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Posers and Puzzles

Posers and Puzzles

  1. Standard member wittywonka
    Chocolate Expert
    05 Dec '06 04:38
    Try these two puzzles out...place eight queens on a board so that no queen can capture another queen in one move. Also, try placing 5 queens on a board so that every space on the board is covered by a queen. These may seem elementary to some but difficult to others!
  2. Standard member wittywonka
    Chocolate Expert
    06 Dec '06 05:52
    Here...I'll post the 5 queens puzzle...

  3. 06 Dec '06 11:50
    Is this the only solution? I remember I had it in that computer game 'Majestic Chess' once and found a different solution. But somehow I cant find it now.
    There are the same puzzles with bishops and other pieces I think, but I can't remember how many you need to cover a board.
  4. Standard member wittywonka
    Chocolate Expert
    07 Dec '06 02:51
    Originally posted by crazyblue
    Is this the only solution? I remember I had it in that computer game 'Majestic Chess' once and found a different solution. But somehow I cant find it now.
    There are the same puzzles with bishops and other pieces I think, but I can't remember how many you need to cover a board.
    There are numerous solutions to both puzzles...I read somewhere that for the 8 queens puzzle there were around 100 possibilities...
  5. 08 Dec '06 23:33
    Has anyone tried proving that five is the minimum number of queens that can cover the whole board?
  6. 09 Dec '06 00:50
    Can you do it in 4?
  7. Standard member wittywonka
    Chocolate Expert
    09 Dec '06 01:33
    Originally posted by dmnelson84
    Can you do it in 4?
    I don't know, actually...I've never tried, but I don't think so...
  8. 11 Dec '06 15:34 / 1 edit
    I'm not sure if 4 is possible, but I thought that I'd try it the math way before I tried to write code to brute force the solution.

    First, I want to see if 64 square coverage is possible with 4 queens. First, if you place one queen on a center square, it covers:

    1 (it's own square, based on original problem)
    + 7 (row)
    + 7 (column)
    + 7 (diagonal 1)
    + 6 (diagonal 2)

    = 28 squares.

    Now, if you could keep placing Queens on the board, each covering 28 squares, then you could cover 103 points with 4 queens, but this is not the case.

    First, we have to look at the Queen as a rook. If you place two rooks on two points on the board, and made sure they were neither in the same row or the same column, together they would double cover one point. Thus, the effectiveness of each subsequent rook is diminished by 2*m, where m = the number of rooks previously placed on the board.

    Next, due to the nature of the Queen as a bishop, we have to look at centralization. The above sum is the result of the Queen being centralized, meaning, in the center 4 squares. The next Queen that is placed CANNOT be in the center four squares, because of what I would like to call 'heavy collision', where:

    collision (c) = the number of squares double protected by any set of two queens on the board.

    Now, collision, for two opposite colored Queens near the center of the board (one at e4, one at c5, if you will), is 10! This is the problem that will kill us in the end. The Second placed Queen will reduce in effectiveness from not only the fact it is in the second row from the center, where it's effectiveness is reduced, but also collision.

    On that previous statement, the effectiveness of a Bishop as it moves away from the center of the board, is diminished by a count of 2. Thus, the function b(x) where x = the distance from the center for x &gt= 0 , x &lt= 3 ranges from 13 to seven. We also have to add that the more centralized bishops of opposite color are in relation to a rook the more collision occurs, in the order of 6 at a maximum (see the collisions of the previous example) and likely 2 at a minimum. Let us just use these values and put in the possibilities from Q1 to Q4:

    Q1 = 28
    Q2 = 16
    Q3 = 16 - 4 - 2 = 10 (I'm assuming collision reduction
    Q4 = 16 - 4 - 2 - 2 = 8 (Assuming Collision reduction)

    28 + 16 + 10 + 8 = 62.

    I would like to see if even this number is possible. I've gotten 59 square coverage with 4 Queens (e4, c5, f6, d2) and a fifth Queen would finish the puzzle (b3).

    While my math is SHAKY and heavily generalized, it is good enough, I think, to prove it is impossible. Even as the collision with the central Queens diminishes, other things subtract enough from the placed Queens to make it impossible.
  9. 11 Dec '06 23:46 / 1 edit
    edit. i'll try again
  10. 11 Dec '06 23:51 / 2 edits
    Here's my solution to the 8 queen puzzle: (just learned FEN a few minutes ago on wikipedia. Had to take one failed attempt to write it right).
  11. 12 Dec '06 02:26
    this is another solution for the 5 queens. sorry for putting kings in there, but im not so familiar with fen code and this chess pad lets me set up only legal positions. so just ignore the kings

  12. Standard member Ari Brenin Cymru
    Siachmat!
    07 Feb '07 22:06
    In a past paper for an university scholarship exam a question asked the candidate to write an algorithm to find a solution to the 8 queens on a board puzzle.

    I'm glad this question didn't come up in the exam!
  13. 08 Feb '07 00:02
    Four queens can't attack all the empty squares, but 4 queens + 1 knight can. Try to find how.