1. Joined
    12 Mar '03
    Moves
    44411
    06 May '07 08:02
    Originally posted by SwissGambit
    Because no piece can observe another.
    my bad, sorry.
  2. In Christ
    Joined
    30 Apr '07
    Moves
    172
    06 May '07 18:21
    Well, I also got 112 with the solution already posted. I wonder though (and this just came to me): What if we put more emphasis on one type of piece? In other words, what if we try to maximize [(number of knights)^2]x(number of bishops), or (number of knights)x[(number of bishops)^2]?
  3. Standard memberSwissGambit
    Caninus Interruptus
    2014.05.01
    Joined
    11 Apr '07
    Moves
    92274
    06 May '07 20:471 edit
    Originally posted by Jirakon
    Well, I also got 112 with the solution already posted. I wonder though (and this just came to me): What if we put more emphasis on one type of piece? In other words, what if we try to maximize [(number of knights)^2]x(number of bishops), or (number of knights)x[(number of bishops)^2]?
    For the first case, the maximum product seems to be 2304.

    For the 2nd, it looks like 900.
  4. Joined
    17 Mar '07
    Moves
    207
    06 May '07 23:10
    Yeah, variations could be interesting. Changing the pieces involved doesn't seem to work (pieces other than bishops and knights work too simply -- for instance, a rook always blocks off exactly 14 squares), nor does changing from product to sum (I seriously doubt you could do better than 32). What happens if pieces of the same kind don't threaten each other?

    SwissGambit, you make a good case for 112. Ideally we'd test it with a computer or something, but I'll buy that it's maximal.
  5. Joined
    06 Mar '07
    Moves
    1476
    06 May '07 23:54
    Originally posted by SwissGambit
    For the first case, the maximum product seems to be 2304.

    For the 2nd, it looks like 900.
    I also got 2304 for the knight^2. However I only managed to get 891 for the bishop^2 version. I assume to get 900 you used 10 bishops and 9 knights but I can't find a way to get that, only 10 bishops and 8 knights or 9 bishops and 11 knights.
  6. Standard memberSwissGambit
    Caninus Interruptus
    2014.05.01
    Joined
    11 Apr '07
    Moves
    92274
    07 May '07 00:31
    Originally posted by PaddyB
    I also got 2304 for the knight^2. However I only managed to get 891 for the bishop^2 version. I assume to get 900 you used 10 bishops and 9 knights but I can't find a way to get that, only 10 bishops and 8 knights or 9 bishops and 11 knights.
    Unfortunately, I didn't save the position with 10 B and 9 N. I can't seem to reproduce it, so I wonder if I simply mis-counted the first time.
  7. In Christ
    Joined
    30 Apr '07
    Moves
    172
    07 May '07 17:04
    A recent situation has inspired a whole new set of puzzles. What if we allow the pieces to threaten or observe each other, but try to maximize the ratio (number of pieces on the board)/(number of unthreatened unoccupied spaces on the board)? This could even be a challenge with queens or rooks, but bishops or knights could be just as fun. And if you want to use two different pieces, then just maximize (number of one piece on the board)x(number of the other piece on the board)/(number of unthreatened unoccupied spaces on the board). These are just as new to me as they are to you, so I don't have any answers at this point. If you come up with an answer for a certain piece (or pair, trio, quartet, or even pentad of pieces), just post it here. That last one would probably be the most interesting: using kings, queens, rooks, bishops, and knights all at the same time. That would even be tough under the previous premise: trying to make none of them threaten (or observe) each other.
  8. Standard memberSwissGambit
    Caninus Interruptus
    2014.05.01
    Joined
    11 Apr '07
    Moves
    92274
    08 May '07 19:111 edit
    Originally posted by Jirakon
    A recent situation has inspired a whole new set of puzzles. What if we allow the pieces to threaten or observe each other, but try to maximize the ratio (number of pieces on the board)/(number of unthreatened unoccupied spaces on the board)? This could even be a challenge with queens or rooks, but bishops or knights could be just as fun. And if you want to u tough under the previous premise: trying to make none of them threaten (or observe) each other.
    This might attract more interest if it were narrowed down a bit. I recommend that you do the preliminary grunt work of testing some of these cases. If there are any that make for good puzzles, those can be shared with the group one at a time (it's more interesting when a few people are trying to solve the same puzzle, and people are more likely to try one clearly-defined problem rather than a research project).
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