Why regularization controls overfitting in PyTorch - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add L2 regularization (weight decay) to the optimizer.

PyTorch

optimizer = torch.optim.SGD(model.parameters(), lr=0.01, weight_decay=[1])

Drag options to blanks, or click blank then click option'

A0.001

B0

C1

D-0.01

Attempts:

3 left

💡 Hint

Common Mistakes

Setting weight_decay to 0 disables regularization.

Using negative values causes errors.

L2 regularization is added via the weight_decay parameter in PyTorch optimizers. A small positive value like 0.001 helps control overfitting by penalizing large weights.

2fill in blank

medium

Complete the code to apply dropout with 50% probability during training.

PyTorch

dropout_layer = torch.nn.Dropout(p=[1])

Drag options to blanks, or click blank then click option'

A0.25

B0.5

C1.0

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1.0 disables all neurons, which breaks training.

Using 0 disables dropout, so no regularization.

Dropout randomly disables neurons during training to prevent overfitting. A dropout probability of 0.5 means half the neurons are dropped each time.

3fill in blank

hard

Fix the error in the training loop to correctly apply dropout only during training.

PyTorch

model.train()
for data, target in train_loader:
    optimizer.zero_grad()
    output = model(data)
    loss = criterion(output, target)
    loss.backward()
    optimizer.step()
model.[1]()

Drag options to blanks, or click blank then click option'

Aeval

Btrain

Cdropout

Ddisable

Attempts:

3 left

💡 Hint

Common Mistakes

Calling train() again keeps dropout active.

Using a non-existent method disables dropout incorrectly.

After training, calling model.eval() sets the model to evaluation mode, disabling dropout and batch norm effects for consistent predictions.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that stores squared values for even numbers only.

PyTorch

squares = {x: x[1]2 for x in range(1, 11) if x [2] 2 == 0}

Drag options to blanks, or click blank then click option'

A**

B%

C//

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using // instead of % for modulus.

Using + instead of exponentiation.

The ** operator squares the number. The % operator checks if the number is even by testing remainder 0 when divided by 2.

5fill in blank

hard

Fill all three blanks to create a filtered dictionary with uppercase keys and values greater than 5.

PyTorch

filtered = [1]: [2] for k, v in data.items() if v [3] 5}

Drag options to blanks, or click blank then click option'

Ak.upper()

Bv

C>

Dk.lower()

Attempts:

3 left

💡 Hint

Common Mistakes

Using k.lower() instead of uppercase.

Using < instead of > for filtering.

Keys are converted to uppercase with k.upper(). Values are kept as v. The filter keeps items where value is greater than 5 using >.

Practice

1. Why does regularization help prevent overfitting in a PyTorch model?

easy

A. It keeps the model weights small by adding a penalty to the loss.

B. It increases the size of the training dataset automatically.

C. It removes layers from the neural network during training.

D. It speeds up the training process by skipping some data points.

Solution

1. Step 1: Understand what overfitting means

   Overfitting happens when a model learns the training data too well, including noise, causing poor performance on new data.

2. Step 2: Explain how regularization affects model weights

   Regularization adds a penalty to large weights, encouraging smaller weights that generalize better to new data.

3. Final Answer:

   It keeps the model weights small by adding a penalty to the loss. -> Option A

4. Quick Check:

   Regularization = penalty on weights = less overfitting [OK]

Hint: Regularization adds penalty to weights to reduce overfitting [OK]

Common Mistakes:

• Thinking regularization increases data size
• Believing regularization removes layers
• Assuming regularization speeds training

2. Which PyTorch code snippet correctly applies L2 regularization (weight decay) during optimizer setup?

easy

A. optimizer = torch.optim.SGD(model.parameters(), lr=0.01, momentum=0.1)

B. optimizer = torch.optim.SGD(model.parameters(), lr=0.01, dropout=0.1)

C. optimizer = torch.optim.SGD(model.parameters(), lr=0.01, weight_decay=0.1)

D. optimizer = torch.optim.SGD(model.parameters(), lr=0.01, decay=0.1)

Solution

1. Step 1: Identify correct parameter for L2 regularization in PyTorch

   PyTorch uses weight_decay in optimizers to apply L2 regularization.

2. Step 2: Check the code options for correct usage

   Only optimizer = torch.optim.SGD(model.parameters(), lr=0.01, weight_decay=0.1) uses weight_decay=0.1, which is the correct way to add L2 regularization.

3. Final Answer:

   optimizer = torch.optim.SGD(model.parameters(), lr=0.01, weight_decay=0.1) -> Option C

4. Quick Check:

   weight_decay = L2 regularization in PyTorch [OK]

Hint: Use weight_decay param for L2 regularization in PyTorch optimizers [OK]

Common Mistakes:

• Using dropout parameter in optimizer
• Confusing momentum with regularization
• Using decay instead of weight_decay

3. Consider this PyTorch training loop snippet with L2 regularization applied:

optimizer = torch.optim.Adam(model.parameters(), lr=0.001, weight_decay=0.01)
for data, target in dataloader:
    optimizer.zero_grad()
    output = model(data)
    loss = loss_fn(output, target)
    loss.backward()
    optimizer.step()

What effect does the weight_decay=0.01 have during training?

medium

A. It adds a penalty to large weights, helping reduce overfitting.

B. It increases the learning rate by 0.01 each step.

C. It drops 1% of neurons randomly during training.

D. It stops training early when loss is below 0.01.

Solution

1. Step 1: Understand weight_decay in optimizer

   The weight_decay parameter adds L2 regularization, penalizing large weights during training.

2. Step 2: Identify the effect on training

   This penalty helps the model avoid overfitting by keeping weights smaller and more generalizable.

3. Final Answer:

   It adds a penalty to large weights, helping reduce overfitting. -> Option A

4. Quick Check:

   weight_decay = L2 penalty = less overfitting [OK]

Hint: weight_decay adds penalty to weights, not learning rate or dropout [OK]

Common Mistakes:

• Confusing weight_decay with learning rate changes
• Thinking weight_decay is dropout
• Assuming weight_decay controls early stopping

4. You have this PyTorch code snippet intended to apply L2 regularization:

optimizer = torch.optim.SGD(model.parameters(), lr=0.01)
for data, target in dataloader:
    optimizer.zero_grad()
    output = model(data)
    loss = loss_fn(output, target) + 0.01 * torch.sum(model.parameters())
    loss.backward()
    optimizer.step()

What is wrong with this code regarding regularization?

medium

A. It uses SGD optimizer which does not support regularization.

B. It forgets to call optimizer.zero_grad() before backward.

C. It applies regularization after optimizer.step(), so no effect.

D. It incorrectly sums parameters instead of their squares for L2 penalty.

Solution

1. Step 1: Check how L2 regularization is computed

   L2 regularization requires summing the squares of parameters, not just their values.

2. Step 2: Analyze the code's regularization term

   The code sums parameters directly with torch.sum(model.parameters()), which is incorrect for L2 penalty.

3. Final Answer:

   It incorrectly sums parameters instead of their squares for L2 penalty. -> Option D

4. Quick Check:

   L2 penalty = sum of squares, not sum of values [OK]

Hint: L2 regularization sums squares of weights, not weights themselves [OK]

Common Mistakes:

• Summing parameters instead of squared parameters
• Thinking SGD can't use regularization
• Misplacing optimizer.zero_grad() call

5. You train two PyTorch models on the same dataset: Model A uses no regularization, Model B uses L2 regularization with weight_decay=0.05. After training, Model A has training accuracy 98% but test accuracy 70%, while Model B has training accuracy 90% and test accuracy 85%. What explains this difference?

hard

A. Model A's higher training accuracy means it will always perform better on test data.

B. Model B's regularization reduced overfitting by keeping weights smaller, improving test accuracy.

C. Model B used a larger learning rate, causing better generalization.

D. Model A trained longer, so it has better test accuracy.

Solution

1. Step 1: Compare training and test accuracies

   Model A fits training data very well but performs poorly on test data, indicating overfitting.

2. Step 2: Understand effect of L2 regularization on Model B

   Model B has lower training accuracy but better test accuracy because regularization keeps weights smaller, improving generalization.

3. Final Answer:

   Model B's regularization reduced overfitting by keeping weights smaller, improving test accuracy. -> Option B

4. Quick Check:

   Regularization = smaller weights = better test accuracy [OK]

Hint: Better test accuracy with regularization means less overfitting [OK]

Common Mistakes:

• Assuming higher training accuracy means better test accuracy
• Confusing learning rate with regularization effect
• Ignoring the role of weight size in generalization