Nesterov momentum looks ahead and often converges faster than classical
momentum.
6406537043612.
large fixed learning rate always speeds convergence.
6406537043613.
Adam uses only a first-moment estimate of gradients.
6406537043614.
None of these.
Published solution
**1. Evaluate each optimization statement:**
- **Option A:** Nesterov Accelerated Gradient (NAG) evaluates gradients at the look-ahead point \( \theta_t - \gamma v_{t-1} \), providing an anticipatory correction that typically converges faster than classical momentum. True.
- **Option B:** An excessively large fixed learning rate causes divergence rather than faster convergence. False.
- **Option C:** Adam uses both first-moment (mean) and second-moment (uncentered variance) gradient estimates. False.
- **Option D:** False since Option A is true.
**Answer: A — Nesterov momentum looks ahead and often converges faster than classical momentum.**
A: **Correct:** Nesterov momentum computes the gradient at the look-ahead position, reducing overshoot and accelerating convergence.
B: **Incorrect:** A large fixed learning rate can cause oscillation and divergence.
C: **Incorrect:** Adam tracks both first- and second-moment estimates of the gradients.
D: **Incorrect:** Option A is true.
**1. Recall the four stages of the SIFT pipeline:**
1. **Scale-space Extrema Detection** (using Difference-of-Gaussians)
2. **Keypoint Localization** (sub-pixel refinement and edge/low-contrast filtering)
3. **Orientation Assignment** (dominant local gradient orientation)
4. **Keypoint Descriptor** (128-D local gradient histogram)
**Answer: B — Extrema Detection → Localization → Orientation → Descriptor.**
A: **Incorrect:** Lists the stages in reverse order.
B: **Correct:** Matches Lowe's four-stage SIFT pipeline.
C: **Incorrect:** The descriptor is computed last, after localization and orientation assignment.
D: **Incorrect:** Option B is the exact sequence.
**1. Test each \( 3 \times 3 \) kernel for separability (rank 1):**
A 2D kernel \( K \) is separable if \( K = u v^T \) for 1D vectors \( u, v \) (rank 1).
- **Option C** is the \( 3 \times 3 \) Gaussian smoothing filter:
\[
\frac{1}{16} \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} \frac{1}{4}\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}
\]
which has rank 1 and is separable. Options A, B, and D have rank \( \ge 2 \) and are not separable.
**Answer: C.**
A: **Incorrect:** Row 3 (`[1, -50, -2]`) is not a scalar multiple of Row 1 (`[1, 0, -1]`), so the matrix is not rank 1.
B: **Incorrect:** The 4-neighbor Laplacian kernel has rank 2 and is not separable.
C: **Correct:** Factors as the outer product \( \frac{1}{4}[1, 2, 1]^T \times \frac{1}{4}[1, 2, 1] \).
D: **Incorrect:** The negative Laplacian kernel has rank 2 and is not separable.
The degrees of freedom for affine transformations in 3D is __
Published solution
**1. Count the parameters of a 3D affine transformation:**
A 3D affine transformation \( \mathbf{x}' = A\mathbf{x} + \mathbf{t} \) has a \( 3 \times 3 \) linear matrix \( A \) (9 entries) and a \( 3 \times 1 \) translation vector \( \mathbf{t} \) (3 entries), giving:
\[
9 + 3 = 12 \text{ degrees of freedom}
\]
**Answer: 12.**
**1. Recall the 8-neighbor discrete Laplacian mask:**
With \( -1 \) weights on all 8 surrounding neighbors, the center weight must equal **8** so that the sum of all mask coefficients is zero:
\[
\begin{bmatrix} -1 & -1 & -1 \\ -1 & 8 & -1 \\ -1 & -1 & -1 \end{bmatrix}
\]
**Answer: 8.**
