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Close Neighbors (Posted on 2004-02-05) Difficulty: 3 of 5
We can label the squares of an 8 x 8 chess board from from 1 to 64 in 64! different ways.

For each arrangement we find D, the largest difference between the labels of two squares which are adjacent (orthogonally or diagonally).

What is the smallest possible D (and how would you prove it)?

  Submitted by DJ    
Rating: 4.5714 (7 votes)
Solution: (Hide)
9

Consider the straightforward ordering:

 1  2  3  4  5  6  7  8
 9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48
49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64
Adjacent squares in the same row have difference 1, adjacent squares in different rows have difference 7, 8 or 9.
So, for this ordering, D = 9.

If we take any two squares on a chessboard there is a path from one to the other of length at most 7, where each step of the path is to an adjacent square [if the squares are at opposite corners of an m x n rectangle with m >= n, then take m - 1 steps, n - 1 of them diagonal and the rest along the longest side.] So there is a path of at most 7 steps from 1 to 64. At least one step of that path must have a difference of at least 9 (since 7 x 9 = 64 - 1).

Thus, we always have D ≤ 9, and the minimal D is 9.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Puzzle ThoughtsK Sengupta2023-09-05 21:38:52
SolutionAnswerMath Man2012-07-05 09:06:37
No Subjectpaul2004-03-08 20:13:32
QuestionFollowup Questionsstan2004-02-06 14:10:06
re(2): possible solution: proofSilverKnight2004-02-05 14:20:01
Solutionre: possible solution: proofBrian Wainscott2004-02-05 13:59:06
Solutionpossible solutionSilverKnight2004-02-05 13:29:19
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