# Tag Info

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You're

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The 'hardest' possible Irregular Sudoku has and it looks like this:

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Glorfindel's answer is sufficient for the main question. To answer the bonus question: Here is an example: To construct this example, As for a starting position,

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Of course we need to use Pythagoras. This leads to the following solution: Here is another more compact solution.

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My position has a fourteenth check too, but that's probably ok. :-) These are the moves: And here's the whole solution uploaded to Lichess. EDIT: Looks like 15 is also doable: The final check feels like such a waste of a piece though; the first fourteen checks are possible with white having only two pieces and a king: Again, here's the Lichess link.

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This solution works: (Places where the wires are close together can be pushed apart a bit - there aren't any crossed wires, though.) My strategy:

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I think you could make cuts as follows

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I must be missing something because I'm getting a lot more than 54 mates-in-two for the position below? I have 116 listed, although I'm doing this by hand so there may be some errors included. Is there some rule I haven't considered?

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For Queen: For Knight: For Rook: For Bishop: EDIT: A winning knight underpromotion I found. All other promotions lead to wins, but underpromoting to a knight is the fastest mate (mate in 1). EDIT: Thanks to RosieF for catching a huge error in the bishop part.

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EDIT: Managed to pare my score down to 10 pieces (incidentally also worth 10 points in the usual piece value reckoning) by entirely locking down the position, thus removing any last remnants of fun the solver might have found in the earlier versions: If white starts with 1. c3, black can thwart white's suicidal plans by promoting at b1, which causes an ...

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Here's a situation where promoting to queen is the only winning move: 1 h8=Q! wins 1 h8=R? only draws: Here's a situation where promoting to bishop is best: Promoting to bishop Promoting to queen Promoting to knight Promoting to rook As luck would have it, this morning, YouTube recommended to me a video discussing a win-study by Selezniev which ...

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Here is the answer for practice ones: 1: 2: 3: To construct b=3; As show below for i=9. The general idea is actually the same as above, if you want to increase b for example, if you want b=5 and i=9; or for example, if you want b=20 and i=9; So you should so you can create polygons with every $i\geq0$ and $b\geq3$ values with the method above.

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A slightly different approach, drawn terribly

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I drew it! Everything below is my original post: @Bass had a very similar idea to mine. Funny how that works. I was going to wait and try to actually draw it, but I feel like it isn't worth the effort, now. Edit

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OK, so here's my track design, or at least the functional part of it. I call it The Candle of Chicken. Unlike the examples above, I've marked the spots that are not walls, that is, there's only a very narrow path available from the start at the bottom. The starting points are symmetrical, and there's only one choice to make: Either stop after the first ...

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Ok. I have a solution with inhabitants. One question I wasn't sure how to answer was, are we allowed to have more clubs than inhabitants? If so then what would you call the extra clubs, just random names? Anyway, from all the rules I have concluded that Next Fore ease of explanation I will name the inhabitants now: Ok, here is the breakdown of ...

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The diagonal symmetry with opposing colours on opposing sides of the axis isn't really possible because there are pieces on the symmetry axis too. Since OP agreed that this observation presents a problem, here's the "most symmetrical" finished position with 4 empty spots I could find. The checkered spots can be any colour you want (as long as they are all ...

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Consider the following solution: The moves proceed as follows (with all moves forced for black): The notation in the last line just indicates that white can promote to either a knight, bishop, or rook. The reasoning is as follows:

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