The 'Chinese' Theory: For and Against

[Jorge_Stolfi]

For N  = 30, K = 8, M =15, J = 8, this formula gives P = ~0.0011 = ~0.11%; that is, less than one chance in 900.  So getting all 8 coins by picking 15 out of 30 boxes IS a remarkable result.

Why N = 30 in this post?

Those numbers were primarily intended to show that the probability computation must take N and K into account, and that the fact that comb(15,8) is a large number does not mean anything by itself.

But indeed the numbers were chosen to be roughly adequate for the Rooster recipe case.

If the claim had been that the mapping was exact, the number N of boxes in the box and coin analogy would be the number of Chinese characters, 72.  

But the gap sizes are allowed to differ from the predicted values, with average deviation of 2-3 EVA letters, which is about half a Chinese character.  Thus, to make the analogy valid, the boxes must be wider than just one character, and then there will be fewer boxes.  Hence the use of N = 30 in that example.

 Can you provide the precise expression you used to arrive at [the gap error scores]?

The formulas I am currently using are given below.  But they are not particularly special.  The only important aspect is that the score for each gap is proportional to the square of the relative error between the actual length in EVA letters and the length predicted from the hanzi count in the corresponding gap of the recipe.  Specifically, the square of the difference of the logs. 


  pppp.txt (Size: 5.13 KB / Downloads: 8)

The other thing I'm going to want to know is what the badness cutoff is---when is that badness too high to count as a match?

Since the scale of the badness score is somewhat arbitrary, there is no cutoff.  I currently use the score only to rank the candidate parags.

The decision of whether the best-ranked (minimum-score) candidate is probably the real translation of the recipe is still done by eye, looking at whether the identified Voynichese keywords were the "canonical" ones (like daiin instead of dain or kaiin) and how big really are the gap errors (they should all be small, mostly 2-3 EVA letters at most).  Thus the mappings between recipes and parags are only tentative.  

That said, with the current formula, when the best candidate has badness score greater than 1.0, it is usually not a real match.  Sometimes I am lucky and there is only parag that matches all the cribs well, with score below 1, and all other parags have mush larger scores.  Sometimes there are a few parags with scores below or just above 1, with no clear gap.  Sometimes the best parag is obviously not it.

Eventually I intend to replace the ad-hoc badness score above with a proper statistical one (log of 1/prob of the pairing being true according to Bayes).  Then one could list the pairs that have 80% or 99% confidence etc.

For the Rooster recipe, there is no need to choose among the ranked candidates.   There is only one candidate paragraph to consider, because both are outliers in the recipe/parag length distributions.  Thus any other parag, being much shorter, would have at least one gap that is way too small.

--stolfi