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Chess problem

  1. Apr 30, 2013 #1
    Imagine a 8x8 chess board, is it possible to place 8 pawns such that they are of different distances from one another?
  2. jcsd
  3. May 1, 2013 #2


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    Asuming diagonal distances are also included, it doesn't seem like it should be too difficult with only 8 pawns.
  4. May 1, 2013 #3


    Staff: Mentor

    What are your thoughts on this?
  5. May 1, 2013 #4
    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...
  6. May 1, 2013 #5
    D wasn't really meant to be a function by the way, just a way of signifying the concept
  7. May 1, 2013 #6


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    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?
  8. May 1, 2013 #7
    jbriggs444 has my interpretation.

    With that interpretation, no pawns should lie on the perpendicular bisector of any two pawns already placed.
  9. May 1, 2013 #8
    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?
  10. May 1, 2013 #9
    ^I remember that problem. It's possible if I remember correctly
  11. May 2, 2013 #10
    please prove and explain to me
  12. May 2, 2013 #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.
  13. May 2, 2013 #12
    well come to think of it, each queen eliminates two diagonals, so I'm not sure it would ever work
  14. May 2, 2013 #13
    Wikipedia says that there are 92 ways of solving the problem though, so I don't know.
  15. May 2, 2013 #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)
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