This is a repost of a puzzle I created in 2014. I used to post puzzles on my blog, a long time ago!

Rules:

- Go from start to finish without crossing any lines.
- The maze contains four copies of itself, except for the blue lines, which are not copied. Pretend that these are perfect copies, even though in the image I cannot copy them perfectly.
- I’ve labeled the four copies as A, B, C, and D, and numbered the 12 entrances. These should be useful for notating your solution. For example, one possible path goes: 7, C10, C11, C12, CA9, CAD6.

This maze was inspired by an article in Mathpuzzle waaaay back, which established the idea of a fractal maze. I designed three of these fractal mazes, and this is the third one. Perhaps I will repost the other two later, although I think the first two were harder to read.

Oscar Cunningham says

Are solutions to the maze supposed to only go finitely deep? Or can out path pass through points like CADCADCAD… ?

Siggy says

@Oscar Cunningham,

All solutions must have a finite number of steps and go finitely deep.

cubist says

I put together a similar maze a few years ago as the cover image of a netzine I edited. Start at any of the “plus” signs, and end at the corresponding “minus” sign.

Orion says

Wait, did you close the discussion on that 2014 post to prevent spoilers? Can I comment on partial paths like how to get from 9 to 5?

Orion says

Er, for some reason my earlier comment didn’t appear? can we describe partial paths like A5 leading to a dead end? also, “solutions”? does this maze have multiple solutions?

Siggy says

@Orion,

1. No, I closed the comments because I closed ALL the comments on my old blog.

2. Comments from new commenters get sent to moderation automatically.

3. You are free to post partial or complete solutions here. The solution will be a string of letters and numbers so I don’t think anyone will be spoiled unless they make an effort to read your solution.

4. There are an infinite number of solutions.

Orion says

If 7 can reach 6, then 7A and 6A can loop an infinite number of times lol

Orion says

9, 7, 5, and 2 have easy paths connected by the center of the maze. 7-C10-C11 which is connected to the center. 2 and 5 are directly connected to the center. 9-D5-D2 connects to the center.

The center also connects to B8 (8 always connects to D4, 4 always connects to B7) , C12 (12 connects to A9 and A8), and D3 (3 connects to B12 and B11)

Orion says

C, being a corner maze, has only 6 connections to the main maze.

C1 and C2 are linked together outside C,

C3 is linked to 6,

C10 and C11 are linked together inside C and to 7, the maze center, and CD1 (1 is linked to A4, thus AB7),

C12 (12 links to A9 and A8) links the maze center, and by extension, 7, to CA9 and CA8

Orion says

Also of note is that A, 1, and 12 are most obviously completely separated from the rest of the maze (including the center) and thus so are 11 and 10

Orion says

Getting to the center of CA is easy enough, going to CA4 and CAB7 will be more difficult

Orion says

The path from the center to 4 begins at B8 and ends at B7, and the path from 8 to 7 begins at D4-DB7 and connects to the center at DB8-D2.

Orion says

We can’t find the path to 7 to 8 if we don’t know 8 -> 7, but there’s another way into DB.

Remember that D3 connects to the center? The path from 4 to 3 goes from B7 to B10-B11.

And B10 is connected to B9 😉

We already know how 9 and 7 are connected to the maze center, and now we know how 8 is as well:

The path from 8 to D3 is D4-DB7-DBC10-DBC11-DBD2-DBD5-DB9-DB10-DB11-D3

Orion says

Now that we know that 8 is connected to the center through D3, We can also connect 8 to 7 through D2 using up to an infinite number of recursions of ((8-DB7)-(7-DB8))-(8-D3)-(D3-7)-((DB7-8)-(DB8-D2-7)), thus making infinite solutions.

The most important info taken from 8’s connection to the center, is that B8 is connected to B’s center, which means that 3-B11-B10-B9 and 4-B7 connect to B8 which connects to the main center.

Orion says

Now we know how to get to CA4, and we can get to the center through C1-C2.

And from there, we know how to get to C3 and thus 6

The shortest solution I can think of is now 7-C10-C11-C12-CA(9-D5-D2-B(8-D4-DB7-DBC10-DBC11-DBD2-DBD5-DB9-DB10-DB11-D3-C11-C10-7)-4)-C1-C2-CB(8-D4-DB7-DBC10-DBC11-DBD2-DBD5-DB9-DB10-DB11-D3-D2-D5-9)-CB10-CB11-C3-6

Siggy says

@Orion, that solution is correct. Congrats!