正在加载视频...

视频加载失败

Graph Convolutional Network by hand ✍️ ~ 12 steps walkthrough below Graph Convolutional Networks (GCNs), introduced by Thomas Kipf and Max Welling in 2017, are the tool for data shaped like a graph: social networks, recommendations, biological networks, drug discovery, molecular chemistry. I drew and calculated a simple GCN...

16,956 次观看 • 3 个月前 •via X (Twitter)

3 条评论

Dispatch Desk 的头像
Dispatch Desk3 个月前

Raw dogging ann with a pencil like a champ.🔥🔥

A.W.E.S.O.M.-O 4000 的头像
A.W.E.S.O.M.-O 40003 个月前

Message passing is the secret sauce behind scalable graph reasoning and representation learning today

cipcioch 的头像
cipcioch3 个月前

can u do st-gcn next ?

相关视频

Backpropagation by hand ✍️ ~ 11 steps walkthrough below Backpropagation is the algorithm that actually trains a neural network, and it is where most people stop following along. It is not calculus you cannot do. It is matrix multiplication, working backward, one layer at a time. So I drew and calculated one entirely by hand. Goal: push the loss gradient back through a 3-layer network and land on a new value for every weight and bias. = 1. Given = A 3-layer perceptron, an input X, predictions Ypred = [0.5, 0.5, 0], and the truth Ytarget = [0, 1, 0]. = 2. Backprop gradient cells = Let us draw empty cells for every gradient we are about to compute. The shape of the answer comes first. = 3. Layer 3 softmax = We get dL/dz3 straight from Ypred minus Ytarget = [0.5, -0.5, 0]. No chain rule needed, and that shortcut is the whole reason softmax and cross-entropy are paired. = 4. Layer 3 weights and biases = Let us multiply dL/dz3 by [a2 | 1]. One multiplication gives the gradient for W3 and b3 together. = 5. Layer 2 activations = We multiply dL/dz3 by W3 to get dL/da2. The gradient moves back across a layer the same way the signal moved forward. = 6. Layer 2 ReLU = Let us pass it through the gate: keep the gradient where the activation was positive, zero it everywhere else. = 7. Layer 2 weights and biases = We multiply dL/dz2 by [a1 | 1]. The same figure as step 4, one layer up. = 8. Layer 1 activations = Let us multiply dL/dz2 by W2. = 9. Layer 1 ReLU = We apply the same gate again, now on a1. = 10. Layer 1 weights and biases = Let us multiply dL/dz1 by [x | 1], and every weight in the network now has a gradient. = 11. Update = We subtract, and the network has learned. In practice a learning rate scales this step. The gradients: dL/dz3 = [0.5, -0.5, 0] dL/da1 = [1, -2, 2, -1] dL/dz1 = [0, -2, 2, -1] The takeaway: matrix multiplication is all you need. Just like the forward pass, backpropagation is matrix multiplications end to end. You can do every one by hand, slowly and imperfectly, which is exactly why a GPU's ability to do them fast mattered so much to deep learning. 💾 Save this post!

Tom Yeh

963,459 次观看 • 3 个月前

ResNet by hand ✍️ ~ 10 steps walkthrough below "Deep Residual Learning for Image Recognition" (Kaiming He, CVPR 2016) is among the most cited papers in all of deep learning. Why does it matter so much? It fixed the exploding and vanishing gradients that kept deep networks from being deep, and made thousands of layers possible. How simple was the fix? An identity matrix. Goal: push three input vectors through a residual block, then through a transformer encoder block, filling in every cell yourself. = 1. Given = A mini batch of three input vectors, 3D, and the weights of the layers ahead. = 2. Linear layer = Let us multiply by the weights, add the bias, and apply ReLU so negatives become 0. Three feature vectors out. This is F(X). = 3. Concatenate = Now the trick. Stack an identity matrix beside the second layer's weights, and stack the input vectors under the features. Draw the lines between rows and columns: those are the skip connections. The identity is the residual. = 4. Linear layer + identity = We multiply the two stacked matrices. The identity carries X straight through while the weights transform it, so a single multiplication computes F(X) + X. Apply ReLU and hand it to the next block. Now watch the same trick inside a transformer, first in attention. = 5. Attention = Let us take three input vectors in 2D, compute the attention matrix, and multiply to get attention weighted vectors. = 6. Concatenate = We stack two identities this time, two residuals, which is how you get 1 + 1, and stack the input vectors with the attention weighted ones. = 7. Add = Multiply the stacked matrices. The identity adds attention to its own input, across the columns, which is how positions get combined. And again in the feed forward layer. = 8. First layer = Let us multiply by the feed forward weights and bias, then ReLU. Three feature vectors. = 9. Concatenate = Stack and link exactly as in step 3: the residual again. = 10. Second layer + identity = We multiply, apply ReLU, and pass the result to the next encoder block. This identity adds across the rows, combining features rather than positions. Takeaway: one simple "add" is what made really deep networks possible. 💾 Save this post!

Tom Yeh

18,152 次观看 • 3 个月前

Dropout by hand ✍️ ~ 10 steps walkthrough below Dropout is the simplest trick in deep learning that actually works: during training you randomly switch neurons off, so the network cannot lean on any one of them. It is two lines of code and almost nobody has worked through what those lines do to the numbers. So I drew and calculated one entirely by hand. Goal: train one pass through a small network with two dropout layers, then run inference with dropout switched off. The network: Linear(2,4), ReLU, Dropout(0.5), Linear(4,3), ReLU, Dropout(0.33), Linear(3,2). = 1. Given = A training set of two examples, X1 and X2, and the weight matrices for all three linear layers. = 2. Draw the first random numbers = Let us draw 4 random numbers, one per neuron in the first hidden layer. Above 0.5 we keep (◯), below we drop (╳). Here that gives [◯, ╳, ◯, ╳]. = 3. Build the first dropout matrix = We turn that pattern into a diagonal matrix. The scaling factor is 1/(1-p) = 2, so a kept neuron gets 2 and a dropped one gets 0. Multiplying by it does both jobs at once: it deletes the 2nd and 4th neurons and doubles the two that survive. = 4. Draw the second random numbers = Let us do it again for the 3 neurons in the next layer, this time against p = 0.33. The result is [◯, ◯, ╳]. = 5. Build the second dropout matrix = We set the diagonal to 1.5 where kept and 0 where dropped. Only the 3rd neuron goes. = 6. Feed forward = Let us run the whole thing top to bottom: one matrix multiplication per layer, ReLU setting the negatives to zero, and the two dropout matrices doing their work in between. The outputs Y come out at the bottom. = 7. MSE loss gradients = We compare Y against the targets Y', subtract, and multiply each element by 2. That is the whole gradient of the mean squared error. = 8. Update the weights = Let us push those gradients back through the network and update the weights (marked in light red). = 9. Deactivate dropout = Training is over, so we set both dropout matrices to the identity. Every neuron is back, and nothing is scaled. = 10. Feed forward again = One more pass, this time on unseen data, to make the prediction. You have just trained and run a network with dropout by hand. ✍️ The outputs: Training outputs Y = [-6, 9; 13, 4] Loss gradients = [-4, 4; 6, -2] Inference outputs = [13, 13; 4, 3] 💾 Save this post! #AIbyHand #Dropout #DeepLearning #NeuralNetworks

Tom Yeh

14,442 次观看 • 3 个月前

Self Attention by hand ✍️ ~ 9 steps walkthrough below Self-attention is what enables LLMs to understand context. How does it work? So I drew and calculated one entirely by hand. Goal: turn four 6D features into four 3D attention weighted features, filling in every cell yourself. = 1. Given = Four feature vectors, six dimensions each, one per position. = 2. Query, key, value = Let us multiply the features by WQ, WK and WV. Queries, keys and values all come out of the same four features, and that is what the word "self" is doing in self-attention. = 3. Prepare for MatMul = We copy the queries across the top and the transposed keys down the side. Lining the two up is half the work. = 4. MatMul = Let us multiply K transpose by Q. Every cell is the dot product of one key with one query, which we use as a matching score. That works because the dot product is the numerator of cosine similarity: it is how alike two vectors are, before anyone divides by their lengths. = 5. Scale = We divide by the square root of dk, the dimension of a key vector, here 3. Without it the scores grow with the dimension and a 64-wide head would swamp the softmax. To keep the page doable in pen, the drawing approximates dividing by root 3 with halving. = 6. e to the power = Let us raise e to the power of each score. This is the first half of softmax, and the drawing uses 3 in place of e, which is close enough to do in your head. = 7. Sum = We add up each column: 16, 6, 7 and 12. = 8. Normalize = Let us divide every cell by its column sum. That gives the attention weight matrix in yellow, and each of its four columns is now a probability distribution over the four positions. The decimals are nudged as they are rounded, so every column still sums to exactly 1. = 9. MatMul = We multiply the value vectors by those weights. Each output is a blend of all four values, mixed in the proportion the attention matrix just decided, and it goes to the position-wise feed forward network in the next layer: the FFN box at the bottom of the page. The outputs: Attention weights (A), by column = [.2, .6, 0, .2], [.2, .4, .2, .2], [.4, .2, 0, .4], [.1, .7, .1, .1] Attention weighted features (Z) = [8, 2, 6], [8, 4, 4], [16, 4, 2], [4, 2, 7] The takeaway: attention is a weighted average, and everything before step 9 exists to decide the weights. Compare every position with every other, turn the scores into one distribution per position, then blend. 💾 Save this post!

Tom Yeh

27,383 次观看 • 3 个月前

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

Tom Yeh

13,602 次观看 • 1 个月前

Discrete Fourier Transform by hand ✍️ ~ 12 steps walkthrough below Here is a little-known secret about the DFT and the inverse DFT: it is just matrix multiplication in both directions, one the transpose of the other, exactly like the forward pass and backpropagation I drew in other examples. Goal: recover which cosine waves a signal is made of, using nothing but multiplication and addition. = 1. Given = Three signals written as sums of cosines, and a fourth, X, that we do not know yet. = 2. Frequency matrix F = Let us write the coefficients as a matrix. Each signal is a row, each frequency a column, so A = cos(w) + 2cos(2w) becomes [1, 2, 0, 0]. = 3. Sample the waves = We read the four cosine waves at ten discrete time points. That word "discrete" is the whole difference between this and the continuous transform. = 4. Cosine matrix W = Let us write those samples as a matrix: each frequency a row, each time point a column. = 5. Frequency to time = We multiply F by W. That combines the four cosine waves in the proportions F specifies, and the result T is the three signals as they would look in time. = 6. Transpose = Let us stand each signal up as a column. = 7. Time to frequency = We multiply W by that transpose. Every cell is the dot product of one signal with one cosine wave, which measures how much of that wave the signal contains. Zero means none of it. = 8. Scale = Let us multiply by 2/n, with n = 10. The projections come out five times too large, and this is the correction. = 9. Transpose back = We turn it back around, and it is F again, exactly. That is the check: the transform recovered the coefficients we started from. = 10. Now solve for X = Let us run the same multiplication on the one signal whose recipe we never knew. = 11. Scale = We divide by 5 again. = 12. Transpose back = And X reads [0, 0, 3, 2], which says X = 3cos(3w) + 2cos(4w). Note: I originally drew this to show that the DFT is a special case of a convolution layer, its filters fixed to sine and cosine waves rather than learned. No wonder, then, that a convolution layer free to learn its own filters can be trained to process signals. 💾 Save this post!

Tom Yeh

25,684 次观看 • 3 个月前

FIVE LAYERS OF AGENT ENGINEERING, EACH ONE WRAPS THE ONE BELOW IT. IF YOU SKIP LAYER 2, YOUR LAYER 5 WILL LOOK BROKEN WHEN IT IS ACTUALLY JUST STANDING ON NOTHING. for weeks i debated harness vs loop vs graph like they were competing choices. then a stack diagram made the shape obvious. they are not choices. they are floors. 01 | prompt engineering. the message. unit of work: one input. inputs are role, instructions, examples, format. output is a single raw response. 02 | context engineering. the memory. unit of work: what stays in the window. a curator selects, compresses, and drops from query, docs, memory, prior turns, and tool outputs before the prompt runs. 03 | harness engineering. the machine. unit of work: the machine itself. gather (context + prompt) → LLM → tools or sub-agents → verifier → final response. the article calls this the operating environment. 04 | loop engineering. the system. unit of work: the run. goal + success criteria + max iterations + budget + completion check wrap around one harness pass. failed pass appends results to context and retries. 05 | graph engineering. the topology. unit of work: the graph run. goal + nodes + edges + state schema. graph routes to agent nodes, tool nodes, or human approval. a reviewer node with a different model and fresh context checks the final answer. the wrapping is the whole point. layer 5 assumes layer 4 works. layer 4 assumes layer 3 works. skip layer 2 and layer 3's verifier keeps failing without a clear reason. this is why swapping the model is a one-day project and swapping the stack is a quarter. the model is the commodity. the five layers around it are the engineering. full three-layer breakdown of the top of the stack (harness, loop, graph) in the post below.

kocer

31,198 次观看 • 2 个月前

Variational Autoencoder by hand ✍️ ~ 11 steps walkthrough below A VAE learns the structure of your data, the mean and variance of its hidden features, and then generates new data from that structure. A GAN only learns to fool a discriminator. It can make convincing fakes without ever knowing what the data is really made of. That is the difference, and it is the whole reason VAEs matter. In 2024 ICLR gave its first ever Test of Time Award to the VAE paper, "Auto-Encoding Variational Bayes" by Diederik Kingma and Max Welling, ten years on. How does it work? Goal: encode three inputs into a distribution, sample from it, decode it back, and read every loss gradient off the page. = 1. Given = Three training examples X1, X2, X3, copied to the bottom as their own targets. Reconstructing your own input is what puts the "auto", meaning self, in autoencoder. = 2. Encoder, layer 1 = Let us multiply the inputs by weights and biases, then apply ReLU, crossing out every negative. = 3. Mean and standard deviation = We multiply the features by two more weight sets. The first predicts the means μ of the latent distributions, the second their standard deviations σ. = 4. A random offset = Let us sample ε from a standard normal, mean 0 and variance 1, and multiply it by σ. This is a random step away from the mean, scaled by how uncertain each feature is. = 5. Mean plus offset = We add the offset back onto μ, and these become the decoder's inputs. Keeping the randomness out in ε is the reparameterization trick: it lets gradients flow straight through the sampling. = 6. Decoder, layer 1 = Let us multiply by weights and biases and apply ReLU again. Here -4 is crossed out. = 7. Decoder, layer 2 = We multiply once more. The output Y is the decoder's attempt to rebuild X from the sampled distribution. = 8. Gradient for the mean = Let us push μ toward 0. A lot of math, the SGVB estimator, collapses the KL gradient to simply μ itself. = 9. Gradient for the standard deviation = We want σ to approach 1. = 10. And its formula = That same math simplifies the gradient to σ minus 1/σ. = 11. Reconstruction gradient = We want the reconstruction Y to match the input X. Mean squared error simplifies its gradient to Y minus X. Takeaway: the two gradients you just calculated each sit at the heart of a modern method, so one VAE teaches you both. The KL divergence is the penalty RLHF like GRPO uses to keep a fine-tuned model from drifting off its base. The reconstruction loss, plain mean squared error, is exactly what trains a diffusion model to denoise. Draw one VAE by hand and you have quietly learned the core of both. 💾 Save this post!

Tom Yeh

17,011 次观看 • 3 个月前

SORA by Hand ✍️ OpenAI’s #SORA took over the Internet when it was announced earlier this year. The technology behind Sora is the Diffusion Transformer (DiT) developed by William Peebles and Shining Xie. How does DiT work? 𝗚𝗼𝗮𝗹: Generate a video conditioned by a text prompt and a series of diffusion steps [1] Given ↳ Video ↳ Prompt: "sora is sky" ↳ Diffusion step: t = 3 [2] Video → Patches ↳ Divide all pixels in all frames into 4 spacetime patches [3] Visual Encoder: Pixels 🟨 → Latent 🟩 ↳ Multiply the patches with weights and biases, followed by ReLU ↳ The result is a latent feature vector per patch ↳ The purpose is dimension reduction from 4 (2x2x1) to 2 (2x1). ↳ In the paper, the reduction is 196,608 (256x256x3)→ 4096 (32x32x4) [4] ⬛ Add Noise ↳ Sample a noise according to the diffusion time step t. Typically, the larger the t, the smaller the noise. ↳ Add the Sampled Noise to latent features to obtain Noised Latent. ↳ The goal is to purposely add noise to a video and ask the model to guess what that noise is. ↳ This is analogous to training a language model by purposely deleting a word in a sentence and ask the model to guess what the deleted word was. [5-7] 🟪 Conditioning by Adaptive Layer Norm [5] Encode Conditions ↳ Encode "sora is sky" into a text embedding vector [0,1,-1]. ↳ Encode t = 3 to as a binary vector [1,1]. ↳ Concatenate the two vectors in to a 5D column vector. [6] Estimate Scale/Shift ↳ Multiply the combined vector with weights and biases ↳ The goal is to estimate the scale [2,-1] and shift [-1,5]. ↳ Copy the result to (X) and (+) [7] Apply Scale/Sift ↳ Scale the noised latent by [2,-1] ↳ Shifted the scaled noised latent by [-1, 5] ↳ The result is "conditioned" noise latent. [8-10] Transformer [8] Self-Attention ↳ Feed the conditioned noised latent to Query-Key function to obtain a self-attention matrix ↳ Value is omitted for simplicity [9] Attention Pooling ↳ Multiply the conditioned noised latent with the self-attention matrix ↳ The result are attention weighted features [10] Pointwise Feed Forward Network ↳ Multiply the attention weighted features with weights and biases ↳ The result is the Predicted Noise 🏋️‍♂️ 𝗧𝗿𝗮𝗶𝗻 [11] ↳ Calculate MSE loss gradients by taking the different between the Predicted Noise and the Sampled Noise (ground truth). ↳ Use the loss gradients to kick off backpropagation to update all learnable parameters (red borders) ↳ Note the visual encoder and decoder's parameters are frozen (blue borders) 🎨 𝗚𝗲𝗻𝗲𝗿𝗮𝘁𝗲 (𝗦𝗮𝗺𝗽𝗹𝗲) [12] Denoise ↳ Subtract the predicted noise from the noised latent to obtain the noise-free latent [13] Visual Decoder: Latent 🟩 → Pixels 🟨 ↳ Multiply the patches with weights and biases, followed by ReLU [14] Patches → Video ↳ Rearrange patches into a sequence of video frames.

Tom Yeh

238,485 次观看 • 2 年前

SVM by hand ✍️ ~ 19 steps walkthrough below (Linear vs RBF) Support Vector Machines reigned supreme in machine learning before the deep learning revolution. An SVM predicts with dot products, the same matrix multiplication every model uses. What it does not do is train by backpropagation: it is fitted by convex optimization, so there is no matrix-multiplication backward pass for a GPU to accelerate. I drew and calculated two SVMs by hand: a linear one (top) and an RBF one (bottom), classifying the same two test vectors. Goal: turn six training vectors and their learned coefficients into a prediction, and see what changing the kernel actually changes. = 1. Given = Six training vectors, their labels, and the coefficients and bias already learned. A coefficient of zero means that vector is not a support vector: too far from the boundary to matter. = 2. Linear kernel, test vector 1 = Let us take the dot product of the test vector with every training vector. The dot product stands in for cosine similarity, and the column of results is the first column of the kernel matrix K. = 3. Linear kernel, test vector 2 = We do the same for the second, and K is complete. = 4. Signed weights = Let us multiply each coefficient by its label. The second training vector drops out here, because its coefficient is 0. = 5. Weighted combination = We multiply the signed weights through K and add the bias b. The result is a signed distance to the decision boundary: 17 and 5. = 6. Classify = Let us take the sign. Both are positive. = 7 to 11. RBF kernel, test vector 1 = Now the same picture with a different kernel, in five moves: square the differences, sum them, take the square root for the L2 distance, multiply by minus gamma, and raise e to that power. The negation is what turns a distance into a similarity, and gamma controls how far a single training vector's influence reaches. = 12 to 16. RBF kernel, test vector 2 = We repeat all five. The numbers change, the moves do not. = 17 to 19. Decision boundary, again = Signed weights, weighted combination, sign. Identical arithmetic to steps 4 through 6, on a K that was built a completely different way. The outputs: Linear K, first column = [13, 25, 12, 15, 19, 27] Linear decision values = 17 and 5, both positive RBF decision values = -2 and 1, so negative and positive The takeaway: the kernel is the only thing that changed, and it changed the answer. The linear SVM calls both test vectors positive; the RBF one splits them. Everything after the kernel matrix, the signed weights and the weighted combination and the sign, is the same page of arithmetic twice. 💾 Save this post!

Tom Yeh

16,916 次观看 • 3 个月前

[RLHF] by Hand ✍️ Yesterday, Jan Leike (Jan Leike) announced he is joining #Anthropic to lead their "super-alignment" mission. He is the co-inventor of Reinforcement Learning with Human Feedback (#RLHF). How does RLHF work? [1] Given ↳ Reward Model (RM) ↳ Large Language Model (LLM) ↳ Two (Prompt, Next) Pairs 🟪 TRAIN RM Goal: Learn to give higher rewards to winners [2] Preferences ↳ A human reviews the two pairs and picks a "winner" ↳ (doc is, him) Embeddings ↳ This prompt has never received human feedback directly ↳ [S] is the special start symbol [11] Transformer ↳ Attention (yellow) ↳ Feed Forward (4x2 weight and bias matrix) ↳ Output: 3 "transformed" feature vector, one per position ↳ More details in my previous post 8. Transformer [] [12] Output Probabilities ↳ Apply a linear layer to map each transformed feature vector to a probability distribution over the vocabulary. [13] Sample ↳ Apply the greedy method, which is to pick the word with the highest score ↳ For output 1 and 2, the model accurately predicts the next word ↳ For 3rd output position, the model's predicts "him" [14] Reward Model ↳ The new pair (CEO is, him) is fed to the reward model ↳ The process is same as [3]-[6] ↳ Output: Reward = 3 [15] Loss Gradient ↳ We set the loss as the negative of the reward. ↳ The loss gradient is simply a constant -1. ↳ Run backpropagation and gradient descent to update LLM's weights and biases (red border)

Tom Yeh

79,916 次观看 • 2 年前

Switch Transformer by hand ✍️ ~ 13 steps walkthrough below The Switch Transformer, by Fedus, Zoph, and Shazeer in 2022, is one of the papers that made sparse Mixture of Experts practical at scale. Today, frontier models use MoE to pack enormous parameter counts while activating only a small slice per token: GPT-4, Claude, DeepSeek-V3, and Kimi all follow this pattern. If you want to understand how those models can be huge to store yet still cheap to run, this paper is a good place to start. How does it work? Goal: run five input features through attention, route each one to a single best expert, and read the output off the page. = 1. Given = Input features X1-X5 arrive from the previous block. = 2. Attention matrix = Feed all five features to a query-key attention module to get an attention weight matrix A. = 3. Pooling = Multiply the input features by A to get attention-weighted features Z1-Z5. The effect is to combine features across positions. = 4. Visualize pooling = Z4 is X4 + X5 because the fourth column of A is [0,0,0,1,1]. = 5. Gate values = Multiply the weighted features by the switch matrix. Each gate value says how well expert A, B, or C can probably handle the feature. = 6. Top expert = Pick the row with the highest gate value. Sparse means only the top expert is selected, not all of them. = 7. Routing = Route each Z to its best expert. Every expert has a fixed capacity of 2, so one feature may overflow. = 8. Expert A, linear = Apply the linear layer to the features routed to Expert A. The effect is to combine features across feature dimensions. = 9. Expert A, aggregate = Send the combined feature to the corresponding output column. = 10. Expert B, linear = Apply the linear layer, as in step 8. = 11. Expert B, aggregate = Send the result to the corresponding output column, as in step 9. = 12. Expert C, linear = Apply the linear layer, as in step 8. = 13. Expert C, aggregate = Send the result to the output column. Since one feature exceeded Expert C's capacity, it passes through as-is. Takeaway: a Switch Transformer keeps attention unchanged, then replaces the dense feed-forward network with a sparse set of experts. Most of the parameters sit in the experts, but only a small fraction are used for any one input. That is how GPT-4, Claude, DeepSeek-V3, and Kimi can be enormous to store and still cheap to run. 💾 Save this post!

Tom Yeh

37,410 次观看 • 2 个月前