Tag → logic puzzle

I’m not really satisfied with the execution, but eh, what the hell. My brain can only function at so much of its full capacity when it’s a few kilometers up in the sky.

This is a Triple Back, variant on MellowMelon’s Double Back. Briefly, draw a closed loop through all square centers visiting each bold-outlined area exactly three (= ⌊π⌋) times. Shaded cells do not influence solving, only aesthetics.

At least one person wants me to post. I’m not even going to try do a life summary. It’s too hard. Let’s just say:

1. right now, my blog drafts contain a backlog of ~7500 words and counting;
2. I was not accepted as an MIT Admissions blogger, which is bad because my blogging will continue to not reach a large audience, but good because my blogging will continue to not reach a large audience. Maybe it had something to do with the fact that, because the application form wouldn’t let me submit without any media, I panickedly cranked out the following puzzle in an hour or so to attach.

CLICKBAIT PERSONALITY TEST THAT YOU CAN DO WITHOUT SOLVING THE PUZZLE: What do you see in the puzzle image below? I have my own thoughts but I won’t bias you by posting them yet. Sound out your thoughts in the comments below! (I don’t expect this to work but I’d love to be proven wrong)

Okay so apparently how puzzles work is I go nearly a year without posting one and then when I post a terrible one, I feel guilty and obligated to post a legitimate one soon after. Testsolved by chaotic_iak.

This is a Fillomino (write a number in every empty cell so that every group of cells with the same number that is connected through its edges has that number of cells) where each tetromino has had their 4s replaced by one of L, I, T, or S describing their shape, and they obey the rules of LITS — they can touch if they are not congruent, they must all be connected, and their squares cannot form a 2×2 block. In addition, cells separated by a thick border may not contain the same number or letter.

5:27 PM phenomist: do you use gridderface to make pretty puzzles?

5:52 PM phenomist: actually nvm excel is probably easier lol

Okay I’m sorry this is a horrible puzzle where the rules don’t make sense and I didn’t even get it testsolved. I just wanted an image to concisely demonstrate the capabilities of gridderface, my puzzle marking and creation program, for the project homepage, after somebody expressed interest in using the program to write a puzzle. Then I got tremendously carried away.

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).