Author: Larry Coon
Date: 09:13:26 11/12/98
(Long question follows -- sorry.) I know I saw SEE talked about recently, but I couldn't locate the discussion here, on rgcc or anybody's web page (I don't remember where I saw it). Can somebody tell me if I'm thinking about this the right way? I'll start with a contrived position: 1kq5/3n2p1/5p2/8/6N1/5R2/5RK1 w Would a SEE do the following? 1. Find all attackers on the square in question (f5). 2. For each side, order the attackers in value, lowest to highest. Include the piece that is initially captured to start off the exchange. Put them into two....whatever, I'll say stacks just to make it easy to talk about. The lowest valued pieces will be at the top of the stack. Only the values of the pieces need to be pushed onto the stack. 3. Start with a "trade balance" of zero. 4. Pop the next value for the side NOT to move (black's stack if it's white's move) and SUBTRACT it from the trade balance, then do the same for the other stack. (This works because at least in my program opposing pieces have opposite signs). 5. When all values have been popped off either one of the stacks, then the exchange is over and the trade balance lets you know where you stand. 6. However, neither side has to follow the exchange through to its completion. So if either of the following happen: a. You pop the next value of "your" stack and subtract it from the trade balance and it ends up negative (for white, positive for black) b. You do the same for the opponent's stack and the result is positive (if black is the opponent, otherwise negative) Then you can assume that the side to move will not continue the exchange, exit the loop now, and the current trade balance will be the final balance. So in my example position, the stacks would look like (top of stack to the left): white: 3, 5, 5 black: -1, -1, -3, -9 And the process would be as follows: Step stack value balance ---- ----- ----- ------- 0 0 1 b -1 1 2 w 3 -2 3 b -1 -1 * 4 w 5 -6 ** 5 b -3 -3 6 w 5 -8 7 b -9 1 So at the point I marked *, the value is negative with black to move and the loop can actually stop here. Black doesn't have to play NxR, and he's won the exchange. Therefore, the initial capture is bad for white. However, if black DOES play NxR the following move takes him to an even better position (**). Maybe I keep track of "most positive after an odd-numbered step" and "most negative after an even numbered step"? Then the "most negative" would be the assumed jumping-out point for black, and the "most positive" would be the jumping-out point for white. Or is it like beta-cutoff in alpha-beta where you don't care how big the cutoff is, as long as you find -a- cutoff? So in this case, it doesn't matter if the next exchange takes black from -1 to -6, because the -1 by itself means the initial capture is bad for white. I think I'm missing something obvious here about the essence of this problem. Can anybody help? Thanks, Larry Coon
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