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by ericpauley
2588 days ago
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log32n/log2n is always 5, highlighting why this is such a meaningless comparison. O(logn) (basically) refers to a constant multiple clogn, by which logic O(log32n) = O(5log2n) = O(log2n) = O(logn). It's all the same. Case in point, if your algorithm takes log32n operations but each OP takes 5x longer its exactly the same as log2n. This is true for any value n, not just large values. |
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