Wednesday, March 21, 2018

mathematics - Professor Halfbrain and the 9x9 chessboard (Part 1)



Professor Halfbrain has spent has spent the last few days with placing pawns on a 9×9 chessboard; each of the 81 squares on the chessboard had side length 1. Halfbrain always started with an empty board, and then one by one placed (point-sized) pawns onto it. Every pawn was placed precisely in the middle of one of the little chessboard squares. Whenever Halfbrain placed a new pawn, its distance to each of the pawns placed before was at least 2.


Professor Halfbrain has proved two extremely deep theorems on such placements of pawns.



Professor Halfbrain's first theorem: It is possible to place three pawns according to the above rules.


Professor Halfbrain's second theorem: It is not possible to place 81 pawns according to the above rules.



This puzzle asks you to improve the two theorems of professor Halfbrain and to make them even deeper. Find an integer x, so that "three pawns" in the first theorem may be replaced by "x pawns", and so that "81 pawns" in the second theorem may be replaced by "x+1 pawns" (again yielding true statements, of course).



Answer



There is a simple way to see that manshu's answer of x=25 is correct.


Rather than consider a 9×9 board and pawns, consider a 10×10 board and 2×2 square pieces. We can associate to each pawn configuration on the 9×9 board a unique configuration of squares in the 10×10 board; consider the image below.



From pawns to 2x2 squares


The blue squares are the squares we added to our board to make it 10×10. The 2×2 square on the side is a piece for our new board, and the red square marks where the pawn lies for our previous board. It's easy to see this relation is a bijection.


Now, our new board has a total of 100 squares, and each piece uses up 4 squares. It's thus clear that no configuration can go beyond 25 pieces.


EDIT: Added an example. In the image below, to the left we have the 9×9 board in which red squares mark pawn positions. To the right is the corresponding 10×10 board with 2×2 square pieces; notice the position of red squares is preserved and no pieces overlap.


An example of the bijection


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