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Here is an example of a maze I'm calling the beehive clock maze:

maze example

Rules. Let's call the hexes TL, TR, L, M, R, BL, and BR (for top-left, top-right, etc).

  1. You start outside and to the left and step into the L hex facing right.
  2. The L hex has the numbers 4, 8, 10, and 12 on it, referring to the respective clock positions. 12 is straight ahead, 6 is behind you, 2 is to your right and slightly in front, 4 is to your left and slightly behind. You get the idea.
  3. Since you start facing right, the 12 on the L hex tells you that you can move to the M hex (and still be facing right). And the 10 on the L hex tells you that you can move to the TL hex, in which case you'll be facing up and a little to the right.
  4. Suppose you move to the TL hex. You now have the option of turning around and going back to the L hex (the 6 o'clock move); or moving to the TR hex (the 2 o'clock move (you'd be facing right if you did this)); or moving to the M hex (the 4 o'clock move (you'd be facing the bottom right of the screen if you did this)).

In this case, you could solve the maze in 7 steps: maze example solution

Here's a simple example that takes 13 steps and doesn't use the bottom two hexes (in fact it's a simple pattern to get through):

maze example 2

The challenge is to find the maze with the most number of steps.

Clarifications of the challenge:

  1. You must provide the o'clocks for each of the hexagons shown. Your maze will have the same start and end as above and the same hexagon configuration. Only the numbers will change.
  2. The solution will always be the shortest route through the maze. A maze that cannot be solved will not count. A maze with multiple solutions will have only its shortest route as the score.
  3. It doesn't matter whether the maze is "difficult" or not.
  4. For benchmarking purposes I have a solution with 21 steps. I suppose 26 is an upper bound since there are 25 states (bonus challenge for the comments: check that statement! - Answered by @athin who realized my original statement of 25 upper bounds and 24 states was off by one since I forgot to count the very first state).
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    $\begingroup$ I suppose the upper bound should be 26 as there are 25 states? (There is an extra state from the starting cell.) $\endgroup$ – athin Apr 17 at 0:03
  • $\begingroup$ Thanks, and updated! $\endgroup$ – Dr Xorile Apr 17 at 0:05
  • $\begingroup$ I believe 21 steps is maximal. I could not get further than that, either. $\endgroup$ – Cloudy7 Apr 17 at 3:36
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    $\begingroup$ Both the L hex and the R hex have a different number of entrances and exits. Since the path endpoints are not inside those hexes it means both hexes will have at least one possible connection unused, and 24 moves cannot be exceeded. $\endgroup$ – Bass Apr 17 at 10:36
  • $\begingroup$ @cloudy7, post your solution. $\endgroup$ – Dr Xorile Apr 17 at 14:04
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EDIT: I think this one hits the theoretical maximum:

enter image description here

It takes

24 moves

to solve, and it uses every possible path except M-L and R-M. That is the best way to place the unavoidable deficiencies of the L and R hexes, which have, respectively, an extra entrance and an extra exit.

Here's what the solution path looks like:

enter image description here

This is, by no means, the only maximally long maze; for example, any "2,8" can be independently replaced with a "4,10" (or vice versa) without affecting the length of the shortest path. Interestingly, this particular maze is also reversible: if you start at the "End", you can reach "Start", but only by a (slightly different) maximally long path.


EARLIER ANSWER:

Here's my best so far:

enter image description here

Unless I'm mistaken, it takes

23 moves

to solve it. Here's the only solution path:

enter image description here

As you can see, in addition to the unavoidable L and R hex deficiencies (those hexes have an odd number of entrances/exits, so one path will not be used), there's exactly one unused path, from BR to BL. This means that this solution is one move away from the theoretical maximum. That maximum can only be achieved by "connecting" both the unavoidable unused paths to the middle hex, but I haven't found a way to do so.

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  • $\begingroup$ Excellent, and confirmed that it takes 23 steps. $\endgroup$ – Dr Xorile Apr 17 at 15:27

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