# Tag Info

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Being a chess composer myself, one motivation came to my mind: self-stalemate. White cannot win, so they must secure a draw via underpromotion. While there are plenty of problems that shows it, here are three problems where the rook, bishop, and knight promotion are maximized. I will leave to you to figure out which is which, so you may try to solve it own ...

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Rook White to move: Explanation:

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We can achieve the stated goals if Knight: Bishop (this one was harder): Doing a similar trick with a rook promotion is harder because

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First of all, antichess has been weakly solved in 2016 : White can force a win with 1.e3 Moreover there are 3 main 'losing lines' for white : d3, d4 and e4. The majority of the experienced antichess players can force a win against these 3 moves with black. I learned the e4 counter ~1 year ago in order to counter inexperienced Lichess players : I obviously ...

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I'll take a guess and say : Why (not a great explanation. This puzzle is complex) : Fun fact : if it's black to move in the final position, there is a checkmate in two moves : Qg3+ Rxg3 Bxg3#

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Here's my answer: Reasoning : But I might be wrong.

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Here is an alternate solution with 47 mates that I found, originally posted here: https://www.chess.com/forum/view/more-puzzles/diams-totally-puzzled-213-page-66-ndash-mate-in-one?page=2 Hmmm... I also just found out at matplus this alternate solution was first found in 1882... oh well! There's nothing new under the sun I guess...

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I found

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TL;DR: The minimum number of moves is Let's start by finding the allowed move pairs: Obviously, a bishop's move can be matched by an equal and opposite bishop's move, and so on. There are also some "special couples" where black and white can move different pieces. Given the weights King=10g. Queen=8g. Rook=5g. Bishop=4g. Knight=4g. Pawn=2g. we can go ...

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My best attempt : Here it is :

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Here's my solution to the problem, which takes There probably is a shorter solution, but I haven't found it yet.

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I found an answer im satisfied with : Reasoning and variations : This puzzle was fun to solve, I really enjoyed it. I still don't know if this is the best solution. With less pieces, checkmating gets really hard. PS : My solution is computer-free, and im not good enough at programming to be able to verify my solution. If I missed something or if there is ...

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I believe the answer is The reason being Maybe this is wrong, but I'd love to hear your thoughts.

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Maybe I'm missing something, but it seems that But I'm not certain.

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The question doesn't say that the queen has to go onto an unattacked square. This means that the upper bound is because

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Horst can place That's because Note

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Here is the solution I had in mind:

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The answers for $M$ are somewhat surprising! I expected smaller numbers. $k=0$ Minimum Maximum $k=1$ Minimum Maximum $k=2$ Minimum Maximum

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You asked for 6 queens, but the maximum is at least

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I have improved the queens solution or some more

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The most moves possible for both sides in a position where all 32 units are still on the board is 164. This answer was stolen from @GloriaVictis's answer to this CSE question. Nenad Petrović, The Fairy Chess Review 1946

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Via integer linear programming, the maximum for knights is The maximum for queens is at least Other maxima are

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Here are my own solutions I put together. With the exception of the Rooks solution, they have already been equaled or surpassed by Daniel's answer, so I'm just posting them for comparison. Bishops: Rooks: Knights: Queens: Kings:

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Bishops Knights Rooks Queens Kings

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Initial thoughts Update 1

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Clues Putting it all together

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The last episode was difficult, with a few twists and turns, but this selfmate left little doubt: the answer is 18 moves. Explanation It's clear that we must manipulate Black's pawn structure somehow, because there's no way a mate can come out of the current structure. The fact that such a manipulation must be forced, i.e. Black's only legal move, ...

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Observations A few basic facts I notice: The breakthrough threat of ...b4, cxb4, and ...c3 prevents the White king from leaving the square a1-d4, namely we must rule out a winning plan of running to the kingside and gobbling up the black pawns on g4 etc. The a6-b7 structure keeps Black's king confined, both spatially and temporally, to b8-a7 (or c7). ...

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Bishops, N=0 Bishops, N=1 Bishops, N=2 Queens, N=2 Queens, N=3 PSE link: Queens, N=0 Lichess links for new or improved monochrome solutions: Knights, N=2 Queens, N=2 Queens, N=4

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For kings, $N=1$ yields a maximum of

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This post is a collection of all the answers. Single Color N = 0 N = 1 N = 2 N = 3 N = 4 Two Colors Pawns, N = 0 Pawns, N = 1 Pawns > 1 is impossible without a cylindrical board. Bishops, N = 0 Bishops, N = 1 Bishops, N = 2 Bishops > 2 is impossible. Knights, N = 0 Knights, N = 1 Knights, N = 2 Knights, N = 3 Knights, N = 4 Rooks, N = 0 ...

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I used a computer program to find solutions for Knights and for Kings. I am assuming that there have to be an equal number of pieces of each colour. KNIGHTS: For N=0, For N=1, see this answer to one of the other questions, which uses 24 of each colour. For N=2: The N=3 case was tackled in another question. For N=4: N=5 or larger is not possible. Any ...

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I found 3 other solutions with 16 knights. Enjoy! 1. 2. 3.

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There are solutions for number of knights on each side equal to 4, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16. For example, here's one for 16: I used integer linear programming, with a binary decision variable $x_{i,j,k}$ to indicate if cell $(i,j)$ contains a knight of color $k$. For each cell $(i,j)$, let $N_{i,j}$ be the set of neighboring cells (one knight's ...

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For the 14 v 14 solution: lichess link

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Here is a solution with 15 knights: I could not find a solution with 14 knights.

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First observation: the colour of the knights corresponds to the colour of the squares - the black knights must be all on one colour and the white knights all on the other colour, since a knight's move always changes colour. So in my pictures I'm just going to put all knights in red, and aim for overall 4-fold rotational symmetry with 28 knights each ...

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Here is a solution with pieces. The colouring of the empty squares ($2\times2$ in white, $4\times4$ in yellow, $6\times6$ in orange, $8\times8$ in red) shows how the pattern generalises to any $2k\times2k$ board. For $8\times 8$ this solution is optimal.

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