedes four subsequent verses marked by a red star.
There are reasons to be optimistic:
1) the title is shorter than the other lines in both Voynichese and Latin.
2) the average length of Voynichese words per verse increase or decrease as the average length of the corresponding Latin words increase or decrease.
One way, and possibly the only way without external reference, that the Voynich script can reduce in size is if each glyph represented a number. Add up a large sequence of numbers, then add the resulting sum of that, and so on, until the result is a low number matching the number of the Latin letter.
Although the number of numeric possibilities are enormous, they are not infinite. In other words, a very good computer programmer should be able to break the code with the Rosetta Stone at hand as this gives him the target result. It could be only a matter of finding patterns, where, for example, the same group of glyphs (numbers) in any order produce a certain letter or punctuation mark in different places of the Latin text.
I’m confident there is someone out there willing to give it a try. For sure, people have already wasted lots of time on efforts a lot less promising than this.
There is a logical flaw in gemetria encryption, as I have pointed out before on many occasions and JKP pointed out above: it's impossible to decrypt.
Gematria works on the premise that the letters of the alphabet can also be used as numbers, and therefore words and phrases acquire distinctive numerical values. Those words and phrases can thus be considered to be "equal".
Let us assume an English equivalent, where each letter of the alphabet is assigned a numeric value corresponding to its position in the alphabet (a=1,b=2....z=26) and encode a word to see how this works.
MR
13 +18 =
31
V O Y N I C H
22+15+25+14+9+3+8 =
96
Now we have several different ways of "encoding" this.
Traditional gemetria would use the value 96 to find other words that have the value of 96 and substitute them for VOYNICH.(Another way is to add 96+31 together and cast about for a word that equals 127, but I'll take the first route). So, KNOWLEDGE, FREEMASON, TURTLE and APHRODITE all total 96.
And CHIEF, FALL, BELL, MALE all total 31.
So..... we can "encode" MR VOYNICH into CHIEF TURTLE or MALE FREEMASON as all three phrases are "equivalent" under the rules of gemetria.
OK, that's encoded with gemetria. I leave the task of decrypting ALABAMA, I CHANGE NOT into the plain text as an exercise for the reader.*
* SOLUTION: Mr Voynich. But how can you be sure I didn't mean CHIEF TURTLE or MALE FREEMASON?
(30-01-2018, 06:43 AM)davidjackson Wrote: You are not allowed to view links. Register or Login to view.There is a logical flaw in gemetria encryption, as I have pointed out before on many occasions and JKP pointed out above: it's impossible to decrypt.
Gematria works on the premise that the letters of the alphabet can also be used as numbers, and therefore words and phrases acquire distinctive numerical values. Those words and phrases can thus be considered to be "equal".
Let us assume an English equivalent, where each letter of the alphabet is assigned a numeric value corresponding to its position in the alphabet (a=1,b=2....z=26) and encode a word to see how this works.
MR
13 +18 = 31
V O Y N I C H
22+15+25+14+9+3+8 = 96
Now we have several different ways of "encoding" this.
Traditional gemetria would use the value 96 to find other words that have the value of 96 and substitute them for VOYNICH.(Another way is to add 96+31 together and cast about for a word that equals 127, but I'll take the first route). So, KNOWLEDGE, FREEMASON, TURTLE and APHRODITE all total 96.
And CHIEF, FALL, BELL, MALE all total 31.
So..... we can "encode" MR VOYNICH into CHIEF TURTLE or MALE FREEMASON as all three phrases are "equivalent" under the rules of gemetria.
OK, that's encoded with gemetria. I leave the task of decrypting ALABAMA, I CHANGE NOT into the plain text as an exercise for the reader.*
* SOLUTION: Mr Voynich. But how can you be sure I didn't mean CHIEF TURTLE or MALE FREEMASON?
I think you are ignoring the 3 to 1 ratios that I indicated. Closest are the 125 glyphs to 41 letters of the first verse and the 151 glyphs to 50 letters of the 3rd verse.
For the 151, let’s suppose the first glyph merely indicates the beginning of a verse, giving us 150 to 50. Therefore, moving across from left to right, every three glyphs converts to one letter (or blank space or punctuation mark).
As you say, 13 is M. There are dozens of ways of arriving at 13 with three glyphs (eg. 10+2+1, 2+10+1, 5+5+3, 4+8+1, etc. etc) but as long as you know the numerical value of each glyph, it can be done quickly.
Let’s now suppose the second letter is a A. There are three possibilities, 0+0+1, 0+1+0, and 1+0+0, to get 1. So there you have, in a matter of seconds, we have MA and continue on.
The entire sequence of 150 glyphs can probably be converted into the 50 Latin letters in a matter of minutes as long as you know the numerical value of each glyph.
Look, I’m not pretending to be an expert on this stuff. I’m old and weak now, so I’m counting on you guys to figure it out. That’s why I’m posting here.
I wonder what leads to most information loss, gematria or abjad anagramming
Quote:I think you are ignoring the 3 to 1 ratios that I indicated.
OK, let us try the thought experiment that you suggest:
Quote:moving across from left to right, every three glyphs converts to one letter (or blank space or punctuation mark).
As you say, 13 is M. There are dozens of ways of arriving at 13 with three glyphs (eg. 10+2+1, 2+10+1, 5+5+3, 4+8+1, etc. etc) but as long as you know the numerical value of each glyph, it can be done quickly.
Let’s now suppose the second letter is a A. There are three possibilities, 0+0+1, 0+1+0, and 1+0+0, to get 1. So there you have, in a matter of seconds, we have MA and continue on.
Some maths gives us the answer. Let a,b,c be natural numbers >= 0. How many combinations of a,b,c be there so that a+b+c = 13?
Since a,b,c can be repeated (ie, 9+2+2) then this is just a combination problem. The solution is given by C(13,3) =
455 combinations of glyphs.
First off, let me note that there are fewer than 30 Voynich glyphs - there are aprox (ignoring the weirdoes) as many glyphs as letters in a European alphabet.
So we need to assume that there is a one on one correspondence going on; that is, that both languages have the same gemetria system.
Voynichese glyphs are
very highly positional aware - they tend to appear in the same positions and correspondences. I would say that it is impossible for you to find more than 400 different combinations to write the glyph which is equivalent to M.
(30-01-2018, 01:31 PM)davidjackson Wrote: You are not allowed to view links. Register or Login to view.Quote:I think you are ignoring the 3 to 1 ratios that I indicated.
OK, let us try the thought experiment that you suggest:
Quote:moving across from left to right, every three glyphs converts to one letter (or blank space or punctuation mark).
As you say, 13 is M. There are dozens of ways of arriving at 13 with three glyphs (eg. 10+2+1, 2+10+1, 5+5+3, 4+8+1, etc. etc) but as long as you know the numerical value of each glyph, it can be done quickly.
Let’s now suppose the second letter is a A. There are three possibilities, 0+0+1, 0+1+0, and 1+0+0, to get 1. So there you have, in a matter of seconds, we have MA and continue on.
Some maths gives us the answer. Let a,b,c be natural numbers >= 0. How many combinations of a,b,c be there so that a+b+c = 13?
Since a,b,c can be repeated (ie, 9+2+2) then this is just a combination problem. The solution is given by C(13,3) = 455 combinations of glyphs.
First off, let me note that there are fewer than 30 Voynich glyphs - there are aprox (ignoring the weirdoes) as many glyphs as letters in a European alphabet.
So we need to assume that there is a one on one correspondence going on; that is, that both languages have the same gemetria system.
Voynichese glyphs are very highly positional aware - they tend to appear in the same positions and correspondences. I would say that it is impossible for you to find more than 400 different combinations to write the glyph which is equivalent to M.
The number of combinations from your example is probably C(15,2)
Ah yes, sorry, it's been a while
. That's still a lot of combinations!
(30-01-2018, 03:43 PM)davidjackson Wrote: You are not allowed to view links. Register or Login to view.Ah yes, sorry, it's been a while . That's still a lot of combinations!
I see no requirement to make use of all the available possibilities.
But yes, the script does look somewhat rigid at places, such as within the following line:
ySheedy okeedy oteedy qokeedy okeedy okeedy chedal okair qoteedar aty
The first six words all end in eedy, and half of those are entirely the same. But if you take six at a time (as it looks like we should do for our second verse), they are all different and therefore convert into different letters:
ySheed y_okee dy_ote edy_qo keedy_ okeedy _okeed y_ched …
The letters change just before a repeat would have occurred.
It looks like Voynichese is able to vary, at will, the number of glyphs used to represent each Latin letter in any particular passage. In our Rosetta Stone of five lines, four of them are at a ratio of 3 to 1 and one is at a ratio of 6 to 1.
Elsewhere, the ratio could be different but, as in our Rosette Stone, the number of Voynich words might reflect the character ratio.
Thus, if a red star passage has 49 words, in could mean that the underlying Latin has six words and the character ratio is 8 to 1, or perhaps that the underlying Latin has seven words and the character ratio is 7 to 1.
Note that the ranges are limited on the Latin end: we can infer from the literal French translation that the underlying Latin should have approximately six words per verse, maybe sometimes five or seven words.
Note that I'm not scoffing at the length of your system (the number of words to encode a letter) as I know of several Renaissance encoding systems that used even longer cipher text, amongst them one of Kircher's universal language attempts.
However, unless you can find a flag indicating the 'ratio' all you're doing is introducing noise; and changing the rules of the game on the fly to meet objections.