The 'Chinese' Theory: For and Against

[Jorge_Stolfi]

This account leaves out that the "butchering freedom", as we're calling it, over-generates potential Voynichese cribs. The Rooster example has 17 cribs in the SPS and 10 in Shennong's Classic.

The longest entry of the SBJ, the "Red Rooster", has 8 (not 10) occurrences of 主.  Excluding the fields that apparently were omitted by the Author in all entries, it is 72 hanzi long:

丹雄鸡主崩中漏下赤白沃补虚温中止血通神杀毒辟不祥头主杀鬼肪主耳聋肠主遗溺肶胵裹黄皮主泄利屎白主消渴伤寒寒热翮羽主下血闭鸡子主除热火疮痫痉可作虎魄

The longest single-star parag of the SPS is f105v.32.  It has 361 EVA letters in my transcription.  Note that 361/72 = ~5, a ratio that seems to be approximately the same for all recipes.  Ignoring spaces, there are 15 occurrences of daiin = daiin and its allowed variations, marked by brackets below:

<f105v.32>  poarkeeo[daiin]qoairaracphheyqoeedeodyqo[kaiin]qote[dair]aporairapylsheodytairoteeyoteeoolotaiinokeeyqo[kaiin]oraiiraldalsheeo[daiin]chsdqokeeey[dair]o[kaiin]otaiinche[daiin]olkallkl[dain]doeeokcheeoltaiinotcheedychoraiino[daiin]chedyotaiinalkaishd[laiin]sheodokeeodyqoaiinytaiinotairchdaldy[daim]ch[daiin]ockhhyysheyckhysheoqoeeol[kaiin]chsokoltchdysheeeyo[kaiin]araildycheodyoaiirainokshey

That's  5 [daiin], 5 [kaiin], 2 [dair], 1 [daim], 1 [dain], 1 [laiin].

I explained before why I allow these but not [taiin], [raiin], [chaiin] etc. Briefly, m is probably an abbreviation of iin or maybe in, and in sloppy cursive handwriting iin may look like ir, and d may look like an l or k (but not a t or other glyphs).

Besides those 15, there are 13 additional occurrences of the strings *aiin, *ain, *aim and *air, that I do not consider acceptable variations of daiin.

t turns out that, mapping the positions of the 主 in the SBJ recipe to positions in the SPS parag, 4 of these fall close to a  [daiin], and 4 to variations thereof.  (More precisely, the gaps between these 8 nearest keys in the parag are very close to the gaps between the 主, scaled by 5.)

Selecting 10 items from 17 where the order of the selection doesn't matter (because we are going to keep them in the order they appear in the SPS) is 17C10 = 19,448 potential matches.

Imagine that you have a row of 30 boxes, and 8 of them have a coin inside,while the others are empty.  You pick 15 of those boxes at random.  You find that 8 of those 15 boxes have a coin inside.

There are comb(15,8) = 6435 subsets of 8 boxes out of the 15.  That is the number of ways you could have got all the 8 coins that way.  That is a big number. Does it mean that this outcome is unremarkable?

Of course not.  That number alone does not mean anything.  To answer that question one must consider also the total number of boxes and how many of them have coins.

The probability P that picking M of N boxes at random will yield J of the K coins they contain is comb(K,J)*comb(N-K,M-J)/comb(N,M).   

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.

How is this relevant to the "Rooster" matching? Hint: the coins in the boxes have a "主" stamped on them.

Sure, the computation of the "null hypothesis" P for the Rooster=f105r.32 claim is more complicated than for that analogy.  

For one thing, I don't compare the absolute positions of daiin-like strings and 主 characters, but the lengths of the 9 gaps before, between, and after the eight 主 characters and the gaps between the eight assigned daiin-like strings, scaled by the 1:5 ratio.  

The discrepancies seen in these gap sizes are almost all less than 5 EVA letters, which is 1 hanzi; but two of them are -7 letters (in the title of the recipe, which is 3 hanzi but only 8 EVA letters) and +9 letter (in a gap that is 46 letters long).  These two discrepancies are equivalent to -1.4 and +1.8 hanzi.  For comparison, the average gap size is about (361-8*5)/9 = ~35 EVA letters.  This tolerance in the gap lengths affects the number N of "boxes" that are available to be "picked" by the M = 15 daiin-like strings.

But anyway you should be able to see that the null-hypothesis P will be quite small,  even though comb(M,J) is a large number.

All the best, --stolfi