Archive for May, 2018

24 Hour Puzzle Competition 2018 (aka Puzzles 4-15)

I did not participate in the 24 Hour Puzzle Competition (24HPC), as I’m nowhere near good enough at solving logic puzzles to really justify spending so much money to fly out and spend several days at an international (even if somewhat light-hearted) puzzling event. However, I was involved in helping write some puzzles, at the behest of chaotic_iak, the leader of our writing team (Puzzlers Club, a particular Discord server that I’m in). I’ve not really written that many logic puzzles, to be honest – there’s, what, 3 logic puzzles on this blog, I’ve written about 1 logic puzzle for a puzzlehunt once, ok I guess I wrote a decent number of “mini 5×5 logic puzzles” back at MOP (but that doesn’t really count, plus most of them were kind of trivial as intended), and I’ve written maybe 10 or so logic puzzles for various events at our Discord.

For this event, I wrote 12 (out of 33) puzzles, so more or less doubling my puzzle writing total. (More than 12, if you count the puzzles that didn’t make the test, plus some easy examples in the instruction booklet.) Surprisingly a lot, given that initially I had asked to “just be a testsolver”. (Turns out that (usually) I am far worse at testsolving for speed (especially since I wrote the puzzles for my strongest types), though I got some useful uniqueness checks in on certain hard puzzles.)

I guess now I’ll just talk about the puzzles that I’ve written. In order of the booklet:

Puzzle 3. Easy as ABC (12 points)

Place a letter in each cell so each row and column has the given letters exactly once. Letters outside the grid indicate the first letter seen when looking from that direction, ignoring empty cells.


In the end, we settled on all point values divisible by 3 (except the first observational puzzle, which was needed to end up with a total of 1000 points, not divisible by 3). Furthermore, chaotic_iak wanted the minimum point value for puzzles to be 10 points, so upon this change, the minimum point value is now 12 points. It probably overly weighs the easy puzzles now (I think a minimum of 9 would be fine, honestly), but that’s fine, the other solvers need all the boost that they can get against Ken Endo šŸ˜› (Is Ken even competing this year? I hope he is, since a design goal was so that he wouldn’t necessarily finish the test. [He is not. RIP] If not… lol @ highest score (though one fairly strong full-test testsolver scored 480 points, so I think 600 points for non-Ken’s is probably doable. [After competition note: top score was 627. Still low, but within my expected range.] Probably 700-800 for Ken, which is about right I’d say.))

With that said, this Easy as ABC isn’t *that* easy, and 12 points is perfectly fine for this puzzle. The 3 theme only vaguely shows up here, you know, 3 letters and all. I don’t often see too many Easy as ABC’s that are 6×6 but still use 3 letters (most 6×6’s seem to go up to 4 letters). As such, the large amounts of empty space – 3 spaces per row/column! – may be fun to play with. Also, I tried to minimize the number of clues used, and the solvepath should be fairly tight and narrow. I’m happy with how this puzzle turned out, especially for my first construction attempt at this genre.

Puzzle 5. FillominoĀ (15 points)

Divide the grid into regions. A number describes the size of its region. A region may have zero, one, or more given numbers. Two regions with the same size may not touch orthogonally.


After several attempts at making a 3-themed fillomino, which ended up being far too easy, this was a fairly satisfactory attempt, noting how the diagonal strands of 3’s interacted with each other. This ends up being a little bit of counting at the beginning to determine how the center-bottom line of 3’s resolves, and then a funny global deduction with some tweaking to get everything right. So yeah – overall, very global fillomino in a genre that’s typically centered around local deductions. I’m happy that the tweaks didn’t end up being the immediately obvious solution, and also happy that none of this compromised the uniqueness of the solution. 20 and 18 are indeed to reference the year number; 18 and 23 are just for stabilization for a maximally interesting solution.

Puzzle 10. Walled HeterominoĀ (21 points)

Divide the grid into triminoes. Triminoes of the same shape and orientation may not touch orthogonally. Some borders have been given to you.


It turns out that the base figure (without the 3 in the middle) is already quite difficult to tile, so I was initially frustrated when my larger-scale changes wouldn’t work out. Eventually, through bifurcation, I was able to find that the 3 (plus the edge to the side) could work to force uniqueness. Later, after it was testsolved, I was able to find a better deduction than simply bifurcating, so I’m happy with this puzzle.

As it turns out, walls aren’t actually standard in Heteromino (not that Heteromino is a terribly standard genre), but I think they are a natural addition.

Puzzle 11. Triopia (24 points)

Place some triminoes in the grid. No two placed triminoes may be identical, counting rotation/reflection as different. (That is, two congruent triminoes with different orientation may touch.) Triminoes may not touch, not even diagonally. If a cell has some directions, the directions point to all nearest directions where there are squares occupied by triminoes. (In other words, for each other direction, either there is no trimino in that direction, or the closest trimino is farther.)


What happens when you tweak the ruleset of Pentopia? You get this far less interesting variant. The decrease from 12 to 6 pieces and the loss of changing orientation really makes it hard to construct an interesting puzzle. The obvious thing to do here is indeed the correct thing – case work all the cases. Fortunately they resolve fairly quickly in all but two cases, one which is the mirror image of the other (disambiguated by the corner clues).

Puzzle 14. TriceptĀ (30 points)

Draw a single closed loop in the grid, passing through all cells. The loop may only go horizontally and vertically, and may only turn on centers of cells. Among any three consecutive squares of the loop, the loop must turn at least once and must go straight at least once. Black squares are inside the loop. White squares are outside the loop.


I think the path mechanic originates from a Puzzling Stack Exchange post by TheGreatEscaper. I couldn’t think of a way to make the puzzle interesting (without clue overload) in the incomplete loop version (i.e. you don’t need to pass through every square), which was the original. And thus, I was able to make this pretty nice antisymmetric grid. In/out makes antisymmetric grids pretty easy to make – draw a grid, notice which places are inside and which places are outside, and tweak things to make unique. I was quite happy to be able to get rid of some of the clues, and I think this puzzle is now clue-optimal.

Puzzle 22. AkariĀ (12 points)

Place some lightbulbs in the grid such that every cell is lit up. Lightbulbs light up all cells in orthogonal directions, until hitting a black cell or the border of the grid. Lightbulbs may not light each other. Each number indicates the number of lightbulbs in orthogonally adjacent cells.


Quick Akari that I whipped up after settling on the 3 formation. It was difficult to force uniqueness but with the addition of quite a few more blocks + some 0 clues, that was doable.

Puzzle 23. All-Three AkariĀ (21 points)

Follow standard Akari rules (puzzle 22). AllĀ instances of 3 on black cells have been givenĀ to you.


This puzzle demonstrates the various clueless deductions that you can make in this variant (which I don’t think I’ve seen before). I’m a little miffed that I had to introduce some clues in the end, though. Ideally, I would have not needed to use any numbered clues at all, nor 1-size islands, but alas.

Puzzle 24. Slitherlink (12 points)

Draw a single closed loop in the grid. The loop may only go horizontally and vertically, and may only turn on the dots. Each number in a cell indicates the number of edges of the cells that are part of the loop.


I started with a bunch of 3’s in a pinwheel formation. Eventually I abandoned my idea of 000-111-222-333 in favor of antisymmetry adding up to 4 instead. Then it was a matter of stabilizing the puzzle with some more clues (including a fifth 3-triple). I like this moreso for the design than the solvepath, which is pretty straightforward, and yeah, some of the clues are unnecessary (except for design).

Puzzle 25. All-Three SlitherlinkĀ (45 points)

Follow standard Slitherlink rules (puzzle 24). All instances of 3 in the grid have been given to you.


This puzzle is also influenced by design. And All-Three puzzles without any 3’s are funny.Ā  (also see – the All-Three Skyscrapers by chaotic_iak) I had to add three 2 clues in the end to force a unique solution, hence the big 2 looks a little weird. Unfortunately I could not find a purely logical solution for the right half of the puzzle besides caseworking.

Puzzle 29. NonogramĀ (36 points)

Shade some cells black. In each row and column, the list of numbers matches, in order, the lengths of consecutive blocks of shaded squares. If there are multiple blocks, each block is separated by at least one unshaded square.


Eh, this puzzle could be better. There is a theoretical global deduction that can be made exploiting the structure of the row clues, but this puzzle does not use that. Furthermore, the picture that you get at the end is a VERY dubious 24 HPC + additional junk in the left corner. Oh well.

Puzzle 31. Star BattleĀ (60 points)

Place some stars into cells. Each row, column, and outlined region has the indicated number of stars. Stars may not touch, not even diagonally.

puzz31Whew. Some neat deductions here, I’d say. I also liked the layout and the idea of making the P and C double regions. (one interpretation – PC not only means Puzzle Competition, but also Puzzlers Club!) Region counting makes the two halves of the puzzle obvious. The bottom half should be relatively straightforward (although the fact that you immediately lose the left column plus an appreciable amount of the right lends to some decent deductions), and then the top half is harder – essentially how I constructed the puzzle, solve and tweak the regions. At the end I was frustrated at all the contradictions I was getting so I drew the last few regions to accommodate the stars, but I think it worked out well.

Puzzle 33. Kropki (84 points)

Place a number from 1-N into each cell, where N is the length of the grid. Each number appears exactly once in each row and column. If a white dot appears between two numbers, the two numbers must be consecutive. If a black dot appears between two numbers, the larger number must be twice the smaller. (There can be either dot between a 1-2 pair.) All possible dots are given.


This puzzle was basically constructed (after placing the initial 3’s, which leads to a decent number of digits being placed) by looking at all of the possibles, and doing a bunch of Sudoku strategies to ensure uniqueness. Then I had to add some more dots, spoiling the design a little. Ah well. I also thought I had an insane pointing quad but alas that was not to be (I accidentally erased a possible). Fortunately I had a spare rectangle flip (like swap the values of x/y;y/x) which could be easily clued with a few additional dots, and all wasĀ  good again. At least there should be a decent logical path now. Plus, I do enjoy the general sparseness of the clues, including the two empty columns side by side.