**This is an open question. There is currently no known answer.**

On a chessboard of any size, is there a configuration of kings and knights such that each king attacks exactly 2 kings and 2 knights, and each knight attacks exactly 2 kings and 2 knights?

(Sub-problem -- if it is possible, what's the smallest board needed?)

WIKI PAGE DEDICATED TO THE PROBLEM; WE CAN COLLECT OUR DATA HERE

Details: Kings can attack any adjacent square, including diagonally. Knights move in an L shape as shown below.

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## Comments

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TopNewestIf we assume the chessboard is infinite, it could be done this way... I guess...

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That's extremely neat! This is a legitimate discovery (while I think the people tackling the other problem were all restricting to the finite case, the infinite case is still quite interesting, and I'm curious if it could be used to form a finite case).

I don't suppose you could post a problem to the group that asks the above question "is it possible on an infinite grid?" yes/no (with the answer yes) and then post this answer?

https://brilliant.org/groups/open-problems-group/

Put "Related to Open Problem #1" somewhere in the text of the title.

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Thanks! I'll post the problem, and I think there might be even more such patterns which we can follow to make other infinite solutions.

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If we assume some smallest, finite board exists, we know the following about that board: 1. The board will have an edge on the left, on the top, on the right, and on the bottom 2. To guarantee the board can’t get shrunken, the board needs at least one piece on each edge

With those two ideas I started looking at a corner. I assumed a piece is put in the corner and kept extrapolating until I arrived at a contradiction. All of my work is over here: https://docs.google.com/document/d/1fEALESsS72xeT1e8YamxNqrrOEg6qAn-3mTHk9Rgxnk/edit?usp=sharing

Assuming I made no mistakes, the conclusion I reached was: There are no pieces on the squares A1, A2, A3, B1, and C1.

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Just proved (visually) that A4 and D1 cannot have knights on them, and therefore B2 is empty. (Assuming there are no pieces on the squares A1, A2, A3, B1, and C1) See https://docs.google.com/spreadsheets/d/1yuFJck60D0HHRelXdyzsC79sqiBwOUpBzIryzvpVZ3Y/edit#gid=1329517548

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However, I wasn't able to exclude kings from A4 and D1, and upon further consideration I realized that B2 is not necessarily empty. Sorry.

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Proof that B2 is empty: I am still assuming that A1, A2, A3, B1, and C1 are all empty and that A4 and D1 are not knights. Assume that B2 is not empty. It cannot be a king. This it must be a knight. Thus A4, C4, D1, D3 must be filled, with A4 and D1 being kings, by assumption. However, that forces A5, B3, B4, B5 to be filled. Note B4 is touching five different squares that must be filled. Thus it is not a king. Say it is a knight, then B3 is a king, which is touching more than five squares that must be filled. C! Thus B2 must be empty.

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Is there a way this could be summarized as a posted question to the group that you then answer yourself, the same way we did with an infinite grid?

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@Stefan van der Waal That was to you. Thanks!

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Can you prove your point after #3 a bit more rigorously? "Since the K in A2 has the same going on as the K on B1, A3 should also be a K and B3 should be a N." It feels like this assumes that any solution would be symmetric. I might be missing something, but as a reader I'm going to question it unless proven otherwise.

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Now I look back at what I wrote, I realize that step made sense in my head, but I worded it awfully (and later on I made a leap based on a similar argument that was about 5 times as big). I just added a bit of clarification, which hopefully makes it a bit more clear, even though I'm still not sure if it's rigorous enough.

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If that's true, then you can't place a king on B2.

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Ok, here is my shoot on solving the following related problem. What are the configurations where all kings are attacking 2 other kings? We know that we can divide the board into 2x2 boards. All the possible configurations for Kings on 2x2 boards are shown below, ignoring rotations and reflections: \(\)

Now, we can discard the configuration 5 since all the kings are attacking 3 other kings. Configuration 4 is already complete by itself, all the kings are attacking 2 other kings, it is one solution as well as combining it with other solutions such that no king on this configuration attacks a king on the other solution is also a solution. Configuration 1 will only form a solution if connected with the others such that the king is adjacent to another king, which can be broken down on one of the other configurations as shown in the following image: \(\)

Configuration 1 \(\)

The other two configurations, 2 and 3, are a bit more interesting. On both of them, there are 2 kings that each needs to attack 1 more king. One can concatenate another of those pieces to make it attack another of those kings, this will satisfy the conditions to one of the previous kings and one of the added kings, leaving us in the same situation as before. The only way to satisfy the conditions for all kings one a finite board then is if the added piece attacks the previous pieces on 2 different points closing a loop like the images shown here : \(\)

One condition I found is that for any 2x2 grid on your pattern, the configuration 4 cannot appear on the loop, for example, loop 1 is invalid. \(\)

loop

I don't know if my proof is rigorous enough, sorry, I'm not a mathematician but those are what I found, hopefully, it helps someone come up with a more rigorous proof. Here another more complicated pattern I created just following these rules. \(\)

pattern

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I have put this data on a new wiki page dedicated the problem, which we can use to store such things: https://brilliant.org/wiki/open-problems-group-open-problem-1/

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I don't think that's a proof, but I don't think that you're trying to prove anything. You're just exploring the question and came to some cool conclusions.

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Oh well, I was trying to find the ways one can find the patterns that where the kings attack only 2 other kings, hope it's something useful the conclusions I got to.

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If there were a solution to this, would it have more kings than knights, or more knights than kings?

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There would probably be more knights, as they can be more spread out. However, this is simply a conjecture.

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Good question! Maybe you can work on that? (They're certainly not necessarily equal - the kings-and-rooks case, for instance, is on a 4 by 4 board with 4 rooks and 8 kings.)

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Which one do you think is needed more? (Just to give me a place to start working.)

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If there were more kings than knights, what conditions must hold?

If there were more knights than kings, what conditions must hold?

If the number of kings and knights were equal, what conditions must hold?

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I have found 2 more patterns for the infinite chessboard and added them to the wiki.

As for the finite cases, some restrictions could maybe help. Since all the pieces must attack exactly 4 out of 8 tiles we can conclude:

Having this in mind, we can see that there is no arrangement of kings and knights on a 4x4 board.

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I am going to assume that we are overlooking the trivial case with 0 kings and 0 knight, right?

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Can we also reword the problem to exclude infinite cases? Otherwise, we're done.

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We aren't, because: (Sub-problem -- if it is possible, what's the smallest board needed?)

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So, I found out that there are 21 different arrangement of pieces or lack thereof on a two by two grid. This is where I go before I call it a night.

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Maybe this can help a little. 2x2 3x3 4x3 5x5 6x5 6x6 8x7

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isn't 4x4 possible with kings on all edges

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True, like this right? 4x4

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Reducing down to the just-kings case is a great way to start. Will all king configurations look like this, though? (Is it possible to make a "winding snake" with a larger setup, that is? And if not, can you prove it?)

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Found a few more patterns with the kings, but I'm just getting a grasp of the problem right now. I'm not sure yet, but I think that if the king is not on the 2x2 pattern I showed, it will have to form a loop, I'll try to prove that and get from there.

5x3 6x3 7x3 8x3

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How about if we start with a much simplified version of the problem here?

And then attempt to solve the 4x4 case, 5x5, case, and work toward the general nxn solution (for just 2 kings)?

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I'm inclined to think so too with about 80-20 confidence. The problem is to prove it!

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I think the answer is

no.Each king attacks 2 kings and 2 knights.

Each knight attacks exactly 2 kings and 2 knights.

Considering a \(4 \times 4\) chessboard, where it doesn't satisfies the condition.

In other words, it could be possible for a certain number of knights and kings. But not possible for each knights and kings.

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This looks like a good start, but it doesn't exhaustively cover every single configuration of kings and knights. To begin with, you might want to think of what configurations with the kings will meet "every king must attack two other kings" condition before adding on the knights.

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of any sizeAdmittedly, one approach that might get some traction is to start with specific board sizes (that is, if you can eliminate 4x4 very thoroughly, then 5x5, that might be useful and suggest a general method).

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3by3 board (A3+B3+B1+C1)"knights. "(A1+A2+C2+C3)"kings.

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King on A1 touches only 1 Knight and 1 King. Each piece needs to attack two of each piece.

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Looks like each knight is only attacking one other knight.

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Show where you're I'm confused because I see two knights attacking two knights.

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That means if you have four knights, each of the four knights will attack 2 knights each, for a total of 8 knight-attacks.

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