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by anchovy_
1107 days ago
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>However, an O(n^3) algorithm will always loose out to an O(n^2*log(n)) algorithm, when the input gets big enough. Sure, this might be the case from a theoretical point of view (as per definition) but this completely disregards the hidden constants that come to light when actually implementing an algorithm. There's a reason why for instance a state-of-the-art matrix multiplication algorithm [0] can be completely useless in practice: The input data will never become large enough in order to amortize the introduced overhead. [0] https://arxiv.org/abs/2210.10173 |
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