Answer to Question 1-1
The activation function in a deep architecture needs to be non-linear because it introduces non-linearity into the model, allowing the network to learn and represent complex relationships between input and output data. Without non-linearity, deeper layers in the neural network would simply be able to learn linear combinations of the previous layer's activations, which would limit the network's expressive power. Non-linear activation functions enable the network to effectively learn and distinguish a wide range of features at different levels of abstraction, making deep architectures more capable of handling tasks like image recognition, natural language processing, and other complex problems.





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Answer to Question 1-2
LayerNorm, BatchNorm, and InstanceNorm are normalization techniques used in deep learning to improve the stability of neural network training.

1. **BatchNorm (Batch Normalization):**
   - It normalizes the activations across a batch of data, meaning it applies to each feature channel for all samples in a mini-batch.
   - The mean and variance are calculated over the entire batch, and then the normalization is applied to each element in the batch.
   - BatchNorm helps in reducing internal covariate shift by making the input to each layer have a similar distribution across different batches, which speeds up training.

2. **InstanceNorm (Instance Normalization):**
   - InstanceNorm normalizes each feature channel for every sample individually, not considering the entire batch.
   - It calculates mean and variance for each image or instance separately, ensuring that each input has zero mean and unit variance.
   - This technique is commonly used in image generation and style transfer tasks where preserving local contrast is important.

3. **LayerNorm (Layer Normalization):**
   - LayerNorm normalizes the activations across all samples within a mini-batch but for each individual feature, meaning it applies to each element of the feature vector.
   - It calculates mean and variance over the feature dimensions, not considering the batch dimension.
   - LayerNorm is useful in recurrent neural networks (RNNs) where hidden states need to have consistent statistics across time steps.

These normalization layers help stabilize training by:
- Reducing internal covariate shift: By normalizing inputs, the distribution of activations remains constant throughout training, making it easier for the network to learn.
- Accelerating convergence: Normalized inputs lead to faster weight updates and can result in quicker convergence during optimization.
- Improving generalization: By reducing the sensitivity to initialization and input statistics, these layers can improve a model's ability to generalize across different datasets.

In summary, BatchNorm focuses on mini-batch statistics, InstanceNorm operates per-instance, and LayerNorm is applied across all samples within a layer. Each technique has its specific use cases and benefits depending on the architecture and task at hand.





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Answer to Question 1-3
The figure "graphics/logistic_regression.jpg" seems to be a depiction of a logistic regression model rather than a recurrent network. However, given the context of the question, we can infer that it's describing an RNN (Recurrent Neural Network) with a final output unit and Sigmoid activation function.

Since no specific details about the recurrent network architecture are provided (such as whether it's an LSTM, GRU, or simple RNN), we'll assume it's a basic RNN. The question asks to determine the function computed by the output unit at the final time step.

In a standard RNN with Sigmoid activation, the output at each time step \( t \) is calculated as:

\[ h_t = \sigma(W_{ih}x_t + W_{hh}h_{t-1} + b_h) \]

where:
- \( x_t \) is the input at time step \( t \).
- \( h_{t-1} \) is the hidden state from the previous time step.
- \( W_{ih} \) and \( W_{hh} \) are weight matrices for input-to-hidden and hidden-to-hidden transitions, respectively.
- \( b_h \) is a bias (which we know is 0 in this case).
- \( \sigma \) is the Sigmoid function.

At the final time step, say \( T \), the output unit computes:

\[ h_T = \sigma(W_{ih}x_T + W_{hh}h_{T-1}) \]

The function computed by the output unit at the final time step would be this Sigmoid transformation of the combination of the last input and the previous hidden state. The exact form depends on the learned weights \( W_{ih} \) and \( W_{hh} \), which are not provided.

In summary, the network computes a sequence of Sigmoid-transformed activations based on the inputs and the evolving hidden state. At the final time step, this computation is solely represented by the output unit's activation, which is a function of the last input and the previous hidden state.





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Answer to Question 1-4
a. At the training phase, the inputs to the RNN language model using teacher forcing consist of the current word (from the ground truth sequence) along with the hidden state from the previous time step. The network is fed the entire known context from the training data, allowing it to learn dependencies between words.

b. At the test phase, when teacher forcing is not used, the inputs are different. The input at each time step is the predicted word from the previous time step, along with the hidden state from the previous time step. In this case, the model generates its own predictions and uses them as input for the next time step, instead of relying on ground truth data like during training.





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Answer to Question 1-5
For the given convolutional neural network/2D time delay neural network, let's fill in the shape and the receptive field's size of the output volume and the number of parameters at each layer.

1. Input (32x32x3):
   - Output Volume Dimension: Not applicable since it is the input layer.
   - Number of Parameters: 0
   - Size of Receptive Field: 1

2. CONV3-8:
   - Output Volume Dimension: (32, 32, 8) (height=32, width=32, channels=8)
   - Number of Parameters: (8 filters * 3 * 3 + 8 biases) = (72 + 8) = 80
   - Size of Receptive Field: 3

3. Leaky ReLU:
   - Output Volume Dimension: Same as the previous layer, (32, 32, 8)
   - Number of Parameters: 0 (since it's an activation function and has no learnable parameters)

4. POOL-2 (2x2 max-pooling):
   - Output Volume Dimension: (16, 16, 8) (height=16, width=16, channels=8)
   - Number of Parameters: 0 (since it's a pooling layer and has no learnable parameters)
   - Size of Receptive Field: Not applicable for pooling layers

5. BATCHNORM:
   - Output Volume Dimension: Same as the previous layer, (16, 16, 8)
   - Number of Parameters: 24 (since it has learnable parameters for mean and variance per channel)

6. CONV3-16:
   - Output Volume Dimension: (16, 16, 16) (height=16, width=16, channels=16)
   - Number of Parameters: (16 filters * 3 * 3 + 16 biases) = (144 + 16) = 160
   - Size of Receptive Field: 3

7. Leaky ReLU:
   - Output Volume Dimension: Same as the previous layer, (16, 16, 16)
   - Number of Parameters: 0

8. POOL-2 (2x2 max-pooling):
   - Output Volume Dimension: (8, 8, 16) (height=8, width=8, channels=16)
   - Number of Parameters: 0
   - Size of Receptive Field: Not applicable for pooling layers

9. FLATTEN:
   - Output Volume Dimension: Flattened to a single dimension with size (8 * 8 * 16) = 1024
   - Number of Parameters: 0 (since it's just reshaping the volume)

10. FC-10 (fully-connected layer):
    - Output Volume Dimension: Not applicable since it is the final layer, but we can say there are 10 neurons.
    - Number of Parameters: (1024 inputs * 10 weights + 10 biases) = (10240 + 10) = 10250

Now let's summarize:

Layer | Output Volume Dimension | Number of Parameters | Size of Receptive Field
Input | (32, 32, 3) | 0 | 1
CONV3-8 | (32, 32, 8) | 80 | 9
Leaky ReLU | (32, 32, 8) | 0 | -
POOL-2 | (16, 16, 8) | 0 | -
BATCHNORM | (16, 16, 8) | 24 | -
CONV3-16 | (16, 16, 16) | 160 | 9
Leaky ReLU | (16, 16, 16) | 0 | -
POOL-2 | (8, 8, 16) | 0 | -
FLATTEN | 1024 | 0 | -
FC-10 | 10 neurons | 10250 | -





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Answer to Question 2-1
True options: 
- [] Vanishing gradient causes deeper layers to learn more slowly than earlier layers
- [] Leaky ReLU is less likely to suffer from vanishing gradients than sigmoid
- [] Xavier initialization can help prevent the vanishing gradient problem

Explanation:
- The statement "tanh is usually preferred over sigmoid because it doesn’t suffer from vanishing gradients" is false. While tanh has a more limited output range [-1, 1] compared to sigmoid's [0, 1], both can still experience vanishing gradients.
- Vanishing gradient does indeed cause deeper layers to learn more slowly than earlier layers, as the gradients become smaller and might not effectively propagate through the network.
- Leaky ReLU is designed to address the issue of vanishing gradients by allowing a small non-zero gradient for negative inputs, unlike the standard ReLU which has a gradient of zero for negative inputs.
- Xavier initialization helps prevent the vanishing gradient problem by adjusting the scale of weight initializations, ensuring that the variance of activations remains consistent across layers, thus avoiding overly small or large gradients.

So, the true options are 2, 3, and 4.





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Answer to Question 2-2
True options: 
- The size of every convolutional kernel
- The number of channels of every convolutional kernel

Explanation:
The receptive field of a neuron in a convolutional neural network (CNN) is the region in the input space that influences its output. Increasing the size of the convolutional kernel or the number of channels can increase the size of the receptive field, as it requires more spatial information to be processed. The activation function does not directly affect the receptive field size, and neither does the size of the pooling layer; instead, the pooling layer typically reduces the spatial dimensions, effectively reducing the receptive field for the neurons in the following layers.





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Answer to Question 2-3
True





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Answer to Question 2-4
The valid activation functions to train a neural net in practice are:
- [] f(x) = 3x + 1
- [] f(x) = min(x, 0.5x) if x < 0; f(x) = min(x, 0.5x) if x >= 0

These functions meet the criteria because they are differentiable, which is necessary for backpropagation in neural networks. The first function is a linear function, and the second function is a piecewise linear function that is also differentiable at the point where the two pieces meet (x = 0). The third option is not suitable because it is not strictly increasing in the region x >= 0, which could lead to vanishing gradients. The fourth option has a discontinuous gradient at x = 0, making it unsuitable for backpropagation.





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Answer to Question 2-5
True: Data augmentation, Dropout, Batch normalization
False: Using Adam instead of SGD





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Answer to Question 2-6
During backpropagation, as the gradient flows backward through a sigmoid function, it will always decrease in magnitude while maintaining polarity. Therefore, the correct answer is:

[] Decrease in magnitude, maintain polarity





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Answer to Question 3-1
a. For the attention layer $l$ head $h$ to attend to the previous token position $n$ the most at decoding step $n+1$, the dot product between the query vector $q^l_h = W^{l, h}_Q x^l_{n+1}$ and the key vector $k^l_n = W^{l, h}_K x^l_n$ should be maximized. This is because in self-attention, attention weights are computed using dot products of query vectors with key vectors, and softmax is applied over these values to normalize them into probabilities. If this dot product is the highest among all positions, then position $n$ will receive the most attention.

b. The self-attention mechanism requires a masked attention mechanism in the Transformer architecture to be able to fulfill this condition for arbitrary sequences. This masking ensures that at each decoding step, the model cannot attend to future tokens, allowing it to focus on past tokens and the current context without seeing information from future positions.





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Answer to Question 3-2
a. In greedy decoding, the next token $t_{k+1}$ is predicted as the one with the highest probability at position $n+1$. The condition for this can be expressed as:
$$ p(t_{k+1} | t_1, \\dots, t_n) = \max_{\forall t'} p(t' | t_1, \\dots, t_n) $$

b. For the layer $l$ attention head $h$ to attend from position $n+1$ to position $k+1$ the most, we can express this as the dot product between the query vector $q^l_h = W^{l, h}_Q(x^l_{n+1})$ and the key vector $k^l_h = W^{l, h}_K(x^l_{k+1})$ being maximized:
$$ q^l_h \cdot k^l_h = \max_{\forall j} (q^l_h \cdot k^l_j) $$

c. A single attention layer cannot fulfill the condition in part (b) for arbitrary sequences and any $k < n$ where $t_k = t_n$, because it lacks the ability to capture long-range dependencies across multiple positions. The self-attention mechanism in a single layer can only compare the current position with the previous positions within that one layer, which may not be sufficient to identify repeating patterns across distant positions.

d. - For attention heads of the same layer, their communication channel is the shared input and output spaces since they operate on the same residual stream $X^l$.
- For successive layers, the communication channel is the output from the previous layer's self-attention, which becomes the input to the next layer's self-attention. The self-attention mechanism in each layer decides what information to pass forward by transforming the input through its own key, value, and query matrices, while the feed-forward network (FFN) contributes to this process by further processing the output.

e. In a two-layer Transformer model with attention layers only:
1. In the first layer $l=1$, design an attention head $h$ that computes the difference between position embeddings of positions $n+1$ and $k+1$. This can be achieved by using a linear transformation (rotation) to subtract the position embedding of $k+1$ from the position embedding of $n+1$.
2. Pass the output of layer 1 through a feed-forward network, which will help propagate this difference information.
3. In the second layer $l=2$, design an attention head $h$ that attends to positions based on this difference. The query vector can be constructed to favor attending to positions where the position embedding matches the computed difference from the previous layer.

By doing so, the second-layer attention head $h$ will attend to position $k+1$ because it is the one whose position embedding closely aligns with the transformed position information from the first layer.





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Answer to Question 4-1
The input vector for the $k$-th token in the sequence $S$, when $s_k$ is the $i$-th vocabulary word, can be represented as a one-hot encoded vector. In this representation, all elements are zero except for the $i$-th element, which is one. This indicates that the input corresponds to the $i$-th word in the vocabulary.

The layer $E$ performs a matrix-vector multiplication where the input vector (the one-hot encoded representation of $s_k$) is multiplied by the word embedding matrix $W_E$. The word embedding matrix $W_E$ has dimensions $|V| \times d$, with each column corresponding to the vectorial representation of a vocabulary word. When the input is the one-hot encoded form of the $i$-th word, the multiplication effectively selects the $i$-th column of $W_E$. The result of this multiplication is the vectorial representation (or embedding) of the $i$-th vocabulary word, which we denote as $x_k$. This embedding captures the semantic information of the word in a lower-dimensional space.





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Answer to Question 4-2
a. The function $g$ must be differentiable, and its gradient should exist for all points in the domain $\mathbb{R}^d$, so that we can compute the gradients necessary for gradient descent.

b. The gradient of the loss function with respect to the word embedding matrix $W_E$ is given by:
$$\nabla l = \frac{\partial l}{\partial W_E} = \frac{\partial g}{\partial E(w)} \cdot \frac{\partial E(w)}{\partial W_E}$$
Since $\frac{\partial E(w)}{\partial W_E}$ is simply the identity matrix with a size of $|V| \times d$, because each column of $W_E$ corresponds to one input word, we have:
$$\nabla l = \frac{\partial g}{\partial E(w)} \cdot I_{|V| \times d}$$

c. For an $i \neq k$, the gradient $\frac{\partial l}{\partial w_{ij}}$ is zero because the embedding vector for word $w$ (the $k$-th column of $W_E$) is not involved in the computation of the loss with respect to any other word's embedding. Thus, only the derivative of $g$ with respect to the correct input ($\frac{\partial g}{\partial E(w)}$) contributes to the gradient.

d. The insight from part (c) signifies that:
- For the forward pass, only the embedding vector corresponding to the current input word is used, so the computational complexity is linear in $d$, the dimension of the embeddings. Memory complexity is also linear in $d$, as we need to store one output vector per input.
- For the backward pass, when computing the gradient with respect to the embedding matrix, only the row corresponding to the current input word will have a non-zero contribution, so the computational complexity for updating each entry in the embedding vector is $\mathcal{O}(d)$. The memory complexity remains linear in $d$, as we need to store the gradients for all dimensions of the embeddings.





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