1.5  Automatic Differentiation

Recall from Section 1.4 that derivatives drive all the optimization algorithms that we will use to train deep networks. While the calculations are straightforward, working them out by hand can be tedious and error-prone, and these issues only grow as our models become more complex.

Fortunately all modern deep learning frameworks take this work off our plates by offering automatic differentiation (often shortened to autograd). As we pass data through each successive function, the framework builds a computational graph that tracks how each value depends on others. To calculate derivatives, automatic differentiation works backwards through this graph applying the chain rule. The computational algorithm for applying the chain rule in this fashion is called backpropagation.

Autograd has a long history: the earliest references date back over half a century (Wengert 1964), and reverse mode, the variant that powers modern backpropagation, was developed by Linnainmaa (1970) . Section 24.3 recounts this history in full. Before exploring methods, let’s first master the autograd package.

import torch
import tensorflow as tf
from jax import numpy as jnp
from mxnet import autograd, np, npx
npx.set_np()

1.5.1 Mechanics

We begin with the basic workflow (attach a gradient, record a computation, run the backward pass), first on a scalar-valued function, then on vector-valued ones.

1.5.1.1 A Simple Function

Let’s assume 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, we assign x an initial value.

x = torch.arange(4.0)
x
tensor([0., 1., 2., 3.])
x = tf.range(4, dtype=tf.float32)
x
<tf.Tensor: shape=(4,), dtype=float32, numpy=array([0., 1., 2., 3.], dtype=float32)>
x = jnp.arange(4.0)
x
Array([0., 1., 2., 3.], dtype=float32)
x = np.arange(4.0)
x
array([0., 1., 2., 3.])

Before we calculate the gradient of \(y\) with respect to \(\mathbf{x}\), we need a place to store it. In general, we avoid allocating new memory every time we take a derivative because deep learning requires successively computing derivatives with respect to the same parameters a great many times, and we might risk running out of memory. Note that the gradient of a scalar-valued function with respect to a vector \(\mathbf{x}\) is vector-valued with the same shape as \(\mathbf{x}\).

Before we calculate the gradient of \(y\) with respect to \(\mathbf{x}\), we need a place to store it. In general, we avoid allocating new memory every time we take a derivative because deep learning requires successively computing derivatives with respect to the same parameters a great many times, and we might risk running out of memory. Note that the gradient of a scalar-valued function with respect to a vector \(\mathbf{x}\) is vector-valued with the same shape as \(\mathbf{x}\).

Before we calculate the gradient of \(y\) with respect to \(\mathbf{x}\), we need a place to store it. In general, we avoid allocating new memory every time we take a derivative because deep learning requires successively computing derivatives with respect to the same parameters a great many times, and we might risk running out of memory. Note that the gradient of a scalar-valued function with respect to a vector \(\mathbf{x}\) is vector-valued with the same shape as \(\mathbf{x}\).

# Can also create x = torch.arange(4.0, requires_grad=True)
x.requires_grad_(True)
x.grad  # The gradient is None by default
x = tf.Variable(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.])

We now calculate our function of x and assign the result to y.

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>
y = lambda x: 2 * jnp.dot(x, x)
y(x)
Array(28., dtype=float32)
# Our code is inside an `autograd.record` scope to build the computational
# graph
with autograd.record():
    y = 2 * np.dot(x, x)
y
array(28.)

Recording the operations gives the framework a computational graph, shown in Figure 1.5.1. Its nodes are operations and its edges carry intermediate values.

Figure 1.5.1: The computational graph for \(y = 2\mathbf{x}^\top\mathbf{x}\). The forward pass (black) flows from \(\mathbf{x}\) to \(y\); reverse-mode automatic differentiation walks the same graph backward (blue), multiplying the local derivative at each node via the chain rule to accumulate \(\partial y / \partial \mathbf{x} = 4\mathbf{x}\).

The forward pass evaluates the graph from \(\mathbf{x}\) to \(y\); to obtain the gradient, automatic differentiation then traverses it in reverse, multiplying the local derivatives along the way. We unpack computational graphs and backpropagation in full in Section 4.3, and the underlying mathematics (both modes of automatic differentiation and their costs) is developed in Section 24.3; for now we simply use the resulting gradients.

We can now take the gradient of y with respect to x by calling its backward method. Next, we can access the gradient via x’s grad attribute.

We can now calculate the gradient of y with respect to x by calling the gradient method.

We can now take the gradient of y with respect to x by passing through the grad transform.

We can now take the gradient of y with respect to x by calling its backward method. Next, we can access the gradient via x’s grad attribute.

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)>
from jax import grad
# The `grad` transform returns a Python function that
# computes the gradient of the original function
x_grad = grad(y)(x)
x_grad
Array([ 0.,  4.,  8., 12.], dtype=float32)
y.backward()
x.grad
array([ 0.,  4.,  8., 12.])

We already know that the gradient of the function \(y = 2\mathbf{x}^{\top}\mathbf{x}\) with respect to \(\mathbf{x}\) should be \(4\mathbf{x}\). We can now verify that the automatic gradient computation and the expected result are identical.

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])>
x_grad == 4 * x
Array([ True,  True,  True,  True], dtype=bool)
x.grad == 4 * x
array([ True,  True,  True,  True])

Now let’s calculate another function of x and take its gradient. Note that PyTorch does not automatically reset the gradient buffer when we record a new gradient. Instead, the new gradient is added to the already-stored gradient. This behavior comes in handy when we want to optimize the sum of multiple objective functions. To reset the gradient buffer, we can call x.grad.zero_() as follows:

Now let’s calculate another function of x and take its gradient. Note that TensorFlow resets the gradient buffer whenever we record a new gradient.

Now let’s calculate another function of x and take its gradient. Note that MXNet resets the gradient buffer whenever we record a new gradient.

x.grad.zero_()  # Reset the gradient
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)>
y = lambda x: x.sum()
grad(y)(x)
Array([1., 1., 1., 1.], dtype=float32)
with autograd.record():
    y = x.sum()
y.backward()
x.grad  # Overwritten by the newly calculated gradient
array([1., 1., 1., 1.])

1.5.1.2 Backward for Non-Scalar Variables

When y is a vector, the most natural representation of the derivative of y with respect to a vector x is a matrix called the Jacobian that contains the partial derivatives of each component of y with respect to each component of x. Likewise, for higher-order y and x, the result of differentiation could be an even higher-order tensor.

While Jacobians do show up in some advanced machine learning techniques, more commonly we want to sum up the gradients of each component of y with respect to the full vector x, yielding a vector of the same shape as x. For example, we often have a vector representing the value of our loss function calculated separately for each example among a batch of training examples. Here, we just want to sum up the gradients computed individually for each example.

Because deep learning frameworks vary in how they interpret gradients of non-scalar tensors, PyTorch takes some steps to avoid confusion. Invoking backward on a non-scalar elicits an error unless we tell PyTorch how to reduce the object to a scalar. More formally, we need to provide some vector \(\mathbf{v}\) such that backward will compute \(\mathbf{v}^\top \partial_{\mathbf{x}} \mathbf{y}\) rather than \(\partial_{\mathbf{x}} \mathbf{y}\). This argument is named gradient because the vector \(\mathbf{v}\) is the gradient arriving from the rest of a larger computation, as will become clear when we study backpropagation in Section 4.3. For a more detailed description, see the PyTorch documentation on the gradient argument to Tensor.backward.

By default, TensorFlow returns the gradient of the sum. In other words, rather than returning the Jacobian \(\partial_{\mathbf{x}} \mathbf{y}\), it returns the gradient of the sum \(\partial_{\mathbf{x}} \sum_i y_i\).

MXNet handles this problem by reducing all tensors to scalars by summing before computing a gradient. In other words, rather than returning the Jacobian \(\partial_{\mathbf{x}} \mathbf{y}\), it returns the gradient of the sum \(\partial_{\mathbf{x}} \sum_i y_i\).

x.grad.zero_()
y = x * x
y.backward(gradient=torch.ones(len(y)))  # Equivalently: 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)>
y = lambda x: x * x
# grad is only defined for scalar output functions
grad(lambda x: y(x).sum())(x)
Array([0., 2., 4., 6.], dtype=float32)
with autograd.record():
    y = x * x  
y.backward()
x.grad  # Equals the gradient of y = sum(x * x)
array([0., 2., 4., 6.])

1.5.2 Controlling the Graph

Sometimes the graph the framework records is not the graph we want to differentiate. The next two subsections show how to prune it by detaching individual intermediate results, and how to switch recording off altogether.

1.5.2.1 Detaching Computation

Sometimes, we wish to move some calculations outside of the recorded computational graph. For example, say that we use the input to create some auxiliary intermediate terms for which we do not want to compute a gradient. In this case, we need to detach the respective computational graph from the final result. The following toy example makes this clearer: suppose we have z = x * y and y = x * x but we want to focus on the direct influence of x on z rather than the influence conveyed via y. In this case, we can create a new variable u that takes the same value as y but whose provenance (how it was created) has been wiped out. Thus u has no ancestors in the graph and gradients do not flow through u to x. Now consider z = x * u. Because u is treated as a constant equal to \(x^2\), the gradient is \(\partial z / \partial x = u = x^2\). Had we not detached, so that z = x * (x * x) \(= x^3\), we would instead have obtained \(\partial z / \partial x = 3x^2\).

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 preserve the compute graph. 
# This lets us 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])>
import jax

y = lambda x: x * x
# jax.lax primitives are Python wrappers around XLA operations
u = jax.lax.stop_gradient(y(x))
z = lambda x: u * x

grad(lambda x: z(x).sum())(x) == u
Array([ True,  True,  True,  True], dtype=bool)
with autograd.record():
    y = x * x
    u = y.detach()
    z = u * x
z.backward()
x.grad == u
array([ True,  True,  True,  True])

Note that while this procedure detaches y’s ancestors from the graph leading to z, the computational graph leading to y persists and thus we can calculate the gradient of y with respect to x.

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])>
grad(lambda x: y(x).sum())(x) == 2 * x
Array([ True,  True,  True,  True], dtype=bool)
y.backward()
x.grad == 2 * x
array([ True,  True,  True,  True])

1.5.2.2 Turning Off Gradient Tracking

Recording operations for a backward pass costs time and memory. When we only need a value (at prediction time, or while updating parameters by hand), we can skip the bookkeeping entirely.

Wrap the computation in a torch.no_grad() block (or decorate a function with @torch.no_grad()). The result still shares data with x, but it is not attached to the graph, so no gradient can flow through it.

TensorFlow only records operations executed inside a tf.GradientTape (a tape is the recorded list of executed operations), so any computation outside a tape is already untracked. To pause recording within a tape, use tape.stop_recording().

JAX never records gradients implicitly: nothing is tracked until you apply a transform such as grad. There is simply nothing to switch off: you opt in to differentiation rather than out of it.

MXNet only builds a graph inside an autograd.record() block, so ordinary computation already carries no gradient bookkeeping. To suspend tracking within a recording scope, wrap the code in autograd.pause().

with torch.no_grad():
    y = 2 * torch.dot(x, x)
y.requires_grad  # False: y is detached from the graph
False
# Outside any GradientTape, nothing is recorded
y = 2 * tf.tensordot(x, x, axes=1)
y
<tf.Tensor: shape=(), dtype=float32, numpy=28.0>
# No graph is built unless we ask for it via a transform like `grad`
y = 2 * jnp.dot(x, x)
y
Array(28., dtype=float32)
with autograd.record():
    with autograd.pause():
        y = 2 * np.dot(x, x)  # not recorded: no gradient will flow through y
y
array(28.)

This untracked mode is the default for inference and evaluation throughout the rest of the book.

1.5.3 Beyond the Basics

Automatic differentiation is more general than the fixed formulas we have differentiated so far: it handles arbitrary control flow, derivatives of derivatives, and even lets us choose the direction in which the graph is traversed.

1.5.3.1 Gradients and Python Control Flow

So far we reviewed cases where the path from input to output was well defined via a function such as z = x * x * x. Programming offers us a lot more freedom in how we compute results. For instance, we can make them depend on auxiliary variables or condition choices on intermediate results. 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. To illustrate this, consider the following code snippet where 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 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
def f(a):
    b = a * 2
    while jnp.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 np.linalg.norm(b) < 1000:
        b = b * 2
    if b.sum() > 0:
        c = b
    else:
        c = 100 * b
    return c

Below, we call this function, passing in a random value, as input. Since the input is a random variable, we do not know what form the computational graph will take. However, whenever we execute f(a) on a specific input, we realize a specific computational graph and can subsequently run 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=4096.0>
from jax import random
a = random.normal(random.key(1), ())
d = f(a)
d_grad = grad(f)(a)
a = np.random.normal()
a.attach_grad()
with autograd.record():
    d = f(a)
d.backward()

Even though our function f is, for demonstration purposes, a bit contrived, its dependence on the input is quite simple: it is a linear function of the scalar a with piecewise defined scale. As such, f(a) / a is a constant and, moreover, it needs to match the gradient of f(a) with respect to a.

a.grad == d / a
tensor(True)
d_grad == d / a
<tf.Tensor: shape=(), dtype=bool, numpy=True>
d_grad == d / a
Array(True, dtype=bool)
a.grad == d / a
array(True)

Dynamic control flow is very common in deep learning. For instance, when processing text, the computational graph depends on the length of the input. In these cases, automatic differentiation is necessary for statistical modeling since it is impossible to compute the gradient a priori.

1.5.3.2 Higher-Order Derivatives

Occasionally we need the derivative of a derivative: the curvature of a function, or the Hessian–vector products (products of the matrix of second derivatives, the Hessian, with a vector) used by some optimizers. Autograd can differentiate through a gradient computation. Take \(f(x) = x^3\), for which \(f'(x) = 3x^2\) and \(f''(x) = 6x\).

Pass create_graph=True so the first gradient is itself a differentiable function of x, then differentiate again.

Nest two GradientTapes: the outer tape differentiates the gradient computed under the inner tape.

grad returns a function, so we just apply it twice.

Higher-order gradients in MXNet require explicitly retaining the graph of the first derivative. The mathematics is framework-agnostic and developed in Section 24.3.

x3 = torch.tensor(2.0, requires_grad=True)
dy = torch.autograd.grad(x3 ** 3, x3, create_graph=True)[0]  # 3x^2 = 12
d2y = torch.autograd.grad(dy, x3)[0]                          # 6x  = 12
dy, d2y
(tensor(12., grad_fn=<MulBackward0>), tensor(12.))
x3 = tf.Variable(2.0)
with tf.GradientTape() as outer:
    with tf.GradientTape() as inner:
        y = x3 ** 3
    dy = inner.gradient(y, x3)   # 3x^2 = 12
d2y = outer.gradient(dy, x3)     # 6x  = 12
dy, d2y
(<tf.Tensor: shape=(), dtype=float32, numpy=12.0>,
 <tf.Tensor: shape=(), dtype=float32, numpy=12.0>)
f = lambda x: x ** 3
dy = grad(f)(2.0)            # 3x^2 = 12
d2y = grad(grad(f))(2.0)    # 6x  = 12
dy, d2y
(Array(12., dtype=float32, weak_type=True),
 Array(12., dtype=float32, weak_type=True))

1.5.3.3 Forward versus Reverse Mode

Automatic differentiation can traverse the computational graph in either direction. Reverse mode, the variant we have used so far and the engine behind backpropagation, sweeps from the output back to the inputs, yielding the gradient of a single scalar with respect to all inputs in one pass. Forward mode sweeps the other way, propagating derivatives from one input outward to every output.

The choice is about cost, and a counting argument settles it. For a function with \(n\) inputs and \(m\) outputs, filling the full matrix of derivatives takes one reverse sweep per output (\(m\) sweeps) or one forward sweep per input (\(n\) sweeps), each sweep costing about as much as one evaluation of the function. A training loss is a single scalar (\(m = 1\)) depending on millions of parameters (\(n\) huge), so reverse mode delivers the entire gradient for the price of roughly one extra forward pass. Forward mode wins in the opposite regime (few inputs, many outputs), and it is also the tool of choice for Hessian–vector products and per-input sensitivities, as in the Julia package ForwardDiff.jl (Revels et al. 2016). The exercises explore this trade-off further, and Section 24.3 derives both modes and their costs in full.

1.5.4 Discussion

Automatic differentiation frees practitioners from deriving gradients by hand, and it makes it practical to train models for which pen and paper gradient computations would be prohibitively time consuming. While we use autograd to optimize models (in a statistical sense), the optimization of autograd libraries themselves (in a computational sense) is a rich subject that matters to framework designers. Here, tools from compilers and graph manipulation are used to compute results quickly and with modest memory.

For now, try to remember these basics: (i) attach gradients to those variables with respect to which we desire derivatives; (ii) record the computation of the target value; (iii) execute the backpropagation function; and (iv) access the resulting gradient.

1.5.5 Exercises

  1. Why is the second derivative much more expensive to compute than the first derivative?
  2. After running the function for backpropagation, immediately run it again and see what happens. Investigate.
  3. In the control flow example where we calculate the derivative of d with respect to a, what would happen if we changed the variable a to a random vector or a matrix? At this point, the result of the calculation f(a) is no longer a scalar. What happens to the result? How do we analyze this?
  4. Let \(f(x) = \sin(x)\). Plot the graph of \(f\) and of its derivative \(f'\). Do not exploit the fact that \(f'(x) = \cos(x)\) but rather use automatic differentiation to get the result.
  5. Let \(f(x) = ((\log x^2) \cdot \sin x) + x^{-1}\). Write out a dependency graph tracing results from \(x\) to \(f(x)\).
  6. Use the chain rule to compute the derivative \(\frac{df}{dx}\) of the aforementioned function, placing each term on the dependency graph that you constructed previously.
  7. Given the graph and the intermediate derivative results, you have a number of options when computing the gradient. Evaluate the result once sweeping from \(x\) to \(f\) (forward mode) and once from \(f\) tracing back to \(x\) (reverse mode).
  8. For the graph of exercise 5, count the operations that forward mode and reverse mode each perform, and the intermediate values each must store. How would the comparison change for a function with many inputs, or with many outputs?