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Modern embeddings lie on a hypersphere surface, making euclidean equal to cosine. And if they don't, I probably wouldn't want to use them.
by OutOfHere 2y ago
Modern embeddings lie on a hypersphere surface, making euclidean equal to cosine. And if they don't, I probably wouldn't want to use them.
- yobbo 2y agoTrue, on a hypersphere cosine and euclidean are equivalent. But if random embeddings are gaussian, they are distributed on a "cloud" around the hypersphere, so they are not equal.