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@hdevalence.bsky.social

Reading §d.2, understand that there's eff gains from sparse computations and why, but it would be helpful to have a mapping between the math formulas used in the definitions of the autoencoders and the specific kernels described in this section ie what are the formulas and for each where does the sparsity come from

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Claude 4 Opus (T) via @hdevalence.bsky.social
Thinking Summary

The user has shared a link to a paper about sparse autoencoders and then asked about Section D.2, which discusses kernel optimizations for computational efficiency. They want to understand the mapping between the mathematical formulas in the autoencoder definitions and the specific kernels described, particularly where the sparsity comes from. Looking at the paper content, I can see: 1. The autoencoder definitions are in Section 2.2 and 2.3: - Encoder: z = TopK(W_enc(x - b_pre)) - Decoder: x̂ = W_dec z + b_pre 2. Section D.2 mentions two main kernels: - DenseSparseMatmul: multiplication between dense and sparse matrix - MatmulAtSparseIndices: multiplication of two dense matrices evaluated at sparse indices 3. Five optimizations are listed: - Decoder forward pass uses DenseSparseMatmul - Decoder gradient uses DenseSparseMatmul - Latent gradient uses MatmulAtSparseIndices - Encoder gradient uses DenseSparseMatmul - Pre-bias gradient uses a trick The user wants to understand how these connect to the math formulas and where sparsity comes from in each case.

Looking at Section D.2, I can help map the mathematical operations to the kernels. Let me break down where the sparsity comes from in each operation:

Forward Pass

Encoder: z=TopK(Wenc(x−bpre))z = \text{TopK}(W_{\text{enc}}(x - b_{\text{pre}}))

  • This produces a sparse vector zz where only kk out of nn latents are non-zero
  • The encoder computation itself is dense (full matrix multiply)

Decoder: x^=Wdecz+bpre\hat{x} = W_{\text{dec}} z + b_{\text{pre}}

  • Uses DenseSparseMatmul: WdecW_{\text{dec}} (dense, d×nd \times n) multiplied by zz (sparse, only kk non-zeros)
  • Only need to compute weighted sum of kk columns of WdecW_{\text{dec}} instead of all nn columns

Backward Pass

Let's denote δx=∂L∂x^\delta_x = \frac{\partial \mathcal{L}}{\partial \hat{x}} as the gradient w.r.t. reconstruction.

1. Decoder gradient: ∂L∂Wdec=δxzT\frac{\partial \mathcal{L}}{\partial W_{\text{dec}}} = \delta_x z^T

  • Uses DenseSparseMatmul: outer product of dense δx\delta_x with sparse zTz^T
  • Only updates kk columns of the weight gradient

2. Latent gradient: ∂L∂z=WdecTδx\frac{\partial \mathcal{L}}{\partial z} = W_{\text{dec}}^T \delta_x

  • Uses MatmulAtSparseIndices: only compute gradients for the kk active latents
  • No need to compute gradients for zero latents (which would be zeroed anyway by TopK)

3. Encoder gradient: ∂L∂Wenc=∂L∂z(x−bpre)T\frac{\partial \mathcal{L}}{\partial W_{\text{enc}}} = \frac{\partial \mathcal{L}}{\partial z} (x - b_{\text{pre}})^T

  • Uses DenseSparseMatmul: only the kk non-zero rows of ∂L∂z\frac{\partial \mathcal{L}}{\partial z} contribute
  • Results in updates to only kk rows of encoder weights

4. Pre-bias gradient: The trick mentioned involves summing gradients across batch before multiplication, reducing memory bandwidth.

The key insight is that TopK creates a sparse bottleneck where only k≪nk \ll n latents are active, allowing all downstream computations to exploit this sparsity for up to 6x speedup in the limit.

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@hdevalence.bsky.social

Ok. I was unsure about MatmulAtSparseIndices since it wasn't obvious when you would ever want to do that mathematically. But it sounds like the answer is that because of the TopK zeroing, we can know in advance which part of the matrix <> matrix computation we don't need?

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Claude 4 Opus (T) via @hdevalence.bsky.social
Thinking Summary

The user is asking about MatmulAtSparseIndices and when you'd want to mathematically compute only certain entries of a matrix multiplication. They correctly understand that it's because TopK tells us in advance which latents are active. Let me explain this clearly - the key insight is that when computing the gradient w.r.t. z, we have: - $\frac{\partial \mathcal{L}}{\partial z} = W_{\text{dec}}^T \delta_x$ - This would normally be a full matrix-vector multiply producing an n-dimensional vector - But we know that only k entries of z were non-zero (the ones selected by TopK) - The chain rule means gradients for the zero entries don't matter - they'll be zeroed anyway - So we only need to compute the k entries of the gradient vector corresponding to the k active latents This is exactly what MatmulAtSparseIndices does - it computes only specific entries of a matrix multiplication result, not the full result.

Exactly right! The key insight is that TopK gives us the sparsity pattern before we need to compute gradients.

When computing ∂L∂z=WdecTδx\frac{\partial \mathcal{L}}{\partial z} = W_{\text{dec}}^T \delta_x:

  • Normally this would produce an nn-dimensional gradient vector (one entry per latent)
  • But we know from the forward pass that only kk specific latents were non-zero
  • By the chain rule, gradients for the zero latents will be zeroed out anyway when backpropagating through TopK
  • So why compute them at all?

MatmulAtSparseIndices exploits this by only computing the kk entries we actually need:

  • Instead of computing all nn dot products in WdecTδxW_{\text{dec}}^T \delta_x
  • We only compute the kk dot products corresponding to the indices where zz was non-zero
  • This turns an O(n⋅d)O(n \cdot d) operation into O(k⋅d)O(k \cdot d)

It's like having a recipe that says "multiply these two matrices" but adding "...oh, but I only need elements [17, 42, 156, ...] of the result" - so you just compute those specific dot products rather than the full matrix multiply.

This pattern appears specifically because TopK creates a deterministic sparsity pattern during the forward pass that we can exploit during backpropagation. Without that known sparsity structure, you'd have to compute the full matrix product first.

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