Adam R. Wood (zotmeister) wrote,
Adam R. Wood

Puzzle 33: Ariadne's Lament

You've probably seen one of these before in a toy store:

They're universally called "Labyrinth", and they are a game of skill: rotating the knobs on the sides tilt the playfield, the task being to maneuver the marble from "Start" to "Finish" in only that manner and without the marble falling into any of the myriad of holes along the way. They're manufactured in many sizes and difficulty levels by a number of different companies. Microsoft Windows XP Plus! comes with an excellent computerized version.

Anyway, Nikoli has this clever and very entertaining puzzle (my absolute favorite among their offerings!) named Yajilin - "arrow link", or Arrow Ring for those who "attended" last year's United States Puzzle Championship - whose solutions look so much like one of these to me that I decided to break the ring and put in a "start" and "finish" to get the logic-puzzle version of the classic game. Throw in a title that alludes to 'labyrinth', and the metamorphosis is complete - without further ado, I bring you Ariadne's Lament:

Unsolved sample on left, unique solution on right.

Condensed instructions would, like The One Ring, not be terribly condensed, so I'll skip straight to the list:

1) The objective is twofold:
- 1a) determine for each dot if it is a "pit" or if it's "open" (I suggest shading in pits and marking open dots with a smaller dot);
- 1b) to create a linear path of "edges", each edge connecting two orthogonally adjacent open dots, that travels from the 'S' (start) to the 'F' (finish) and uses ALL open dots EXACTLY ONCE and NEVER uses a pit. I call this path the "thread", and it's essentially the way out of the maze - like the black path on the labyrinth game, it connects start to finish without encountering any holes.
2) The dots marked 'S' and 'F' are open (natch).
3) The thread may not touch or cross itself at any point. This means the 'S' and 'F' dots will have exactly one edge incident on them, and all other open cells will have exactly two edges.
4) All the edges must be unit-length (so you can't draw an edge through spots where dots are "missing" from the otherwise evenly-spaced grid).
5) Number/arrow pairs - "signs" - appear where dots are missing. Each number is the quantity of pits pointed at by, and directly aligned with, its arrow. These signs, not unlike the arrows in Stargazers, "see" right through each other, straight on through to the edge of the grid.
6) No two pits may be orthogonally adjacent to each other (that is, one edge-length apart).

The sample serves the job of introduction rather nicely; if you're finding it a bit tricky, I'll show you what to look for:

(Start of sample puzzle solution)
The key to solving these is to note that any dot that isn't part of the thread is a pit, and vice versa. Working back and forth from marking dots and connecting open ones is typically necessary. It also helps to keep an open mind regarding the grid as a whole; although these can be solved purely deductively, they decidedly reward intuitive solvers by giving up where to apply the logic faster.

I start with the bottom-center sign. It says there's one pit to its left. Well, there are two dots to its left, and one is the start, which I know must be open, so I shade in the dot to the right of the 'S', marking it as a pit. Then, since two pits can't be next to each other, I put a small pencil mark in the dot above it.

Now I start to roll out the thread. From the 'S', the next dot in the thread must be up, since there's a pit on the right. I draw in an edge connecting the 'S' to the dot above it. Since it's part of the thread, it must be open - meaning the sign in its column now has only one cell left that could be the pit it's referring to. I mark that pit, and the open dot to its right. The thread must now turn right; I draw the edge so as to overlap the pencil mark I placed there earlier (the edge now serves the same purpose).

Hm - the finish is right there. What's keeping us from connecting the thread to the 'F' and calling it a day? Well, two things - firstly, we have an open dot just ahead, and ALL open dots must be part of the thread. Secondly, any dot that isn't open must be a pit - and that'd be a LOT of pits right next to each other! (Also note that the puzzle grid doesn't have any walls in it - but that doesn't mean they aren't there! Just like the pits, they're invisible to begin with. Walking into one would hurt. Especially if it was lined with spikes. Or electrified. Or something. But I digress.) So the thread must now turn upward.

But what happens now? The signs aren't of any use at this point... or are they? Not what's written on the signs, sure, but the positions of the signs themselves are useful. Look at the top-left corner. That dot isn't the start or finish, and it has only one way in or out. It can't possibly be part of the thread - it's a dead end! But if it's not part of the thread, it's a pit, meaning the dot to its right must be part of the thread. And how convenient that this dot is now in a little corner! It needs edges connecting to two different open dots, and there are only two alternatives, each of which only has a single option from there. I draw edges for two dots down from there - which links up with the previously laid thread - and for two dots right.

The next clue is the right-pointing sign. There are two dots there, neither of which is marked as open. Can you tell which is the pit? I can: if the pit were on the right, it would simultaneously force the upper-right-corner dot to be open, since it'd be next to a pit... and a pit, since it'd be in a dead end! That can't be, so the pit is right next to the sign, and the thread works around it. The up-pointing sign can now be satisfied, which then forces a leftward step for the thread.

To finish, the remaining unmarked dot on the bottom row - between the signs - must be a pit, since it's a dead end; the open cell above it can only look up and left for its thread buddies, and that completes the path from start to finish. But wait... what about the center dot? Easy - it has to be a pit. It has to be a pit because it isn't part of the thread, and it can't be part of the thread because the thread is finished! Yes, that is perfectly valid logic. Think about it.
(End of sample puzzle solution)

This is my first attempt at creating one of these - I actually made this before the sample - and I must say, I really like how it came out. It's fairly challenging. I ran into a few snafus putting it together, but it was educational. I'm glad I can pull these off. Any questions or comments about the rules or design can be posted here; email me your solution and whether or not you own a working NES and I'll check your solution and maybe offer a prize. Maybe. - ZM

Tags: ariadne, puzzles
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