[Part 1] PSNR vs SNR
In signal processing, the Signal-to-Noise Ratio (SNR) is defined as:
\[\mathrm{SNR} = 10 \log_{10} \left( \frac{\mathrm{Var}(\text{signal})}{\mathrm{Var}(\text{noise})} \right)\]- Unit: dB
- Measures the ratio between signal power (variance of the signal) and noise power (variance of the error).
The Peak Signal-to-Noise Ratio (PSNR) replaces signal variance with the squared peak signal value:
\[\mathrm{PSNR} = 10 \log_{10} \left( \frac{\mathrm{MAX}^2}{\mathrm{MSE}} \right)\]- Unit: dB
- \(\mathrm{MAX}\) is the maximum representable signal value.
- \(\mathrm{MSE}\) is the mean squared error between reference and reconstructed data.
For point cloud geometry:
- \(\mathrm{MAX}\) = diagonal length of the 3D bounding box of the reference point cloud.
- \(\mathrm{MSE}\) is computed using nearest-neighbor distances.
Procedure:
- For each point \(p_i\) in the reference point cloud, find its nearest neighbor \(\hat{p}_i\) in the reconstructed cloud.
- Compute:
Colored PSNR: For attribute (color) distortion:
- \(\mathrm{MAX} = 255\) (for 8-bit color channels).
- MSE is computed across RGB channels and averaged:
where:
- \((R_i, G_i, B_i)\) are original colors,
- \((\hat{R}_i, \hat{G}_i, \hat{B}_i)\) are reconstructed colors.
PSNR is then computed as:
\[\mathrm{PSNR} = 10 \log_{10} \left( \frac{255^2}{\mathrm{MSE}} \right)\][Part 2] Loss - Chamfer Distance
A common geometry loss for point clouds is the Chamfer Distance (CD).
Given two point sets:
- Reference point cloud: \(P = \{p_i\}_{i=1}^N\)
- Reconstructed point cloud: \(\hat{P} = \{\hat{p}_j\}_{j=1}^M\)
The (symmetric) point-to-point Chamfer distance is:
\[\mathcal{L}_{\text{CD}}(P, \hat{P}) = \frac{1}{N} \sum_{i=1}^{N} \min_{\hat{p} \in \hat{P}} \| p_i - \hat{p} \|^2 + \frac{1}{M} \sum_{j=1}^{M} \min_{p \in P} \| \hat{p}_j - p \|^2\]This is essentially the average squared nearest-neighbor distance in both directions.
Is Chamfer Distance Differentiable? Short answer:
Yes, almost everywhere — and it is widely used in gradient-based optimization.
- The squared Euclidean distance \(\|p - \hat{p}\|^2\) is fully differentiable.
- The only non-smooth operation is the nearest neighbor selection (the \(\min\)).
Step 1: Forward Pass Nearest-neighbor (NN) assignment
Define the (squared-distance) NN index for each reference point:
\[j^*(i) \;=\; \arg\min_{j} \|p_i - \hat p_j\|^2.\]Then the loss can be written with the discrete assignment made explicit:
\[\mathcal{L}_{P\to \hat P} = \frac{1}{N}\sum_{i=1}^{N} \|p_i - \hat p_{j^*(i)}\|^2.\]This is the key idea: during the backward pass we treat the current NN assignments as fixed (they are recomputed each forward pass, but within a given forward/backward evaluation they are constants). You can compute the Jacobian with respect to point positions:
\[\frac{\partial}{\partial \hat{p}} \| p - \hat{p} \|^2 = 2(\hat{p} - p)\]So in the graph?, the jacobian of the final cost w.r.t a contributing point is:
\[\frac{\partial \mathcal{L}_{P\to \hat P}}{\partial \hat p_k} = \frac{1}{N}\sum_{i\in S_k} 2(\hat p_k - p_i).\]Step 2: Chamfer Distance — Gradient w.r.t. Reconstructed Points
The reconstructed point cloud \(\hat{P} = \{\hat{p}_j\}_{j=1}^M\) is the output of the network, so during backpropagation we compute gradients with respect to these points.
The symmetric Chamfer Distance between the original point cloud \(P = \{p_i\}_{i=1}^N\) and the reconstructed point cloud \(\hat{P}\) is:
\[\mathcal{L}_{\text{CD}}(P,\hat P) = \frac{1}{N}\sum_{i=1}^{N} \min_{j} \|p_i - \hat{p}_j\|^2 + \frac{1}{M}\sum_{j=1}^{M} \min_{i} \|\hat{p}_j - p_i\|^2.\]To compute the gradient with respect to a reconstructed point \(\hat{p}_k\), we consider the second term (reconstructed → original). Let \(S_k\) be the set of original points for which \(\hat{p}_k\) is the nearest neighbor. Then the gradient is:
\[\frac{\partial \mathcal{L}_{\hat{P} \to P}}{\partial \hat{p}_k} = \frac{1}{M} \sum_{i \in S_k} 2(\hat{p}_k - p_i).\]This gradient pushes each reconstructed point toward the original points for which it is the closest match, ensuring the predicted cloud aligns with the ground truth during training.
Step 3: Backward Pass — Jacobian View (Aggregation)
Stack reconstructed points into a single vector:
\[\hat{\mathbf{x}} = [\hat p_1^\top,\hat p_2^\top,\dots,\hat p_M^\top]^\top \in \mathbb{R}^{3M}.\]For a single term \(\|p_i - \hat p_{j^*(i)}\|^2\), the gradient w.r.t. \(\hat{\mathbf{x}}\) is zero everywhere except in the block corresponding to \(j^*(i)\):
- For block \(k = j^*(i)\):
- For blocks \(k \neq j^*(i)\):
Thus, the Jacobian is block-sparse: each reference point contributes to exactly one reconstructed point (for this one-way term).
Step 4: Aggregating Over All Points
Now summing over all reference points in the \(P \to \hat P\) term,
\[\mathcal{L}_{P \to \hat P} = \frac{1}{N} \sum_{i=1}^{N} \|p_i - \hat p_{j^*(i)}\|^2,\]the gradient for a reconstructed point \(\hat p_k\) becomes:
\[\frac{\partial \mathcal{L}_{P \to \hat P}}{\partial \hat p_k} = \frac{1}{N} \sum_{i:\, j^*(i)=k} 2(\hat p_k - p_i).\]Define the index set
\[S_k = \{\, i \mid j^*(i) = k \,\}.\]Then we can write compactly:
\[\frac{\partial \mathcal{L}_{P \to \hat P}}{\partial \hat p_k} = \frac{2}{N} \sum_{i \in S_k} (\hat p_k - p_i).\]So aggregation simply amounts to summing all contributions from reference points that selected \(\hat p_k\) as their nearest neighbor.
Q & A: Where non-differentiability comes from (and why it’s “almost everywhere”)
The only non-smooth part is:
\[j^*(i) = \arg\min_j \|p_i-\hat p_j\|^2.\]As you move \(\hat p_j\)’s continuously, most of the time the identity of the minimizer stays the same, so \(j^*(i)\) is constant locally and the loss is a smooth quadratic in that region.
A change happens only when two candidates tie:
\[\|p_i-\hat p_a\|^2 = \|p_i-\hat p_b\|^2,\]which defines a boundary of measure zero in continuous space. That’s the “assignment switch” event.
At those boundaries:
- the gradient is not uniquely defined (there are multiple minimizers),
- but optimization still works because you almost never land exactly on the boundary, and even if you do, any small perturbation picks a side.
