Hidden state management in PyTorch - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the hidden state for an RNN with batch size 1 and hidden size 5.

PyTorch

hidden = torch.zeros(1, 1, [1])

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A7

B10

C3

D5

Attempts:

3 left

💡 Hint

Common Mistakes

Using batch size or sequence length instead of hidden size.

The hidden state size must match the RNN's hidden size, which is 5 here.

2fill in blank

medium

Complete the code to detach the hidden state from the computation graph to avoid backpropagating through entire history.

PyTorch

hidden = hidden.[1]()

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Adetach

Bzero_

Crequires_grad_

Dclone

Attempts:

3 left

💡 Hint

Common Mistakes

Using clone() does not detach gradients.

Using zero_() resets values but does not detach.

Using detach() breaks the computation graph, preventing gradients from flowing backward beyond this point.

3fill in blank

hard

Fix the error in the code to correctly initialize hidden state for a 2-layer LSTM with batch size 3 and hidden size 4.

PyTorch

hidden = (torch.zeros([1], 3, 4), torch.zeros([1], 3, 4))

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A2

B3

C1

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using batch size or hidden size as the first dimension.

The first dimension is the number of layers, which is 2 here for a 2-layer LSTM.

4fill in blank

hard

Fill both blanks to correctly update hidden state inside a training loop for an RNN, detaching it to avoid backpropagating through entire history.

PyTorch

for input in inputs:
    output, hidden = rnn(input, [1])
    hidden = hidden.[2]()

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Ahidden

Binput

Cdetach

Dclone

Attempts:

3 left

💡 Hint

Common Mistakes

Passing input instead of hidden as RNN state.

Using clone() instead of detach().

The hidden state from the previous step is passed as input to the RNN, then detached to stop gradient flow beyond this step.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that stores the length of each word only if the length is greater than 3.

PyTorch

lengths = {word: [1] for word in words if len(word) [2] 3 and word.[3]('a')}

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Alen(word)

B>

Cstartswith

Dendswith

Attempts:

3 left

💡 Hint

Common Mistakes

Using endswith instead of startswith.

Using '<' instead of '>'.

This comprehension stores the length of each word if its length is greater than 3 and it starts with 'a'.