# Chess numbers

Arjun Shastry

Ranch Hand

Posts: 1893

posted 4 years ago

- 0

I am trying to solve this problem but not able to.(Taken from Mathematical Circles(Russian experience)).

On 8x8 chess board, numbers from 1 to 64 are written . Each number is written exactly once.

Prove that you can always find one pair of adjacent squares(which share common edge) such that difference in their numbers is at least 5.

On 8x8 chess board, numbers from 1 to 64 are written . Each number is written exactly once.

Prove that you can always find one pair of adjacent squares(which share common edge) such that difference in their numbers is at least 5.

MH

Mike Simmons

Ranch Hand

Posts: 3028

10

posted 4 years ago

- 0

Consider two squares (not necessarily adjacent) - one at (x1, y1), and the other at (x2, y2). The first point has number N1 and the second has number N2. The shortest path between them has length abs(x2-x1) + abs(y2-y1), if we only use horizontal moves or vertical moves. If we consider how N changes as we move from square to square, we know that

Now, somewhere on the 64 squares there's a 1 and a 64. Let those be N1 and N2. What's the smallest possible value of avg? The numerator is clearly 63. The largest possible value for the denominator is if N1 and N2 are on opposite corners, which are 7 squares apart vertically, and 7 horizontally. So the smallest possible value of avg is 63/14 = 4.57. Since all individual changes in N must be integers, we know that at least one change

*average*rate of change on a particular path must be avg = (N2 - N1)/(abs(x2-x1) + abs(y2-y1)).Now, somewhere on the 64 squares there's a 1 and a 64. Let those be N1 and N2. What's the smallest possible value of avg? The numerator is clearly 63. The largest possible value for the denominator is if N1 and N2 are on opposite corners, which are 7 squares apart vertically, and 7 horizontally. So the smallest possible value of avg is 63/14 = 4.57. Since all individual changes in N must be integers, we know that at least one change

*must*be 5 or greater. And at least one must be 4 or less.
Ryan McGuire

Ranch Hand

Posts: 1048

4

posted 4 years ago

Nice insight!

I had worked out a proof that starts with some number in the middle of the board and proves that the in placing the next so many numbers, you are forced to end up with two adjacent numbers that are more than 4 apart. I was dreading typing in all the verbiage for the proof.

- 0

Mike Simmons wrote:[minimum average difference between 1 and 64 must be > 4.]

Nice insight!

I had worked out a proof that starts with some number in the middle of the board and proves that the in placing the next so many numbers, you are forced to end up with two adjacent numbers that are more than 4 apart. I was dreading typing in all the verbiage for the proof.

I agree. Here's the link: http://aspose.com/file-tools |