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Grimy
Grimy
9:42 AM
-10 on (?*) version of "find the largest square-free factor":

^(?=((?*(x+))(?=((x*)(?=\2\4*$)x)(\3*$))\5(?=\2*$)|x*)*)\1
 
 
1 hour later…
Grimy
Grimy
10:52 AM
174 on consecutive-prime-constant-exponent:

# N = smallest root of N
^(?=(x+?)(((\1(x+))(?=(\4*)\1*$)\4*(?=\5$\6))*)x$)\2

# Assert that there exists no trio of prime numbers such that N is divisible by the
# smallest and largest prime but not the middle prime.
(?!(((x+)(?=\9+$)(x+))(?!\8+$)(x+))\7*(?=\7$)(?!(\11\10?)?((xx+)\14+|x?)$))

# Assert that N is square-free (its prime factors all have single multiplicity)
((?=(xx+?)\16*$)(?=(x+)(\17+$))\18(?!\16*$))*x$

The overall algo is mine, but all I did was copy three of your regexes, change a + to a * in perfect powers, and adjust the back…
 
 
2 hours later…
Grimy
Grimy
1:06 PM
Never mind, I got 155 by doing something significantly different (special-casing multiples of 12, can you believe that?)
 
 
2 hours later…
Grimy
Grimy
3:01 PM
137 and considerably faster, still room for improvement
 
 
3 hours later…
Grimy
Grimy
5:57 PM
I posted my totient regex: https://codegolf.stackexchange.com/questions/83533/calculate-eulers-totient-function/180255#180255

This time I polished the explanation a bit more, but feel free to improve it if you want to.
 
Deadcode
Deadcode
6:46 PM
@Grimy Just gave it a quick look and looks great... but you accidentally put the same TIO link for both versions. There's no TIO for the number-of-matches version.
 
Grimy
Grimy
Whoops
Fixed
 
Deadcode
Deadcode
@Grimy Did you realize that Martin Ender's solution does basically the exact same thing, and with the same length of regex?
 
Grimy
Grimy
Oh, I didn’t
Guess that makes mine redundant, might as well remove it
 
Deadcode
Deadcode
Well, there is one interesting thing about it
That it works in ECMAScript 2018
(And the one from the Retina program also does)
And it is interesting that you independently came up with it, and that it's the same length
Something I'd like to look into is using Japt for some of the TIO links instead of (or in addition to) Node.js. I wonder if the number-of-matches solution could be well golfed using Japt.
 
 
1 hour later…
Deadcode
Deadcode
8:26 PM
Oops, I mean Leaky Nun's solution
 
Deadcode
Deadcode
8:40 PM
@Grimy (x+x) is more efficient in a lookbehind than (xx+)
 

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