Inverted dropout

Problem

Dropout sets a random part of the activations to 0 during training. Inverted dropout also divides the kept activations by 1 - p, so the expected value of each activation does not change. At inference time the layer then does nothing. You write the forward pass for both modes and the backward pass.

Task

Write two functions.

1. dropout_forward(x, p, rng, training=True) returns (out, mask).
2. dropout_backward(grad_output, mask, p) returns the gradient of the loss with respect to x.

In training mode, build the mask with exactly one call:

True in the mask means kept. Each entry is dropped with probability p. The output is

In inference mode (training=False), return a copy of x and a mask that is True everywhere. Do not draw from rng in inference mode.

The backward pass uses the mask from the forward pass: \partial L / \partial x = \text{grad\_output} \cdot \text{mask} / (1 - p).

Input and output

• dropout_forward returns a tuple (out, mask). out is a float array and mask is a bool array. Both have the shape of x.
• dropout_backward returns a float array with the shape of x.

Constraints

• x has 1 to 4 axes and at most 40000 entries.
• 0 <= p < 1.
• Do not change the input arrays.
• Use numpy only.

Examples

pythonCopyx = np.array([[1.0, 2.0, 3.0, 4.0], [5.0, 6.0, 7.0, 8.0]])
out, mask = dropout_forward(x, 0.5, np.random.default_rng(0))
grad = dropout_backward(np.ones((2, 4)), mask, 0.5)

textCopyout
[[ 2.  0.  0.  0.]
 [10. 12. 14. 16.]]
mask
[[ True False False False]
 [ True  True  True  True]]
grad
[[2. 0. 0. 0.]
 [2. 2. 2. 2.]]

pythonCopyout, mask = dropout_forward(x, 0.5, np.random.default_rng(0), training=False)

textCopyout
[[1. 2. 3. 4.]
 [5. 6. 7. 8.]]
mask
[[ True  True  True  True]
 [ True  True  True  True]]