[HoustonHokie]

How many rooks could you place on a chess board so that none of them are able to attack the others?

Make the following assumptions:

1. They have to move according to the normal rules

2. Any rook can attack any other rook - i.e., color is not important

3. The rooks can be placed anywhere on the board

4. There are only rooks on the board

5. The board is a normal 8x8 chess board

How about bishops? Or knights? Or queens?

Finally, suppose you placed a single queen on the board in a location of your choosing. Now you try to place as many rooks, bishops, or knights as possible so that they can neither attack nor be attacked by the queen or each other. What are your new answers, and how do they vary with the location chosen for the queen?

[Guest]

8- one in each square of a diagonal

[Guest]

15 - every square on two adjacent edges (8+7=15)

[Guest]

15 - every square on two adjacent edges (8+7=15)

i don't think that would work because every bishop except for the one in the corner could be attacked.

[Guest]

i don't think that would work because every bishop except for the one in the corner could be attacked.

Right, diagonals go both ways.

Yikes.

Ho

14. 8 along one edge, 6 along the opposite edge (not in the corners).

w about

Edited August 27, 2008 by Chuck Rampart

[Guest]

How many rooks could you place on a chess board so that none of them are able to attack the others?

Make the following assumptions:

1. They have to move according to the normal rules

2. Any rook can attack any other rook - i.e., color is not important

3. The rooks can be placed anywhere on the board

4. There are only rooks on the board

5. The board is a normal 8x8 chess board

How about bishops? Or knights? Or queens?

Finally, suppose you placed a single queen on the board in a location of your choosing. Now you try to place as many rooks, bishops, or knights as possible so that they can neither attack nor be attacked by the queen or each other. What are your new answers, and how do they vary with the location chosen for the queen?

Rooks: 8

Bishors: 14

Knights: 24

Pawns: 32 (unless using the weird pawn rule, em-passent ( spelt wrong))

no matter where you place the queen it takes up 29 spaces..

so you can have only 6 rooks now,

Edited August 27, 2008 by taliesin

[Guest]

you could put 8 all along one side and then 6 in the 6 the middle squares of the opposite edge so 14

edit: darn Chuckie beat me to it.

Edited August 27, 2008 by lemonymelon

[Guest]

edit: darn Chuckie beat me to it.

you could put 8 all along one side and then 6 in the 6 the middle squares of the opposite edge so 14

Mine doesn't count, because I made a laughable error on my first attempt.

So taliesin beat you to it.

[Guest]

Rooks: 8

Bishors: 14

Knights: 24

Pawns: 32 (unless using the weird pawn rule, em-passent ( spelt wrong))

no matter where you place the queen it takes up 29 spaces..

so you can have only 6 rooks now,

I think I can do better on the 1 queen, x rooks:

If you put the queen in a corner, she takes up 22 spaces - 7 along one edge, 7 along the adjacent edge, 7 along the diagonal, 1 that she's on. I think I can get 7 rooks in there:

Number the chess board from (1,1) to (8,8) because I don't like normal chess notation. Put the queen on (1,1). Now (1,x), (y,1) and (z,z) are out for all x, y and z. Put one rook at (2,3), another at (3,4), etc up to (7,8). That's 6. Now you can put the 7th at (8,2)

Am I wrong here?

[Guest]

I think I can do better on the 1 queen, x rooks:

If you put the queen in a corner, she takes up 22 spaces - 7 along one edge, 7 along the adjacent edge, 7 along the diagonal, 1 that she's on. I think I can get 7 rooks in there:

Number the chess board from (1,1) to (8,8) because I don't like normal chess notation. Put the queen on (1,1). Now (1,x), (y,1) and (z,z) are out for all x, y and z. Put one rook at (2,3), another at (3,4), etc up to (7,8). That's 6. Now you can put the 7th at (8,2)

Am I wrong here?

I think you have proven me wrong here, *crawls back in to my cave*

[HoustonHokie]

8- one in each square of a diagonal

Rooks: 8

Bishors: 14

Knights: 24

Pawns: 32 (unless using the weird pawn rule, em-passent ( spelt wrong))

no matter where you place the queen it takes up 29 spaces..

so you can have only 6 rooks now,

Good job - we've got rooks, bishops, and knights solved. taliesin even took a stab at pawns (kudos for the attempt). I didn't include them originally because pawns are color dependent - they can't go backwards. Plus the en passant rule does make things more difficult, as does the fact that they're allowed to move two spaces initially and one space afterwards - too confusing.

What about queens?

And I question taliesin's assertion about how many spaces a queen takes up...

[HoustonHokie]

I think I can do better on the 1 queen, x rooks:

If you put the queen in a corner, she takes up 22 spaces - 7 along one edge, 7 along the adjacent edge, 7 along the diagonal, 1 that she's on. I think I can get 7 rooks in there:

Number the chess board from (1,1) to (8,8) because I don't like normal chess notation. Put the queen on (1,1). Now (1,x), (y,1) and (z,z) are out for all x, y and z. Put one rook at (2,3), another at (3,4), etc up to (7,8). That's 6. Now you can put the 7th at (8,2)

Am I wrong here?

that's more like it...

[Prime]

I think I can improve the numbers just a bit.

And offer an algorithm.

Bishops and Rooks are easy:

Rooks: 8 (you can set them on a long diagonal).

Bishops: 14 as shown:

Here is a formal proof:

Rook occupies one square and attacks another 14, no matter where you place it.

Each square may be attacked by at the most 2 rooks without those attacking each other.

Therefore the upper limit for the rooks can be found from inequality: x + 14x/2 <= 64; Yielding x less or equal 8.

For the bishop the most advantageous placement is at the edge of the board where it occupies one square and attacks another 7.

A square may be attacked by 2 bishops at the most without them attacking each other.

Thus, upper limit with best placement: x + 7x/2 <= 64; Yielding x <= 14 and 2/9. And there are 14 squares available for best placement. (See the picture above.)

With the knights and queens it is more difficult. There are different placements with different number of squares attacked. Knight in the corner attacks only 2 square, in the middle - 8; a square may be attacked by 8 knights without any of them attacking another.

So here is an example of 25-knight placement:

Can there be more than 25? I don't know.

Here is an example of 7-Queen placement:

Can we place more? I doubt.

I did all the placements without aid of a computer. Nonetheless, here is a secret algorithm that I found:

Spoiler for Secret Algorithm:

After placing first Queen somewhere at the edge, place consequtive queens so as to minimize the number of "clean" (yet unattacked)squares that the newly placed queen attacks.

With knights, it's a bit more tricky. But same criterion -- minimize number of "clean" squares that each new placement attacks.

[Guest]

I can get 32 knight's on the board at the same time

Knights are Light Red

Pawns?

32

Pawns as grey

Kings?

16

Pawns as grey

[Prime]

I can get 32 knight's on the board at the same time

Knights are Light Red

Now, that knight placement is going to be hard to imporve upon!

There I go making algorithms when it was as simple as:

A knight placed on a white square can only attack black squares!

[Guest]

Now, that knight placement is going to be hard to imporve upon!

There I go making algorithms when it was as simple as:

A knight placed on a white square can only attack black squares!

I stumbled upon that by accident, it wan't my plain goinging into the drawing..

[Prime]

As far as Queen co-existing peacefully with other pieces:

7 Rooks can coexist with one Queen, as Chuck Rampart already noted. (Except, I don't understand his notation.)

(Queen is red)

It seems, we need to sacrifice one Bishop (from 14) to place Queen:

As for the Knights, we can use Taliesin's solution, placing Queen onto opposite color diagonal, into the corner, where it would be attacked only by 2 Knights:

To sum up:

7 Rooks, 13 Bishops, 22 Knights.

[Guest]

8, running corner to corner

How many rooks could you place on a chess board so that none of them are able to attack the others?

Make the following assumptions:

1. They have to move according to the normal rules

2. Any rook can attack any other rook - i.e., color is not important

3. The rooks can be placed anywhere on the board

4. There are only rooks on the board

5. The board is a normal 8x8 chess board

How about bishops? Or knights? Or queens?

Finally, suppose you placed a single queen on the board in a location of your choosing. Now you try to place as many rooks, bishops, or knights as possible so that they can neither attack nor be attacked by the queen or each other. What are your new answers, and how do they vary with the location chosen for the queen?

[HoustonHokie]

As far as Queen co-existing peacefully with other pieces:

7 Rooks, 13 Bishops, 22 Knights.

nice... If you moved the queen, would it reduce the number of other pieces you could fit on the board?

[Prime]

nice... If you moved the queen, would it reduce the number of other pieces you could fit on the board?

As far as Queen with Rooks, or Bishops -- there are several similar placements for the queen, giving the same result. Think of the postion in terms of lines and diagonals.

With Queen and Knights, I can see only one other such placement -- Queen in the opposite corner. (Or you could put knights on the white squares and Queen at the end of black diagonal.) Either way, all those are symmetrical variations.

[soop]

Just an interesting aside; I was in the pub one night, and one of the tables had a chess board printed on it. As far as I remember, I put empty glasses on the board in a way that:

1. There were as many glasses on the table that there could possibly be given that

2. If the glasses could have the same moves as every chess piece, no one piece could take another. It was some time ago, but I believe there were about 7 or 8 glasses.

(bearing in mind that this was late on a Friday night, years ago, but it could make for an interesting puzzle).

*edit* so far, I can only get 6

Edited August 28, 2008 by soop

[Guest]

I refrained from answering the all queens puzzle, because I already knew the answer. It's actually a well-known math / computer science problem. I won't spoil the solution though. You can read about it on Wikipedia, though, if you'd like (I'd provide a link, but the title would give away the answer).

[Guest]

Now, that knight placement is going to be hard to imporve upon!

There I go making algorithms when it was as simple as:

A knight placed on a white square can only attack black squares!

that was funny - coz its as easy as the rooks - not sure i can see the pawns yet - en passent is only the first move, does it count if you are 'placing' pawns, which means not a true first move unless on its natuaral atarting square. My head cant take it in

edit; en passent/placing clarity - it that is clear?

Edited August 28, 2008 by Lost in space

[HoustonHokie]

As far as Queen with Rooks, or Bishops -- there are several similar placements for the queen, giving the same result. Think of the postion in terms of lines and diagonals.

With Queen and Knights, I can see only one other such placement -- Queen in the opposite corner. (Or you could put knights on the white squares and Queen at the end of black diagonal.) Either way, all those are symmetrical variations.

I guess what I was going for was - if you moved the queen out of the corner with the rooks, would you still be able to place 7 rooks on the board? Or would it be fewer than that? Can you place the queen in such a way to reduce it even further? I don't know the answer myself; I'm just asking questions.

[Guest]

that was funny - coz its as easy as the rooks - not sure i can see the pawns yet - en passent is only the first move, does it count if you are 'placing' pawns, which means not a true first move unless on its natuaral atarting square. My head cant take it in

edit; en passent/placing clarity - it that is clear?

I have added a solution of pawns, which solves including em-passent ( it is really the 2nd move) and doesn't matter whether the pawns move up or down.. It has 32 and I cant see much improvement on that.

[Guest]

I guess what I was going for was - if you moved the queen out of the corner with the rooks, would you still be able to place 7 rooks on the board? Or would it be fewer than that? Can you place the queen in such a way to reduce it even further? I don't know the answer myself; I'm just asking questions.

I think 7 has to be the maximum for one queen + rooks. A queen does everything a rook does (and more, of course). We know for sure that every piece must be in its own row and its own column, and there are only 8 rows and 8 columns.

I am also pretty sure that there can be 7 rooks for any queen placement, and multiple 7-rook arrangements at that. I haven't found a general algorithm to place the rooks, but I can't find a counter-example either.