# Automatic Differentiation

## Introduction

Derivatives show up everywhere in data science, and most notably in optimization, because of gradient descent (and its variants). Writing derivatives by hand is error-prone and annoying, especially for complicated expressions. Automatic differentiation (“autodiff”) is an algorithm that solves all of these problems: evaluating gradients is usually only as expensive as evaluating the functions themselves, and since it’s the computer doing it, you’re never going to miss a term again. Autodiff is also the backbone of libraries like PyTorch and TensorFlow.

Automatic differentiation works by traversing a data structure representing the value we want to compute. People call this “expression graph”, “expression tree”, or an “computation graph”: they’re the same thing. Usually, we ask Python to compute the value of an expression directly:

>>> 3 + (5 * 6)
33


Instead, we can build a data structure representing that, and ask Python to derive the value:

class Variable:
def __init__(self, value):
self.v = value
def evaluate(self):
return self.v

class Sum:
def __init__(self, left, right):
self.l = left
self.r = right
def evaluate(self):
return self.l.evaluate() + self.r.evaluate()

class Mult:
def __init__(self, left, right):
self.l = left
self.r = right
def evaluate(self):
return self.l.evaluate() * self.r.evaluate()

>>> x = Sum(Variable(3), Mult(Variable(5), Variable(6)))
>>> x.evaluate()
33


(This is, in fact, pretty much how Python actually evaluates expressions internally.)

## Given a computation graph, derivatives are local

The first important bit of intuition on how derivatives can be computed comes from the way the chain rule works. You’ve probably seen the chain rule written in this way:

$\frac{df}{dx} = \frac{df}{du} \times \frac{du}{dx}$

This version of the chain rule says that if $f$ depends on $x$ only through $u$, and $u$ depends on $x$, then the derivative $df/dx$ is the product of the “immediate derivative” from $f$ to $u$, $df/du$, and the “immediate derivative” from $u$ to $x$, $du/dx$.

This lets us compute the derivative of $f = \sin (\cos x)$ knowing only the base case rules that if $u = \sin x$, then $du/dx = \cos x$, and if $v = \cos x$, then $dv/dx = -\cos{x}$:

$df/dx = \cos (\cos x) \times - \sin (x)$

# Forward-mode autodiff

If we organize our computation in terms of nodes that know how to evaluate their own derivatives, that computation will only need the derivatives of the immediate neighbors. This is because, as we saw above, the chain rule means that “dependencies propagate one node at a time”.

Imagine the computation of the example above, $3 + (5 \times 6)$, except that we create named variables: $a + bc$, $a = 3, b = 5, c = 6$. Say we wanted to compute the derivative of that expression with respect to $b$: $d { a + bc }/db = c = 6$. Can we come up with an algorithmic way to do this? The easiest way is called “forward-mode automatic differentiation”.

## Just carry the derivatives with the values

If you inspect the source code for the Python evaluation library we wrote above, the evaluation of each node needs only to know the values of its immediate neighbors. Forward-mode autodiff simply takes this idea one step further, and carries the derivatives together with the values:

class Variable:
def __init__(self, value, derivative):
self.v = value
self.d = derivative
def evaluate(self):
return (self.v, self.d)

class Sum:
def __init__(self, left, right):
self.l = left
self.r = right
def evaluate(self):
(left_v, left_d) = self.l.evaluate()
(right_v, right_d) = self.r.evaluate()
return (left_v + right_v, left_d + right_d)

class Mult:
def __init__(self, left, right):
self.l = left
self.r = right
def evaluate(self):
(left_v, left_d) = self.l.evaluate()
(right_v, right_d) = self.r.evaluate()
return (left_v * right_v,
left_v * right_d + left_d * right_v) # product rule

>>> a = Variable(3, 0)
>>> b = Variable(5, 1)
>>> c = Variable(6, 0)
>>> x = Sum(a, Mult(b, c))
>>> x.evaluate()
(33, 6)


There is a bit of a trick going on, which is how we encode which variable we are taking the derivative over. We simply ask variables to store the values of their derivatives with respect to whatever variable we’re interested in. In our example, we want to take a derivative over $b$, and so $da/db = 0, db/db = 1, dc/db = 0$.

Of course, you have to teach your library about other expressions and derivative rules as well:

class Sin:
def __init__(self, v):
self.v = v
def evaluate(self):
(v, d) = self.v.evaluate()
return (math.sin(v), math.cos(v) * d)

class Cos:
def __init__(self, v):
self.v = v
def evaluate(self):
(v, d) = self.v.evaluate()
return (math.cos(v), -math.sin(v) * d)

# class Exp, Log, etc.


This is called forward-mode autodiff because the derivatives “flow forward” with the computation of values. You can see the algorithm in action here, with uniformly random values between 0 and 1 assigned to the variables:

Expression:

This works well for simple expressions. However, consider the case of computing the gradient of a function: the vector of partial derivatives with respect to all variables. If we want to use forward-mode autodiff to compute the gradient, we end up having to evaluate the same expression over and over again, just changing the values of the d fields of the variables a, b, and c. That’s quite inefficient, so let’s try to identify where the inefficiency comes from.

Pay attention to the partial derivatives we are computing with forward-mode autodiff. For every invocation, we choose one variable to compute the derivative over. We are keeping the “denominator” of the derivative fixed, and varying the “numerator” over all possible computation nodes. But a gradient does the opposite. It keeps the “numerator” fixed, and varies the “denominator” over all variables of the function: $\nabla f(x, y, z, w) = [ \partial f / \partial x, \partial f / \partial y, \partial f / \partial z, \partial f / \partial w ]$. So if we use forward-mode autodiff to compute gradients, all the computations of the derivatives of the intermediate nodes are wasted.

There is a better way to do this! It’s called reverse-mode autodiff.

# Reverse-mode Autodiff

Let’s start with that same original expression, $f = a + bc$. First, you should convince yourself that the gradient of $f$, $\nabla f = [\partial f/\partial a, \partial f/\partial b, \partial f/\partial c]$ is equal to $[1, c, b]$, and then break $f$ into simple expressions:

$\begin{eqnarray*}d &=& b \times c\\ f &=& a + d\end{eqnarray*}$

If we wanted to compute the gradient of $f$, we could replace this with a slightly more general thing: let’s think about computing “all possible denominators of the fraction $\partial f/\partial *$”. At the very start, we only know how to compute one of these: $\partial f/\partial f = 1$. It’s a bit weird to think about taking a derivative of a function with respect to itself. But if you think of it as a variable that has some constraints, then it makes sense: as long as you find a way to change the value of the variable, the rate of change of $f$ with respect to $f$ itself is just $1$.

Ok, so we found $\partial f/\partial f = 1$. How does that help us at all? Well, we can use that to percolate the derivative information “backwards” up the expression tree.

Say we want to find the derivative of $f$ with respect to $d$. Because $d$ (and $a$) define $f$ directly, this case is “easy”: we take the derivative of the whole expression $f = a + d$ with respect to $d$, and get $\partial f/\partial d = \partial a / \partial d + \partial d / \partial d$, or $\partial f / \partial d = \partial d / \partial d = 1$. The same trick works for $a$ (and so $\partial f / \partial a = 1$). However, for $b$ and $c$ we need something different.

Let’s start with $b$. We want $\partial f / \partial b$, so somewhere along the way we will need to take a derivative over $b$. Since the only simple expression we have involving $b$ is $d = b \times c$, we might as well take a derivative over $b$, and get $\partial d / \partial b = c$. And now, just like in the forward-mode case, we can use the chain rule: $\partial f / \partial b = \partial f / \partial d \times \partial d / \partial b$. Since we know from before that $\partial f / \partial d = 1$, we get $\partial f / \partial b = c$. The same argument works for a derivative with respect to $c$, and we get that the gradient will be $[1, c, b]$, as expected.

In forward-mode autodiff, the chain rule lets us “push derivatives down the tree along with the values”; in reverse-mode autodiff, the chain rule lets us “push derivatives up the tree”: notice how $\partial f / \partial b$ is defined in terms of $\partial f / \partial d$, a derivative that is “towards the bottom”. This really is working in the “reverse” direction of evaluation (and that’s why this is called “reverse-mode” autodiff), but the chain rule is symmetric, so it all works out.

I’ve described to you in words how reverse-mode autodiff works, but haven’t really given you an algorithm that’s easy to implement.

## “Pushing derivatives up”

There’s a trick that highly simplifies the implementation of reverse-mode autodiff. The main issue with the above explanation is that the base case needs to be handled differently for every kind of expression, and there can be very many different expression kinds to handle: sums, products, sines, cosines, etc.

It’s much easier to write the code if every base case behaves the same way. The trick to make everything uniform is simply to posit that the “reverse-mode pass” always starts at a special variable (let’s call it $g$), and that $g$ is defined to be equal to the expression you care about (in our case $g = f$).

We’re almost ready to write the algorithm. But we need two reminders. First, in forward-mode autodiff, the derivative values of a node $v$ stored $\partial v/\partial x$, where $x$ was the chosen variable of interest. In reverse-mode autodiff, the derivative values of a node $v$ will store $\partial f/\partial v$, where $f$ is the overall expression. Second, the version of the chain rule which we need for reverse-mode autodiff is a little more general than what we’ve seen before. Specifically, we need to handle a situation when $f$ depends on $x$ through multiple “intermediate paths”. In that case, the chain rule sums over all of terms of each independent path:

$\frac{\partial f}{\partial x} = \sum_v \frac{\partial f}{\partial v} \frac{\partial v}{\partial x}$

With that in mind, here’s the algorithm for reverse-mode autodiff:

1. Evaluate the expression tree for the values as you would do in forward-mode autodiff, but without computing derivatives. This is the “forward pass”.

2. Initialize the derivative values of all nodes to 0, except $g$, which is initialized to 1.

3. Traverse the computation graph in some topological order, from the node of the final expression (the “root” if it were a tree) up. The invariant we seek here is that by traversing the graph in this way, for every node we visit, the derivative of $g$ with respect to that node will have been fully computed by the time we visit it.

4. When we visit a node, we “push derivatives up” the tree, adding the appropriate derivative values to the derivative of the node’s parents.

5. When this “backward pass” is finished, each node’s derivative values will hold the value of the derivative of $g$ with respect to the node itself.

Let’s make things more concrete. Consider, for example, the case where we visit the node $d = c \times b$. By our invariant, at that point we will have fully computed $\partial f / \partial d$. By taking the two possible derivatives, we see that $\partial d / \partial b = c$, and $\partial d / \partial c = b$. The chain rule says $\partial f / \partial b = \sum_v (\partial f / \partial v) (\partial v / \partial b)$. Setting $v = b$ gives us one term of the sum, so we need to increment the currently-stored value of $\partial f / \partial b$ by $(\partial f / \partial d) \times c$. We increment (as opposed to storing) because there could be other nodes in the graph that also use $b$, and those will eventually have to add their own contribution to the chain rule sum.

As a result, reverse-mode autodiff “pushes derivatives” up the tree based on specific rules for each kind of expression, and you can derive each one of them by taking derivatives with respect to the possible parameters. Here are a few examples of the behavior of the algorithm when visiting nodes of different types, always assuming that $g$ is the variable representing the entire expression:

• $a = b + c$ adds $\partial g / \partial a$ to both $\partial g / \partial b$ and $\partial g / \partial c$,
• $a = b - c$ adds $\partial g / \partial a$ to $\partial g / \partial b$, and subtracts $\partial g / \partial a$ from $\partial g / \partial c$,
• $a = \sin b$ adds $(\partial g / \partial a) \times \cos b$ to $\partial g / \partial b$,
• $a = \log b$ adds $(\partial g / \partial a) \times (1 / b)$ to $\partial g / \partial b$, etc.

You can see this new algorithm in action below, again with uniformly random values between 0 and 1 assigned to the variables. First, the algorithm performs a forward pass to compute the expression values, and then the algorithm performs a backward pass, to propagate the derivatives up the tree back to the variables. Note that the derivatives with respect to the variable nodes here are represented in text field separate from the variable nodes. We do this so that different branches can refer to the same variable, and the derivative with respect to that variable can accumulate correctly.

Expression:

## Backpropagation?

In neural networks, “reverse-mode autodiff” is often referred to as “error back-propagation”, from the paper that made it popular for neural networks. The general algorithm for reverse-mode automatic differentiation was known before the specific use-case was published for neural networks, and was developed in a 1970 master’s thesis (!).

# Why does this matter?

Although forward-mode autodiff is more straightforward than the reverse-mode varient, the performance gains of using reverse-mode autodiff in the case of gradient calculations are big. Consider, for example, the notorious VGG19 neural network, which in 2014 set the state of the art for classification accuracy in a 1000-class image classification problem. It has 144 million parameters, so evaluating the gradient using forward-mode autodiff would have taken 144 million times as long, turning a problem which in 2014 took about 3 weeks to train into a problem that would have taken over 8 million years!

# Acknowledgments

Before I built this (and also during!), I extensively referenced Rufflewind’s AD tutorial. I still highly recommend it.

Thanks to Matthew Conlen for spotting mistakes on this.