Author: Robert Hyatt
Date: 07:48:50 05/31/00
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On May 31, 2000 at 01:10:52, Tom Kerrigan wrote: >On May 30, 2000 at 21:48:52, Robert Hyatt wrote: > >>>All I'm saying is that the numbers you gave (e.g. 25%) are (hash hits/hash >>>probes) and not (hash hits/nodes). The latter is what's more important in this >>>case. >>> >>>-Tom >> >> >>That number would be meaningless. Of course you can't "hit" if you don't >>"probe". Who would care? I only want to know what percentage of the time I >>get a hit after doing a probe... which seems like the only reasonable measure >>of anything. If a program probes in the q-search, those numbers would match >>mine _exactly_ since in his case, probes == nodes. > >No, that number would NOT be meaningless. > A number computed from dividing one independent value into another independent value is meaningless. such a hit percentage would be useless unless you compare two programs with identical q-searches. Because you are factoring in q-search nodes which has absolutely nothing to do with the number of hash hits and misses I get. Since I neither hit nor miss in the q-search. >Sure, if you are trying to gague the effectiveness of a hash entry replacement >scheme, it makes more sense to measure hits/probes. But in this case, "we" need >to find out how many times the hash move can short-circuit move generation. So >unless you don't generate moves in qsearch() either, the number is important. > >-Tom Why? The question is does not generating moves save time? Not "does not generating moves only in the normal search save time?" or not "does not generating moves only in the q-search save time?" Rather than trying to develop a mathematical model for everything, it is far easier to get a much better number by just computing it within the program. In simple terms, if you do less work by not calling movgen every time, you win. If you do more work when you avoid calling movgen, you lose. For me it is a 10% win, roughly.
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