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01:17
3
Q: What do I receive when I get a drop for a character whose constellation is already complete?

LemonOn the last couple 10 Wishes I have pulled, I have obtained Noelle on every single pull, so I have received 5 of Noelle's Stella Fortuna. Once I have Noelle's constellation complete, what do I get the next time the character drops on a wish?

 
2 hours later…
03:23
1
Q: What do the Letters and Numbers above the ISSN called, and what do they mean?

MorganWhat do the letters and numbers above the ISSN mean and what are them called? HAC P ALZAA TRA-HAC-ALZAA-AUS-1 I'm presuming the "AUS" part is because it's the Australian version of the game.

 
1 hour later…
04:44
Ah yes Microsoft, thanks for letting me play the hottest new game of this season - Space Warlord Organ Trading Simulator
 
2 hours later…
07:05
Holy snap, crackle and pop. The bonus yo u get for doing thos epuzzles within an hour or two is just silly.
Here I am, unable to sleep, so I decided to do the puzzle... thinking I'll do well
but NOPE
still 100 points behind @badp
oh wait that's for all days
man I'm sleepy
anyway, day 12 done. Another fun one. Good ol' recursion.
I was worried I wouldn't have tiome for day 12 since I have plans tomorrow (uh, today!)... but then I couldn't sleep sooo VICTORY!...?
Anyway now I'm going back to tossing and turning in bed.
 
1 hour later…
08:31
@Wipqozn what bonus?
@Wipqozn hooray, we're bad sleep buddies!!
 
5 hours later…
13:53
@badp Yay!
@badp The leaderboard gives points based on how quickly you complete it
 
2 hours later…
16:02
1
Songwriting

Proposed Q&A site for a community of songwriters, to help with lyrics, melodies, judging each other's songs or just for general songwriting-based discussion!

Currently in definition.

 
2 hours later…
17:52
My solution to today's puzzle was way overengineering and ran both parts in a total of 1.378 seconds. But on the upside, I did so much extra work for part 1 that part 2 took about 30 seconds to write
@Sterno I ended up rewriting my part 1 after I started part 2. Realized there was a smarter way to make sure I didn't visit the same cave twice, that also made it easier to visit only one twice
So my part 1 took me 1h11m and my part 2 took me 5 minutes lol
Part 1 runs in .043s and part 2 runs in .418s
Mine was pretty brute-forcey. Had classes and objects for "Cave", "Cave Path", "Cave Segment"(the thing from the input file that tells you how two things connect), and I was using recursing and cloning CavePath over and over again and just walking everything until I hit the end or a dead end. Dumb, but fun to write!
I'm willing to bet that starting at the end instead of the beginning, building up partial strings of paths to the end, and just concatenating those as you keep walking back to the start would have been way faster
My way of walking everything and starting at the start meant many paths were repeatedly going down the same dead ends
Their scoring system is pretty cool. Someone could do the day 1 puzzle first, never do another one, and still pretty much be guaranteed more than the 0 points I'll ever have because I do them the next morning
Wish instead of 100 - your placement it was #peopleregistered - your placement or something
18:33
the global leaderboard is kinda dumb yeah
I did crack the top 1,000 this morning though
19:07
@Sterno My key algorithm was just:
Whoops
Well I'm not typing this on mobile
But recursion
 
4 hours later…
22:52
Lol
Yeah, I had a recursive algorithm for it
I first built a dictionary that consisted of {cave: [caves it links to]}
Then the recursive function was just:
def path(loc, connections, trail=[]):
    if loc.islower() and loc in trail:
        return
    global paths
    if loc == 'end':
        paths.append(trail)
        return
    for conn in connections[loc]:
        path(conn, connections, trail[:] + [loc])
Checks if we're in a small cave that we've already visited, if so that potential path is dead
if we're at the end that's a successful path so we plunk that onto the global array of paths
Otherwise we add the current location onto the trail and recurse onto each of the caves that connects to the current one
Part 2 just required a minor modification that tracked whether we had visted a small cave twice yet
And if we hadn't, it would, instead of ending when it detected a duplicate, set the flag to say we'd already visited a small cave twice
23:35
def Explore(self, path):
        if self.isend:
            self.counter += 1
            return

        if self.IsMinor and self.visited:
            return

        self.visited = True
        for p in self.paths:
            p.Explore(path)

        self.visited = False
@SaintWacko You don't even need to actually track the paths outside of debugging purposes.
that's just part 1 of course

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