Quote from: robotrebellion on August 06, 2013, 07:29:14 PM

Quote from: Meni Rosenfeld on August 06, 2013, 07:10:36 PM

Quote from: robotrebellion on August 06, 2013, 03:24:59 PM

Is it possible to calculate in advance how many shares will be included in the window? I'm thinking in terms of an implementation using a shares database (e.g., SQL). Before I begin collecting shares, can I compute approximately how many I will need to look at?

Yes, it will be approximately X * D. Depending on how you intend to use the approximation, a better one can be found.

My goal is to limit the number of rows fetched from the table to reduce server overhead. I would of course want to avoid the possibility of under-estimating, else the total payout wouldn't equal the block reward. Would I need to add anything to X*D to avoid underestimation?

If D1 is the current difficulty and D2 is the previous difficulty, an upper bound would be X * max (D1, D2) + 3. Taking the max makes sure that if the difficulty has just decreased, you won't miss the shares needed at the previously high difficulty; the +3 term helps with rounding, off-by-one errors etc., it can probably be tightened but it costs very little so I wouldn't bother.

Note that if the window spans more than one difficulty adjustment it may not work (you might need the maximum over previous difficulties as well) but that's not a situation I believe should happen.

If miners submit shares with difficulty greater than 1 this can be decreased. If d is the minimal difficulty that they can submit (assuming that throughout the window you only have shares with this difficulty or higher), and upper bound is X * max (D1, D2) / d + 3.