## A chess board problem

Posted by Chris on December 28, 2012 – 7:35 am

If no row, column or diaganol can have more than five pieces on it, what is the maximum number of squares that can be occupied on a chess board?

If you want to show a solution, use “_” and “X” for empty an occupied. Whatever you use, I’ll make sure it aligns (within 24 hours of your posting).

December 28th, 2012 at 11:45 am

I can get 39 with a quick, brute-force approach

X X _ X X _ _ X

X X X _ X _ _ X

X X _ _ X X X -

X _ _ X X _ X X

_ X X _ _ X X X

_ X X X X _ _ X

X _ X X _ X X _

_ _ X X _ X X _

I (very arbitrarily) started with an “every 3rd square” approach, then filled in from there.

There’s probably a more efficient or mathematical way to fill them in, but it’s a base-point for the next guy!

December 28th, 2012 at 6:45 pm

I can get to 40 with a slight modification to DP’s solution (see rows 6 and 8):

X X _ X X _ _ X

X X X _ X _ _ X

X X _ _ X X X _

X _ _ X X _ X X

_ X X _ _ X X X

_ X X X _ X _ X

X _ X X _ X X _

_ _ X X X X X _

I believe 40 is the entitlement since each of the eight rows and columns cannot hold more than 5 pieces. In addition, there are ten diagonals to be concerned with: five heading upwards of 6, 7, or 8 spaces, and five heading downwards of 6, 7, or 8 spaces. All 8 columns, 8 rows, 5 up diagonals, and 5 down diagonals meet the criterion.

December 29th, 2012 at 7:42 am

My dad came up with the following (which is nice, as my son gave me the problem). Like Chuck H, he noticed that 5*8 = 40 is the max conceivable number.

He started with a tidy 32 squares solution, shown with X’s, then added 8 more pieces, shown with x’s.

X_X_X_Xx

_XxX_X_X

XxX_X_X_

xX_X_X_X

_X_X_XxX

X_X_XxX_

_X_XxX_X

X_XxX_X_

The only other thing that I had realised was that there are 8!/(5! 3!) = 56

ways of arranging a line with 5 counters in it. That made me believe that there was a very good chance that 40 squares could be occupied.

The recent hats problems made me think of the following: If a particular square in a row does contain a piece, then there are 7!/(4! 3!) = 35 ways of having 4 occupied squares on the rest of the line. If 2 particular squares are occupied, then there are 6!/(3! 3!) = 20 ways for the remaining squares to be occupied. etc. However, I’ve not gone beyond such ruminations.