I should probably have asked this before the last Puzzle GP round, but I don't think I understand how to approach these puzzles (3 in the last GP and one recently uploaded to the LMI expo).
I think I understand the idea that in a directed loop all pieces on the edge of the grid will go round in the same clockwise or anti-clockwise direction, and that adjacent pieces inside will go the opposite way, but I can't see how to apply this, or even if this is how I'm meant to start the puzzles.
Anyone got any solving tips?
Traffic Loop
Re: Traffic Loop
I'm not sure where to start either, to be honest. There might be some mileage in thinking about the vertices of the grid, but there might also be a better way of looking at it.
Depending on how the loop goes next to the border of the grid, it can be thought of as either right-handed (ie clockwise) or left-handed. I *think* that a right-turn cell that is part of a right-handed loop will have 3 of its vertices outside the loop and 1 inside, and vice versa for left-turn cells of that loop.
That seems to give a few mini-theorems:
- In 2026 GP R3 Q28 (the second one) the vertices on the border cells of the grid are all outside, and so the right-hand turn cells on the edge force the loop to be clockwise.
- In Q27 (the first puzzle), there are pairs of opposite facing turn cells. Considering the edge between those two cells, if that edge is either fully inside, or fully outside the loop, then it would break one of the clues (because you can't then have 3 vertices inside the loop for one of the clues whilst also having 3 vertices outside the loop for the other clue). So the edge between opposite turn clues must be crossed by the loop. This lets you immediately put line segments in joining adjacent opposite-facing turn clues.
There's probably some other easy steps that I've not figured out yet, or different ways of thinking about it, so I'd also be interested to hear of any tips
Depending on how the loop goes next to the border of the grid, it can be thought of as either right-handed (ie clockwise) or left-handed. I *think* that a right-turn cell that is part of a right-handed loop will have 3 of its vertices outside the loop and 1 inside, and vice versa for left-turn cells of that loop.
That seems to give a few mini-theorems:
- In 2026 GP R3 Q28 (the second one) the vertices on the border cells of the grid are all outside, and so the right-hand turn cells on the edge force the loop to be clockwise.
- In Q27 (the first puzzle), there are pairs of opposite facing turn cells. Considering the edge between those two cells, if that edge is either fully inside, or fully outside the loop, then it would break one of the clues (because you can't then have 3 vertices inside the loop for one of the clues whilst also having 3 vertices outside the loop for the other clue). So the edge between opposite turn clues must be crossed by the loop. This lets you immediately put line segments in joining adjacent opposite-facing turn clues.
There's probably some other easy steps that I've not figured out yet, or different ways of thinking about it, so I'd also be interested to hear of any tips
Re: Traffic Loop
Haven’t solved any actual puzzles, but I guess there’s a bit of topology going on. One way of thinking of a directed loop is that the inside of the loop is always going to be to the left of the loop or it is always going to be the right of the loop.
Seems to me the key pattern is Maurice’s 2nd lemma (where you have adjacent RL turns you draw the segment) as that gives you segments. From there I imagine it’s just a case of keeping outside the loop from touching the inside? - even when you aren’t totally sure which is which (I guess some sort of parity you might also call it…) I guess I need to print off the GP and have a go myself
Seems to me the key pattern is Maurice’s 2nd lemma (where you have adjacent RL turns you draw the segment) as that gives you segments. From there I imagine it’s just a case of keeping outside the loop from touching the inside? - even when you aren’t totally sure which is which (I guess some sort of parity you might also call it…) I guess I need to print off the GP and have a go myself
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Re: Traffic Loop
I’ve just worked my way through the puzzles from the GP, having ignored them entirely during the contest.
Maurice’s two rules seem pretty solid to me. I solved the puzzle similarly to the way I solve Inturnals, by identifying vertices in or out of the loop. Right and left turns must have either 3 in and 1 in or 1 in and 3 in, depending on the orientation of the loop, as explained above.
You can’t have two consecutive “out” vertices along the outermost perimeter which isn’t the external perimeter - that’s terrible notation, but hopefully a helpful rule if you’re trying to solve these.
It took me forever to get into the 10 point puzzle, but having adopted this in/out notation of vertices, the 41 and 113 point puzzles took about the right amount of time for me to solve (given that it’s a loop puzzles took about, which are always slow for me), so I’m satisfied that these steps are about enough to be getting on with for me.
Maurice’s two rules seem pretty solid to me. I solved the puzzle similarly to the way I solve Inturnals, by identifying vertices in or out of the loop. Right and left turns must have either 3 in and 1 in or 1 in and 3 in, depending on the orientation of the loop, as explained above.
You can’t have two consecutive “out” vertices along the outermost perimeter which isn’t the external perimeter - that’s terrible notation, but hopefully a helpful rule if you’re trying to solve these.
It took me forever to get into the 10 point puzzle, but having adopted this in/out notation of vertices, the 41 and 113 point puzzles took about the right amount of time for me to solve (given that it’s a loop puzzles took about, which are always slow for me), so I’m satisfied that these steps are about enough to be getting on with for me.