Andrei Alexandrescu wrote: > Wait, doesn't a benchmark always measure an algorithm with the same > input?
The fact that you formulate as a question indicates that you are unsure about the wright answer---me too, but 1) surely one can define a benchmark to have this property. But if one uses this definition, the used input would belong to the benchmark as a description. I have never seen a description of a benchmark including the input, but because I am more interested in theory I may have simply missed such descriptions. 2) if a probilistic algorithms is used, the meaning of input becomes unclear, because the state of the machine influences T. 3) if a heuristic is used by the benchmarked algorithm, then a made up family of benchmarks can "prove" T= O(n*n) for quick sort. -manfred
