wow, so many regex gurus here... codegolf.stackexchange.com/q/179747/17602 seems like it might be their next target

heh

@Neil In a pure regex though, taking unary input, it'd have to return 0 for 0 and 1/n for everything else

perhaps no-match for 0 would be more appropriate

codegolf.stackexchange.com/a/180263/78849 Is a spicy interpretation of arbitrary bases

@Deadcode I was wondering whether it would be more efficient to divide by all primes of the form 4k+1 and then check whether the result is a square or twice a square

@Neil Great idea, thanks! Did it.

I was wondering if were merging k sorted lists. Our best time would Onlogn using a heap. Which is the same as sorting (quick sort) . and that's because we have to look up values in our heap for every value we come across.

How about instead of using a heap to look up and insert, we use a binary tree to break down the list into intervals using right and left pointers. left = lo to mid and hi = mid+1 to hi. until our leaf is a single pointer on a single list. Then we just merge every subsequent list into the furthest left node as we traverse the tree. Would this be O(nk) or O(nlogk)

I came across this paper which says it's nlogn but I don't know if it's right.

web.cs.dal.ca/~whidden/CS3110/assignments/a5_solution.pdf

@Deadcode checking for a square is still eye-watering if you don't have recursive backreferences though

@Deadcode since I got it so badly wrong last time, would you mind explaining why |))\12(?!^) can't be ))\12?

I'm teaching my friend 05AB1E, any suggestions on explaining what a stack is?

Another good example is those plate stacks that you'll sometimes find at self-serve bars/restaurants/whatever where there's a spring at the bottom so when you put a plate on top, it pushes the stack lower, and when you take a plate off, the stack lifts up.

Super useful dice trick: instead of rolling 6d8, you can simply roll 4d4 + 2d6 + 1d10 + 1d12 - 2 and the results are almost indistinguishable.

@MilkyWay90 how much prior programming experience does this person have?

@PhiNotPi Beginner in Javascript (forgot most things about Javascript)and experienced in python

@Neil Now that you mention it, it seems to work, but I didn't try it originally, because I thought it shouldn't work. If the perfect square test fails, it should backtrack the loop that divides by ≡1 mod 4 primes, and that should be able to make the perfect square test succeed when it shouldn't. So it seems I have to explain why it does work when it seemingly shouldn't...

Do you know why it works?

Oh, silly me. Yes, backtracking will never change a failure to a success. It'll only make the regex slower, but that's fine, this is golf.

joys of the burden of proof... :P