Puzzle : A heavy chessboard!

Well,

I have a thing for dramatic titles! The puzzle that I am about to post here is actually very simple. Without further ado, here it is.

Problem :  There is a standard 8×8 chessboard. Each of its 64 square is assigned a weight. These weights are assigned in such a manner that weight of a square is an average of the weight of the square that it is surrounded by. That is : w(i,j) = w(i-1,j-1) + w(i-1,j) + w(i-1,j+1) + w (i,j-1) + w(i,j+1) + w(i+1,j-1) + w(i+1,j) + w(i+1,j+1)/(num of squares surroinding (i,j)). Of course, the intelligent reader can deduce what would be the formula for the squares lying at the boundary. Now, prove that all the squares have equal weight!

Solution : The solution would require lesser space than the problem description :-). Assume that not all the squares have equal weight. Then there has to be a square whose weight is the least. Weight of this square can not be the average of the weight of the squares surrounding it! Please observe that if there are multiple squares with the least weight, the argument still holds as you will be able to find at least one square which has a neighbour with a higher weight. If not, then all of the squares have the least weight!!

Thanks to Deepanjan for this puzzle!

–Saurabh Joshi

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2 Responses to “Puzzle : A heavy chessboard!”

  1. Dhruva Sagar Says:

    You mentioned w(i,j) = w(i-1,j-1) + w(i-1,j) + w(i-1,j+1) + w (i,j-1) + w(i,j+1) + w(i+1,j-1) + w(i+1,j) + w(i+1,j+1), but it should be w(i,j) = (w(i-1,j-1) + w(i-1,j) + w(i-1,j+1) + w (i,j-1) + w(i,j+1) + w(i+1,j-1) + w(i+1,j) + w(i+1,j+1))/(number of surrounding squares)

    It is supposed to be the average of the weights, not sum.

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