Author: David Paulowich
Date: 10:50:56 02/15/00
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On February 15, 2000 at 09:12:25, Shep wrote: >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 > SKIP >--- >Shep I am still trying to shake off a winter cold, but I will give this a try. THERE IS NO SOLUTION FOR k = 2 (OR GREATER). Suppose the slow machine plays game/m hours and the k-times as fast machine plays game/1 hour. Following your calculations, with 50% success rate for predictions, we obtain effective reflection times of: R(slow) = m + 0.5 hours = 28.5 hours when m = 28 R(fast) = k(1 + 0.5m) hours = 28.5 hours when m = 28 and k = 1.9 Solving the equation: m + 0.5 = k(1 + 0.5m) gives the formula: m = (2k -1)/(2 - k) for values of "k" between 1 and 2. For example, m = 298 when k = 1.99 No matter how large "m" gets, you cannot handle a speed advantage "k" greater than 2, assuming the machines have a 50% success rate in pondering.
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