# Tag Info

26

This seems to work: And the position looks like this: Apart from the symmetrical solution, this might very well be unique:

16

As far as I know, the only way to figure this out is by letting a computer run through all the possibilities. It is a small puzzle, so this does not take long. First I will assume that you want the final solved position to have the blank in the bottom right corner, with the tiles in numerical order: 123 456 78. (See further below for the results with the ...

15

12

Suppose you can set your pair of compasses to length 1. Then Then we know that the length of the line that would join those two points is The simplest construction would use:

12

Here are my first idea (both sides are essentially the same answer, so the hint fits too): Both positions seem to be independently reachable by a legal game. It might be possible to find a legal game leading to the whole position too, but that would take a bit of time. Before that, I'm going to double check for any simpler solutions. :-) Since OP ...

11

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.

9

9

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:

8

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?

8

I think you could make cuts as follows

7

7

How about I think that's an alternative 6+1-piece solution.

7

Edited: added a drawing for the first step. Edited again: Dr Xorile very clever solution eliminates the second step.

7

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.

6

The general idea is Here's a solution with 14 pieces total.

6

My guess would be Because

6

Here's one way to do it Strategy

6

I've managed to reduce it to 34 moves. I can't prove this is optimal but I can't seem to get any lower using this strategy. Apronus link

6

I haven't seen this puzzle before. I cannot tell you the name or who produced it. But assuming your goal is to reassemble it, I can help. It seems the problem is to fit all 5 pieces in a 4x4x4 cube. In the solution there are visible holes left. Edit: As Weather Vane noted, a missing 6th piece can complete the puzzle. The goal was to build a solid ...

5

Following the excellent answer from @FlorianF here is a set of images showing his missing piece (four angles). This is Florian's solution, not mine, I just added a graphic. (Earlier) I have not seen this exact puzzle. I wondered if it is complete. Edit: So I am as mystified as ever by the puzzle, but perhaps the centre must be filled too.

5

A slightly different approach, drawn terribly

5

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.

5

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 ...

4

I'm reasonably sure this is the best possible solution. The most recent move must be

4

I believe (without proof) that the only solution up to symmetry is: This is solvable since:

4

Here's one way of doing it. (Without pawns on the promotion squares, and with bishops on different colours.)

4

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 ...

4

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

3

Here's a way to do I'm not sure that this is optimal, though:

3

I would release: Because: A third shape to release:

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