Equation 8 · What Interpretability Actually Costs to Do at Scale
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Work out what that architecture actually spends compute on per token, because the two halves of it behave differently. The encoding step needs a score for every one of the n candidate latents before it can select the top k , so it is an unavoidably dense matrix multiply: roughly 2dn floating-point operations. The decoding step only touches the k latents that survived, so it is sparse: roughly 2dk operations. Summed and multiplied across T training tokens, a first-order compute model for training the dictionary is
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