Author: Andrew Dados
Date: 13:02:59 05/09/01
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On May 09, 2001 at 15:49:26, Dann Corbit wrote: >Suppose that I have a collection of sorting algorithms. > >I know that the count of vector elements to be put into sorted order will never >exceed a size_t. > >I have a counting sort that can sort in O(n*log(element_length)) > >I have a relaxed-quick-heapsort that can sort in O(log(n!)) > >I have an shell sort that can sort in O(n^(1 + epsilon/n)) > >Now, since we know that the count will never exceed a size_t, have all these >calculations really simplified to O(1)? Is there really no way to choose >between them? If they really do retain their original behaviors, then what is >the utility of declaring them to be O(1)? In chess, no matter what position, it is guaranteed to be solved after less then 2^100 steps. Now find me such a sorting algorithm that will require less then Constant steps * no matter what size_t *. Can't you see that simple difference? -Andrew- -Andrew-
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