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## 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).

This post is under “Logic” and has 3 respond so far.

### 3 Responds so far- Add one»

1. 1. DP Said：

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!

2. 2. Chuck H Said：

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.

3. 3. Chris Said：

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.

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