Author: Enrique Irazoqui
Date: 06:36:18 02/15/00
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On February 15, 2000 at 09:12:25, Shep wrote: Hi Shep, Your math is interesting, but don't you think that the easiest and most accurate procedure in the case of uneven hardware is the simple swapping of machines? Each program would play half the match on the fast computer and half on the slow one. Enrique >Hi all, > >I was trying to write this post earlier, but found a flaw in my calculations, so >here goes the fixed one: > >Suppose you want to play a match on two machines with PB and only have two >machines of different speed. (Obviously, just adjusting the time controls to >reflect the speed difference won't make it fair.) > >Assume that machine A is the faster one and that machine B is slower by a factor >of k>1. The natural impulse would be to give machine B k-times as much >reflection time. >For simplicity, let's assume that both machines predict precisely 50% of their >opponent's moves. Then the effective reflection times for both side are: > > R_a(k) = 1+k/2 > R_b(k) = k+1/2 > >Both reflection time functions are linear in k and intersect at k=1, thus it's >easy to see that > > R_a(k) < R_b(k) for all k>1 > >Thus > > l := R_b(k)/R_a(k) > 1 for all k>1 > >is the factor by which B's reflection time is too big to be fair. > >Consequently, to have a fair match, the multiplier for B's reflection time needs >to be k/l instead of k. >A quick calculation yields the final formula: > > m := k/l = (k^2+2k)/(2k+1) > >Thus, if machine B is given m-times the reflection time of machine A, the match >is fair (under the assumption that both sides predict 50% which is not precise, >but rather realistic - the final formula is slightly more complex for uneven >prediction rates). > >Example table: > > k m > -------- > 2 ~1.6 > 3 ~2.1 > 4 ~2.7 > 5 ~3.1 > 10 ~5.8 > >--- >Shep
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