Answer to Question 1-1
The activation function in a deep architecture needs to be non-linear for several reasons:

Non-linear activation functions allow the network to compute non-trivial problems using a less complex model. Linear activation functions, on the other hand, limit the network's ability to represent complex mappings from inputs to outputs, regardless of the number of layers or the number of neurons in each layer. 

If we use linear activation functions, the entire network can be reduced to a single-layer network because the composition of linear functions is also a linear function. This defeats the purpose of deep architectures, which are designed to exploit the hierarchical structure of data and to model complex relationships.

Non-linear functions help in introducing complexity by allowing the activation thresholds to vary nonlinearly. This gives deep learning models the flexibility to learn from highly intricate data by creating complex mappings from the input data to the output predictions.

The use of non-linear activation functions introduces the concept of decision boundaries in classification problems, allowing the model to create non-linearly separable boundaries, which is necessary for categorizing data points that are not linearly separable.

Therefore, using non-linear activation functions in deep architectures is crucial for enabling the model to solve complex problems that require non-linearity for better representation and prediction.





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Answer to Question 1-2
1. LayerNorm (Layer Normalization):
Layer normalization works by computing the normalization statistics (mean and variance) across the features for a single example in a batch. This normalization is applied independently for each individual example, and the same operation is performed at training and test time. LayerNorm is particularly useful in recurrent neural networks (RNNs) where batch normalization can be difficult to apply due to variable input sequence lengths.

2. BatchNorm (Batch Normalization):
Batch normalization, on the other hand, computes the normalization statistics (mean and variance) over the entire mini-batch of data. The statistics are calculated separately for each feature. It helps to stabilize the learning process by normalizing the inputs to layers within the network, reducing internal covariate shift (which loosely describes the change in the distribution of network activations due to the update of weights).

3. InstanceNorm (Instance Normalization):
Instance normalization is similar to layer normalization, but it performs the normalization for each individual channel (in the field of computer vision, this would correspond to each color channel of an image) in each data sample independently. It's used primarily in style transfer applications where the contrast of the image is normalized for each channel, which guarantees stylization independence from the contrast of the content image.

Why these kinds of normalization layers help neural network trained more stable:
Normalization layers stabilize the training of neural networks by fixing the distribution of the layer inputs. This reduces the amount of shift in the distribution of activations within the network, which is known as internal covariate shift. By doing so, the learning process becomes faster and more stable. It enables the use of higher learning rates, accelerates the convergence of the training process, and also helps to reduce the sensitivity to the initial starting weights. Furthermore, it delivers some form of regularization effect, which can reduce the problem of overfitting to some extent.





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Answer to Question 1-3
Based on the provided recurrent network diagram and the given information, we can identify the function computed by this network at the final time step as follows:

1. The input at each time step is multiplied by 1 before being sent to the linear hidden unit.
2. The hidden unit also receives input from its previous output multiplied by -1 (due to the recurrent connection).
3. The output of the hidden unit is then multiplied by 10 before being passed to the logistic output unit.
4. The activation function of the output layer is the Sigmoid function, which has the general form: \(S(x) = \frac{1}{1 + e^{-x}}\).
5. All biases are 0, so they do not influence the activations.

Assuming the initial hidden state (h0) is 0 (since it is not specified), let's compute the hidden state after processing an input sequence of even length:

At the first time step (t1), the input unit receives some integer \(x_1\). The hidden unit would then compute \(h_1 = (-1) * h_0 + 1 * x_1 = x_1\), since \(h_0 = 0\).

At the second time step (t2), the input unit receives another integer \(x_2\). The hidden unit would then compute \(h_2 = (-1) * h_1 + 1 * x_2 = -x_1 + x_2\).

This pattern continues with the hidden unit computing \(h_t = -h_{t-1} + x_t\) at each subsequent time step.

Because the length of the input sequence is even, after processing the entire input sequence, the hidden unit ultimately computes the alternating sum of the inputs. If the sequence length is 2n, the final hidden state before being passed to the output unit would be:

\[ h_{2n} = x_1 - x_2 + x_3 - x_4 + ... + x_{2n-1} - x_{2n} \]

This sum is then multiplied by 10 and passed through the Sigmoid function to compute the output at the final time step:

\[ output_{final} = S(10 * h_{2n}) = S(10 * (x_1 - x_2 + x_3 - x_4 + ... + x_{2n-1} - x_{2n})) \]

Hence, the function computed by the output unit at the final time step is the Sigmoid of ten times the alternating sum of the input sequence.





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Answer to Question 1-4
a: The inputs to the network at the training phase of an RNN language model using teacher forcing are the ground truth previous token (word or character based on the granularity of the model) and the current hidden state. Essentially, at each time step of training, the model receives the actual previous token from the training data as input along with the computed hidden state from the previous time step.

b: The inputs to the network at the test phase are the predicted token from the previous time step and the current hidden state. Here, instead of the ground truth, the model uses its own predictions from the preceding time step as the input for the subsequent step. The hidden state from the previous time step is also input into the model, as in the training phase.





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Answer to Question 1-5
CONV3-8:
- Output volume dimension: (32, 32, 8) as given
- Number of Parameters: For each filter, there are 3(height) * 3(width) * 3(depth) weights + 1 bias, and there are 8 filters. So, the total number of parameters is (3*3*3 + 1)*8 = 224.
- Size of receptive field: 3 x 3 = 9 as given

Leaky Relu:
- Output volume dimension: (32, 32, 8) (same as CONV3-8, activation layers don't change the dimensions)
- Number of Parameters: 0 (activation functions do not have parameters)
- Size of receptive field: 3 x 3

POOL-2:
- Output volume dimension: (16, 16, 8) (A 2x2 max pooling with stride 2 will reduce the size by a factor of 2)
- Number of Parameters: 0 (pooling layers don't have parameters)
- Size of receptive field: Since the pooling is 2x2, the receptive field size would be 4 as it represents the area from which the max value will be taken.

BATCHNORM:
- Output volume dimension: (16, 16, 8) (Batch normalization does not change the dimensions)
- Number of Parameters: For batch normalization, the number of parameters is twice the depth of the layer - one for scaling (gamma) and one for shifting (beta), so it's 8*2 = 16.
- Size of receptive field: 4 (remains the same as the last layer because batch normalization does not affect the receptive field)

CONV3-16:
- Output volume dimension: (16, 16, 16) (since CONV3 has padding 1 and stride 1, the output volume remains the same height and width as the input)
- Number of Parameters: For each filter, there are 3*3*8 weights + 1 bias, and there are 16 filters. So, the total number of parameters is (3*3*8 + 1)*16 = 1168.
- Size of receptive field: 7 x 7 (since it takes into account the pooling and previous convolutional layers)

Leaky ReLU:
- Output volume dimension: (16, 16, 16) (same as CONV3-16)
- Number of Parameters: 0 (no parameters for activation functions)
- Size of receptive field: 7 x 7

POOL-2:
- Output volume dimension: (8, 8, 16) (A 2x2 max pooling with stride 2 will reduce each of the volume's height and width by a factor of 2)
- Number of Parameters: 0 (no parameters for pooling layers)
- Size of receptive field: Computed based on previous layers and the pooling operation

FLATTEN:
- Output volume dimension: (1024) (Since 8*8*16 = 1024, flattening the output of the previous layer results in a volume of 1024 elements)
- Number of Parameters: 0 (Flatten layers don't have parameters)
- Size of receptive field: Same as the last layer

FC-10:
- Output volume dimension: (10) (since this layer is fully connected with 10 neurons)
- Number of Parameters: (Input features * Neurons) + Biases. The input features are the flattened output from the previous layer (1024) and the biases are equal to the number of neurons, so (1024*10 + 10) = 10250.
- Size of receptive field: Not applicable for fully connected layers

Indexing the answers:

1. CONV3-8: Output (32, 32, 8), Parameters 224, Receptive Field 9
2. Leaky Relu: Output (32, 32, 8), Parameters 0, Receptive Field 9
3. POOL-2: Output (16, 16, 8), Parameters 0, Receptive Field 4
4. BATCHNORM: Output (16, 16, 8), Parameters 16, Receptive Field 4
5. CONV3-16: Output (16, 16, 16), Parameters 1168, Receptive Field 7x7
6. Leaky ReLU: Output (16, 16, 16), Parameters 0, Receptive Field 7x7
7. POOL-2: Output (8, 8, 16), Parameters 0, Receptive Field (depends on earlier layers)
8. FLATTEN: Output (1024), Parameters 0, Receptive Field (same as the last layer)
9. FC-10: Output (10), Parameters 10250, Receptive Field N/A





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Answer to Question 2-1
1. false - While tanh is usually preferred over sigmoid because it is zero-centered, it can still suffer from vanishing gradients.
2. true - Vanishing gradient does cause deeper layers to learn more slowly than earlier layers because the gradients become increasingly small as they propagate back through the layers, leading to negligible updates to the weights.
3. true - Leaky ReLU is less likely to suffer from vanishing gradients than sigmoid because it allows for a small, non-zero gradient when the unit is not active.
4. true - Xavier initialization, also known as Glorot initialization, can help prevent the vanishing gradient problem by maintaining the variance of the gradients across layers.





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Answer to Question 2-2
1. The size of every convolutional kernel - True. Increasing the size of the convolutional kernel increases the area of the input image that is considered at once, hence increasing the receptive field.

2. The number channel of every convolutional kernel - False. The number of channels of a convolutional kernel does not affect the size of the receptive field; it relates to the depth of the kernel, which corresponds to the number of feature maps created, not the area of the input it encompasses.

3. The activation function of each layer - False. The choice of activation function does not influence the receptive field size. Activation functions are applied after convolution operations and merely introduce non-linearities into the model.

4. The size of pooling layer - True. Pooling layers reduce the spatial size of the representation, which indirectly increases the receptive field. A larger pooling layer will cover a greater area of the input, thus enlarging the receptive field size collectively with subsequent layers.





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Answer to Question 2-3
False

Dividing the weight vector W by 2 effectively scales the outputs of the logistic function. However, since the logistic function is a non-linear function, this change can potentially alter the decision boundary. If the decision boundary changes, the classification of some data points could change, which in turn could lead to a change in the test accuracy (Acc) on a given dataset.





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Answer to Question 2-4
- f(x) = min(2,x) - False. This function outputs 2 for all x greater than 2, which means it has a constant gradient in this range, leading to issues with gradient-based optimization methods.

- f(x) = 3x + 1 - False. This function is linear and therefore does not introduce non-linearity into the neural network, which is required to solve non-linear problems.

- f(x) = min(x, 0.5x) if x < 0; f(x) = min(x, 0.5x) if x >= 0 - False. This function is essentially the same as f(x) = 0.5x for all x. It is a linear function just with a different slope and hence does not introduce non-linearity.

- f(x) = min(x, 0.1x) if x < 0; f(x) = max(x, 0.1x) if x >= 0 - True. This function is non-linear and resembles the behavior of leaky ReLU, which is a known activation function. It provides a small gradient for negative inputs and maintains the gradient for positive inputs, which is beneficial during backpropagation as it does not saturate or kill gradients.





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Answer to Question 2-5
1. Data augmentation: true
2. Dropout: true
3. Batch normalization: false, this technique helps with model's training convergence and stability, but on its own it doesn't reduce overfitting.
4. Using Adam instead of SGD: false, replacing SGD with Adam optimizer does not necessarily reduce overfitting; it's more about improving the training process.





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Answer to Question 2-6
1. Increase in magnitude, maintain polarity: False
2. Increase in magnitude, reverse polarity: False
3. Decrease in magnitude, maintain polarity: True
4. Decrease in magnitude, reverse polarity: False

The gradient often decreases in magnitude due to the nature of the sigmoid function's derivative, which scales the gradient by a factor of y * (1-y). This factor is always between 0 and 0.25 for any input (since 0 < y < 1 for sigmoid), thus it cannot increase the magnitude of the gradient, nor can it change the sign of the gradient (reverse its polarity).





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Answer to Question 3-1
a. The condition that must be met for the attention layer \(l\) head \(h\) to attend to the previous token position \(n\) the most at decoding step \(n+1\) is that the attention weight \(a^{l,h}_{(n+1,n)}\) for the previous position \(n\) is maximized relative to the other weights at that attention head. This can be formulated as:

\[ a^{l,h}_{(n+1,n)} > a^{l,h}_{(n+1,m)} \quad \forall m \in \{1, \dots, n-1, n+1\} \]

where \(a^{l,h}_{(n+1,n)}\) is the attention weight calculated using the dot product of the query vector and the key vector, followed by a softmax function:

\[ a^{l,h}_{(n+1,n)} = \text{softmax}\left( (x^{l}_{n+1} W^{l,h}_Q) (x^l_n W^{l,h}_K)^T \right) \]

The softmax function is applied across all positions up to the current one, ensuring that the sum of the attention weights is 1.

b. The self-attention mechanism requires the query vector \(x^{l}_{n+1} W^{l,h}_Q\) and the key vectors \(x^l_n W^{l,h}_K\) for all \(n\) to be able to fulfill this condition for arbitrary sequences. To specifically ensure that the attention head attends to the previous token, the self-attention mechanism must create a scenario where the query vector has the highest degree of similarity with the key vector corresponding to the previous token's position. This might require the model to learn during training to generate query and key vectors in such a way that this similarity (prior to softmax) is maximized for the previous token. The ability to do this for arbitrary sequences depends on the capability of the Transformer to learn complex patterns in data during training and then apply those during inference.





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Answer to Question 3-2
a. For the induction heads to predict \( t_{k+1} \) as the next token at decoding step \( n+1 \) using greedy decoding, the condition that must be satisfied is that the probability of \( t_{k+1} \) being the next token, given the currently decoded token sequence \( T \), must be the highest among all possible next tokens. Mathematically, the condition can be expressed as:

\[ P(t_{k+1} | T) > P(t_i | T) \quad \forall i \neq k+1 \]

where \( P(t_i | T) \) represents the probability of the token \( t_i \) being the next token, given the current token sequence \( T \).

b. For the layer \( l \) attention head \( h \) to attend from position \( n+1 \) to position \( k+1 \) the most, the condition can be formalized in terms of the attention score. Specifically, the dot-product attention score between the query vector at position \( n+1 \) and the key vector at position \( k+1 \) has to be maximal in comparison with all other key vectors at positions \( j \neq k+1 \). The attention score is computed using the query matrix \( W_Q \), the key matrix \( W_K \), and their respective input vectors \( x \) as follows:

\[ \text{Score}(x^l_{n+1}, x^l_{k+1}) = \left(W^{l, h}_Q \cdot x^l_{n+1}\right) \cdot \left(W^{l, h}_K \cdot x^l_{k+1}\right)^\top \]

The condition for maximal attention from position \( n+1 \) to position \( k+1 \) is:

\[ \text{Score}(x^l_{n+1}, x^l_{k+1}) > \text{Score}(x^l_{n+1}, x^l_j) \quad \forall j \neq k+1 \]

c. A Transformer model with only a single attention layer can potentially fulfill the condition for attending from position \( n+1 \) to position \( k+1 \) for arbitrary sequences and any \( k < n \) where \( t_k = t_n \), depending on its learned parameters. If the attention weights are learned in such a way that they maximize the attention score for similar tokens, then the attention mechanism could "focus" on the matching token \( t_k \) when predicting the next token given \( t_n \). However, the model could struggle if it also needs to deal with a large number of competing tokens that could distract from the desired behavior, as the single layer has to encode all necessary relationships between tokens, making it less flexible compared to models with multiple layers.

d. In a task where multiple attention heads cooperate:
- For attention heads within the same layer (the trick question part), they do not have a direct means of communication with each other as each head operates independently. Their outputs are typically concatenated or combined in some way after all heads have processed their input, and only then is the information effectively "shared" as it passes on to subsequent layers or output layers.
- For attention heads of successive layers, the communication channel consists of the output of one layer serving as the input to the next layer. The part of the self-attention that determines what to write to this communication channel is the output transformation matrix \( W^{l, h}_O \), while the query matrix \( W^{l+1, h}_Q \) for the attention heads in the subsequent layer decides what to read from it by influencing the dot product with the keys of that layer.

e. For the two-layer Transformer model with only attention layers, we can design a sequence of operations such that the layer \( l > 1 \) attention head \( h \) attends to position \( k+1 \):
1. In the first attention layer, learn parameters that allow the model to identify tokens that have previously occurred (like an induction head). The attention mechanism can learn to assign high attention scores to token positions that match the current input token.
2. The output of the first layer (after attention scores have been applied) acts as input to the second layer, carrying forward the information about the previously occurring token's position.
3. The second layer's attention heads can then learn to use the information about the position of the previous occurrence to predict the next token based on the context established by the first layer. This could be done, for example, by learning to align the query vector for the current position (\( n+1 \)) with the key vector that corresponds to the position immediately following the previously attended token (\( k+1 \)) from the first layer.





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Answer to Question 4-1
**The layer $E$ as matrix-vector multiplication:**

**Input vector for $s_k$ being the $i$-th vocabulary word:**
- When $s_k$ is the $i$-th vocabulary word, the input vector is a one-hot encoded vector $v \in \mathbb{R}^{|V|}$. This vector $v$ has a 1 at the $i$-th position and 0s everywhere else.

**Multiplication:**
- This input vector $v$ is then multiplied with the word embedding matrix $W_E$. The multiplication can be described as: $x_k = W_E \cdot v$.
- Since $v$ is one-hot encoded, the multiplication effectively picks the $i$-th column from $W_E$, which represents the vectorial representation of the $i$-th word in the vocabulary.





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Answer to Question 4-2
a. The function \( g \) must be differentiable with respect to its input if we want to perform training via gradient descent. This is because gradient descent relies on the ability to compute the gradient of the loss function with respect to the parameters, which in turn requires the functions involved to be differentiable.

b. The gradient of the loss \( l = \mathcal{L}(f(w), t) \) with respect to the word embedding matrix \( W_E \) can be given by the chain rule as:

\[
\nabla l = \frac{\partial \mathcal{L}}{\partial f(w)} \frac{\partial f(w)}{\partial W_E} = \frac{\partial \mathcal{L}}{\partial f(w)} \frac{\partial g}{\partial E(w)} \frac{\partial E(w)}{\partial W_E}
\]

where \( \frac{\partial E(w)}{\partial W_E} \) is the indicator matrix with a 1 at the position corresponding to the input word \( w \) and zeros elsewhere. The only unresolved gradient term in the full form is \( \frac{\partial g}{\partial E(w)} \) as requested.

c. The gradient \( \frac{\partial l}{\partial w_{ij}} \) for an \( i \neq k \) is 0. This is because the loss \( l \) for a particular input word is only affected by the corresponding row in the word embedding matrix \( W_E \), not by other rows corresponding to other vocabulary words.

d. The insight from part (c) signifies the following for the computational and memory complexity of the embedding layer:

- Forward pass: The embedding layer only needs to look up the vector corresponding to the input word \( w \). This operation is very efficient because it requires no computation over the entire vocabulary and is therefore of very low computational complexity. The memory complexity is also low because only the current word's vector is used from \( W_E \) without the need to load the entire matrix. 

- Backward pass: During the gradient update, only the row of the embedding matrix corresponding to the current input word \( w \) needs to be updated. This means that for a single input word, the changes are sparse, and we do not need to compute or store gradients for the entire matrix \( W_E \), leading to low computational and memory complexity. Only the specific row for the input word \( w \) in \( W_E \) is updated based on the gradient. 

In both the forward pass and the backward pass, the computation and memory required are proportional to the dimensionality of the word vectors \( d \) and not to the size of the vocabulary \( |V| \), which significantly reduces complexity.





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