[Guest]

What is the maximum number of Bishops (that can not attack each other) that can be placed on a four by four grid? And in how many ways?

Edited June 4, 2009 by adiace

[Guest]

Maximum of 6, 4 different ways?

[Guest]

6 bishops.

If you call each place as a1,a2..,b1,b2,, as in standard chess, 12 different combinations are possible, but most of them are symmetric.

[Glycereine]

6 bishops, 3 ways, each rotateable 90 degrees for 12 total:

1) a1, a2, a3, a4, d2, d3

2) a1, a2, a3, d1, d2, d3

3) a1, a2, b4, c4, d1, d2

Edited June 5, 2009 by Glycereine

[Guest]

6 bishops, 3 ways, each rotateable 90 degrees for 12 total:

1) a1, a2, a3, a4, d2, d3

2) a1, a2, a3, d1, d2, d3

3) a1, a2, b4, c4, d1, d2

In way 3, a2 and c4 can attack one another

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[Guest]

What is the maximum number of Bishops (that can not attack each other) that can be placed on a four by four grid? And in how many ways?

Rotate the square 45 degrees into a diamond and look at all possible left-to-right sweeps that can be covered by a bishop. There should be 2n-1 of them in an (n x n) board, so 7 of them in this case. Since the one at the top and bottom corners can reach each other, you can't put a bishop in both... but there are ways to put bishops in all the other ones by staggering them appropriately. For two rows of the same width, one bishop must be on the left and the other bishop on the right. So, 2n-1 minus 1 more for one of the corners is 2n-2 for 6 maximum bishops. An (8 x 8) board would have 14 maximum bishops for the same reasons.

From this same perspective, with a little bit of experimentation, you can see the following:

1. You'll always place each bishop at a left or right edge of any given row, never in a middle square.

2. You'll always place one bishop at either the top or bottom row.

3. You'll always place one bishop in the left or right edge of the widest row (the one that is n columns wide).

4. There are exactly 2 rows of each width from 2 to n-1, and you'll place one bishop on the left of one, and one on the right of the other.

If you are convinced of the truth of the above four statements, you can arrive at a formula for calculating the total number of arrangements of the maximum number of non-attacking bishops in an (n x n) square.

2 * 2 * ( 2 ^ (n - 2) ) which reduces to 2 ^ n

For example, in the (4 x 4) square, you can arrange your bishops in the following combos:

Bishop 1: A1 or D4

Bishops 2 and 6: (A2,D3) or (B1,C4)

Bishops 3 and 5: (A3,D2) or (C1,B4)

Bishop 4: A4 or D1

So, your 6 bishops can be arranged 16 different ways.

1. A1,A2,D3,A3,D2,A4

2. A1,A2,D3,A3,D2,D1

3. A1,A2,D3,C1,B4,A4

4. A1,A2,D3,C1,B4,D1

5. A1,B1,C4,A3,D2,A4

6. A1,B1,C4,A3,D2,D1

7. A1,B1,C4,C1,B4,A4

8. A1,B1,C4,C1,B4,D1

9. D4,A2,D3,A3,D2,A4

10. D4,A2,D3,A3,D2,D1

11. D4,A2,D3,C1,B4,A4

12. D4,A2,D3,C1,B4,D1

13. D4,B1,C4,A3,D2,A4

14. D4,B1,C4,A3,D2,D1

15. D4,B1,C4,C1,B4,A4

16. D4,B1,C4,C1,B4,D1

[Guest]

6 bishops, 3 ways, each rotateable 90 degrees for 12 total:

1) a1, a2, a3, a4, d2, d3

2) a1, a2, a3, d1, d2, d3

3) a1, a2, b4, c4, d1, d2

In way 3, a2 and c4 can attack one another

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Really, since I had found the result as 12, I agreed this solution and didn't check it.

As phat impressed, way 3 is wrong.

In my solution it would be

a2, a3, a4, d2, d3, d4

This is very similar to way 2 but not the same, even it's rotated 90⁰ four times.

So the result is still 12, I suppose.

Opps. I saw phat's recent result . I'll check it.

Edited June 8, 2009 by nobody