oshfdk > 20-09-2025, 11:46 PM
(20-09-2025, 11:36 PM)SherriMM Wrote: You are not allowed to view links. Register or Login to view.A few samples of vertical patterns within the starting characters of the text (with an example attached):
You are not allowed to view links. Register or Login to view.
RobGea > 21-09-2025, 01:57 AM
Rafal > 21-09-2025, 01:11 PM
RobGea > 21-09-2025, 06:59 PM
(21-09-2025, 01:11 PM)Rafal Wrote: You are not allowed to view links. Register or Login to view.Good spot! For me it is not a coincidence. Notice that it happens mostly at neighbour pages.I dunno, man, it's tricky.
Koen G > 22-09-2025, 06:18 AM
SherriMM > 29-09-2025, 07:26 PM
Rafal > 29-09-2025, 09:15 PM
Jorge_Stolfi > 29-09-2025, 09:34 PM
(20-09-2025, 11:36 PM)SherriMM Wrote: You are not allowed to view links. Register or Login to view.A few samples of vertical patterns within the starting characters of the text (with an example attached): You are not allowed to view links. Register or Login to view.
SherriMM > 29-09-2025, 10:50 PM
(29-09-2025, 09:34 PM)Jorge_Stolfi Wrote: You are not allowed to view links. Register or Login to view.(20-09-2025, 11:36 PM)SherriMM Wrote: You are not allowed to view links. Register or Login to view.A few samples of vertical patterns within the starting characters of the text (with an example attached): You are not allowed to view links. Register or Login to view.
The blogpost cited by Rafal is too simplistic. The first letter of each line is not a random letter of the Voynichese alphabet with equal probability. The line-initial letters have a highly skewed distribution. That greatly increases the probability of certain three-letter combinations.
(The skewed probabilities themselves are a feature that requires explanation. I posted some possible explanations somewhere on this forum, a few days ago.)
let P(x) be the relative frequency of EVA letter x in line-initial position. Namely, if 35% of the lines start with y, then P(y) = 0.35, and so on.
If you have a string S of n letters randomly and independently generated with those probabilities, and you pick any index k in 0 to n-3, the probability that the substring S[k]S[k+1]S[k+2] is qoy is Z = P(q)P(o)P(y). In that string, you should expect to find about (n-2)Z such strings. (A bit less because the strings cannot overlap, but that should be close enough).
Can you compute the expected number of qoy substrings with this method?
Note that even if the letters are drawn independently, some three-letter combinations will show up more often than expected by the above formulas. A three-letter combination would have to be a lot more frequent than predicted in order to be "notable".
All the best, --jorge