30-01-2018, 07:45 AM
(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.