A reader responded to Daniel Eastwood’s efforts (click here to see the exertions) to solve ‘The Rodriguez File’:
“Mrs Rodriguez may have meant ‘straight vertical lines of 5 or 6 spaces’. You have drawn lines that have different angles. Would that make a difference?”
Naturally, Dan Eastwood, again dug into the file and replied:
“Yes it does. I interpreted her question to mean the diagonal line ending at about the extra ‘e’ before lucubrations. Rereading this now, I think I misunderstood her intent. Vertical lines are much easier to count though:
(Length, # of): 2, 23; 3, 2; 4, 0; 5, 1; 6, 1. This is an average length of 0.33, but a Poisson distribution (my original hypothesis) is most certainly not correct. Unfortunately, this makes the math harder.This isn?t a complete answer, but it?s the best I can do now:There are 36 cases where a space on one line is followed by a space on the subsequent line (this counts longer lines several times) out of 15 subsequent rows. This is an average of 36/15 = 2.4 per pair of (subsequent) rows. In an 80 character line, the probability that a space will be followed in the next row by another space is 2.4/80 = 0.03.
Assuming the characters are random and the rows independent, then the probability a line of length 2 being followed by a third subsequent space is (0.03)^2 = 0.0009, a fourth is (0.003)^4 = 0.000027, and so on.
So the probability of getting straight lines of spaces 5 or 6 rows long is pretty small. Even without the math, I would guess that my assumptions of randomness and independence are probably not true. […] Short of creating a bootstrap simulation of a random distribution of words in the paragraph and the ‘lines’ that result, I don?t see any way to get at this problem. The distribution of words isn?t really random, so maybe it?s not too surprising that those lines appear.”
