# Tag Info

## New answers tagged chess

1

Maximizing number of accesories Number of accessories: 19 pieces. Regicide score: 19 - 3x4 = 7 points. Maximizing the regicide score Number of accessories: 11 pieces. Regicide score: 11 points.

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For good measure here is my intended answer: Q1: Q2:

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As I understand it, this should do

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My interpretation of the question is that you want a checkmate position where none of the pieces can be removed and it remain a checkmate position. In that case I present the following: In this setup the black king is in check by the white knights which cannot be captured and the king has no valid moves due to being blocked by black pieces or the white ...

4

If I understand the question correctly, Thanks to @aguy for spotting a mistake (now fixed).

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4: (can probably be improved by a ton) Chris in the comments claims to describe an 11-piece solution but I don't fully understand his comment so I hope he posts a solution with his answer. I'll update this answer in the future if I find a better answer (which I may soon)...

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An alternate phrasing of the movement rules, which took a lot of toil to get a decent description of the Bishop: (Movements ending at the starting square are implicitly forbidden) A bonk is a unit move (knight leap for the Knight, diagonal step for the bishop), except if this move would take them to a square off the board, the piece moves to the closest on-...

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We can also have rotational (180° only, though) and mirror symmetry at the same time: Please note that only four squares really use Swiss rules (labeled s in the diagram). All others are covered by plain chess moves.

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I think this arrangement works

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From what I can surmise from the Bishops rule, the following configuration of Swiss Queens should work More Explanation

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Well, for starters, you need Here's one rather quick way to do that: The game can be replayed here, this is the final position:

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I used integer linear programming (and, sorry, a computer): The solution is unique up to rotation and reflection.

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Explanatory 'showcase' answer provided by OP: All three of the existing answers here made great contributions to solving the puzzle (@Graylocke, @samm82, and @berkeleybross - to whom I have awarded the checkmark for being first to find the final answer), but none of them quite described the intended solution path in full (without making any assumptions), ...

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EDIT: All solutions up to mirror symmetry, rotation, colour permutation found by brute force: End of EDIT. One possibility (I write X,O,+ for 0,1,2): A bit on how this was found:

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It feels like this can be improved but the best I've been able to do so far is As follows

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Using a modified version of Albert Lang's method on the previous question, the best I've managed so far is As follows

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One possible answer is It doesn't look very symmetric at first glance, but if you look more closely, I think it's likely that there isn't any answer which does not follow this pattern. I'm pretty sure this property holds for all answers, with the following reasoning: Let's classify the cells into four categories. A B C D C D A B B A D C D C B A And let's ...

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I finally wrote a heuristic solver that finds good solutions. I found many solutions that achieve the longest chain length of Here are some example solutions

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Surely must be enough, and also required. After white has two unavoidable captures coming up: on white's next move, the Black can prevent only one of those moves, so the first objective is completed. Continuing from there, white will play next. It threatens black's F-pawn, and also the queen's rook or some other piece in front of it. All those pieces ...

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Another proof of Albert's lemma (and one that I believe is much more elegant than the others presented): I will prove a stronger lemma instead. Namely: Proof: (A proof of the reduced statement follows. This proof is similar to Gareth's answer to the same question, but does not rely on an arbitrary choice of "leftmost".)

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For guessing a formula on integers, generally we want to start at combinations of inputs that result in "guessable nice numbers". In this case, I look at: f(100, -100) = 60 = 100 * 3/5 f(100, -50) = 75 = 100 * 3/4 f(100, 0) = 100 = 100 * 3/3 f(100, 50) = 150 = 100 * 3/2 f(100, 100) = 300 = 100 * 3/1 f(200, -100) = 120 = 200 * 3/5 f(200, 50) = 300 =...

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Here's (what I think is) a simpler proof of Albert's lemma than the one in loopy_wall's answer. We'll find either a king-path of 0-squares connecting N and S sides, or a king-path of 1-squares connecting W and E sides. The basic idea is to walk along the boundary between 0-squares and 1-squares until we reach an edge of the board. So here's an example board; ...

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Proof of Albert's lemma (which solves the bonus question; please note that some repetitive details are omitted to keep down overall length to something reasonable):

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Every pair of zeros and every pair of ones are connected via some King chain is confusing. If you mean that every $1$ can be reached from every other $1$ by the traversal rules (similarly for $0$), then I think 8 is a minimum. If we can have disconnected pairs, gets you to 7. As for a proof, there might be fertile ground in looking at what can stop a path (...

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This is just a very minor detail on top of @justhalf's work:

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Here is an obviously true but devilishly hard to actually prove lemma from which optimality of for the bonus question follows: Lemma: No proof :-( Also, alternative solution for the main question:

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As a first attempt, longest king chain of 8: As for the bonus,

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Edit: As a GIF: As a picture The final formation being

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Credit to Lukas Rotter in the comments who corrected some mistakes I had. Assuming we don't have to move all the pieces, here is one way to do it

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Reasoning: More reasoning as to the number of queens:

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(Note: I have treated the question as a classical covering problem, while OP apparently intended that the occupied squares need to be attacked as well. I'm leaving the answer up anyway, since this interpretation yields an interesting puzzle too.) UPDATE: Here's the biggest one I got: It took surprisingly long to fiddle the placements so that everything fit, ...

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I'll get things started with Improvement:

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Here is my answer, with 32 pieces. Can add another couple of objects on a5 & b5 without causing a square to appear.

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Building on samm82's answer, and fixing the French fashion designer answer as Enigma said, you get As Graylocke observed, Removing every other letter reveals i.e. the solution is Edit: just realised:

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This answer was a group effort between @Graylocke, @berkeleybross, and myself, with a little confirmation from @Enigma, so I'm more than happy to make this a community wiki answer (unless there are any objections or @Stiv wants the +2 rep for accepting XD). The answers to the questions: What you should have done is

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I don't know if this is the right answer, but it seems... plausible? So... What does this mean?

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@PaulPanzer made a summary of existing answers here (including some they gave themselves). I want to provide a clearer visualization for each of the existing solutions: Source code for the above program

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The minimum is

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I can manage to get Methodology:

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I used integer linear programming to minimize the number of unattacked squares. Here is one optimal solution (unique up to symmetry), with

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@RobPratt's answer is correct. Here are the alternative arrangements I've found, all of which are only slightly different from each other.

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You can solve the problem via integer linear programming as follows. Define a graph with one node per cell and an edge for each pair of cells that are a knight's move away from each other. For node $i\in N$, let $N_i \subset N$ be the neighbors of $i$, and let binary decision variable $x_i$ indicate whether node $i$ is selected. The problem is to minimize ...

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I think this will work as a solution

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These answers have been very entertaining for me, thanks everyone! Here's my guess: Because:

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It can be done in: Here is the move list. It takes quite some precision of Black moves to do it! Replay it here. Here is proof of optimality.

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There are 6 missing black pieces, and since the black pawns haven't moved, the bishop at c8 must have been captured by white knight. That gives 5 pawn captures, and so by counting, the third white knight must have promoted at g7. That means the white knight at h8 must have been there before black pawn got into g6. Now, consider white knight at h1. It must ...

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