What is the maximum number of knights in an 8 × 8 chessboard, so that no knights can attack each other?
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@Abhay Tiwari This is how to approach the problem.
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Thanks a lot for your concern sir. And now I can see where I made the mistake. But whoever controls that "challenger master note" is a bit rude.
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That was me. I was summarizing the various issues wrong with the solution so that others would not read it thinking that the solution was correct. Given that yours was the only solution, I didn't want the unsuspecting reader to believe that it was fine.
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@Calvin Lin – Ok, sir. I am sorry for posting wrong solution then. Will keep this thing in mind. But sir, you were rude in your last comment.
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@Abhay Tiwari – I'm sorry that you considered the comment as rude. That was not my intention.
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@Calvin Lin – Sir, I guess it was my fault, I should have deleted my solution in the first place.
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@Abhay Tiwari – You are even considering the staff as rude..... so in short if anyone opposes you, you consider it as rude.
@Calvin Lin – Exactly!..... Even I was not being rude and even then Abhay Tiwari considered my words rude here :(
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@Nitesh Chaudhary – I consider the usage of derogatory words rude. Your first deleted comment would have been fine without the last word which is a personal insult. Unfortunately, by using such a word, you end up being targeted even if the rest of your comments are not rude.
Let's phrase things in a positive light and encourage each other in this Brilliant community. E.g.
Can you post a more exciting question than class 5 exponents? Perhaps share some deeper material that you're working on?
Another way we could show that we cannot have more than 32 is by showing that the existence iof the Knight's tour guarantees us that if there are 33 or more knights, 2 of them must lie on consecutive squares on a Knight's tour.
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Great! That's a nice way to establish the pigeonhole :)
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1) First, we show that 32 is possible.
Place the knights on the 32 white squares. Then, the squares that such a knight attacks must be black, hence it doesn't attack any other knight that is on a white square.
2) Now, we show that 32 is an upper bound.
Break up the board into 8 smaller boards of size 2 × 4 . In each board, label the squares as such:
On each of these smaller boards, we see that we can place at most 1 knight in each of the pairs of squares with the same number, since these pairs of squares attack each other. Hence, there are at most 4 knights on this small board.
Since we have 8 different small boards, this shows us that we can have at most 8 × 4 = 3 2 knights on the total chessboard.