The Impossibility of Double Gallows

[ChenZheChina]

Hi ChenZheChina,

Question 1: Counting until the occurrence of any new gallows.
Question 2: No, cTh are two chars and one gallow ( counted one ).
Question 3: 91 is what comes out as result, every char and LF was counted. It is just what the script returns.

Thank you for the explanation.

I checked what nablator found in f89v2 , and it seems that the transcription is well done with no obviously ambiguous characters.

ychey okeey qoeol daiin chor chor cheos qoleeeey dal chody cheor chey qoaiin chody
dair or cheol chom qol cheo lcho l or cheo daiin chkam

In this case, the distance between the first and the last k would be 127, if all spaces and newlines (LF) are counted. Even if we remove all suspicious spaces in the second line (cheo lcho l or cheo daiin -> cheolcholorcheodaiin), the distance would still be 122, which is much larger than 91.

Is it possible that the transcription you used is incomplete, or the bash script is accidentally missing some lines?