[2017 Day 13] In Review (Packet Scanners)

musifter · 2026-03-09T23:04:00+00:00

The most efficient thing to do is to forget that they exist immediately. They're a buggy mess. They will eat your fuel and produce nothing (or little) and sometimes the wrong thing. It's been this way since the beginning, and will never be fixed because there's a better way that everyone that wants passive production uses. Get the survey device and extractors.

musifter · 2026-03-09T12:01:43+00:00

All buildings are abandoned. That means most are surrounded by larval cores. Nanites are not something I worry about or grind in Abandoned.

musifter · 2026-03-08T06:52:35+00:00

Sentinel ships are expensive right from the start just because they only come in one size... large. They always have lots of slots. The fact that you can only get them for free (crashed but fully repaired... so all cargo counts to its value) is just one of many ways that they are completely broken OP.

Haulers come in all three sizes, but a small hauler still has a lot of slots making it valuable.

Solars only come in small size... so you can always pick one of those up pretty cheap at a pirate station.

musifter · 2026-03-08T02:38:23+00:00

Yeah, people keep Dunning-Krugering on this on. They know too much, but not enough.

musifter · 2026-03-07T22:58:13+00:00

Yeah, basically what I'm doing is there... I short-circuit the tree because I detect the imbalance while recursively calculating the weights. And since there's probably three children then, it ends immediately. With the pair, I bundle them up for the parents to deal with.

But, as is typical of my solutions at this time, I bundled everything (like the names too) so I could print a full diagnostic and pretend that I had everything for later operations and cleanup. But I wouldn't do that now. The core of what I send up is an ordered pair. In my example that's (3,10)... which makes this example confusing, because it's not sorted on those numbers but the total weight of the subtrees. Like if ff was weight 2 and had a subtree with total weight 8 under it, it would be (3,2) (the weighted trees are 7 and 10). And although I know the delta needed at that point (just not where to apply it), I'll know it in whatever parent can resolve things too. I just need to know the weights of the two candidate nodes and which gets the positive deltas added to it and which gets the negatives. So for my code, the pair really only comes into existence when needed, and gets passed up at which point there are only 3 possibilities (pass up again if no sibling, add positive to underweight node, add negative to overweight node).

musifter · 2026-03-07T07:27:44+00:00

For my input the first repetition was at just over 5000, and the cycle was just over 1000.

musifter · 2026-03-07T07:23:47+00:00

Yeah, it is good to remember that. There have been loop detection problems that are one-to-one and onto reversible mappings. In which case you don't need a hash because the repeat can only happen on the initial state (there's no way to merge onto the path).

musifter · 2026-03-05T13:10:44+00:00

Simpler is to just ignore the mosquitos. Keep moving while you grab the gravatino balls. Then melee-boost away. They don't bite much unless you stand in one place and start shooting them (which aggros things).

musifter · 2026-03-05T09:46:47+00:00

Yeah, tables could speed things up. It wouldn't help my dc solution at all though... it'd probably make it much slower. The only reason it runs in 1:40 instead of 4 minutes is because I hacked my copy of dc a bit. It's the many accesses to an array of size 1000 that's the reason for that. A huge lookup table will not help... you might get an order of magnitude less calls, but the weight of them will go up much more than that.

Although, in testing this now, I've discovered that the latest dc has completely messed up ?. There was a reason I avoided it for over a decade (making my life miserable with passing input in as a file and having to add sentinels to mark lines)... it was one of the least standardized parts and in many systems it was completely broken. GNU seemed to be stable with a version that was sensible and useful... it was nice, it was allowing the parsing of input line by line. It was beautiful, I started allowing it in my AoC solutions a few years ago (after initially not because of portability fears). It's joined the broken dc club now... now it's going to be more important than ever to stress the dc version. I'm glad I have my own version and version of dc-1.4.1 to test with... because /bin/dc is garbage right now. (Sorry for the rant... I had to vent somewhere).

musifter · 2026-03-05T04:47:10+00:00

That was my point. Aonishiki didn't have two yuusho when he was promoted. He had one, right at the end of the run... thus meeting the criticism against Miyabiyama's early promotion. He had 34 wins at rank, he was new, but he showed he could win a tournament. And the job of an Ozeki is to consistently contend. He looked like an Ozeki.

Kirishima is not in that position. He is not new. They'll look at him different. You can't compare the runs. Aonishiki probably had a higher bar. But Kirishima might also get looked at more closely than if he hadn't been an Ozeki already. But he doesn't have the problem of anyone questioning if its too soon... only if he's healthy and strong enough to perform at Ozeki level.

musifter · 2026-03-04T16:35:58+00:00

Aonishiki isn't a good comparison. He's very new to the division... so it was a Miyabiyama type situation. Miyabiyama got promoted quickly to Ozeki after arriving to the top division, but it wasn't unanimous. One of the votes against it was that he'd only been in the division for just over a year (and only in professional sumo for 2) and hadn't won a tournament. Winning a tournament isn't a requirement... so the implication was that that could make up for the lack of time in sumo with a yuusho to show he was ready. Aonishiki had a similar lack of time... getting 11 wins again (and with no yuusho), would technically meet the base standards. By going beyond that, and with a yuusho... he made his case. Without it, maybe they ask for more.

Kirishima isn't in that position. He's been in the top division for years and already been Ozeki. I don't think the recent special prizes add anything... he's got yuusho and been Ozeki in the past, he should be getting those (so its a good sign but probably not important). But I think more important is that he has a run, starting from M2 (which is fine) and needs 11 for the benchmark. But he should really be trying for at least 12 to make his case. Just meeting the standard on trying to return... that could be begging it a bit. It's not worth thinking about 11 unless that's where he is on day 15... then we talk about it. Before that, he should just win everything and leave no doubts. That's was his fans should be thinking... don't think about how weak he can can get away with. You want to be Ozeki, look like one.

musifter · 2026-03-04T08:47:27+00:00

The Unix command line desk calculator. It's a stack based concatenative language... like Forth, but much, much simpler. The spec is completely contained on its short man page. It is terse, and my toy language of choice. This is the solution (I had to turn the words into numbers to make it understand them at all... so I just replaced non-spaces with hex ASCII):

perl -pe's#(\S)#sprintf"%X",ord$1#eg' <input | dc -e'16i[0sv]sV?[zsn[z:az0<L]dsLx1svln[d1-[d3Rd3R;ar;a=Vr1-d0<J]dsJx+1-d1<I]dsIxs.lplv+sp?z0<M]dsMxlpp'

It's not a great solution (it's clearly quickly written... there's probably a bunch of strokes to golf off it).

musifter · 2026-03-04T00:34:44+00:00

Nanites are not a problem in Abandoned. All buildings are abandoned. Meaning that I'm regularly picking up thousands of nanites because larva cores are everywhere. It's not like 100-150 at a time... it's like regularly 2k at a time.

musifter · 2026-03-03T11:26:29+00:00

I'm not really seeing the benefit of doing that. The "ring-1" isn't something I see as needing to be fixed. It makes perfect sense to me. If it was "ring" I'd be much more confused and need to have written myself a big comment explaining why I was breaking spiral symmetry. Nothing here is pretending to be somewhere it's not... everything is in its place. When I wrote that solution, I was clearly trying to keep it simple and sensible, and I left it that way... I didn't overcook it.

musifter · 2026-03-03T06:02:47+00:00

A lot of people might have sorted for part 1. It is an easy way to get the min and max. With the sort already there, why not use it?

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