Well, it’s been over a week, which is a long time for blog posts to
be delayed after the event they’re documenting in probably all of the
world except my blog. So.
I guess this post should start with a bit of background. I’ve been
puzzlehunting for… wow, three and a half years now. I was introduced to
puzzlehunts from AoPS, when some fellow members got together a team for
CiSRA
2011, and I think I’ve participated to some degree in every known
internet Australian puzzlehunt since.
But as for my experience with the MIT Mystery Hunt in particular, I
sort of hunted with a decidedly uncompetitive AoPS team in 2012 (I think
we solved one puzzle exactly), but my serious hunting career began when
dzaefn recruited me into the Random team (then Random Thymes)
for the 2013 hunt (and I did
blog obliquely about it). We didn’t win (and I actually didn’t
participate that much because I was traveling with family) but the next
year (as One Fish Two Fish Random Fish Blue Fish (1f-2f-17f-255f (I am
evidently in a parentheses mood today because as you’ve probably
noticed, the amount and depth of parentheses in this sentence are
positively alarming (lol)))) we won.
And I do have a half-written post about that which will never get
posted (and I also didn’t participate that much, because my
family was moving that weekend) but okay, let’s just drop any semblance
of chronological coherence on this blog and dump a short version of the
list of puzzles and parts towards which I contributed solving, as I
wrote them down one year ago:
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.
![Informatix [MUMS Puzzle Hunt 2013]](/img/informatix.jpg?w=300)
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.
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.)