No, you can have any number of tiles, so brute force isn't enough.
Shame.
You can always score zero with the strategy I describe; you can always score 1 with the strategy you describe.
Agreed.
Actually I think you can always score at least ceil(n/2): choose the letter which appears most often above, and lay that down in every place below. If the tiles are alternating then that's the best you can do.
Reckon so. That's a nice strategy - it does well in lots of circumstances. Presumably you want a solution that'll outperform that when possible? I'm not finding any immediately to hand.
no subject
Shame.
You can always score zero with the strategy I describe; you can always score 1 with the strategy you describe.
Agreed.
Actually I think you can always score at least ceil(n/2): choose the letter which appears most often above, and lay that down in every place below. If the tiles are alternating then that's the best you can do.
Reckon so. That's a nice strategy - it does well in lots of circumstances. Presumably you want a solution that'll outperform that when possible? I'm not finding any immediately to hand.