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[Deep RNN] by Hand ✍️ A Deep Recurrent Neural Network (RNN) extends a basic single-layer RNN into multiple layers of hidden states, effectively incorporating deep learning into the RNN architecture. How does a Deep RNN work? [1] Given ↳ A sequence of four inputs X1, X2, X3, X4 ⬛️...

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[LSTM] by Hand ✍️ LSTMs have been the most effective architecture to process long sequences of data, until our world was taken over by the Transformers. LSTMs belong to the broader family of recurrent neural network (RNNs) that process data sequentially in a recurrent manner. Transformers, on the other hand, abandon recurrence and use self-attention instead to process data concurrently in parallel. Recently, there is renewed interest in recurrence as people realized self-attention doesn’t scale to extremely long sequences, like hundreds of thousands of tokens. Mamba is a good example to bring back recurrence. All of a sudden, it is cool to study LSTMs. How do LSTMs work? [1] Given ↳ 🟨 Input sequence X1, X2, X3 (d = 3) ↳ 🟩 Hidden state h (d = 2) ↳ 🟦 Memory C (d = 2) ↳ Weight matrices Wf, Wc, Wi, Wo Process t = 1 [2] Initialize ↳ Randomly set the previous hidden state h0 to [1, 1] and memory cells C0 to [0.3, -0.5] [3] Linear Transform ↳ Multiply the four weight matrices with the concatenation of current input (X1) and the previous hidden state (h0). ↳ The results are feature values, each is a linear combination of the current input and hidden state. [4] Non-linear Transform ↳ Apply sigmoid σ to obtain gate values (between 0 and 1). • Forget gate (f1): [-4, -6] → [0, 0] • Input gate (i1): [6, 4] → [1, 1] • Output gate (o1): [4, -5] → [1, 0] ↳ Apply tanh to obtain candidate memory values (between -1 and 1) • Candidate memory (C’1): [1, -6] → [0.8, -1] [5] Update Memory ↳ Forget (C0 .* f1): Element-wise multiply the current memory with forget gate values. ↳ Input (C’1 .* o1): Element-wise multiply the “candidate” memory with input gate values. ↳ Update the memory to C1 by adding the two terms above: C0 .* f1 + C’1 .* o1 = C1 [6] Candiate Output ↳ Apply tanh to the new memory C1 to obtain candidate output o’1. [0.8, -1] → [0.7, -0.8] [7] Update Hidden State ↳ Output (o’1 .* o1 → h1): Element-wise multiply the candidate output with the output gate. ↳ The result is updated hidden state h1 ↳ Also, it is the first output. Process t = 2 [8] Initialize ↳ Copy previous hidden state h1 and memory C1 [9] Linear Transform ↳ Repeat [3] [10] Update Memory (C2) ↳ Repeat [4] and [5] [11] Update Hidden State (h2) ↳ Repeat [6] and [7] Process t = 3 [12] Initialize ↳ Copy previous hidden state h2 and memory C2 [13] Linear Transform ↳ Repeat [3] [14] Update Memory (C3) ↳ Repeat [4] and [5] [15] Update Hidden State (h3) ↳ Repeat [6] and [7]

Tom Yeh

73,077 views • 2 years ago

[Graph Convolutional Network] by hand ✍️ Graph Convolutional Networks (GCNs), introduced by Thomas Kipf and Max Welling in 2017, have emerged as a powerful tool in the analysis and interpretation of data structured as graphs. This exercise demonstrates how GCN works in a simple application: binary classification. -- Goal -- Predict if a node in a graph is X. -- Architecture -- 🟪 Graph Convolutional Network (GCN) 1. GCN1(4,3) 2. GCN2(3,3) 🟦 Fully Connected Network (FCN) 1. Linear1(3,5) 2. ReLU 3. Linear2(5,1) 4. Sigmoid Simplications: • Adjacent matrices are not normalized. • ReLU is applied to messages directly. -- Walkthrough -- [1] Given ↳ A graph with five nodes A, B, C, D, E [2] 🟩 Adjacency Matrix: Neighbors ↳ Add 1 for each edge to neighbors ↳ Repeat in both directions (e.g., A->C, C->A) ↳ Repeat for both GCN layers [3] 🟩 Adjacency Matrix: Self ↳ Add 1's for each self loop ↳ Equivalent to adding the identity matrix ↳ Repeat for both GCN layers [4] 🟪 GCN1: Messages ↳ Multiply the node embeddings 🟨 with weights and biases ↳ Apply ReLU (negatives → 0) ↳ The result is one message per node [5] 🟪 GCN1: Pooling ↳ Multiply the messages with the adjacent matrix ↳ The purpose is the pool messages from each node's neighbors as well as from the node itself. ↳ The result is a new feature per node [6] 🟪 GCN1: Visualize ↳ For node 1, visualize how messages are pooled to obtain a new feature for better understanding ↳ [3,0,1] + [1,0,0] = [4,0,1] [7] 🟪 GCN2: Messages ↳ Multiply the node features with weights and biases ↳ Apply ReLU (negatives → 0) ↳ The result is one message per node [8] 🟪 GCN2: Pooling ↳ Multiply the messages with the adjacent matrix ↳ The result is a new feature per node [9] 🟪 GCN2: Visualize ↳ For node 3, visualize how messages are pooled to obtain a new feature for better understanding ↳ [1,2,4] + [1,3,5] + [0,0,1] = [2,5,10] [10] 🟦 FCN: Linear 1 + ReLU ↳ Multiply node features with weights and biases ↳ Apply ReLU (negatives → 0) ↳ The result is a new feature per node ↳ Unlike in GCN layers, no messages from other nodes are included. [11] 🟦 FCN: Linear 2 ↳ Multiply node features with weights and biases [12] 🟦 FCN: Sigmoid ↳ Apply the Sigmoid activation function ↳ The purpose is to obtain a probability value for each node ↳ One way to calculate Sigmoid by hand ✍️ is to use the approximation below: • >= 3 → 1 • 0 → 0.5 • <= -3 → 0 -- Outputs -- A: 0 (Very unlikely) B: 1 (Very likely) C: 1 (Very likely) D: 1 (Very likely) E: 0.5 (Neutral)

Tom Yeh

46,779 views • 2 years ago

[VAE] by Hand ✍️ A Variational Auto Encoder (VAE) learns the structure (mean and variance) of hidden features and generates new data from the learned structure. In contrast, GANs only learn to generate new data to fool a discriminator; they may not necessarily know the underlying structure of the data. The International Conference on Learning Representations (ICLR) this year announced its first ever "Test of Time Award" to recognizes the VAE paper, published 10 years ago. This exercise demonstrates how to calculate a VAE by hand. [1] Given: ↳ Three training examples X1, X2, X3 ↳ Copy training examples to the bottom ↳ The purpose is to train the network to reconstruct the training examples. ↳ Since each target is a training example itself, we use the Greek word "auto" which means "self." This crucial step is what makes an autoencoder "auto." [2] Encoder: Layer 1 + ReLU ↳ Multiply inputs with weights and biases ↳ Apply ReLU, crossing out negative values (-1 -> 0) [3] Encoder: Mean and Variance ↳ Multiply features with two sets of weights and biases ↳ 🟩 The first set predicts the means (𝜇) of latent distributions ↳ 🟪 The second set predicts the standard deviation (𝜎) of latent distributions [4] Reparameterization Trick: Random Offset ↳ Sample epsilon ε from the normal distribution with mean = 0 and variance = 1. ↳ The purpose is to randomly pick a offset away from the mean. ↳ Multiply the standard deviation values with epsilon values. ↳ The purpose is to scale the offset by the standard deviation. [5] Reparameterization Trick: Mean + Offset ↳ Add the sampled offset to predicted mean ↳ The result are new parameters or features 🟨 as inputs to the Decoder. [6] Decoder: Layer 1 + ReLU ↳ Multiply input features with weights and biases ↳ Apply ReLU, crossing out negative values. Here, -4 is crossed out. [7] Decoder: Layer 2 ↳ Multiply features with weights and biases ↳ The output is Decoder's attempt to reconstruct the input data X from reparameterized distributions described by 𝜇 and 𝜎. [8]-[10] KL Divergence Loss [8] Loss Gradient: Mean 𝜇 ↳ We want 𝜇 to approach 0. ↳ A lot of math called SGVB simplifies the calculation of loss gradients to simply 𝜇 [9,10] Loss Gradient: Stdev 𝜎 ↳ We want 𝜎 to approach 1. ↳ A lot of math simplifies the calculation to 𝜎 - (1/ 𝜎) [11] Reconstruction Loss ↳ We want the reconstructed data Y (dark 🟧) to be the same as the input data X. ↳ Some math involving Mean Square Error simplifies the calculation to Y - X.

Tom Yeh

48,529 views • 2 years ago

[SVM] by Hand ✍️ Support Vector Machines (SVMs) reigned supreme in machine learning before the ascendancy of the deep learning revolution. This exercise compares Linear vs RBF SVMs---how they classify two test vectors, after SVMs are trained from six training vectors. [1] Given xi: Six training vectors (blue rows 🟦) yi: Labels Using xi and yi, we learned ai and b (red borders): ↳ai: coefficient for each training vector i. • Non-zero: A Support Vector that defines the decision boundary • Zero: Too far from the decision boundary, ignored ↳b: bias (how much the decision boundary should be shifted) x’j: Two test vectors (yellow columns 🟨) (To simplify hand calculation, training and test vectors are not normalized.) ~ 𝗟𝗶𝗻𝗲𝗮𝗿 𝗦𝗩𝗠 ~ 🟩 Kernel Matrix (K) [2]-[3] [2] Test Vector 1 ↳ Take dot product between the test vector 🟨 and every training vector 🟦 ↳ The dot product approximates the “cosine similarity” between two vectors ↳ Output: 1st column of K [3] Test Vector 2 ↳ Similar to [2] ↳ Output: 2nd column of K 🟧 Decision Boundary [4]-[6] [4] Unsigned Coefficients → Signed Weights ↳ Multiply each coefficient with the corresponding label ↳ The 2nd training vector is NOT a support vector because its coefficient is 0. [5] Weighted Combination ↳ Multiply weights and bias with K ↳ Output: “signed” distance to the decision boundary ... X'1: (2)*13+0+(1)*12+(-1)*15+(1)*19+(-1)*27+(2) = 17 X'2: (2)*2+0+(1)*3+(-1)*4+(1)*8+(-1)*8+(2) = 5 ... [6] Classify ↳ Take the sign ... X'1: 1 > 0 → Positive + X'2: 5 > 0 → Positive + ... ~ 𝗥𝗕𝗙 𝗦𝗩𝗠 ~ Given ai: Learned coefficients b: Learned bias 🟩 Kernel Matrix (K) [7]-[15] 𝘛𝘦𝘴𝘵 𝘝𝘦𝘤𝘵𝘰𝘳 (𝘟’1) 🟨 L2 Distance 🟪 [7]-[9] [7] Squared Difference ... i=1: (1-2)^2=1, (2-4)^2=4, (1-3)^2=4 ... [8] Sum [9] Square Root [10] Negative Scaling ↳ Multiply by -1: Note that L2 is a distance metric. The negation converts distance to similarity. ↳ Multiply by gamma γ: The purpose is to control how much influence each training example has. A small gamma means each training example pulls the decision boundary more lightly, resulting in smoother decision boundaries. ↳ The result is “negative scaled L2” [11] Exponentiate ↳ Raise e to the power of the “negative scaled L2” ↳ Use the provided table to look up the value of e^ ↳ Output: The 1st column of K 𝘛𝘦𝘴𝘵 𝘝𝘦𝘤𝘵𝘰𝘳 2 (𝘟’2) 🟨 L2 Distance 🟪 [12]-[14] [12] Squared Difference [13] Sum [14] Square Root [15] Negative Scaling [16] Exponentiate ↳Output: The 2nd column of K 🟧 Decision Boundary [17]-[19] [17] Unsigned Coefficients → Signed Weights [18] Weighted Combination ... X'1: (20)*0.7+(-10)*0.8+(10)*0.9+(-10)*0.7+0+(-10)*0.9+(-1) = -2 X'2: (20)*0.9+(-10)*0.7+(10)*0.8+(-10)*0.8+0+(-10)*0.9+(-1) = 1 ... [19] Classify ... X'1: -2 0 → + ... ↳Output: The 2nd column of K 🟧 Decision Boundary [17]-[19] [17] Unsigned Coefficients → Signed Weights [18] Weighted Combination ... X'1: (20)*0.7+(-10)*0.8+(10)*0.9+(-10)*0.7+0+(-10)*0.9+(-1) = -2 X'2: (20)*0.9+(-10)*0.7+(10)*0.8+(-10)*0.8+0+(-10)*0.9+(-1) = 1 ... [19] Classify ... X'1: -2 0 → + ...

Tom Yeh

35,708 views • 2 years ago

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halleseed

16,120 views • 2 years ago

[Backpropagation] by Hand✍️ [1] Forward Pass ↳ Given a multi layer perceptron (3 levels), an input vector X, predictions Y^{Pred} = [0.5, 0.5, 0], and ground truth label Y^{Target} = [0, 1, 0]. [2] Backpropagation ↳ Insert cells to hold our calculations. [3] Layer 3 - Softmax (blue) ↳ Calculate ∂L / ∂z3 directly using the simple equation: Y^{Pred} - Y^{Target} = [0.5, -0.5, 0]. ↳ This simple equation is the benefit of using Softmax and Cross Entropy Loss together. [4] Layer 3 - Weights (orange) & Biases (black) ↳ Calculate ∂L / ∂W3 and ∂L / ∂b3 by multiplying ∂L / ∂z3 and [ a2 | 1 ]. [5] Layer 2 - Activations (green) ↳ Calculate ∂L / ∂a2 by multiplying ∂L / ∂z3 and W3. [6] Layer 2 - ReLU (blue) ↳ Calculate ∂L / ∂z2 by multiplying ∂L / ∂a2 with 1 for positive values and 0 otherwise. [7] Layer 2 - Weights (orange) & Biases (black) ↳ Calculate ∂L / ∂W2 and ∂L / ∂b2 by multiplying ∂L / ∂z2 and [ a1 | 1 ]. [8] Layer 1 - Activations (green) ↳ Calculate ∂L / ∂a1 by multiplying ∂L / ∂z2 and W2. [9] Layer 1 - ReLU (blue) ↳ Calculate ∂L / ∂z1 by multiplying ∂L / ∂a1 with 1 for positive values and 0 otherwise. [10] Layer 1 - Weights (orange) & Biases (black) ↳ Calculate ∂L / ∂W1 and ∂L / ∂b1 by multiplying ∂L / ∂z1 and [ x | 1 ]. [11] Gradient Descent ↳ Update weights and biases (typically a learning rate is applied here). 💡 Matrix Multiplication is All You Need: Just like in the forward pass, backpropagation is all about matrix multiplications. You can definitely do everything by hand as I demonstrated in this exercise, albeit slow and imperfect. This is why GPU's ability to multiply matrices efficiently plays such an important role in the deep learning evolution. This is why NVIDIA is now close to $1 trillion in valuation. 💡Exploding Gradients: We can already see the gradients are getting larger as we back-propagate up, even in this simple 3-layer network. This motivates using methods like skip connections to handle exploding (or diminishing) gradients as in the ResNet. I did the calculations entirely by hand. Please let me know if you spot any error or have any questions!

Tom Yeh

64,783 views • 2 years ago

[Discrete Fourier Transform] by Hand ✍️ In signal processing, the Discrete Fourier Transform (DFT) is no doubt the most important method. But the math involved is extremely complex, literally, involving a summation over a complex number term e^(-iwt). I developed this exercise to demonstrate that underneath such complexity, DFT is just a series of matrix multiplications you can calculate by hand. ✍️ Once you see that, it should not surprise you that a deep neural network, which is also a series of matrix multiplications, with activation functions in-between, can learn to perform DFT to process and analyze signals so effectively. How does DFT work? [1] Given ↳ Signals A, B, and C in the 🟧 frequency domain: ◦ A = cos(w) + 2cos(2w) ◦ B = cos(w) + cos(3w) + cos(4w) ◦ C = -cos(2w) + cos(3w) ◦ Each signal is a weighed sum of four cosine waves at frequencies 1w, 2w, 3w, and 4w. ◦ We will apply Inverse DFT to convert the signals to time domain representations, and then demonstrate DFT can convert back to their original frequency domain representations. ↳ Signal X in the 🟩 time domain. X is sampled at 10 time points 1t, 2t, …, 10t: ◦ X = [-2.5, -1.8, 3, -0.7, -1.0, -0.7, 3, -1.8, -2.5, 5] ◦ Suppose X is also a weighted sum of the same four cosine waves, but we don’t already know their weights. We will apply DFT to discover them. [2] 🟧 Frequency Matrix (F) ↳ Write the coefficients of A, B, C as a matrix F. Each signal is a row. Each frequency is a column. ↳ A → [1, 2, 0, 0] ↳ B → [1, 0, 1, 1] ↳ C → [0, 1-, 1, 0] [3] Cosine → Discrete ↳ Sample from the continuous cosine waves at discrete time points 1t, 2t, 3t, to 10t. [4] Cosine Matrix (W) ↳ Write the samples as a matrix, Each frequency is a row. Each time point is a column. [5] Inverse DFT: 🟧 Frequency → 🟩 Time ↳ Multiply the frequency matrix F and the cosine matrix W. ↳ The meaning of this multiplication is to linearly combine the four cosine waves (rows in W) into time-domain signals (rows in T) using the weights specified in F. ↳ The result is matrix T, which are signals A, B, C converted to the time domain. Each signal is a row. Each time point is a column. [6] Transpose ↳ Transpose T, converting each signal’s time domain representation from a row to a column. [7] DFT: 🟩 Time → 🟧 Frequency ↳ Multiply the cosine matrix W with the transpose of matrix T. ↳ The purpose of this multiplication is to take a dot-product between each time-domain signal (columns in the transpose of T) and each cosine wave (rows in W), which has the effect of projecting the signal onto a cosine wave to determine how much they are correlated. Zero means not correlated at all. ↳ The result is an intermediate version of the “recovered” frequency matrix where each column corresponds to a signal and each row corresponds to a frequency. ↳ Compared to the original frequency matrix F, this intermediate matrix has non-zero weights in the correct places, but scaled up by a factor of 5 (n/2, n=10). For example, signal A, originally [1,2,0,0], is recovered at [5,10,0,0]. [8] Scale ↳ Multiply each value by 2/n = 1/5 to scale down the intermediate matrix to match the magnitude of the original frequency matrix F. [9] Transpose ↳ Transpose the recovered frequency matrix back to the same orientation of the original frequency matrix F. ↳ Like magic 🪄, the result is identical to the original F, which means DFT successfully recovered the frequency components of signals A, B, C. [10] Apply DFT to X: 🟩 Time → 🟧 Frequency ↳ Now that we have some confidence in DFT’s ability to recover frequency components, we apply DFT to X’s time-domain representation by multiplying W with X. ↳ The result is the an intermediate matrix. [11] Scale ↳ Similarly, we scale down by a factor of 5 to obtain the recovered frequency components of X (a column). [12] Transpose ↳ Similarly, we transpose the recovered column to row to match the orientation of the frequency matrix. ↳ Using the coefficients [0,0,3,2], we can write the equation of X as 3cos(3w) + 2cos(4w). Notes: I hope this by hand exercise helps you understand the essence of DFT. But there is more technical details, such as: • Sine: The complete DFT math also includes sine waves that follow a similar calculation process. • Phase: Here, we assume all the cosine waves are aligned at the origin, namely, phase is 0. If a phase p is added, for example, cos(w+p), we will need to calculate the sine component and use their ratio to figure out what p is. • Magnitude: If phase is not zero, the magnitude will need to be calculated by combining both cosine and sine terms.

Tom Yeh

116,622 views • 2 years ago

MLP in PyTorch by hand ✍️ ~ 7 steps walkthrough below Goal: fill in every blank in the PyTorch code to build a multi-layer perceptron. 1. Given Let us start with a code template on the left and the network it is supposed to build on the right. Every blank in the code can be worked out from the picture. 2. Linear layer We count: 3 features in, 4 features out. So the weight matrix is 4 by 3. There is an extra column for the biases, which means bias = T. 3. ReLU Let us apply the activation. ReLU crosses out the negatives, so -1 becomes 0. 4. Linear layer The input size is 4, because that is what the previous layer put out. The output size is 2. A 2 by 4 weight matrix, and this time no extra column, so bias = F. 5. ReLU We cross out the negatives again. 6. Linear layer Two features in, five out. A 5 by 2 weight matrix, with a bias column, so bias = T. 7. Sigmoid Let us finish. Sigmoid squashes the raw scores (3, 0, -2, 5, -5) into probabilities between 0 and 1. You have just implemented a three-layer deep neural network by hand. ✍️ == Story == Three years ago I gave this exercise to my students, to connect the code to the math. They found it odd. Every other AI course they were taking lived inside a Jupyter notebook, and here I was handing out paper. Three years later, my colleagues are the ones rushing to move their materials to paper. The exercise has not changed. Paper still asks the one thing a notebook lets you skip: do you actually understand what the code is doing? If you can tell me why the weight matrix is 4 by 3, and why bias is F on the second layer, you understand nn.Linear better than someone who has been copy-pasting it for a year. 💾 Save this post! #AIbyHand #PyTorch #DeepLearning

Tom Yeh

13,318 views • 3 months ago

Transformer by hand ✍️ ~ 6 steps walkthrough below Open the hood of a transformer and the parts list is overwhelming: embeddings, positional encoding, attention weighting, self-attention, cross-attention, multi-head attention, layer norm, skip connections, softmax, linear, Nx, shifted right, query, key, value, masking. Which of those actually make the car run? Two of them. Attention weighting and the feed-forward network. Everything else is an enhancement to make it run faster and longer, which is how we got from a car to a truck, and to the word "large" in large language model. So I drew and calculated those two parts entirely by hand. Goal: push five features through one transformer block, filling in every cell yourself. 1. Given Five positions of input features, arriving from the previous block. 2. Attention matrix Let us feed all five features to a query-key module (QK) and read back an attention weight matrix, A. The details of that module are a post of their own. 3. Attention weighting We multiply the input features by A to get the attention weighted features, Z. Still five positions. The effect is to combine features *across positions*, horizontally: X1 becomes X1 + X2, X2 becomes X2 + X3, and so on. 4. First layer Let us feed all five weighted features into the first layer of the FFN. Multiply by the weights and biases. This time the combining happens *across feature dimensions*, vertically, and each feature grows from 3 numbers to 4. Note that every position goes through the same weight matrix. That is what "position-wise" means. 5. ReLU We cross out the negatives. They become zeros. 6. Second layer Let us bring it back down: 4 dimensions to 3. The output feeds the next block, which has a completely separate set of parameters, and the whole thing runs again. You have just calculated a transformer block by hand. ✍️ The takeaway: the two parts are doing two different jobs, and neither one alone is enough. Attention mixes *across positions*, so a feature can see its neighbours. The FFN mixes *across feature dimensions*, so each position can think about itself. Horizontal, then vertical. Then that pattern repeats N times, each block with its own separate set of weights. That is the Nx from the list up top, and that is what makes the transformer run. 💾 Save this post! #AIbyHand #Transformers #DeepLearning

Tom Yeh

26,279 views • 3 months ago

Batch Normalization by hand ✍️ ~ 7 steps walkthrough below Batch normalization is common practice for improving training and achieving faster convergence. It sounds simple. But it is often misunderstood. 🤔 Does batch normalization involve trainable parameters, tunable hyper-parameters, or both? 🤔 Is batch normalization applied to inputs, features, weights, biases, or outputs? 🤔 How is batch normalization different from layer normalization? So I drew and calculated one entirely by hand. Goal: normalize a mini-batch of 4 examples to mean 0 and variance 1, then let the network scale it back. = 1. Given = A mini-batch of 4 training examples, each with 3 features. = 2. Linear layer = Let us multiply by the weights and add the biases. Batch norm sits after this, which answers the second question: what gets normalized is features, not inputs, weights or biases. = 3. ReLU = We apply the activation, and -2 becomes 0. Negative values are suppressed before any statistic is taken. = 4. Batch statistics = Let us compute the sum, mean, variance and standard deviation, one row at a time. A row is a feature and the four columns are the four examples, so every number here measures one feature against the rest of the batch. That is the "batch" in batch normalization, and it is exactly what layer normalization does not do. The statistics are rounded to whole numbers, which is what keeps the rest of the page doable in pen. = 5. Shift to mean 0 = We subtract the mean, in green. The four values in each feature now average to zero. = 6. Scale to variance 1 = Let us divide by the standard deviation, in orange. Each feature now has variance one, whatever scale it arrived at. = 7. Scale and shift = We multiply by a linear transformation and pass the result on. The diagonal and the last column are trainable, so having just forced every feature to mean 0 and variance 1, we hand the network the means to undo it. The outputs: Mean of each feature = [2, 1, 2] Std dev of each feature = [1, 1, 2] To the next layer = [2, -2, 2, 0], [-3, 3, 6, -3], [2, 0, 1, 2] The answers: 🤔 Both. The scale and shift are trainable, the statistics are not. Epsilon and the momentum on the running statistics are the hyper-parameters, and one mini-batch by hand needs neither. 🤔 Features, after the linear layer, not inputs, weights or biases. 🤔 Batch norm measures across the batch, one feature at a time. Layer norm measures across the features, one example at a time. 💾 Save this post!

Tom Yeh

20,848 views • 3 months ago