3 ms·
Yes see this: https://arxiv.org/abs/1507.00687 https://arxiv.org/abs/1507.00687 . There is also this answer ( https://mathoverflow.net/a/421380 https://mathover
by HappyPanacea 1y ago
Yes see this: https://arxiv.org/abs/1507.00687 https://arxiv.org/abs/1507.00687 . There is also this answer ( https://mathoverflow.net/a/421380 https://mathoverflow.net/a/421380 ) in MathOverflow which state that you can't do better than cubic time if you want strong form of numerical stability.
- ttoinou 1y agoWow thank you I do not understand everything written and didn't take time to do so (and can't even find reference 23 paper on Brent stability), but I was thinking we don't necessarily need a new algorithm, we could tweak existing ones. For example swapping columns and rows before applying a product, so that accuracy is best kept (then reswapping back at the end of the operation). It seems to me the Strassen-like algorithm choose arbitrarily some elements of the matrices over others to make their temporary sum-products variables, so we could exploit this asymmetry (not sure if it's the correct word here) to choose which elements are to be chosen
- HappyPanacea 1y agoThis sounds like interesting topic to investigate which I don't remember seeing but I don't think this will gain much. I think asymmetry is completely fine in this context.