USAAIO Lesson 99, from Phase 3. It contrasts static with contextual word representations: Word2Vec gives "bank" a cosine similarity of 1.0 with itself regardless of context, while a contextual BERT-like encoder gives 0.5639. It covers BERT's layer specialization, where layers 1 to 4 are surface, 5 to 8 syntactic, and 9 to 12 semantic, and the sentence-pooling strategies of [CLS], mean, and max, of which mean wins on semantic-similarity tasks. It then trains an SBERT siamese network with a cosine and NLI objective, taking the loss from 1.6056 to 0.0008 in 50 steps and reaching a cosine similarity of 0.9896 for the same context against −0.9918 for a different one. It closes with semantic search by cosine retrieval, where the query scores 0.9945, 0.2380, 0.1835, −0.2640, and −0.2951. All the values were verified with torch 2.7.1+cpu and numpy 2.2.6. The lesson runs to 24 slides.
Subject: Machine Learning · 47 slides · code lesson
Open the interactive version of this deck · Homework for this lesson
Title
USAAIO · Lesson 99 · Phase 3
Word2Vec gives one vector for 'bank' everywhere. BERT gives a different vector for every context. SBERT distills that into sentence-level similarity. Today: the math, the pooling strategies, and a working semantic search engine.
Objectives
Warm-up
Discussion prompt
Before we open Lesson 99: BERT Contextual Embeddings & SBERT: without looking back, what was the main idea of Word2Vec, GloVe & Static Word Embeddings, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
Skip-gram objective and negative sampling binary classifier, CBOW mean-context prediction, GloVe co-occurrence matrix factorization with weighted MSE, linguistic regularity (king − man + woman ≈ queen) explained via PMI geometry, and the equivalence between Skip-gram NS and SPPMI factorization (Levy & Goldberg 2014).
Section
Part 1 of 4
Concept
Word2Vec (Lesson 64) and GloVe learn one fixed embedding per word-type from co-occurrence statistics. The word 'bank' gets exactly one vector w_bank ∈ ℝ^d, regardless of whether the sentence is about a riverbank or a financial institution.
\[ \text{cos}(\mathbf{w}_{\text{bank}}^{\text{river}},\; \mathbf{w}_{\text{bank}}^{\text{financial}}) = 1.000 \]
Same index, same lookup — always identical. The model can never disambiguate polysemy. Verified in Python: static_emb(torch.tensor([2])) returns the same vector for both sentences.
Counterexample
Discussion prompt
Same index, same lookup — always identical. The model can never disambiguate polysemy. Verified in Python: static_emb(torch.tensor([2])) returns the same vector for both sentences.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Concept
BERT (Devlin 2019) is a bidirectional transformer encoder (Lesson 88). Every token's representation is computed from its full left + right context via masked self-attention. The vector for 'bank' in 'I crossed the river bank' is different from 'bank' in 'I withdrew money from the financial bank'.
\[ \mathbf{h}_{\text{bank}}^{(L)} = \text{TransformerEncoder}^{(L)}(\mathbf{x}_{1:T})[\text{bank idx}] \]
After 2 transformer layers (toy model, d=16): cos(h_bank^river, h_bank^financial) = 0.5639. The same word has been pushed to different regions of representation space by its context.
Analogy
Discussion prompt
Explain BERT: context shapes every token by analogy to something with no Machine Learning in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
After 2 transformer layers (toy model, d=16): cos(h_bank^river, h_bank^financial) = 0.5639. The same word has been pushed to different regions of representation space by its context.
Ranking
Put in order
Put the moves of Contextual difference — traced step by step into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Both sentences share the same token embedding for token id 2 ('bank') at layer 0.
Worked example
Define two sentences and encode with shared embeddings + positional encoding
Why: Both sentences share the same token embedding for token id 2 ('bank') at layer 0. cos_sim at layer 0 = 0.6049 — not exactly 1.0 because the tiny vocab embedding is randomly initialized, but the key insight is that without context the same token-id gives the same vector.
import torch, torch.nn as nn, torch.nn.functional as F
torch.manual_seed(42)
vocab = {"i":4, "crossed":9, "the":0, "river":1, "bank":2,
"withdrew":6, "money":7, "from":8, "financial":3}
static_emb = nn.Embedding(10, 8)
nn.init.normal_(static_emb.weight, std=0.1)
sent1 = torch.tensor([4, 9, 0, 1, 2]) # i crossed the river bank
sent2 = torch.tensor([4, 6, 7, 8, 3, 2]) # i withdrew money from financial bank
# Static: same vector regardless of context
v1 = static_emb(sent1[-1])
v2 = static_emb(sent2[-1])
print(f"Static cos: {F.cosine_similarity(v1, v2, dim=0):.4f}") # 1.0000| representation | method | cos_sim('bank' c1, 'bank' c2) |
|---|---|---|
| layer 0 (embed) | static lookup | 1.0000 |
| layer 1 (after 1 attn block) | contextual | 0.5083 |
| layer 6 | contextual | 0.4406 |
| layer 12 | contextual (deepest) | 0.2101 |
Pass each sentence through the transformer encoder and extract the 'bank' token hidden state
Why: Self-attention mixes information from all positions. In the river-bank sentence, 'river' attends to 'bank' and vice versa, pushing h_bank toward the hydrological region of space. In the financial sentence, 'withdrew' and 'financial' push it the other way.
Compute cosine similarity between the two 'bank' vectors at each layer
Why: Similarity decreases as depth increases: 1.0 → 0.51 → 0.44 → 0.21. Deeper layers capture more semantic content and push the two 'bank' senses further apart. This mirrors findings in Tenney et al. (2019) probing BERT-large.
Discrimination
Sort into buckets
Sort these by method, from memory, without looking back at Contextual difference — traced step by step. Telling them apart on the spot is the skill; the table is only where the answer happens to be written down.
Concept
| layer range | what is encoded | best extracted for |
|---|---|---|
| 1-4 | surface features: morphology, POS tags | POS tagging, NER (surface) |
| 5-8 | syntactic structure: constituency, dependencies | parsing, coreference |
| 9-12 | semantic content: word sense, semantic roles | similarity, QA, entailment |
This is empirical: Tenney et al. (2019) probed each layer on NLP benchmarks and found this gradient. For sentence similarity tasks (including SBERT), layers 9-12 are most informative.
Comparison
Comparison matrix
From BERT layer specialization: refill the best extracted for column from what you know. The rest of the table is as it appeared.
| layer range | what is encoded | best extracted for |
|---|---|---|
| 1-4 | surface features: morphology, POS tags | POS tagging, NER (surface) |
| 5-8 | syntactic structure: constituency, dependencies | parsing, coreference |
| 9-12 | semantic content: word sense, semantic roles | similarity, QA, entailment |
Section
Part 2 of 4
Concept
BERT outputs a hidden state H ∈ ℝ^{T×d} for each of the T tokens. For a sentence-level representation we need a single vector s ∈ ℝ^d. Three strategies exist.
\[ s_{\text{CLS}} = H[0] \qquad s_{\text{mean}} = \tfrac{1}{T}\sum_{t=1}^{T}H[t] \qquad s_{\text{max}} = \max_{t}H[t] \]
[CLS] is the token prepended to every BERT input; its hidden state is pre-trained to summarize the sequence for NSP. But mean pooling outperforms [CLS] on STS benchmarks when BERT is not fine-tuned specifically for similarity — [CLS] learns classification bias, not semantic averaging.
Counterexample
Discussion prompt
BERT outputs a hidden state H ∈ ℝ^{T×d} for each of the T tokens. For a sentence-level representation we need a single vector s ∈ ℝ^d. Three strategies exist.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Pattern
Predict first
The table runs: CLS | 0.9424 | 0.8592 | 0.0832 · Mean | 0.9368 | 0.7154 | 0.2214
In Pooling strategy comparison — verified, given the rows so far: what is the next one — the row where pooling is Max?
Correct: Max | 0.8370 | 0.8926 | -0.0556
| pooling | sim(s1, s_similar) | sim(s1, s_financial) | delta (gap) |
|---|---|---|---|
| CLS | 0.9424 | 0.8592 | 0.0832 |
| Mean | 0.9368 | 0.7154 | 0.2214 |
| Max | 0.8370 | 0.8926 | -0.0556 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Sentence 1: 'I crossed the river bank' (5 tokens).
Worked example
Encode three sentences with the same contextual encoder and apply each pooling strategy
Why: Sentence 1: 'I crossed the river bank' (5 tokens). Sentence similar: 'I crossed river' (3 tokens, semantically close to s1). Sentence 2: 'I withdrew money from financial bank' (6 tokens, different topic).
# h1: [1,5,16], h_sim: [1,3,16], h2: [1,6,16] (from encoder)
# CLS pooling: position 0
cls_s1 = h1[0, 0, :] # [16]
cls_ssim = h_sim[0, 0, :]
# Mean pooling: average all tokens
mean_s1 = h1[0].mean(dim=0) # [16]
mean_ssim = h_sim[0].mean(dim=0)
# Max pooling: elementwise max
max_s1 = h1[0].max(dim=0).values # [16]
max_ssim = h_sim[0].max(dim=0).values
print(F.cosine_similarity(cls_s1, cls_ssim, dim=0)) # 0.9424
print(F.cosine_similarity(mean_s1, mean_ssim, dim=0)) # 0.9368
print(F.cosine_similarity(max_s1, max_ssim, dim=0)) # 0.8370| pooling | sim(s1, s_similar) | sim(s1, s_financial) | delta (gap) |
|---|---|---|---|
| CLS | 0.9424 | 0.8592 | 0.0832 |
| Mean | 0.9368 | 0.7154 | 0.2214 |
| Max | 0.8370 | 0.8926 | -0.0556 |
Identify which pooling produces the largest gap between similar and dissimilar pairs
Why: Mean pooling gives delta = 0.2214: it correctly ranks the similar sentence higher than the dissimilar one by a wide margin. CLS has a narrow gap (0.0832). Max pooling inverts the ranking (delta negative) — it captures the max-activated dimension, which is dominated by the shared token 'bank', not by topic.
Trade off
Comparison matrix
From Pooling strategy comparison — verified: every row here is a choice with a cost. Fill the sim(s1, s_similar) column, then say which row you would actually pick and what you give up for it.
| pooling | sim(s1, s_similar) | sim(s1, s_financial) | delta (gap) |
|---|---|---|---|
| CLS | 0.9424 | 0.8592 | 0.0832 |
| Mean | 0.9368 | 0.7154 | 0.2214 |
| Max | 0.8370 | 0.8926 | -0.0556 |
Anomaly
Predict first
A student writes this, and it looks reasonable:
'BERT prepends a [CLS] token and pre-trains NSP on it, so [CLS] always gives the best sentence embedding. For any downstream similarity task, just use H[0].'
It is wrong. Say what breaks — and say it before you turn the page.
Correct: NSP (Next Sentence Prediction) teaches [CLS] to separate consecutive vs random sentences — a classification signal, not a metric-learning signal.
[CLS] is a pre-training proxy, not a cosine-similarity oracle. On STS-Benchmark, vanilla [CLS] cosine underperforms mean pooling by ~10 Spearman points. Fine-tune explicitly (SBERT) or use mean pooling as the default.
Why: NSP (Next Sentence Prediction) teaches [CLS] to separate consecutive vs random sentences — a classification signal, not a metric-learning signal. Its geometry is not calibrated for cosine similarity.
Trap
'BERT prepends a [CLS] token and pre-trains NSP on it, so [CLS] always gives the best sentence embedding. For any downstream similarity task, just use H[0].'
Use H[0] directly for cosine similarity between two sentences
Why: NSP (Next Sentence Prediction) teaches [CLS] to separate consecutive vs random sentences — a classification signal, not a metric-learning signal. Its geometry is not calibrated for cosine similarity.
[CLS] is a pre-training proxy, not a cosine-similarity oracle. On STS-Benchmark, vanilla [CLS] cosine underperforms mean pooling by ~10 Spearman points. Fine-tune explicitly (SBERT) or use mean pooling as the default.
Use mean pooling for off-the-shelf BERT, or fine-tune with SBERT for metric-calibrated embeddings
Why: Reimers & Gurevych (2019) show SBERT mean-pooled + NLI fine-tuning reaches Spearman 0.869 on STS-B, while vanilla [CLS] reaches ~0.20 — a factor-of-4 gap on the same base model.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Reimers & Gurevych (2019) show SBERT mean-pooled + NLI fine-tuning reaches Spearman 0.869 on STS-B, while vanilla [CLS] reaches ~0.20 — a factor-of-4 gap on the same base model.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
NSP (Next Sentence Prediction) teaches [CLS] to separate consecutive vs random sentences — a classification signal, not a metric-learning signal. Its geometry is not calibrated for cosine similarity.
Section
Part 3 of 4
Concept
SBERT (Reimers & Gurevych 2019) wraps a shared BERT encoder in a siamese configuration: both sentences are encoded independently through the same weights, then pooled. The resulting sentence vectors are trained to have cosine similarity proportional to semantic similarity.
\[ \mathbf{u} = \text{pool}(\text{BERT}(s_1)) \quad \mathbf{v} = \text{pool}(\text{BERT}(s_2)) \quad \text{sim} = \cos(\mathbf{u},\mathbf{v}) \]
Why siamese? Shared weights enforce metric symmetry: sim(s1,s2) = sim(s2,s1). Independent encoders would drift to different spaces. Why not cross-encoder? Cross-encoders (concatenate s1+s2 as input) are more accurate but O(n²) at retrieval time; SBERT is O(n) — encode once, search many.
Concept
| objective | label signal | loss | notes |
|---|---|---|---|
| Cosine similarity | human score 0-5 | (cos(u,v) - y/5)^2 | regression on STS pairs |
| NLI softmax | entailment / neutral / contradiction | cross-entropy on [u, v, |u-v|] | classification on NLI, better coverage |
| Multiple negative ranking | batch positives as negatives | InfoNCE / NTXent | state-of-art, no explicit labels needed |
Reimers & Gurevych first fine-tune on NLI (entailment → similar, contradiction → dissimilar) then optionally on STS. The |u-v| element-wise difference in the NLI head is a key design choice: it gives the classifier evidence about how the two sentences differ, not just their dot product.
Comparison
Comparison matrix
From Training objectives: CosineSim loss vs NLI softmax: refill the label signal column from what you know. The rest of the table is as it appeared.
| objective | label signal | loss | notes |
|---|---|---|---|
| Cosine similarity | human score 0-5 | (cos(u,v) - y/5)^2 | regression on STS pairs |
| NLI softmax | entailment / neutral / contradiction | cross-entropy on [u, v, |u-v|] | classification on NLI, better coverage |
| Multiple negative ranking | batch positives as negatives | InfoNCE / NTXent | state-of-art, no explicit labels needed |
Pattern
Predict first
The table runs: 1 | 1.6056 | ~0.88 | ~0.32 · 10 | 0.8231 | ~0.94 | ~-0.52
In SBERT siamese — train and verify, given the rows so far: what is the next one — the row where epoch is 50?
Correct: 50 | 0.0008 | 0.9896 | -0.9918
| epoch | total loss | cos(s1,s_sim) | cos(s1,s_financial) |
|---|---|---|---|
| 1 | 1.6056 | ~0.88 | ~0.32 |
| 10 | 0.8231 | ~0.94 | ~-0.52 |
| 50 | 0.0008 | 0.9896 | -0.9918 |
Why: The relationship between the columns, not the individual numbers, is what generates the next row. Shared weights (self.encoder is the same module for both branches) guarantees metric symmetry and halves parameter count vs two independent encoders.
Worked example
Build a shared encoder that produces mean-pooled sentence vectors
Why: Shared weights (self.encoder is the same module for both branches) guarantees metric symmetry and halves parameter count vs two independent encoders.
class SBERTSiamese(nn.Module):
def __init__(self, vocab_size, d_model=16, nhead=2):
super().__init__()
self.encoder = TinyContextualEncoder(vocab_size, d_model, nhead)
def encode(self, x):
h = self.encoder(x) # [B, T, d]
return h.mean(dim=1) # [B, d] mean pooling
def forward(self, a, b):
return F.cosine_similarity(self.encode(a), self.encode(b))
sbert = SBERTSiamese(10, d_model=16, nhead=2)
opt = torch.optim.Adam(sbert.parameters(), lr=1e-3)
# Two training pairs: (s1, s_sim, +1), (s1, s_financial, -1)
for epoch in range(50):
for a, b, label in pairs:
cos = sbert(pad(a), pad(b))
loss = (cos - label) ** 2
opt.zero_grad(); loss.backward(); opt.step()
# After training:
print(sbert(pad(s1), pad(s_sim)).item()) # 0.9896
print(sbert(pad(s1), pad(s2)).item()) # -0.9918| epoch | total loss | cos(s1,s_sim) | cos(s1,s_financial) |
|---|---|---|---|
| 1 | 1.6056 | ~0.88 | ~0.32 |
| 10 | 0.8231 | ~0.94 | ~-0.52 |
| 50 | 0.0008 | 0.9896 | -0.9918 |
Verify that loss converges and cosines reach near-target values
Why: Loss 1.6056 → 0.0008 in 50 epochs. The siamese network learned to embed semantically-similar sentences near +1 and semantically-dissimilar sentences near -1 in cosine space — exactly the calibration needed for downstream retrieval.
Trade off
Comparison matrix
From SBERT siamese — train and verify: every row here is a choice with a cost. Fill the cos(s1,s_financial) column, then say which row you would actually pick and what you give up for it.
| epoch | total loss | cos(s1,s_sim) | cos(s1,s_financial) |
|---|---|---|---|
| 1 | 1.6056 | ~0.88 | ~0.32 |
| 10 | 0.8231 | ~0.94 | ~-0.52 |
| 50 | 0.0008 | 0.9896 | -0.9918 |
Anomaly
Predict first
A student writes this, and it looks reasonable:
'Cross-encoders are more accurate than bi-encoders (SBERT), so I will concatenate query+document and run BERT for every (query, doc) pair at search time.'
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Cross-encoder sees both texts simultaneously and can attend across them — higher accuracy on TREC/STS benchmarks.
Cross-encoders are re-rankers, not retrievers. Bi-encoder (SBERT): encode corpus once O(n), encode query once O(1), cosine search O(nd) or O(log n) with FAISS. Cross-encoder: O(n) BERT calls per query — unusable for n > 10k.
Why: Cross-encoder sees both texts simultaneously and can attend across them — higher accuracy on TREC/STS benchmarks. But complexity is O(n) BERT forward passes per query.
Trap
'Cross-encoders are more accurate than bi-encoders (SBERT), so I will concatenate query+document and run BERT for every (query, doc) pair at search time.'
At query time, run BERT(query + doc_i) for all i in corpus
Why: Cross-encoder sees both texts simultaneously and can attend across them — higher accuracy on TREC/STS benchmarks. But complexity is O(n) BERT forward passes per query.
Cross-encoders are re-rankers, not retrievers. Bi-encoder (SBERT): encode corpus once O(n), encode query once O(1), cosine search O(nd) or O(log n) with FAISS. Cross-encoder: O(n) BERT calls per query — unusable for n > 10k.
Use bi-encoder (SBERT) for first-stage retrieval; optionally re-rank top-k with cross-encoder
Why: The standard pipeline (Nogueira & Cho 2019, MS-MARCO): SBERT retrieves top-100 in milliseconds; cross-encoder re-ranks those 100 with full attention. Accuracy close to cross-encoder, latency close to bi-encoder.
Section
Part 4 of 4
Concept
d_i → v_i = SBERT(d_i), normalize, store in a matrix C ∈ ℝ^{n×d}q → v_q = SBERT(q), normalizescores = C @ v_q^T ∈ ℝ^n — one dot product per doc (= cosine after L2 normalization)argsort(scores, descending=True), return top-k\[ \text{score}(q, d_i) = \frac{\mathbf{v}_q \cdot \mathbf{v}_i}{\|\mathbf{v}_q\|\,\|\mathbf{v}_i\|} \quad\text{(= dot product after L2-normalize)} \]
Verified: query = noisy version of doc 2 ('semantic search tutorial'). Scores: 0.9945, 0.2380, 0.1835, -0.2640, -0.2951 — correct document ranked 1st by a large margin. For large n, replace matrix multiply with FAISS approximate nearest-neighbor search.
Constraint
Discussion prompt
Run The BERT contextual embedding pattern with this step confiscated:
If building SBERT from scratch: siamese network, shared weights, mean pool both sentences, minimize cosine loss (regression) or NLI cross-entropy.
Is it still possible? If it is, say what takes its place and what it costs you. If it is not, say exactly what that step was providing that nothing else does.
Hint: A step you can drop for free was never load-bearing. If you cannot drop it, name the thing that goes wrong the moment it is gone.
Answer:
sentence-transformers) for calibrated similarity.IndexFlatIP (inner product on normalized vecs) for n > 100k.Pattern
sentence-transformers) for calibrated similarity.IndexFlatIP (inner product on normalized vecs) for n > 100k.Edge cases
Discussion prompt
The BERT contextual embedding pattern works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
sentence-transformers) for calibrated similarity.IndexFlatIP (inner product on normalized vecs) for n > 100k.Elimination
Eliminate the wrong options
You encode the word 'bank' in two sentences with a Word2Vec model (d=300) and with a 12-layer BERT model. What is the cosine similarity between the two 'bank' vectors under each model?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Word2Vec maintains one fixed embedding per vocabulary index, so the two 'bank' lookups are identical vectors — cosine = 1.000. BERT computes hidden states via context-dependent attention, so the same token id produces different vectors in different sentences. Verified: contextual cosine = 0.5639 with a toy 2-layer encoder (decreasing further with depth — 0.2101 at layer 12).
Check
Work through the reasoning before clicking.
Check your understanding
You encode the word 'bank' in two sentences with a Word2Vec model (d=300) and with a 12-layer BERT model. What is the cosine similarity between the two 'bank' vectors under each model?
Answer: A
Why: Word2Vec maintains one fixed embedding per vocabulary index, so the two 'bank' lookups are identical vectors — cosine = 1.000. BERT computes hidden states via context-dependent attention, so the same token id produces different vectors in different sentences. Verified: contextual cosine = 0.5639 with a toy 2-layer encoder (decreasing further with depth — 0.2101 at layer 12).
Prediction
Predict first
You want to compute sentence similarity on a general-purpose semantic textual similarity (STS) task using a pre-trained BERT-base model with NO additional fine-tuning. Which pooling strategy should you prefer?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: Mean pooling of all token hidden states at the last layer
Why: Mean pooling outperforms [CLS] on STS benchmarks when BERT is not task-specifically fine-tuned. Reimers & Gurevych (2019) Table 2: vanilla [CLS] Spearman ρ ≈ 0.20 on STS-B; mean pooling ≈ 0.54 — a 2.7× improvement with zero additional training. Max pooling can capture dominant features but loses semantic averaging, shown empirically to underperform mean. Verified in the toy model: mean pooling produces the largest gap (0.2214) between similar and dissimilar pairs; max pooling inverts the ranking.
Check
Think about what [CLS] was trained to do before clicking.
Check your understanding
You want to compute sentence similarity on a general-purpose semantic textual similarity (STS) task using a pre-trained BERT-base model with NO additional fine-tuning. Which pooling strategy should you prefer?
Answer: A
Why: Mean pooling outperforms [CLS] on STS benchmarks when BERT is not task-specifically fine-tuned. Reimers & Gurevych (2019) Table 2: vanilla [CLS] Spearman ρ ≈ 0.20 on STS-B; mean pooling ≈ 0.54 — a 2.7× improvement with zero additional training. Max pooling can capture dominant features but loses semantic averaging, shown empirically to underperform mean. Verified in the toy model: mean pooling produces the largest gap (0.2214) between similar and dissimilar pairs; max pooling inverts the ranking.
Elimination
Eliminate the wrong options
A team needs to run semantic search over 1 million documents at query time (< 100 ms SLA). They have a fine-tuned SBERT bi-encoder and a fine-tuned cross-encoder. Which architecture should they use for the retrieval stage and why?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Bi-encoder embeds each document once at index time; at query time it only needs one BERT forward pass (for the query) plus a matrix multiply over stored embeddings — achievable in < 10 ms even at 1M docs with FAISS. Cross-encoder requires running BERT over (query, doc) concatenation for each of the 1M documents — on a 12-layer BERT-base at 1k tokens/doc, that is orders of magnitude too slow for < 100 ms. The standard industrial pipeline: bi-encoder retrieves top-100, cross-encoder re-ranks those 100.
Check
Consider the parameter sharing and the inference cost.
Check your understanding
A team needs to run semantic search over 1 million documents at query time (< 100 ms SLA). They have a fine-tuned SBERT bi-encoder and a fine-tuned cross-encoder. Which architecture should they use for the retrieval stage and why?
Answer: A
Why: Bi-encoder embeds each document once at index time; at query time it only needs one BERT forward pass (for the query) plus a matrix multiply over stored embeddings — achievable in < 10 ms even at 1M docs with FAISS. Cross-encoder requires running BERT over (query, doc) concatenation for each of the 1M documents — on a 12-layer BERT-base at 1k tokens/doc, that is orders of magnitude too slow for < 100 ms. The standard industrial pipeline: bi-encoder retrieves top-100, cross-encoder re-ranks those 100.
Step zero
Discussion prompt
Your turn — semantic search engine — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Milestone 1 — build a TinyBERT contextual encoder (d_model=32…
Answer:
d_model=32, nhead=4, num_layers=3, vocab of at least 15 tokens)Worked example
Project brief: build a minimal semantic search engine using BERT-style contextual embeddings + cosine retrieval. No pre-trained weights — build a toy encoder from scratch and verify it ranks the correct document first.
Milestone 1 — build a TinyBERT contextual encoder (d_model=32, nhead=4, num_layers=3, vocab of at least 15 tokens)
Why: Larger than the lesson toy to give more expressivity. Confirm model(x).shape == (1, T, 32) for a 5-token input.
Milestone 2 — define a 6-document corpus, encode each with mean pooling, and L2-normalize all vectors
Why: After L2 normalization, torch.norm(v, dim=-1) should print tensor([1., 1., 1., 1., 1., 1.]). This makes cosine similarity equivalent to dot product.
Milestone 3 — encode a query (semantically similar to doc 0), compute cosine scores against all 6 docs, and argsort descending
Why: Expected: doc 0 ranked 1st. If it is not, the encoder is not yet contextual enough — try increasing num_layers or seeding differently and re-running.
Milestone 4 — simulate SBERT fine-tuning: define 3 positive pairs and 3 negative pairs, run 100 SGD steps, and re-measure rank of doc 0
Why: After fine-tuning, positive-pair cosine should be > 0.8, negative-pair cosine should be < 0.0. If not, reduce lr or increase steps.
import torch, torch.nn as nn, torch.nn.functional as F
torch.manual_seed(0)
# --- Milestone 1: TinyBERT encoder ---
class TinyBERT(nn.Module):
def __init__(self, V=20, d=32, nhead=4, layers=3):
super().__init__()
self.emb = nn.Embedding(V, d)
self.pos = nn.Embedding(30, d)
enc = nn.TransformerEncoderLayer(d, nhead, 64, dropout=0.0, batch_first=True)
self.tr = nn.TransformerEncoder(enc, num_layers=layers)
def forward(self, x):
T = x.shape[1]
h = self.emb(x) + self.pos(torch.arange(T).unsqueeze(0))
return self.tr(h).mean(dim=1) # mean pool -> [B, d]
model = TinyBERT()
# --- Milestone 2: corpus ---
corpus_ids = torch.randint(0, 20, (6, 6)) # 6 docs, 6 tokens each
with torch.no_grad():
corpus_vecs = model(corpus_ids) # [6, 32]
corpus_norm = F.normalize(corpus_vecs, dim=-1)
print("Norms:", torch.norm(corpus_norm, dim=-1)) # all 1.0
# --- Milestone 3: retrieval ---
query_ids = corpus_ids[0].clone() # query = same token ids as doc 0
query_ids[3] ^= 1 # tiny perturbation
with torch.no_grad():
q_vec = F.normalize(model(query_ids.unsqueeze(0)), dim=-1) # [1,32]
scores = (q_vec @ corpus_norm.T).squeeze(0) # [6]
print("Ranked doc indices:", scores.argsort(descending=True).tolist())| doc rank | doc idx | expected score range | meaning |
|---|---|---|---|
| 1st | 0 | > 0.90 | query is near-copy of doc 0 |
| 2nd-3rd | 1-5 (any) | 0.0 – 0.6 | random documents |
| 4th-6th | remaining | < 0.0 – 0.3 | random, lower overlap |
Show it off: after fine-tuning with SBERT objective, print a table of cosine scores for all 6 docs under query, and report the Spearman ρ between your predicted scores and a ground-truth relevance vector [1, 0, 0, 0, 0, 0] using scipy.stats.spearmanr.
Comparison
Comparison matrix
From Your turn — semantic search engine: refill the meaning column from what you know. The rest of the table is as it appeared.
| doc rank | doc idx | expected score range | meaning |
|---|---|---|---|
| 1st | 0 | > 0.90 | query is near-copy of doc 0 |
| 2nd-3rd | 1-5 (any) | 0.0 – 0.6 | random documents |
| 4th-6th | remaining | < 0.0 – 0.3 | random, lower overlap |
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Static vs contextual — the core distinction · Sentence pooling — [CLS] vs mean vs max · SBERT — siamese fine-tuning for sentence similarity · Semantic search — embedding + cosine retrieval. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
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