Post funny things about programming here! (Or just rant about your favourite programming language.)
^.?$|^(..+?)\1+$
Matches strings of any character repeated a non-prime number of times
knowing Matt Parker it only matches prime numbers or multiples of e or something.
looks at <ansewer>
Yeah see?
So, here's my attempt
The first portion (^.?$) matches all lines of 0 or 1 characters.
The second portion (^(..+?)\1+$) is more complicated:
(..+?) is a capture group that matches the first character in any line, followed by a smallest possible non-zero number of characters such that (2) still matches (note that the minimum length of this match is 2)\1+ matches as many as possible (and more than 0) repeats of the (1) groupI think what this does is match any line consisting of a single character with the length
1 (due to the note in (1), so that the repeating portion has to be at least 2 characters long), orTherefore, combined with the first portion, it matches all lines of the same character whose lengths are composite (non-prime) numbers? (it will also match any line of length 1, and all lines consisting of the same string repeated more than one time)
So this is a definite example of "regex" that's not regular, then. I really don't think there's any finite state machine that can track every possible number of string repeats separately.
You got downvoted here but you're absolutely right. It's easy to prove that the set of strings with prime length is not a regular language using the pumping lemma for regular languages. And in typical StackExchange fashion, someone's already done it.
Here's their proof.
Claim 1: The language consisting of the character 1 repeated a prime number of times is not regular.
A further argument to justify your claim—
Claim 2: If the language described in Claim 1 is not regular, then the language consisting of the character 1 repeated a composite number of times is not regular.
Proof: Suppose the language described in Claim 2 is regular if the language described in Claim 1 is not. Then there must exist a finite-state automaton A that recognises it. If we create a new finite-state automaton B which (1) checks whether the string has length 1 and rejects it, and (2) then passes the string to automaton A and rejects when automaton A accepts and accepts when automaton A rejects, then we can see that automaton B accepts the set of all strings of non-composite length that are not of length 1, i.e. the set of all strings of prime length. But since the language consisting of all strings of prime length is non-regular, there cannot exist such an automaton. Therefore, the assumption that the language described in Claim 2 being regular is false.
By now, I have just one, so thanks for the assist. There's always that one (sometimes puzzling) downvote on anything factual.
The pumping lemma, for anyone unfamiliar. It's a consequence of the fact an FSM is finite, so you can construct a repeatable y just by exhausting the FSM's ability to "remember" how much it's seen.
Yeah backreferences in general are not "regular" in the mathematical sense.
You can have states point to each other in a loop, no?
Yeah, but in an FSM all you have are states. To do it the obvious way, you need a loop with separate branches for every number greater than 2, or at the very least every prime number, and that's not going to be finite.
If the set of all strings of composite length is a regular language, you can use that to prove the set of all strings of prime length are also a regular language.
But it's also easy to prove that the set of language of strings of prime length is not regular, and thus the language of strings of composite length also can't be regular.
Thank you for this. I'll review this when I can.
...either an empty string, a single character, or the same sequence of characters repeated more than once?
A non prime number of times... It looks like the string of characters could repeat number of times because the whole capture group repeats. I don't see a prime constraint.
The capture group must be the same each time it repeats, so the number of characters stays the same. So X groups of Y characters = string of length X*Y. X and Y can be anything so any string length that can be made by multiplying two numbers-- which is every non-prime string length-- is matched. 0 and 1 are handled specially at the start.
This is brilliantly disgusting.
Literal interpretation of the regex
The regex matches either a line with a single character or a line with a sequence of two or more characters that's repeated two or more times. For some examples: the regex matches "a", "b", "abab", "ababab", "aaaa", and "bbbbbb", but does not match "aa", "bb", "aaa", "ab", "aba", or "ababa".
Hint for the special thing it matches
For a line with a single character repeated n times, what does matching (or not matching) this regex say about the number n?
You forgot empty line. Since first part is ^.?$ it's one or zero of any character.
I could be wrong but I think the (..+?) portion will either remove a dud or replenish the allowance.
Is there a reason to use (..+?) instead of (.+) ?
Yes, the first one matches only 2 more characters while the second matches 1 or more. Also the +? is a lazy quantifier so it will consume as little as possible.
Ah, didn't know +? was lazy, thanks
I thought, the +? was going to be a syntax error. 🙃
I was like, why specify "one or more" and then make it optional? Isn't that just .*?
For a second I thought I was still in the thread about monkeys face-rolling typewriters until the heat death of the universe not eventually producing Hamlet
It matches for non-primes and doesn't match for primes.
The pipe is throwing me off because usually I have to do parentheses for that to work...
The answer given in the spoiler tag is not quite correct!
Test case
According to the spoiler, this shouldn't match "abab", but it does.
Corrected regex
This will match what the spoiler says: ^.?$|^((.)\2+?)\1+$
Full workup
Any Perl-compatible regex can be parsed into a syntax tree using the Common Lisp package CL-PPCRE. So if you already know Common Lisp, you don't need to learn regex syntax too!
So let's put the original regex into CL-PPCRE's parser. (Note, we have to add a backslash to escape the backslash in the string.) The parser will turn the regex notation into a nice pretty S-expression.
> (cl-ppcre:parse-string "^.?$|^(..+?)\\1+$")
(:ALTERNATION
(:SEQUENCE :START-ANCHOR (:GREEDY-REPETITION 0 1 :EVERYTHING) :END-ANCHOR)
(:SEQUENCE :START-ANCHOR
(:REGISTER
(:SEQUENCE :EVERYTHING (:NON-GREEDY-REPETITION 1 NIL :EVERYTHING)))
(:GREEDY-REPETITION 1 NIL (:BACK-REFERENCE 1)) :END-ANCHOR))
At which point we can tell it's tricky because there's a capturing register using a non-greedy repetition. (That's the \1 and the +? in the original.)
The top level is an alternation (the | in the original) and the first branch is pretty simple: it's just zero or one of any character.
The second branch is the fun one. It's looking for two or more repetitions of the captured group, which is itself two or more characters. So, for instance, "aaaa", or "abcabc", or "abbaabba", but not "aaaaa" or "abba".
So strings that this matches will be of non-prime length: zero, one, or a multiple of two numbers 2 or greater.
But it is not true that it matches only "any character repeated a non-prime number of times" because it also matches composite-length sequences formed by repeating a string of different characters, like "abcabc".
If we actually want what the spoiler says — only non-prime repetitions of a single character — then we need to use a second capturing register inside the first. This gives us:
^.?$|^((.)\2+?)\1+$.
Specifically, this replaces (..+?) with ((.)\2+?). The \2 matches the character captured by (.), so the whole regex now needs to see the same character throughout.
Average Matt Parker code
I seem to remember he wrote something in Python that took hours to run, and his community got it down to milliseconds in C.
What took Matt's code over an entire month to run, viewers optimized so damn hard that the majority of the runtime of the code is just loading the words, so they started optimizing the code to run while the word list was loading. Takes like 4 milliseconds to load the word list, and 2 milliseconds to run the program
People joke about the Parker Square, but he's unironically the most inspiring public figure imo. The king of Doing The Damn Thing
I remember the time he Excel'd himself.
The answer says "any character" not "any characters", so it is still correct.
I upvoted this because I hate it.
Whatever you do, don't get in a time machine back to 1998 and become a Unix sysadmin.
(Though we didn't have CL-PPCRE then. It's really the best thing that ever happened to regex.)
I was a sysadmin with some Linux usage in 1998, does that count?
Just waiting for the oppertunity to hide this in prod.
I'm going to assume the answer is a magic square attempt that just isn't very good
