# Q1c · Skip-gram and CBOW

Subject: Natural Language Processing

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/nlp/#tee/papers/legacy-f24/q1c

Canonical: https://deep-learning-study-room.vercel.app/read/nlp/tee/papers/legacy-f24--q1c/

Previous examination · complete worked answers

**Question. Explain Skip-gram and CBOW.** [individual allocation not printed; Q1 totals 20]

### Answer

**Word2Vec learns dense word embeddings by predicting words from nearby context.** Its two common architectures are CBOW and skip-gram. A vocabulary ID or one-hot vector is the input representation; the learned embedding table converts that identity into a short trainable vector.

| Architecture | Input | Prediction | Memory cue |
| --- | --- | --- | --- |
| CBOW | Surrounding context words | Centre/target word | Context → centre |
| Skip-gram | Centre/target word | Surrounding words | Centre → context |

CBOW: the + sat

→

Average context embeddings

→

Predict: cat

Skip-gram: cat

→

Embedding of cat

→

Predict: the and sat

For vocabulary size 

$$
V
$$

 and embedding dimension 

$$
d
$$

, an input one-hot row is 

$$
1\times V
$$

; the input embedding matrix is 

$$
V\times d
$$

; the hidden row is 

$$
1\times d
$$

; the output matrix is 

$$
d\times V
$$

; output scores are 

$$
1\times V
$$

. Multiplying a one-hot row by the embedding matrix simply selects one row.



$$
P(w_o\mid w_i)=\frac{\exp(u_{w_o}^{T}v_{w_i})}{\sum_{w\in V}\exp(u_w^{T}v_{w_i})}.
$$





$$
v_{w_i}
$$

 is the input word embedding, 

$$
u_w
$$

 an output-context embedding, and the dot product is a compatibility score. Softmax turns all scores into probabilities that sum to one. Training increases the probability of observed context words. Negative sampling or hierarchical softmax can reduce computation in larger vocabularies; use the algorithm specified by the question.

**Question. For the sentence “the cat sat on mat”, generate CBOW and skip-gram training data for centre word sat with a window of one token on each side.**

#### Given

Tokens indexed 1 to 5: the, cat, sat, on, mat. Centre = sat.

#### To find

The CBOW example and the skip-gram pairs.

#### Formula and why it applies

CBOW predicts the centre from its neighbours; skip-gram predicts each neighbour from the centre.

#### Step 1. Identify neighbours

Left neighbour = cat; right neighbour = on. “the” and “mat” are outside this window.

#### Step 2. CBOW example

Input [cat,on] → target sat.

#### Step 3. Skip-gram examples

(sat,cat) and (sat,on), with first item as input and second as predicted context.

**Answer.** CBOW: [cat,on]→sat. Skip-gram: sat→cat and sat→on.

**Question. The embeddings of context words cat and on are [1,2] and [3,0]. Compute the CBOW hidden vector using mean aggregation. Output vectors for [sat,ran] are [1,0] and [0,1]. Compute softmax probabilities.**

#### Given

Context embeddings [1,2], [3,0]; mean aggregation; two candidate output vectors.

#### To find

The hidden vector, logits and probabilities.

#### Formula and why it applies

Mean context vector 

$$
h=(v_1+v_2)/2
$$

. Logit 

$$
z_w=u_w^Th
$$

. Softmax 

$$
P(w)=e^{z_w}/\sum_j e^{z_j}
$$

.

#### Step 1. Add coordinates



$$
[1+3,\,2+0]=[4,2]
$$

.

#### Step 2. Divide by number of words



$$
h=[4/2,2/2]=[2,1]
$$

.

#### Step 3. Compute logits



$$
z_{sat}=1(2)+0(1)=2
$$

. 

$$
z_{ran}=0(2)+1(1)=1
$$

.

#### Step 4. Exponentials



$$
e^2\approx7.389056
$$

, 

$$
e^1\approx2.718282
$$

; sum 

$$
\approx10.107338
$$

.

#### Step 5. Divide



$$
P(sat)\approx0.731059
$$

, 

$$
P(ran)\approx0.268941
$$

.

**Answer.** h=[2,1]; sat is more probable (0.731059).

**Question. A skip-gram centre embedding is v=[1,−1]. Output embeddings for [eat,sleep,drive] are [1,1], [0,1], [1,0]. Calculate the predicted context probabilities.**

#### Given

One centre vector and three output vectors; full softmax over these three words.

#### To find

Three logits and three probabilities.

#### Formula and why it applies



$$
z_j=u_j^Tv
$$

, then 

$$
P(j)=e^{z_j}/\sum_k e^{z_k}
$$

.

#### Step 1. Eat dot product



$$
1(1)+1(-1)=0
$$

.

#### Step 2. Sleep and drive dot products



$$
z_{sleep}=0(1)+1(-1)=-1
$$

. 

$$
z_{drive}=1(1)+0(-1)=1
$$

.

#### Step 3. Exponentiate and sum



$$
e^0=1
$$

, 

$$
e^{-1}\approx0.367879
$$

, 

$$
e^1\approx2.718282
$$

. Sum 

$$
\approx4.086161
$$

.

#### Step 4. Normalize



$$
P(eat)\approx0.244728
$$

; 

$$
P(sleep)\approx0.090031
$$

; 

$$
P(drive)\approx0.665241
$$

.

**Answer.** drive has the highest probability in this toy learned model.

**Limitations:** a standard static Word2Vec vector merges a word's senses; it has limited representation of subword morphology and out-of-vocabulary words; its local prediction objective does not directly fit a full global co-occurrence table. Similarity is learned from data and is not a guarantee of factual or causal understanding.

CBOW = neighbours se missing centre. Skip-gram = centre se neighbours.