# Chess problem

1. Apr 30, 2013

### superconduct

Imagine a 8x8 chess board, is it possible to place 8 pawns such that they are of different distances from one another?

2. May 1, 2013

### rcgldr

Asuming diagonal distances are also included, it doesn't seem like it should be too difficult with only 8 pawns.

3. May 1, 2013

### Staff: Mentor

What are your thoughts on this?

4. May 1, 2013

### hello95

so are we talking a series of pawns (p(1), p(2), .... , p(8)) such that D(p(i)) - D(p(i + 1)) is different for every i? Because something like a conch shell spiral works if that's the case. More generally I think with any nxn board it should be possible to place n pawns at different distances from one another using the same method. (start at square (n -1, n +1), move one place to the right, then two places up, 3 to the left, etc...

5. May 1, 2013

### hello95

D wasn't really meant to be a function by the way, just a way of signifying the concept

6. May 1, 2013

### jbriggs444

My interpretation of the problem is different. For each pair of pawns Pi and Pj, the distance must be unique.

With that interpretation, I'd treat it as a counting problem. How many distinct non-zero distances are there on an 8x8 chessboard? How many distinct pairs of pawns are there in a set of 8?

7. May 1, 2013

### superconduct

jbriggs444 has my interpretation.

With that interpretation, no pawns should lie on the perpendicular bisector of any two pawns already placed.

8. May 1, 2013

### eddybob123

Here's another one: is it possible to place 8 queens on the board such that no queens are "attacking" any other queen, and why?

9. May 1, 2013

### hello95

^I remember that problem. It's possible if I remember correctly

10. May 2, 2013

### superconduct

please prove and explain to me

11. May 2, 2013

### hello95

Actually I'm not sure if it does work now that I'm trying to work it out... each queen eliminates a row, column, and diagonal from future options of where to place the next queen. Eventually the eliminated columns and diagonals amount to eliminating a row no matter how you place the first 4 or 5 queens, which seems to indicate that it would be impossible to put 8 queens on the board, but I'm not sure. I'd have to spend more time on it.

12. May 2, 2013

### hello95

well come to think of it, each queen eliminates two diagonals, so I'm not sure it would ever work

13. May 2, 2013

### hello95

Wikipedia says that there are 92 ways of solving the problem though, so I don't know.

14. May 2, 2013

### superconduct

but is the queen version equivalent to my pawn version? I mean like when you place a queen then another directly next to it then another 1 square away from the 2nd, this is not legit in queen version but is legit in pawn version(not including future pawns)