2.5. Automatic Differentiation¶ Open the notebook in SageMaker Studio Lab
As we have explained in Section 2.4, differentiation is a crucial step in nearly all deep learning optimization algorithms. While the calculations for taking these derivatives are straightforward, requiring only some basic calculus, for complex models, working out the updates by hand can be a pain (and often error-prone).
Deep learning frameworks expedite this work by automatically calculating derivatives, i.e., automatic differentiation. In practice, based on our designed model the system builds a computational graph, tracking which data combined through which operations to produce the output. Automatic differentiation enables the system to subsequently backpropagate gradients. Here, backpropagate simply means to trace through the computational graph, filling in the partial derivatives with respect to each parameter.
2.5.1. A Simple Example¶
As a toy example, say that we are interested in differentiating the
function \(y = 2\mathbf{x}^{\top}\mathbf{x}\) with respect to the
column vector \(\mathbf{x}\). To start, let us create the variable
x and assign it an initial value.
from mxnet import autograd, np, npx
npx.set_np()
x = np.arange(4.0)
x
array([0., 1., 2., 3.])
import torch
x = torch.arange(4.0)
x
tensor([0., 1., 2., 3.])
import tensorflow as tf
x = tf.range(4, dtype=tf.float32)
x
<tf.Tensor: shape=(4,), dtype=float32, numpy=array([0., 1., 2., 3.], dtype=float32)>
Before we even calculate the gradient of \(y\) with respect to \(\mathbf{x}\), we will need a place to store it. It is important that we do not allocate new memory every time we take a derivative with respect to a parameter because we will often update the same parameters thousands or millions of times and could quickly run out of memory. Note that a gradient of a scalar-valued function with respect to a vector \(\mathbf{x}\) is itself vector-valued and has the same shape as \(\mathbf{x}\).
# We allocate memory for a tensor's gradient by invoking `attach_grad`
x.attach_grad()
# After we calculate a gradient taken with respect to `x`, we will be able to
# access it via the `grad` attribute, whose values are initialized with 0s
x.grad
array([0., 0., 0., 0.])
x.requires_grad_(True) # Same as `x = torch.arange(4.0, requires_grad=True)`
x.grad # The default value is None
x = tf.Variable(x)
Now let us calculate \(y\).
# Place our code inside an `autograd.record` scope to build the computational
# graph
with autograd.record():
y = 2 * np.dot(x, x)
y
array(28.)
y = 2 * torch.dot(x, x)
y
tensor(28., grad_fn=<MulBackward0>)
# Record all computations onto a tape
with tf.GradientTape() as t:
y = 2 * tf.tensordot(x, x, axes=1)
y
<tf.Tensor: shape=(), dtype=float32, numpy=28.0>
Since x is a vector of length 4, an dot product of x and x
is performed, yielding the scalar output that we assign to y. Next,
we can automatically calculate the gradient of y with respect to
each component of x by calling the function for backpropagation and
printing the gradient.
y.backward()
x.grad
[19:54:27] src/base.cc:49: GPU context requested, but no GPUs found.
array([ 0., 4., 8., 12.])
y.backward()
x.grad
tensor([ 0., 4., 8., 12.])
x_grad = t.gradient(y, x)
x_grad
<tf.Tensor: shape=(4,), dtype=float32, numpy=array([ 0., 4., 8., 12.], dtype=float32)>
The gradient of the function \(y = 2\mathbf{x}^{\top}\mathbf{x}\) with respect to \(\mathbf{x}\) should be \(4\mathbf{x}\). Let us quickly verify that our desired gradient was calculated correctly.
x.grad == 4 * x
array([ True, True, True, True])
x.grad == 4 * x
tensor([True, True, True, True])
x_grad == 4 * x
<tf.Tensor: shape=(4,), dtype=bool, numpy=array([ True, True, True, True])>
Now let us calculate another function of x.
with autograd.record():
y = x.sum()
y.backward()
x.grad # Overwritten by the newly calculated gradient
array([1., 1., 1., 1.])
# PyTorch accumulates the gradient in default, we need to clear the previous
# values
x.grad.zero_()
y = x.sum()
y.backward()
x.grad
tensor([1., 1., 1., 1.])
with tf.GradientTape() as t:
y = tf.reduce_sum(x)
t.gradient(y, x) # Overwritten by the newly calculated gradient
<tf.Tensor: shape=(4,), dtype=float32, numpy=array([1., 1., 1., 1.], dtype=float32)>
2.5.2. Backward for Non-Scalar Variables¶
Technically, when y is not a scalar, the most natural interpretation
of the differentiation of a vector y with respect to a vector x
is a matrix. For higher-order and higher-dimensional y and x,
the differentiation result could be a high-order tensor.
However, while these more exotic objects do show up in advanced machine learning (including in deep learning), more often when we are calling backward on a vector, we are trying to calculate the derivatives of the loss functions for each constituent of a batch of training examples. Here, our intent is not to calculate the differentiation matrix but rather the sum of the partial derivatives computed individually for each example in the batch.
# When we invoke `backward` on a vector-valued variable `y` (function of `x`),
# a new scalar variable is created by summing the elements in `y`. Then the
# gradient of that scalar variable with respect to `x` is computed
with autograd.record():
y = x * x # `y` is a vector
y.backward()
x.grad # Equals to y = sum(x * x)
array([0., 2., 4., 6.])
# Invoking `backward` on a non-scalar requires passing in a `gradient` argument
# which specifies the gradient of the differentiated function w.r.t `self`.
# In our case, we simply want to sum the partial derivatives, so passing
# in a gradient of ones is appropriate
x.grad.zero_()
y = x * x
# y.backward(torch.ones(len(x))) equivalent to the below
y.sum().backward()
x.grad
tensor([0., 2., 4., 6.])
with tf.GradientTape() as t:
y = x * x
t.gradient(y, x) # Same as `y = tf.reduce_sum(x * x)`
<tf.Tensor: shape=(4,), dtype=float32, numpy=array([0., 2., 4., 6.], dtype=float32)>
2.5.3. Detaching Computation¶
Sometimes, we wish to move some calculations outside of the recorded
computational graph. For example, say that y was calculated as a
function of x, and that subsequently z was calculated as a
function of both y and x. Now, imagine that we wanted to
calculate the gradient of z with respect to x, but wanted for
some reason to treat y as a constant, and only take into account the
role that x played after y was calculated.
Here, we can detach y to return a new variable u that has the
same value as y but discards any information about how y was
computed in the computational graph. In other words, the gradient will
not flow backwards through u to x. Thus, the following
backpropagation function computes the partial derivative of
z = u * x with respect to x while treating u as a constant,
instead of the partial derivative of z = x * x * x with respect to
x.
with autograd.record():
y = x * x
u = y.detach()
z = u * x
z.backward()
x.grad == u
array([ True, True, True, True])
x.grad.zero_()
y = x * x
u = y.detach()
z = u * x
z.sum().backward()
x.grad == u
tensor([True, True, True, True])
# Set `persistent=True` to run `t.gradient` more than once
with tf.GradientTape(persistent=True) as t:
y = x * x
u = tf.stop_gradient(y)
z = u * x
x_grad = t.gradient(z, x)
x_grad == u
<tf.Tensor: shape=(4,), dtype=bool, numpy=array([ True, True, True, True])>
Since the computation of y was recorded, we can subsequently invoke
backpropagation on y to get the derivative of y = x * x with
respect to x, which is 2 * x.
y.backward()
x.grad == 2 * x
array([ True, True, True, True])
x.grad.zero_()
y.sum().backward()
x.grad == 2 * x
tensor([True, True, True, True])
t.gradient(y, x) == 2 * x
<tf.Tensor: shape=(4,), dtype=bool, numpy=array([ True, True, True, True])>
2.5.4. Computing the Gradient of Python Control Flow¶
One benefit of using automatic differentiation is that even if building
the computational graph of a function required passing through a maze of
Python control flow (e.g., conditionals, loops, and arbitrary function
calls), we can still calculate the gradient of the resulting variable.
In the following snippet, note that the number of iterations of the
while loop and the evaluation of the if statement both depend on
the value of the input a.
def f(a):
b = a * 2
while np.linalg.norm(b) < 1000:
b = b * 2
if b.sum() > 0:
c = b
else:
c = 100 * b
return c
def f(a):
b = a * 2
while b.norm() < 1000:
b = b * 2
if b.sum() > 0:
c = b
else:
c = 100 * b
return c
def f(a):
b = a * 2
while tf.norm(b) < 1000:
b = b * 2
if tf.reduce_sum(b) > 0:
c = b
else:
c = 100 * b
return c
Let us compute the gradient.
a = np.random.normal()
a.attach_grad()
with autograd.record():
d = f(a)
d.backward()
a = torch.randn(size=(), requires_grad=True)
d = f(a)
d.backward()
a = tf.Variable(tf.random.normal(shape=()))
with tf.GradientTape() as t:
d = f(a)
d_grad = t.gradient(d, a)
d_grad
<tf.Tensor: shape=(), dtype=float32, numpy=102400.0>
We can now analyze the f function defined above. Note that it is
piecewise linear in its input a. In other words, for any a there
exists some constant scalar k such that f(a) = k * a, where the
value of k depends on the input a. Consequently d / a allows
us to verify that the gradient is correct.
a.grad == d / a
array(True)
a.grad == d / a
tensor(True)
d_grad == d / a
<tf.Tensor: shape=(), dtype=bool, numpy=True>
2.5.5. Summary¶
Deep learning frameworks can automate the calculation of derivatives. To use it, we first attach gradients to those variables with respect to which we desire partial derivatives. We then record the computation of our target value, execute its function for backpropagation, and access the resulting gradient.
2.5.6. Exercises¶
Why is the second derivative much more expensive to compute than the first derivative?
After running the function for backpropagation, immediately run it again and see what happens.
In the control flow example where we calculate the derivative of
dwith respect toa, what would happen if we changed the variableato a random vector or matrix. At this point, the result of the calculationf(a)is no longer a scalar. What happens to the result? How do we analyze this?Redesign an example of finding the gradient of the control flow. Run and analyze the result.
Let \(f(x) = \sin(x)\). Plot \(f(x)\) and \(\frac{df(x)}{dx}\), where the latter is computed without exploiting that \(f'(x) = \cos(x)\).