14 Apr '08 14:01>
In the game Battleship, what's the optimal way to choose targets in order to find the enemy ships fastest?
Originally posted by deriver69I'd have to disagree and say go for the smaller ships first in a similar fashion, but also taking into account the corners and middle first. In that way, you don't have to retrace your steps looking for the 2-hit destroyer.
Not sure it is optimal, but I would go for spreading my shots so that the biggest ship could not escape, so in diagonal lines 4 apart (assuming the biggest ship is 5 long). If I was lucky enough to get smaller ships before I got it then that is even better. there may be a psychological way which takes into account where people are likely to put ships.
I ...[text shortened]... ship using the above method I suppose I would then close in on the next largest ship and so on.
Originally posted by deriver69this seems like it would be a sound strategy, if used in tandem with the mirroring idea i just mentioned... perhaps imagine a checkerboard lattice on the board, and fill it in using spaces from one of each quadrant, and then filling in 4x4 checkers before doing the interior 2x2 spaces? and hope to get lucky with the smaller ships? would be great to see some mathematics, given particular assumptions about ship placement (such as each square is equally likely to contain a boat as any other square... what is optimal then?)
A lot depends on how long the longest ship is. To find every ship may require in its worse case every other square to be marked off. Each targetted square must therefore be part of a checkerboard pattern. I would still start off by spacing these further out then close in though hoping to get that 2 square one luckily.
Originally posted by AetheraelIt seems like a sound strategy, but the question posed is "what's the optimal way to choose targets in order to find the enemy ships fastest?"
this seems like it would be a sound strategy, if used in tandem with the mirroring idea i just mentioned... perhaps imagine a checkerboard lattice on the board, and fill it in using spaces from one of each quadrant, and then filling in 4x4 checkers before doing the interior 2x2 spaces? and hope to get lucky with the smaller ships? would be great to see s ...[text shortened]... as each square is equally likely to contain a boat as any other square... what is optimal then?)
Originally posted by broblutothe only issue i have with this is that there will be many many games in which this method will require going to completion to find all of the ships... imagine a spread of ships all over the board - you would likely have to complete your entire algorithm for the average case where ships are as likely to be in the top left as they are to be in the bottom right quadrant of the board.
It seems like a sound strategy, but the question posed is "what's the optimal way to choose targets in order to find the enemy ships fastest?"
The optimal way would have to be:
a. fastest
b. find ALL ships
c. repeatable results using the same method
The solution fails if you can not find all ships, and leaving it to luck to find the small ships ...[text shortened]... ent question. Luck is a huge factor, especially if both opponents employ the same strategy.🙂
Originally posted by broblutoYour objection to forementioned approach (checkerboard corners of a 5x5 square) is that it leaves alot to luck.
It seems like a sound strategy, but the question posed is "what's the optimal way to choose targets in order to find the enemy ships fastest?"
The optimal way would have to be:
a. fastest
b. find ALL ships
c. repeatable results using the same method
The solution fails if you can not find all ships, and leaving it to luck to find the small ships ...[text shortened]... ent question. Luck is a huge factor, especially if both opponents employ the same strategy.🙂
Originally posted by AetheraelYou are correct. But the method would not take into account the average of the ships, but the smallest ship left.
and after posting that ridiculously long message, another thing occurred to me...
would it still be optimal to look at every other square after you've eliminated the destroyer? surely the checkerboard pattern could be beaten by an "every three" method by the same logic that now our smallest ship is of length 3?
i think maybe the pattern with which o ...[text shortened]... y 3. and removing the battleship makes the average ship length even smaller)
just an idea
Originally posted by strokem1That would work but only until you have to go back over territory you've already covered. Then it becomes inefficient because you'll be hitting squares horizontally or vertically adjacent to other squares.
I don't know if my strategy is optimal but I like to start near a corner and start playing Knight moves each time... 😛 It seems to do a decent job