Solving the 8 Pawns Puzzle on an 8x8 Chess Board

superconduct
Messages
31
Reaction score
1
Imagine a 8x8 chess board, is it possible to place 8 pawns such that they are of different distances from one another?
 
Mathematics news on Phys.org
Asuming diagonal distances are also included, it doesn't seem like it should be too difficult with only 8 pawns.
 
superconduct said:
Imagine a 8x8 chess board, is it possible to place 8 pawns such that they are of different distances from one another?
What are your thoughts on this?
 
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...
 
D wasn't really meant to be a function by the way, just a way of signifying the concept
 
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?
 
jbriggs444 has my interpretation.

With that interpretation, no pawns should lie on the perpendicular bisector of any two pawns already placed.
 
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?
 
^I remember that problem. It's possible if I remember correctly
 
  • #10
please prove and explain to me
 
  • #11
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
well come to think of it, each queen eliminates two diagonals, so I'm not sure it would ever work
 
  • #13
Wikipedia says that there are 92 ways of solving the problem though, so I don't know.
 
  • #14
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)
 

Similar threads

Replies
20
Views
6K
Replies
64
Views
23K
Replies
26
Views
685
Replies
42
Views
4K
Replies
25
Views
17K
Back
Top