Yeah, that's interesting. Your solution and algorithm sound sensible.
I don't think simon's hypothesised equivalence holds, either, assuming the #s are assumed to extend all the way into the corners of their squares.
You're right, I think it works in the case we were thinking about (directly along an edge in a hex grid), but I overgeneralised. I think the same might be relevant for corners in a square grid, though I'm not positive.
In fact, even on a hex grid, the straight line for a 4x1 step
A . O
. O . B
passes through both "O" squares. But I think it's right for my purposes to treat it as the 1x1 step
A .
. B
and require either of the squares to be free. It looks like you ought to be able to see :) But I don't know what rigorous condition, if any, it corresponds to :)
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Date: 2008-09-22 12:53 pm (UTC)I don't think simon's hypothesised equivalence holds, either, assuming the #s are assumed to extend all the way into the corners of their squares.
You're right, I think it works in the case we were thinking about (directly along an edge in a hex grid), but I overgeneralised. I think the same might be relevant for corners in a square grid, though I'm not positive.
In fact, even on a hex grid, the straight line for a 4x1 step passes through both "O" squares. But I think it's right for my purposes to treat it as the 1x1 step and require either of the squares to be free. It looks like you ought to be able to see :) But I don't know what rigorous condition, if any, it corresponds to :)