Well, there had to be something here.

Unfortunately I don’t have time for anything more complex, so here’s a low-effort illogical puzzle for the occasion. (It *has* been testsolved, at least. Thanks, Yoshiap.) It also features a brand new category, so as not to distract the people on LMI.

If you don’t already know what occasion it is, it’s easy to find out by looking through my archives or possibly anywhere else I’ve left a trail online. Or you could solve the puzzle! (Or you could scroll down to read the solution!) If this puzzle had an answer, it would be a nine-letter word, although like most of mathematics, it’s less about the answer than about the path you take to get there…

2017 edit: A warning that this has somewhat linkrotted and is harder to solve, but theoretically still possible.
As requested, a puzzle post! Straight from the WTF-variant department. Quite hard.

This is a Fillomino, with the additional constraint that for each polyomino, there must *not* exist a path (i.e. a sequence of cells, each orthogonally adjacent to the next) that includes each of the polyomino’s cells exactly once (and does not include cells outside the polyomino).

As a degenerate case, 1-ominoes are banned as well.

Edit 11/19 7:23 AM UTC+8: Fixed some transcription errors on right: **R4C17 is D instead of N, and R15C17 is G instead of E.** Neither change should greatly impact solvability. Thanks to ksun for pointing out an error.
Logic puzzles are easy to construct if one doesn’t have some specific pattern or theme in mind. It’s just that, given the increasing number of constructors and puzzles with amazing themes, I don’t think it’s very meaningful for me to just construct more puzzles of the same genres by putting down clues randomly. That’s why, for my seventeenth birthday, I took the puzzlehunt route and made something without instructions that is not completely solved by logical deduction. Still, I’ve provided all the information needed to do this puzzle initially, so I hope my not getting the inductive bits test-solved can be excused.

I’m extremely satisfied — a little incredulous, in fact — with how this puzzle came out. chaotic_iak labels it the “most ridiculous fillomino ever in history”. Apparently, it’s rather tricky.

ETA: Journalistic responsibility compels me to mention that chaotic_iak also added, “might be beaten later”. Oops?
This is a Fillomino combining the Nonrectangular (polyominoes can’t be rectangles) and Walls (polyominoes can’t span thick lines) variant rules. I think the first variant first came from mathgrant; I’m not as sure about the second, but they both appeared in Fillomino-Fillia 2, at least.

Write a number in every empty cell so that every group of cells with the same number that is connected through its edges is a shape that’s not a rectangle with that number of cells. In addition, cells separated by a thick border may not contain the same number.

Oops, I forgot the “puzzles” category was semi-reserved for puzzles I constructed/wrote, because among other things an LMI bot is following it. Anyway, if this makes up for anything, I have a puzzle that I’ve procrastinated posting for very, very long.

This is a Fillomino puzzle. Inequality signs in the grid must be satisfied by the two numbers they touch.

So, I somehow managed to get 25 points all by myself in MUMS Puzzle Hunt 2013. Well, I pestered chaotic_iak a little with 3.3 Diagnosus (.html with animated .gif) but we still didn’t recognize all the Pokémon until hint 3, at which point Google sufficed for me.

This is nowhere near the top, but compared to the usual results of whatever AoPS team I form, it’s amazing. By far the best result of AoPS was on CiSRA in 2010 (46th with 58 points), before I discovered puzzle hunts in AoPS; unfortunately due to people getting older and the influx of younger and younger people to the fora, there are less possible teammates each year and they have less time, so here I am by myself. (Also I could have accepted an invitation from a guy in the some-form-of-Elephant team, but I figure if you can win two MUMS hunts in a row you don’t need any more people.)

All in all: Yay!

I survived midterms.

This is a Slitherlink mutant. Draw a loop through adjacent vertices that cannot intersect itself. Each number indicates how many of the four edges around it are drawn. In addition, each pair of colored squares in corresponding positions (e.g. R1C1 and R6C6, R2C8 and R7C3) must have an equal number of edges drawn around them (i.e. if there were numbers placed there, they would be equal).

Yeah, and there’s this. chaotic_iak rejected this variant for his February sequence in order to get consistent 7x7 dimensions, so I made one. It’s been about a month. I have no idea why I procrastinated posting it until now.

This is a Samurai Fillomino, which means each grid satisfies the constraints on its own. Write a number in every empty cell so that, in each square grid, every group of cells with the same number that is connected through its edges has that number of cells. Note that the two grids must contain the same numbers where they overlap, but the grouping should be considered independently. I’d explain this really carefully if it weren’t the main gimmick of this puzzle.

Yay crazy hybrids! I guess this one is kind of hard.

Draw a loop through adjacent vertices that cannot intersect itself. For each pair of symmetrically placed numbers, one is a Slitherlink clue which indicates how many of the four edges around it are drawn, and one is a Contact clue which indicates the total length of all straight segments adjacent to it where segment length is always measured up to the nearest turns in the loop.

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Please click on the image if it looks weird, which it very likely will.
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(This was a WordPress bug that should no longer be relevant.)

A Naoki Inaba (JP) type, as seen on Para’s Puzzle Site.

Draw a loop through adjacent vertices that cannot intersect itself; each number indicates the total length of all straight segments adjacent to it, where segment length is always measured up to the nearest turns in the loop.

Let’s see how well WordPress’s scheduling works again. Happy Chinese New Year!