How to Match “A B C” where A+B=C: The Beast Tamed (2018) 95 points by tosh 46 days ago | hide | past | web | favorite | 11 comments

 Wonder what happened to the regex101 demo link [1], it just redirects to home page---Edit: Found a newer article [2] which does point a working regex101 link [3]
 Real example of how comodity enterprise software is configured.
 "# Carrying. This is where it gets complicated"Oh, I thought we were past that already...
 This is a crime against sanity
 This could be the best and the worst thing I've seen all year.
 What a fascinatingly bizarre little program. It's like some kind of immutable-data assembly language with tortured syntax. The next level of Brainfuck.
 a b c where a+b=c is not a regular language. What exactly is the constraint on this challenge?
 Whatever the PCRE supports. This essentially means that you can have at least a context-sensitive grammar, which I think a+b=c indeed is.
 Should probably read the articles first... I was really hoping for:`````` a + b = c, where a = b = c `````` I could only find:* a = b = c = 0* a = b = c = k/inf (where k < inf and k > -inf)Off-topic but are there other solutions? I checked in Wolfram Alpha and nothing...
 Not sure where you get those weird infinity results from, but the "a = b = c = 0" solution is easily show:Note that "a + b = c" implies "b = c - a", and given our assumption that "a = b = c", we can rewrite as "a = a - a = 0". This obviously results in "a = b = c = 0".EDIT: Modifying the proof above, it becomes apparent that this is a property of groups (most number systems are groups): Again "a + b = c", iff "a + a = a" iff "a + a + (-a) = a + (-a)" which is "a = e".Note that addition over IEEE754 floats do not constitute a group, (in part) because "inf + inf == inf" evaluates to true.

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