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@justhalf Yes, I tried to exploit that by throwing Bob out of the loop after a failed "RR", but eventually he'll still manage to break out of any "LL" traps/loops and approach the "RR" from the other side and get out! By the way, glad you noticed the "Suck it LL, RR!", I was afraid it was a bit too 'subtle' for people to really look at the random notes on the paper ;) - But if anyone can provide a maze that protects against Bob's strategy, we'll surely hear about it! – Katai 2 days ago
@Katai: Let's discuss more here: chat.stackexchange.com/rooms/33277/edwards-maze =) – justhalf 2 days ago
LRRLRLRLLLRLRLLLRLRLLR can't obey your rule because of the LL towards the end there - every sequence of L's needs to be odd if we're working with primes. I believe that should be LRRLRLRLLLRLRLLLRLRLLLR. – Zerris yesterday
@Zerris: that's true! Thanks for spotting that. Again, I missed a T-junction in Maze 2 between c and d, haha – justhalf yesterday
I found a maze which can trap Bob thanks to Zerris! Will give the image later – justhalf yesterday
@Katai: Poor Bob... – justhalf yesterday
Nice job! I was trying to figure out a way to abuse the parity of primes only being even, but couldn't find it. As an aside, the smallest such maze is only a 6x6: i.imgur.com/3qGXBXB.png – Zerris yesterday
@Zerris You can remove last row to make it 5x6. Yours is even more evil than mine, Bob can't finish from the trap directly, he must go through the start – justhalf yesterday
Righty-o, you can remove the bottom row, which also makes the maze look quite elegant! It's a delicious pretzel of maziness. – Zerris yesterday
Can you link to the improved one? I also want to lay my eyes on the delicious pretzel of maziness. – justhalf yesterday
i.imgur.com/BCAPzsv.png - for just 20 total squares of maze. – Zerris yesterday
@Zerris Nice job you two! It's beautiful - and utterly evil. Well, since I wasn't 100% sure either, I revise my earlier assumption of course! Yes, Edward can (obviously) trap Bob. And when I look at it... yeah, I don't know how I managed to miss THAT shape in all the mazes I made during the last week! – Katai yesterday
Define a Maze homomorphism Fx(y, z) such that F is the transformation function, {x} is the infinite series of turns made under Bob's rule, and y, z are mazes. Then Fx(y,z) returns true IFF the paths through y and z with each intersection labeled consecutively in order of appearance under moves {x} cause Bob to see the exact same finite or infinite ordered list of intersections. By this definition, it's easy to see that my map and @justhalf's are homomorphic, and I claim that all maps utilizing this trap are homomorphic to both of ours under Fx(y, z) for Bob's given {x}. – Zerris yesterday
Of particular note for the difficulty of accidentally encountering such a homomorphic maze or submaze, this means that as soon as you attempt place your trap further than the third intersection, your maze can no longer be homomorphic to ours, and thus can no longer use this trick. There may be other more complicated traps that are not homomorphic to ours, but that at least explains how you missed this one. (I also would not be surprised if you could prove that a trap is only possible with the entering move being a R at even parity, with this and minor variations as the only possible answers) – Zerris yesterday
Yeah, I agree with Zerris that this maze requires the trap to be the third intersection, and therefore hard to find. So, don't worry Katai =D @Zerris: so you define each maze using the orders, like in our case: 12343234343...? – justhalf yesterday
Now, the interesting thing here is, that "simpler" rules like "always turn left" or "always turn right" will work again. So, if your theory is right, Edward's downfall was that he focused on stopping those simple rules first. But I think I can come up with a maze that stops those behaviours as well, I have an idea... When I'm back home. – Katai yesterday
@justhalf - correct. For any given {x}, if those sequences of intersections are the same, then the maze is the same, as far as bob can tell. For any single square trap for {x} = prime, It is always possible to reverse your direction. This is because you have two exits from the trap for R or L, and both must come back by different entrances, ergo there are at least two ways to enter the trap. This cascades for any # entrances/exits > 4, and implies symmetry for # = 4. The only symmetry that can't be retraced in {x} = prime is an even prime, so our trap is the minimal single-square one. – Zerris yesterday
And, although I can't easily prove it, I suspect our trap is the only single square one - that is to say, if you take this trap and add any intersections to it which make it more complicated, I believe it can no longer be contained. The basic concept is that our central trap is the only possible one of its size, and anything that adds to it either allows for multiple possibilities for re-entering the center (which then ceases to be a trap) or all identical possibilities - in which case, we have actually built a second copy of this that somehow doesn't start with {2}, which is a contradiction. – Zerris yesterday
