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Sequel to:

The Hallway and Tele Doors

Note: You need not have to read the above to understand this puzzle, however this puzzle does spoil the above one.


The tele door system:

Each pair of doors, represented by letters, are "linked", you go enter one door on the left side, and you go exit the other in the right. Example hallway:


A1 B1 A2 B2


Prisoners always enter through the leftmost door. So in the example hallway, they'd go enter through the first A1->A2->B2->B1->A2->A1->B1->B2->

Note: There will never be an 3 ( representing any letter).

You, my friend, are a wickid man. You wish to trap someone in a hallway for eternity using nothing but pairs of tele doors. But sadly, you have finally come to understand this is not possible.

But how long CAN you keep them there?

Given 26 pairs of teles, how many tele doors can you force them walk through, before escaping?

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  • $\begingroup$ Given the answers to the first question, it really seems like the answer to this one is trivial. $\endgroup$ – GentlePurpleRain Nov 9 '15 at 15:28
  • $\begingroup$ @Gentle Sadly, Ik. $\endgroup$ – warspyking Nov 9 '15 at 16:30
4
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This answer (which I wrote) to the original question shows that it is impossible to enter any door twice:

[Suppose] that you enter the same teleportation door twice at some point. Consider the first time that this happens, and what happened immediately beforehand. Because you can't go towards the left, you can only possibly enter the leftmost door once. Therefore the repeated door must have another door immediately to its left, and you must have just come out of that one both times. But this contradicts the fact that it's the first time you used a door twice.

Therefore the maximum will involve walking into every door once, for a total of 52.

Given two pairs of doors, it can be arranged for all four to be walked through, like this: A1 B1 A2 B2

Then this can be repeated 13 times to achieve the maximum.

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  • $\begingroup$ Please explaining the reasoning, as I don't want other users to have to read the original question to understand. Make it entirely standalone. $\endgroup$ – warspyking Nov 9 '15 at 1:18
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All of them.

For each pair of two doors, do this pattern:

----[A]---[B]---[A]---[B]----

They'll walk through in this order:

--1-[A]-4-[B]-3-[A]-2-[B]-5--

I'm fairly sure that odd amounts are impossible, but I haven't proven that yet.

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