I’ve been thinking about Martin’s desire for the pretty single backslash
approach. I think the discussion should be about the probability of collisions
(I.e., files that cannot be folded due to false positives)
Assuming 95 printable characters (127-32) plus ‘\n’, a total of 96 characters,
the unconditioned probability of a given of an n-character string occurring on
a line is (1/96)^n.
Here are the 3 options being discussed (please check my math!)
1. The pretty double backslash approach in the current draft
- folds only on any column and supports indents
Scan the text content to ensure no existing lines already end with a backslash
('\') character when the subsequent line starts with a backslash ('\')
character as the first non-space (' ') character.
P(“\\\n[ ]*\\”)
= sum of (1/96)^n for 3<=n<69
= ∑((1÷96)^x; x; 3; 69)
= 0.0000011421783625730994152046783625731
~= 1 / 1,000,000
2. The not pretty single backslash approach in I-D -06
- folds only on max-column with no support for indents
Scan the artwork to ensure no existing lines already end with a '\' character
on the desired maximum column.
P(“.\{$maxcol-1\}\\\n”)
= P( (not ‘\n’ for $maxcol-1 chars), followed by a “\\\n”)
= ((1−(1÷96))^68)×(1÷96)^2
= 0.0000532376463396105463857306859461496
~= 1 / 20,000
3. Martin’s pretty single backslash approach
- folds only on any column and supports indents
Scan the artwork to ensure no existing lines already end with a '\' character
OR that a white space character appears on the max column.
P(“\\\n”) + P(“.\{$maxcol\} “)
= (1/96)^2 + ((1−(1÷96))^69)×(1÷96)
= 0.00516608334670744635108885960932866
~= 1 / 200
Note that each of these assume a long-line. The number of long lines in a
given piece of text is small, 1/100? Thusly, while option #3 is two orders of
magnitude more like than option #2, it may only be detected 1 / 200,000 text
samples, that are themselves detected as needing to be folded (1/5?), so maybe
one in a million text samples?
Maybe an automated folding algorithm could try #3 and, only if detecting the
precondition, switch to option #1?
Kent // contributor
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