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?
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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.
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.
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.
Please click on the image if it looks weird, which it very likely will.
(This was a WordPress bug that should no longer be relevant.)
Nikoli Slitherlink + Masyu hybrid; I don’t know who first put them together but combinations like this aren’t rare.
Draw a loop through vertices that cannot intersect itself; each number indicates how many of the four edges around it are drawn; the loop must pass through all large dots, and it must go straight through white dots while turning either before or after (or both), while it must turn on black dots without turning either before or after.
mathgrant’s hybrid type: 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 2x2 block.