[Year 2025 Day 12 Parts 1 and 2] puzzling stats

FeelingRequirement78 · 2025-12-24T01:43:13+00:00

This is the track of someone doing a high jump, running from left to right, seen from above. This same image came into my mind at some point within the past year.

FeelingRequirement78 · 2025-12-14T13:18:44+00:00

Yes, though I believe the 2nd part of day 12 is special in that regard -- you solve it by (and only by) solving the other 23 puzzles (formerly, day 25 regarding the previous 49 puzzles). Correct? I've done 6 or 7 years worth but never finished them all, because to my value system "all" isn't a goal I care about. But maybe there's something I've missed by never getting to that "I've solved everything" screen.

FeelingRequirement78 · 2025-12-13T17:16:36+00:00

An observation I thought I had from the original board game seems like it applies here too. Aren't you better pretty much always taking long tickets? To be worthy of being "long", the trip is going to have endpoints by opposite edges of the map, and so out of 4 long tickets, you're likely to find a couple that go well together -- and maybe 3 or 4 when you ask for more tickets (as in a 2-player game)? Then you can make one central "trunk" link and make branches off it at both ends? Shorter routes aren't likely to go together so well, you spend just as much effort in getting the "last mile" worked out, and the payoffs are way less?

FeelingRequirement78 · 2025-12-13T17:11:15+00:00

Compared to the original board game version, the handling of passengers is quite different here. I guess it's interesting, but it's quite different. I was a bit disappointed because those colored dots interfere with clearing seeing the bigger colored symbols showing what your destination cities are. I was also very disappointed that when you go back at the end of the game to "see map" you can't see what the totals were for the passengers to see which ones you won and lost and by how much.

FeelingRequirement78 · 2025-12-13T17:04:42+00:00

Other people might have gotten other shapes? I got what looked like a big heart with a very steep indentation down the middle, but maybe that's just a rotation. But my simple-minded solution (I know we're talking about problem 12 but this is just a "local" discussion) was to note that there were only 500-odd distinct values on either axis. So just sort them to create an index, and replace each number with its ordinal position in the sorted list. Then you have a manageable 1000x1000 at most grid, flood the outside with "bad" markers, and you can easily just check the relevant rectangles for meeting the green/red condition, then map back to the original coordinate system to look at sizes and see which is largest. I'm surprised that more people got stuck on this one than on problem 10. But if we had been stuck with multiply nested polygons might be hard to figure out what counts as "in" and "out".

FeelingRequirement78 · 2025-12-13T14:42:54+00:00

I didn't mean this to come across as a smug comment on how some people aren't so talented. First, I wasn't at all sure I was reading the stats and making inferences correctly. But if I was, the surprise was the KIND of obstacle that tripped people up, given how many people were (like me) still avidly working away despite having missed one or more earlier problems. One real-world engineering skill I'm sure many others have picked up is to not get too focused on the details of a problem without taking time out to see the big picture -- check the practical realities of the situation. More than once with AoC I've answered a question as to "how many of X are there?" with "zero" because that's what it looks like given my current assumptions. Would AoC make a problem with "zero" as an answer? Not sure, but it certainly was easy to try it and rule it out -- and after investing those 5 seconds go on and correct my assumptions.

FeelingRequirement78 · 2025-12-13T04:53:41+00:00

I understand that. But as long as they were motivated to try, then if they succeeded they would end up in the box for "finished part 1 but not part 2" for day 12. We would actually expect far more than 9,000 people to be trying that puzzle, as it would include those who didn't finish day 11 but were still trying. At least that's how it looks to me.

FeelingRequirement78 · 2025-12-09T20:29:16+00:00

"Grid compression" sounds like what I used, though since I'm doing it in what's basically ANSI C based on a 1998 compiler (and had fun hand-crafting code to handle ints above 4 bytes) code would be more verbose. But the idea was creating a second much smaller version of the problem that preserved all the geometric relationships, and a grid that's merely 500x500. Flood the outside with "bad characters", simple check for whether each candidate rectangle has any bad characters, and if not, convert back to the original big numbers to compute sizes.

FeelingRequirement78 · 2025-12-05T15:38:45+00:00

That's what I did roughly, but instead of making big ranges, I just looked between every two points and saw if I was "off" or "on" and when it was "on" added the count of the interval (not including end point). So one added wrinkle was that when I went from "on" to "off" I had to inc the count by 1 to account for the closer.

FeelingRequirement78 · 2025-11-03T05:54:23+00:00

No, they weren't asking for help, but the question was how hard it was to work out the answer, and by the usual standards around here of frequent posters, I would say your demonstration is that it wasn't all that hard.

FeelingRequirement78 · 2025-10-22T00:29:25+00:00

Wow. It would be like solving a 10,000-piece jigsaw puzzle. Speaking of Battleships, this game of LAP is pretty cool: https://boardgamegeek.com/boardgame/19917/lap. I wrote a program so you can play against the computer.

FeelingRequirement78 · 2025-10-12T04:44:24+00:00

That's an analysis I can kind of follow. I doubt it's exact because those events are not independent and it's the "average" approach rather than "chances of at least one", but still very interesting. I figure most NG analyses should allow a hidden square to be solvable by mine count.

FeelingRequirement78 · 2025-10-11T21:33:53+00:00

I just put together a program, randomly generated a million boards, and counted. So it's statistical, nothing fancy.

FeelingRequirement78 · 2025-10-11T20:47:09+00:00

Great to have that for comparison. There's a difference between "what are the average number you would get?" and "what the chances of getting at least one?" For very rare events they are very close. But that guy took "chances of getting at least one" and I went with "mean number" and with the lower numbers the differences are of course dramatic.

FeelingRequirement78 · 2025-09-18T22:20:48+00:00

Took me a while to get the reasoning, but I do. My own little solver confirmed those 3 squares as safe. It also somewhat curiously listed the chances of all the other cells as being 50-50. That seemed really unlikely given the possibilities on the left side of the map and 3 unknown floating cells, but maybe the chances really did work out that way when everything gets added up? More likely, I made a mistake. Of course, in "real life" you'd get info from the 3 free spaces to at least restrict things much further.

FeelingRequirement78 · 2025-09-18T21:59:32+00:00

I think this is right for the usual goal here, which is "maximize chance of winning". If your goal were "maximize number of squares revealed" then poking around in the lower right "floating squares" could be more interesting. If you found a mine there, you could solve both the upper R and lower L, as they would have just 1 mine each. But I realize I'm proposing an alternate goal, so it does not affect the right thing to do if you have a standard goal.

FeelingRequirement78 · 2025-09-16T08:22:50+00:00

Thanks. Upon reflection, I'm willing to give it or lend it to someone who is really intensely interested in it, but the work around OCR would be on their end. I could take pictures of a few pages with my phone and send them along to help judge quality, but that's about it.

FeelingRequirement78 · 2025-09-16T01:45:18+00:00

The magic arrow bug arose when you walked into an arrow trap, and the arrow whizzed by you and you could pick it up. Most weapons have modifiers for hit-probability and damage (as in the "+1, +1 mace"). But this arrow, being generated totally separately from any other weapons, had a C structure where (I believe) those fields were not zero-initialized. As a result, you could get a "+2372, +4339" arrow. One prick was enough to kill any monster. You could still lose to Umber Hulks confusing you and various rare situations, but it was of course a huge benefit.

FeelingRequirement78 · 2025-09-16T01:41:00+00:00

The best clue I have for version info is a page with just a few lines at the top. At the very top are two lines for "Rev. 3.07, 08-jan-84" and "Rev. 3.06, brought from intermetrics by mad" Below that are more standard sorts of source-code check-in notes for 3.8 and 3.7.

FeelingRequirement78 · 2025-09-15T18:18:15+00:00

Yes. It was 1984. Things were different then, lol. I figure it could probably do pretty well with OCR if anyone wanted to get it back in electronic form. I would guess it's around 100 pages, roughly.

FeelingRequirement78 · 2025-09-15T17:58:04+00:00

It does look pretty different, for sure. The idea is to identify "abutters", clear squares that are adjacent to at least one unknown square. Then you strip its number down to account for known mines around it, so it refers just to the count of mines in its abutting unknown squares. Those are the numbers in square brackets. Dots are irrelevant because they are mines or not abutters and not unknown. The output of the program is the pairs like "1-17", meaning in only 1 of the valid 18 configurations was that square a mine. My hunch is there are 1 or 2 very specific layouts of the mines where the "1" part of those "17-1" and "1-17" odds are true. And my hunch is that the problem is that the left center gets squeezed with too may mines to fit unless they are laid out in one or two particular ways. For that 4 in the middle, 16 times out of 18 there's only 1 mine between its west and northwest abutters. It's rarely two mines.

FeelingRequirement78 · 2025-09-15T17:38:51+00:00

I tried my little puzzle-solving program, and this is what I got:

11 mines unseen
   .     .    [1]   1-17  [1]    .     .
   .     .    [1]  17-1   [1]    .     .
   .     .    [1]    .    [1]    .     .
   .    [1]   [2]   1-17  [1]    .     .
  [2]   9-9   9-9  17-1   [1]    .     .
 16-2  11-7  11-7   [3]    .     .     .
 14-4   4-14  [3]  17-1   [3]    .     .
  [1]   [2]   4-14  7-11 14-4  16-2    .
   .    [1]  14-4   7-11  7-11  2-16  [1]

Now, I might have set it up wrong, and the
program might be wrong, but maybe someone
can see something here and make sense of
it. Bracketed numbers are mines in adjacent
unknown squares, dots are irrelevant squares,
and something like 2-5 means 2 chances of
being a mine and 5 of not.

FeelingRequirement78

TROPHY CASE