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This maze was inspired by the Deusovi Honeypot .

The rules of this maze are simple. The solid colored triangles inside the bigger triangles are shrunken replicas of the bigger triangles of there same color. and don't forget to rotate them 30 degrees clockwise for each level in you go.

The yellow triangles are portals, portals only work when both of them are exactly the same, by both size, and angle.

So the question is, assuming that you don't make any mistakes which two entrances (A,B,C,D,E,F) are the hardest to move between? enter image description here

To rate which two entrances are correct, use this system.

Moving in a level costs 1 move.

Teleportation costs 1 move.

P.S.(for the colorblind; red inside of blue on top, blue inside red on bottom).

For anyone really fascinated with this type of maze. try getting rid of the portal in "F" or bottom left. the solutions will get so much deeper! :).

Hashbrowns
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    Doesn't that mean that the portals in the blue triangle are always unusable? Also, which is the top level maze? Only one triangle is going to have actual entrances/exits. – Deusovi Dec 22 '17 at 03:45
  • @Deusovi Yes the portals on the blue triangle are useless, but their smaller versions deeper in are used. Both are the top level, they are meant to go side by side, so all six exits are usable. Think of it as two mazes that are only connected by portals. hope that helps :) – Hashbrowns Dec 22 '17 at 03:51
  • Can you go from the portal inside A to the one inside E? If so, doesn't that mean that two of the three blue portals are irrelevant? – Deusovi Dec 22 '17 at 03:52
  • I think the other blue portals may work in the deeper levels, where both triangles' portals will have rotated by a bit. One of the deeper red portals may match one of the deeper blue portals. – votbear Dec 22 '17 at 03:53
  • @Deusovi sorry yes your correct two of the three blue portals are useless at this level. – Hashbrowns Dec 22 '17 at 03:55
  • Not at this level, but at any level, right? After all, you're always turning the same amount. – Deusovi Dec 22 '17 at 03:56
  • @Deusovi I Don't know I haven't checked, but that would seem right. – Hashbrowns Dec 22 '17 at 03:58
  • What exactly do you mean by "hardest to move between"? – Deusovi Dec 22 '17 at 03:58
  • And have you not checked the puzzle before you posted it? :/ – Deusovi Dec 22 '17 at 03:59
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    @Deusovi Hardest to move between means the most moves required to get from one exit to the other. And I did check the puzzle, I just didn't check what you were talking about, But I know it doesn't effect the puzzle. – Hashbrowns Dec 22 '17 at 04:03
  • The center of the blue triangle is supposed to be red, I believe? (Maybe my phone browser has a colour rendering issue?) Edit : Oh, there’s an explanation at the bottom. – Bass Dec 22 '17 at 05:38
  • Am I allowed to flip the triangles? Also gasp I have inspired someone? Squeee! – Bass Dec 22 '17 at 05:44
  • @Bass No, only rotate 30 degree's each time. anything else would break the puzzle. Enjoy :) – Hashbrowns Dec 22 '17 at 06:04
  • @Bass I edited it to fix the color rendering issue. hope that helps :) – Hashbrowns Dec 22 '17 at 06:08
  • I love PuzzlingSE! One starts with a joke and soon after there is not even just one genuinely new and creative puzzle, but a whole family of them... +1 – BmyGuest Dec 23 '17 at 16:16

1 Answers1

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Answer:

B and E, which take 8 steps to move between.

Solution:

As was done in the Deusovi Honeypot, I'll name the outermost exits we see $A_1$, $B_1$, etc., and each subsequent smaller copy with increasing subscripts. I'll denote by $X \Leftrightarrow Y$ that $X$ and $Y$ are connected by a teleport, and $X \leftrightarrow Y$ that they're connected by walking between fractal levels. Then we have for all positive $n$:

$A_n \Leftrightarrow E_n$, $D_n \Leftrightarrow F_n$

$A_n \leftrightarrow E_{n+1}$, $B_n \leftrightarrow F_{n+1}$, $C_n \leftrightarrow D_{n+1}$, $D_n \leftrightarrow A_{n+1}$, $E_n \leftrightarrow B_{n+1}$, $F_n \leftrightarrow C_{n+1}$

At this point we have a graph so we can brute force an answer with a few dozen lines of code and a shortest-paths library. It's not too hard to do by hand either - we can get quick upper bounds for most of them by reaching the nearest A or E and then changing fractal levels "in place": $E_1 \Leftrightarrow A_1 \leftrightarrow E_2 \Leftrightarrow A_2$ etc. Once B-E starts looking like the longest path, you can exhaustively search that tree of possibilities; it never gets too big because nothing branches more than two directions and since you can quickly get an upper bound of 8, you never have to travel to the 4th level or beyond (since any travel to the 4th level will take at least 8 steps - 4 inward and 4 outward). Here is one such path of length 8: $B_1 \leftrightarrow F_2 \Leftrightarrow D_2 \leftrightarrow A_3 \Leftrightarrow E_3 \leftrightarrow A_2 \Leftrightarrow E_2 \leftrightarrow A_1 \Leftrightarrow E_1$. This is the only tree you have to search that thoroughly, since once we have B-E at 8, we do not need to find shortest paths for all the other pairs; merely paths shorter than 8.

bmcfluff
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