In [15]:
pyplot.plot(range(num_iters+1), losses_gd);
pyplot.xlabel("iteration of gradient descent");
pyplot.ylabel("loss");
pyplot.title("Convergence of Gradient Descent on CPUsmall Linear Regression");
import numpy
import scipy
import matplotlib
import time
def hide_code_in_slideshow():
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var cell = p;
cell.find(".input").addClass("hide-in-slideshow")
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display.display_html(html, raw=True)
Suppose we have a dataset $\mathcal{D} = \{ (x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n) \}$, where $x_i \in \mathcal{X}$ is an example and $y_i \in \mathcal{Y}$ is a label. Let $h: \mathcal{X} \rightarrow \mathcal{Y}$ be a hypothesized model (mapping from examples to labels) we are trying to evaluate. Let $L: \mathcal{Y} \times \mathcal{Y} \rightarrow \mathbb{R}_+$ be a non-negative loss function which measures how different two labels are.
The empirical risk is $$ R(h) = \frac{1}{n} \sum_{i=1}^n L(h(x_i), y_i). $$ Most notions of error or accuracy in machine learning can be captured with an empirical risk. A simple example: measure the error rate on the dataset with the 0-1 Loss Function $$ L(\hat y, y) = 0 \text{ if } \hat y = y \text{ and } 1 \text{ if } \hat y \ne y. $$
Loss function:
Regularizers:
We need to compute the empirical risk a lot during training, both during validation (and hyperparameter optimization) and testing, so it's nice if we can do it fast.
Question: how does computing the empirical risk scale? Three things affect the cost of computation.
We don't just want to compute the empirical risk: we also want to minimize it. To do so, we parameterize the hypotheses using some parameters $w \in \R^d$. That is, we assign each hypothesis a $d$-dimensional vector of parameters and vice versa and solve the optimization problem \begin{align*} \mbox{minimize: }& R(h_w) = \frac{1}{n} \sum_{i=1}^n L(h_w(x_i), y_i) = f(w) = \frac{1}{n} \sum_{x \in \mathcal{D}} f(w; x) \\ \mbox{ over }& w \in \R^d \end{align*} where $h_w$ denotes the hypothesis associated with the parameter vector $w$ and $\mathcal{D}$ denotes the dataset. This is an instance of a principle of scalable ML: Write your learning task as an optimization problem, then solve it with an optimization algorithm.
Initialize the parameters at some value $w_0 \in \R^d$, and decrease the value of the empirical risk iteratively by running $$ w_{t+1} = w_t - \alpha_t \cdot \nabla f(w_t) = w_t - \alpha_t \cdot \frac{1}{n} \sum_{x \in \mathcal{D}} \nabla f(w; x) $$ where $w_t$ is the value of the parameter vector at time $t$, $\alpha_t$ is a parameter called the learning rate or step size, and $\nabla f$ denotes the gradient (vector of partial derivatives) of $f$.
Gradient descent is not the only classic optimization method though...
Initialize the parameters at some value $w_0 \in \R^d$, and decrease the value of the empirical risk iteratively by running $$ w_{t+1} = w_t - \left( \nabla^2 f(w_t) \right)^{-1} \nabla f(w_t). $$ Newton's method can converge more quickly than gradient descent, because it is a second-order optimization method (it uses the second derivative of $f$) whereas gradient descent is only a first-order method (it uses the first derivative of $f$).
Suppose that computing gradients of the examples takes $\Theta(d)$ time and computing Hessians takes $\Theta(d^2)$ time, and express your answer in terms $n$ and $d$.
Gradient descent: $ w_{t+1} = w_t - \alpha_t \cdot \nabla f(w_t) = w_t - \alpha_t \cdot \frac{1}{n} \sum_{x \in \mathcal{D}} \nabla f(w; x) $
Time: $O(nd)$
Memory: $O(d)$
Newton's method: $ w_{t+1} = w_t - \left( \nabla^2 f(w_t) \right)^{-1} \nabla f(w_t). $
Time: $O(nd^2 + d^3) = O(nd + nd^2 + d^3 + d^2 + d)$
Memory: $O(d^2)$
...and what does it mean for it to work?
Intuitively, GD decreases the value of the objective at each iteration, as long as the learning rate is small enough and the value of the gradient is nonzero. Eventually, this should result in gradient descent coming close to a point where the gradient is zero. We can prove that gradient descent works under the assumption that the second derivative of the objective is bounded (this is sometimes called an $L$-smoothness bound).
Suppose that for some constant $L > 0$, for all $x$ and $y$ in $\R^d$,
$$ \| \nabla f(x) - \nabla f(y) \| \le L \| x - y \|. $$Here, $\| \cdot \|$ denotes the $\ell_2$ norm: $\|u\| = \sqrt{\sum_{i=1}^d u_i^2}$.
Starting from this condition, let's look at how the objective changes over time as we run gradient descent with a fixed step size.
\begin{align*} f(w_{t+1}) &= f(w_t - \alpha \nabla f(w_t)). \end{align*}Let's consider this expression as a function of $\alpha$. Call this $\rho(\alpha) = f(w_t - \alpha \nabla f(w_t))$. Observe that
$$\rho'(\alpha) = \frac{\partial}{\partial \alpha} f(w_t - \alpha \nabla f(w_t)) = -\nabla f(w_t - \alpha \nabla f(w_t))^T \nabla f(w_t).$$Now, by the fundamental theorem of calculus,
$$ \rho(\alpha) - \rho(0) = \int_0^\alpha \rho'(\eta) \; d \eta. $$If we substitute back in our definition of $\rho$, we get
$$f(w_{t+1}) = f(w_t - \alpha \nabla f(w_t)) = f(w_t) - \int_0^{\alpha} \nabla f(w_t - \eta \nabla f(w_t))^T \nabla f(w_t) \; d \eta.$$We can now simplify this as follows. (Keep in mind: $\| \nabla f(x) - \nabla f(y) \| \le L \| x - y \|$).
\begin{align*} f(w_{t+1}) &= f(w_t) - \int_0^{\alpha} \nabla f(w_t - \eta \nabla f(w_t))^T \nabla f(w_t) \; d \eta. \\&= f(w_t) - \alpha \nabla f(w_t)^T \nabla f(w_t) - \int_0^{\alpha} \left( \nabla f(w_t - \eta \nabla f(w_t)) - \nabla f(w_t) \right)^T \nabla f(w_t) \; d \eta. \\&= f(w_t) - \alpha \| \nabla f(w_t) \|^2 + \int_0^{\alpha} \left( \nabla f(w_t) - \nabla f(w_t - \eta \nabla f(w_t)) \right)^T \nabla f(w_t) \; d \eta. \\&\le f(w_t) - \alpha \| \nabla f(w_t) \|^2 + \int_0^{\alpha} \left\| \nabla f(w_t) - \nabla f(w_t - \eta \nabla f(w_t)) \right\| \cdot \left| \nabla f(w_t) \right| \; d \eta. \end{align*}Where the last line follows from $x^T y \le \| x \| \cdot \| y \|$. Now by our smoothness bound, \begin{align*} f(w_{t+1}) &\le f(w_t)
Note that there is a somewhat easier way of getting to this result using Taylor's theorem rather than FTC; this easier approach is the one I use in the notes. I'm using this more-long-winded way of getting there to make sure that everyone gets the intuition.
So, we got to
$$f(w_{t+1}) \le f(w_t) - \alpha \left(1 - \frac{1}{2} \alpha L \right) \cdot \| \nabla f(w_t) \|^2.$$If we choose our step size $\alpha$ to be small enough that $1 \ge \alpha L$, then this simplifies to
$$f(w_{t+1}) \le f(w_t) - \frac{\alpha}{2} \cdot \| \nabla f(w_t) \|^2.$$That is, the objective is guaranteed to decrease at each iteration! This matches our intuition for why gradient descent should work.
Now, if we sum this up across $T$ iterations of gradient descent, we get
$$ \frac{1}{2} \alpha \sum_{t = 0}^{T-1} \norm{ \nabla f(w_t) }^2 \le \sum_{t = 0}^{T-1} \left( f(w_t) - f(w_{t+1}) \right) = f(w_0) - f(w_T) \le f(w_0) - f^* $$where $f^*$ is the global minimum value of the loss function $f$. From here, we can get
$$ \min_{t \in \{0, \ldots, T\}} \norm{ \nabla f(w_t) }^2 \le \frac{1}{T} \sum_{t = 0}^{T-1} \norm{ \nabla f(w_t) }^2 \le \frac{2 (f(w_0) - f^*)}{\alpha T}. $$This means that the smallest gradient we observe after $T$ iterations is getting smaller proportional to $1/T$. So gradient descent converges...
...
Certainly, we can do this when there is only one global optimum.
The simplest case of this is the case of convex objectives.
A function $f: \R^d \rightarrow \R$ is convex if for all $x, y \in \R^d$, and all $\eta \in [0,1]$
$$f(\eta x + (1 - \eta) y) \le \eta f(x) + (1 - \eta) f(y).$$What this means graphically is that if we draw a line segment between any two points in the graph of the function, that line segment will lie above the function. There are a bunch of equivalent conditions that work if the function is differentiable. In terms of the gradient, for all $x, y \in \R^d$,
$$(x - y)^T \left( \nabla f(x) - \nabla f(y) \right) \ge 0,$$and in terms of the Hessian, for all $x \in \R^d$ and all $u \in \R^d$
$$u^T \nabla^2 f(x) u \ge 0;$$this is equivalent to saying that the Hessian is positive semidefinite.
...
An even easier case than convexity: strong convexity.
A function is strongly convex with parameter $\mu > 0$ if for all $x, y \in \R^d$,
$$f(y) \ge f(x) + \nabla f(x)^T (y - x) + \frac{\mu}{2} \norm{x - y}^2$$or equivalently, for all $x \in \R^d$ and all $u \in \R^d$
$$u^T \nabla^2 f(x) u \ge \mu \norm{u}^2.$$One (somewhat weaker) condition that is implied by strong convexity is
$$\norm{\nabla f(x)}^2 \ge 2 \mu \left( f(x) - f^* \right);$$this is sometimes called the Polyak-Lojasiewicz condition.
As before, let's look at how the objective changes over time as we run gradient descent with a fixed step size.
Now applying the Polyak-Lojasiewicz condition condition to this gives us
$$ f(w_{t+1}) \le f(w_t) - \alpha \mu \left( f(w_t) - f^* \right).$$Subtracting the global optimum $f^*$ from both sides produces
$$ f(w_{t+1}) - f^* \le f(w_t) - f^* - \alpha \mu \left( f(w_t) - f^* \right) = (1 - \alpha \mu) \left( f(w_t) - f^* \right). $$So we got that
$$f(w_{t+1}) - f^* \le (1 - \alpha \mu) \left( f(w_t) - f^* \right).$$But this means that
\begin{align*} f(w_{t+2}) - f^* &\le (1 - \alpha \mu) \left( f(w_{t+1}) - f^* \right) \\&\le (1 - \alpha \mu) \left( (1 - \alpha \mu) \left( f(w_t) - f^* \right) \right) \\&\le (1 - \alpha \mu)^2 \left( f(w_t) - f^* \right). \end{align*}Applying this inductively, after $K$ total steps
$$f(w_K) - f^* \le (1 - \alpha \mu)^K \left( f(w_0) - f^* \right) \le \exp(-\alpha \mu K) \cdot \left( f(w_0) - f^* \right).$$This shows that, for strongly convex functions, gradient descent with a constant step size converges exponentially quickly to the optimum. This is sometimes called convergence at a linear rate.
If we use the largest step size that satisfies our earlier assumption that $1 \ge \alpha L$ (i.e. $\alpha = 1/L$), then this rate becomes
$$ f(w_K) - f^* \le \exp\left(-\frac{\mu K}{L} \right) \cdot \left( f(w_0) - f^* \right).$$Equivalently, in order to ensure that $f(w_K) - f^*$ for some target error $\epsilon > 0$, it suffices to set the number of iterations $K$ large enough that
$$K \ge \frac{L}{\mu} \cdot \log\left( \frac{f(w_0) - f^*}{\epsilon} \right).$$The main thing that affects how large $K$ needs to be here is the ratio $L / \mu$, which is a function just of the problem itself and not of how we initialize or how accurate we want our solutions to be. We call the ratio
$$\kappa = \frac{L}{\mu}$$the condition number of the problem. The condition number encodes how hard a strongly convex problem is to solve.
When the condition number is very large, the problem can be very hard to solve, and take many iterations of gradient descent to get close to the optimum. Even if the cost of running gradient descent is tractible as we scale, if scaling a problem causes the condition number to blow up, this can cause issues!
import numpy
import scipy
import matplotlib
from matplotlib import pyplot
import http.client
matplotlib.rcParams['figure.figsize'] = (10, 7)
matplotlib.rcParams['font.size'] = 16
def process_libsvmdataset_line(line, num_features):
(label_str, features_string) = line.split(" ",maxsplit=1)
label = float(label_str)
features = numpy.zeros(num_features)
for f in features_string.split(" "):
(i,v) = f.split(":");
features[int(i)-1] = float(v);
return (label,features)
def get_cpusmall_dataset():
num_features = 12
conn = http.client.HTTPSConnection("www.csie.ntu.edu.tw")
conn.request("GET", "/~cjlin/libsvmtools/datasets/regression/cpusmall")
r = conn.getresponse()
assert(r.status == 200)
cpusmall_dataset_bytes = r.read()
cpusmall_dataset_str = cpusmall_dataset_bytes.decode("utf-8")
cpusmall_dataset_lines = cpusmall_dataset_str.strip().split("\n")
num_examples = len(cpusmall_dataset_lines)
labels = numpy.zeros(num_examples)
features = numpy.zeros((num_examples, num_features+1)) #add one extra feature that's always 1, for the constant offset
for i in range(num_examples):
(label,feats) = process_libsvmdataset_line(cpusmall_dataset_lines[i], num_features)
labels[i] = label
features[i,0:num_features] = feats
features[i,num_features] = 1.0
return (labels, features)
# load the CPUsmall dataset from the libsvm datasets website
(cpusmall_labels, cpusmall_examples) = get_cpusmall_dataset();
This is a linear regression task, which has empirical risk $$ R(w) = \frac{1}{n} \sum_{i=1}^n (x_i^T w - y_i)^2. $$ The gradient is $$ \nabla R(w) = \frac{2}{n} \sum_{i=1}^n x_i (x_i^T w - y_i). $$
# loss of linear regression task
def linreg_loss(w, Xs, Ys):
return numpy.mean((Xs @ w - Ys)**2)
# gradient of linear regression task
def linreg_grad(w, Xs, Ys):
return (Xs.T @ (Xs @ w - Ys)) * (2/len(Ys))
def linreg_gradient_descent(Xs, Ys, alpha, w0, num_iters):
w = w0;
losses = numpy.zeros(num_iters + 1)
for k in range(num_iters):
losses[k] = linreg_loss(w, Xs, Ys)
w = w - alpha * linreg_grad(w, Xs, Ys)
losses[num_iters] = linreg_loss(w, Xs, Ys)
return (w, losses);
# we can find the exact solution as follows
w_optimal = numpy.linalg.inv(cpusmall_examples.T @ cpusmall_examples) @ cpusmall_examples.T @ cpusmall_labels;
loss_optimal = linreg_loss(w_optimal, cpusmall_examples, cpusmall_labels);
print(f"optimal loss: {loss_optimal}");
optimal loss: 96.36575268592969
w0 = numpy.zeros(13)
alpha = 1e-13 # setting the step size larger results in divergence!
num_iters = 50
(w_gd, losses_gd) = linreg_gradient_descent(cpusmall_examples, cpusmall_labels, alpha, w0, num_iters)
pyplot.plot(range(num_iters+1), losses_gd);
pyplot.xlabel("iteration of gradient descent");
pyplot.ylabel("loss");
pyplot.title("Convergence of Gradient Descent on CPUsmall Linear Regression");
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pyplot.plot(range(num_iters+1), losses_gd, label="gradient descent");
pyplot.plot(range(num_iters+1), 0 * losses_gd + loss_optimal, linestyle=":", c="k", label="optimal");
pyplot.xlabel("iteration of gradient descent")
pyplot.ylabel("loss")
pyplot.title("Convergence of Gradient Descent on CPUsmall Linear Regression")
pyplot.legend();
cpusmall_examples[1,:]
array([1.000000e+00, 0.000000e+00, 2.165000e+03, 2.050000e+02,
1.010000e+02, 4.000000e-01, 1.200000e+00, 4.310700e+04,
4.413900e+04, 3.000000e+00, 1.300000e+02, 1.131931e+06,
1.000000e+00])
# normalize the examples in two ways
# mean = 0, variance = 1
cpusmall_mean = numpy.mean(cpusmall_examples, axis=0)
cpusmall_mean[12] = 0.0
cpusmall_var = numpy.mean((cpusmall_examples - cpusmall_mean)**2, axis=0)
cpusmall_normalized1 = (cpusmall_examples - cpusmall_mean) / numpy.sqrt(cpusmall_var)
# scale by max all features lie in [-1,1]
cpusmall_max = numpy.max(numpy.abs(cpusmall_examples), axis=0)
cpusmall_normalized2 = cpusmall_examples / cpusmall_max
# we can find the exact solution as follows
w_opt_normalized1 = numpy.linalg.inv(cpusmall_normalized1.T @ cpusmall_normalized1) @ cpusmall_normalized1.T @ cpusmall_labels
loss_opt_normalized1 = linreg_loss(w_opt_normalized1, cpusmall_normalized1, cpusmall_labels)
w_opt_normalized2 = numpy.linalg.inv(cpusmall_normalized2.T @ cpusmall_normalized2) @ cpusmall_normalized2.T @ cpusmall_labels;
loss_opt_normalized2 = linreg_loss(w_opt_normalized2, cpusmall_normalized2, cpusmall_labels)
# should all be the same
print(f"optimal loss (without normalization): {loss_optimal}");
print(f"optimal loss (with normalization 1): {loss_opt_normalized1}");
print(f"optimal loss (with normalization 2): {loss_opt_normalized2}");
optimal loss (without normalization): 96.36575268592969 optimal loss (with normalization 1): 96.3657526859297 optimal loss (with normalization 2): 96.3657526859297
w0 = numpy.zeros(13)
alpha = 0.1 # setting the step size larger results in divergence!
(w_gdn, losses_gdn1) = linreg_gradient_descent(cpusmall_normalized1, cpusmall_labels, alpha, w0, num_iters)
w0 = numpy.zeros(13)
alpha = 0.5 # setting the step size larger results in divergence!
(w_gdn, losses_gdn2) = linreg_gradient_descent(cpusmall_normalized2, cpusmall_labels, alpha, w0, num_iters)
pyplot.plot(range(num_iters+1), losses_gdn1, label="GD (var-normalized)");
pyplot.plot(range(num_iters+1), losses_gdn2, label="GD (max-normalized)");
pyplot.plot(range(num_iters+1), losses_gd, label="GD (unnormalized)");
pyplot.plot(range(num_iters+1), 0 * losses_gd + loss_optimal, linestyle=":", c="k", label="optimal");
pyplot.xlabel("iteration of gradient descent");
pyplot.ylabel("loss");
pyplot.title("Convergence of Gradient Descent on CPUsmall Linear Regression");
pyplot.ylim((-100,2000))
pyplot.legend();
Can we interpret this in terms of the condition number?
For linear regression, $$\nabla^2 R(w) = 2 \frac{1}{n} \sum_{i=1}^n x_i x_i^T.$$
D2R_unnormalized = (cpusmall_examples.T @ cpusmall_examples) * 2 / len(cpusmall_labels);
D2R_normalized1 = (cpusmall_normalized1.T @ cpusmall_normalized1) * 2 / len(cpusmall_labels);
D2R_normalized2 = (cpusmall_normalized2.T @ cpusmall_normalized2) * 2 / len(cpusmall_labels);
def condition_number(H):
# make sure H is symmetric for the eigendecomposition
H = (H + H.T) / 2;
ev = numpy.linalg.eigvals(H);
mu = numpy.min(ev);
L = numpy.max(ev);
kappa = L / mu;
return kappa;
print(f"unnormalized, kappa = {(condition_number(D2R_unnormalized))}")
unnormalized, kappa = 49943068262314.78
print(f"var normalized, kappa = {(condition_number(D2R_normalized1))}")
var normalized, kappa = 43.80530335460704
print(f"max normalized, kappa = {(condition_number(D2R_normalized2))}")
max normalized, kappa = 14128.211688196658