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Gradient Descent

Follow gradient descent on smooth objectives and apply it to image restoration. Changing the step size reveals the relationship between curvature, stability, and the speed at which the objective decreases.

Run this tour

Run the cells in order with a Python 3 kernel. The first cell locates the companion data and toolbox and installs missing dependencies when needed. All worked examples include their implementation directly in this notebook. Random seeds make comparisons reproducible; you can change them to explore other samples.

# Locate the companion toolbox locally, or fetch it for a standalone/Colab copy.
from pathlib import Path
import importlib.util
import os
import subprocess
import sys

working = Path.cwd()
candidates = [working, working / "python", working.parent / "python"]
python_dir = next((p for p in candidates if (p / "nt_toolbox").is_dir()), None)
if python_dir is None:
    checkout = working / "numerical-tours-support"
    if not checkout.exists():
        subprocess.run(
            [
                "git",
                "clone",
                "--depth",
                "1",
                "--branch",
                "master",
                "https://github.com/gpeyre/numerical-tours.git",
                str(checkout),
            ],
            check=True,
        )
    python_dir = checkout / "python"
os.chdir(python_dir)
if str(python_dir) not in sys.path:
    sys.path.insert(0, str(python_dir))
requirements = python_dir / "requirements.txt"
if any(
    importlib.util.find_spec(name) is None
    for name in [
        "numpy",
        "scipy",
        "matplotlib",
        "skimage",
        "sklearn",
        "pywt",
        "ipywidgets",
        "cvxpy",
        "skfmm",
        "autograd",
        "progressbar",
        "celer",
    ]
):
    subprocess.run(
        [sys.executable, "-m", "pip", "install", "-r", str(requirements)], check=True
    )

import numpy as np
import matplotlib.pyplot as plt

np.random.seed(0)
plt.rcParams.update(
    {
        "figure.figsize": (8, 4),
        "figure.dpi": 100,
        "axes.spines.top": False,
        "axes.spines.right": False,
        "font.size": 11,
        "image.cmap": "gray",
    }
)
%matplotlib inline

This tour explores the use of gradient descent method for unconstrained and constrained optimization of a smooth function

$\newcommand{\dotp}[2]{\langle #1, #2 \rangle}$ $\newcommand{\pd}[2]{ \frac{ \partial #1}{\partial #2} }$ $\newcommand{\umin}[1]{\underset{#1}{\min}\;}$ $\newcommand{\qandq}{\quad\text{and}\quad}$ $\newcommand{\qwhereq}{\quad\text{where}\quad}$ $\newcommand{\qifq}{ \quad \text{if} \quad }$ $\newcommand{\ZZ}{\mathbb{Z}}$ $\newcommand{\RR}{\mathbb{R}}$ $\newcommand{\pa}[1]{\left(#1\right)}$ $\newcommand{\si}{\sigma}$ $\newcommand{\Nn}{\mathcal{N}}$ $\newcommand{\Hh}{\mathcal{H}}$ $\newcommand{\Bb}{\mathcal{B}}$ $\newcommand{\EE}{\mathbb{E}}$ $\newcommand{\norm}[1]{\|#1\|}$ $\newcommand{\abs}[1]{\left|#1\right|}$ $\newcommand{\choice}[1]{ \left\{ \begin{array}{l} #1 \end{array} \right. }$ $\newcommand{\al}{\alpha}$ $\newcommand{\la}{\lambda}$ $\newcommand{\ga}{\gamma}$ $\newcommand{\Ga}{\Gamma}$ $\newcommand{\La}{\Lambda}$ $\newcommand{\Si}{\Sigma}$ $\newcommand{\be}{\beta}$ $\newcommand{\de}{\delta}$ $\newcommand{\De}{\Delta}$ $\newcommand{\phi}{\varphi}$ $\newcommand{\th}{\theta}$ $\newcommand{\om}{\omega}$ $\newcommand{\Om}{\Omega}$

%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt

Gradient Descent for Unconstrained Problems

We consider the problem of finding a minimum of a function $f$, hence solving $$\umin{x \in \RR^d} f(x)$$ where $f : \RR^d \rightarrow \RR$ is a smooth function.

Note that the minimum is not necessarily unique. In the general case, $f$ might exhibit local minima, in which case the proposed algorithms is not expected to find a global minimizer of the problem. In this tour, we restrict our attention to convex function, so that the methods will converge to a global minimizer.

The simplest method is the gradient descent, that computes $$ x^{(k+1)} = x^{(k)} - \tau_k \nabla f(x^{(k)}), $$ where $\tau_k>0$ is a step size, and $\nabla f(x) \in \RR^d$ is the gradient of $f$ at the point $x$, and $x^{(0)} \in \RR^d$ is any initial point.

In the convex case, if $f$ is of class $C^2$, in order to ensure convergence, the step size should satisfy $$ 0 < \tau_k < \frac{2}{ \sup_x \norm{Hf(x)} } $$ where $Hf(x) \in \RR^{d \times d}$ is the Hessian of $f$ at $x$ and $\norm{\cdot}$ is the spectral operator norm (largest eigenvalue).

Gradient Descent in 2-D

We consider a simple problem, corresponding to the minimization of a 2-D quadratic form $$ f(x) = \frac{1}{2} \pa{ x_1^2 + \eta x_2^2 } ,$$ where $\eta>0$ controls the anisotropy, and hence the difficulty, of the problem.

Anisotropy parameter $\eta$.

eta = 4

Function $f$.

f = lambda x: (x[0] ** 2 + eta * x[1] ** 2) / 2

Background image of the function.

tx = np.linspace(-0.3, 1, 101)
ty = np.linspace(-0.6, 0.6, 101)
[v, u] = np.meshgrid(ty, tx)
F = (u**2 + eta * v**2) / 2

Display the function as a 2-D image.

plt.contourf(tx, ty, F.transpose(), 10);
No description has been provided for this image

Gradient.

Gradf = lambda x: np.array([x[0], eta * x[1]])

The step size should satisfy $\tau_k < 2/\eta$. We use here a constant step size.

tau = 1.6 / eta

Initial point for the descent.

x0 = np.array([0.9, 0.3])

Exercice 1:

Perform the gradient descent using a fixed step size $\tau_k=\tau$. Display the decay of the energy $f(x^{(k)})$ through the iteration. Save the iterates so that X(:,k) corresponds to $x^{(k)}$.

x = x0
niter = 20
E = np.zeros((niter, 1))
X = np.zeros((2, niter))
for i in np.arange(0, niter):
    X[:, i] = x
    E[i] = f(x)
    x = x - tau * Gradf(x)

plt.clf()
h = plt.plot(np.log10(E))
plt.axis("tight")
plt.title("$log_{10}f(x^{(k)})$")
Text(0.5, 1.0, '$log_{10}f(x^{(k)})$')
No description has been provided for this image

Display the iterations.

plt.contourf(tx, ty, F.transpose(), 10)
plt.plot(X[0, :], X[1, :], "k.-")
plt.axis("equal");
No description has been provided for this image

Exercice 2:

Display the iteration for several different step sizes.

niter = 20

plt.clf()
plt.contourf(tx, ty, F.transpose(), 10)
tau_list = np.array([0.5, 1, 1.5, 1.9]) / eta

for itau in np.arange(0, tau_list.size):
    tau = tau_list[itau]
    x = x0
    X = np.zeros((2, niter))
    for i in np.arange(0, niter):
        X[:, i] = x.flatten()
        x = x - tau * Gradf(x)
    plt.plot(X[0, :], X[1, :], ".-")
plt.axis("equal")
(np.float64(-0.3), np.float64(1.0), np.float64(-0.6), np.float64(0.6))
No description has been provided for this image

Exercice 3:

Implement the optimal step size selection (exact linesearch).

niter = 20

x = x0

X = np.zeros((2, niter))
for i in np.arange(0, niter):
    X[:, i] = x.flatten()
    # optimal step size
    tau = (x[0] ** 2 + eta**2 * x[1] ** 2) / (x[0] ** 2 + eta**3 * x[1] ** 2)
    x = x - tau * Gradf(x)


plt.clf()
plt.contourf(tx, ty, F.transpose(), 10)
plt.plot(X[0, :], X[1, :], "k.-")
plt.axis("equal")
(np.float64(-0.3), np.float64(1.0), np.float64(-0.6), np.float64(0.6))
No description has been provided for this image

Gradient and Divergence of Images

Local differential operators like gradient, divergence and laplacian are the building blocks for variational image processing.

Load an image $x_0 \in \RR^N$ of $N=n \times n$ pixels.

from nt_toolbox.general import clamp, np, plt
from nt_toolbox.signal import div, grad, imageplot, load_image, np, plt
import numpy as np
import matplotlib.pyplot as plt
from numpy import (
    abs,
    concatenate,
    equal,
    linspace,
    maximum,
    meshgrid,
    newaxis,
    ones,
    r_,
    size,
    sqrt,
    sum,
    tile,
)
from matplotlib.pyplot import axis, matplotlib, np, step
from numpy.linalg import norm
from numpy.random import randn
n = 256
name = "nt_toolbox/data/lena.png"
x0 = load_image(name, n)

Display it.

imageplot(x0)
No description has been provided for this image

For a continuous function $g$, the gradient reads $$\nabla g(s) = \pa{ \pd{g(s)}{s_1}, \pd{g(s)}{s_2} } \in \RR^2.$$ (note that here, the variable $d$ denotes the 2-D spacial position).

We discretize this differential operator on a discrete image $x \in \RR^N$ using first order finite differences. $$(\nabla x)_i = ( x_{i_1,i_2}-x_{i_1-1,i_2}, x_{i_1,i_2}-x_{i_1,i_2-1} ) \in \RR^2.$$ Note that for simplity we use periodic boundary conditions.

Compute its gradient, using finite differences.

def grad(x):
    hdiff = x - x[r_[len(x) - 1, : len(x) - 1], :]
    vdiff = x - x[:, r_[len(x) - 1, : len(x) - 1]]
    return concatenate((hdiff[..., newaxis], vdiff[..., newaxis]), axis=2)

One thus has $\nabla : \RR^N \mapsto \RR^{N \times 2}.$

v = grad(x0)

One can display each of its components.

imageplot(v[:, :, 0], "d/dx", (1, 2, 1))
imageplot(v[:, :, 1], "d/dy", (1, 2, 2))
No description has been provided for this image

One can display its magnitude $\norm{(\nabla x)_i}$, which is large near edges.

imageplot(sqrt(sum(v * v, axis=2)))
No description has been provided for this image

The divergence operator maps vector field to images. For continuous vector fields (v(s) \in \RR^2), it is defined as $$\text{div}(v)(s) = \pd{v_1(s)}{s_1} + \pd{v_2(s)}{s_2} \in \RR.$$ (note that here, the variable (s) denotes the 2-D spacial position). It is minus the adjoint of the gadient, i.e. $\text{div} = - \nabla^*$.

It is discretized, for $v=(v^1,v^2)$ as $$\text{div}(v)_i = v^1_{i_1+1,i_2} - v^1_{i_1,i_2} + v^2_{i_1,i_2+1} - v^2_{i_1,i_2} .$$

div = lambda x: (
    x[r_[1 : len(x), 0], :, 0] - x[:, :, 0] + x[:, r_[1 : len(x), 0], 1] - x[:, :, 1]
)

The Laplacian operatore is defined as $\Delta=\text{div} \circ \nabla = -\nabla^* \circ \nabla$. It is thus a negative symmetric operator.

delta = lambda x: div(grad(x))

Display $\Delta x_0$.

imageplot(delta(x0))
No description has been provided for this image

Check that the relation $\norm{\nabla x} = - \dotp{\Delta x}{x}.$

print("Should be 0:\n", (grad(x0) * grad(x0)).sum() + (delta(x0) * x0).sum())
Should be 0:
 0.0

Gradient Descent in Image Processing

We consider now the problem of denoising an image $y \in \RR^d$ where $d = n \times n$ is the number of pixels ($n$ being the number of rows/columns in the image).

Add noise to the original image, to simulate a noisy image.

sigma = 0.1
y = x0 + sigma * randn(n, n)

Display the noisy image $y$.

imageplot(clamp(y))
No description has been provided for this image

Denoising is obtained by minimizing the following functional $$\umin{x \in \RR^d} f(x) = \frac{1}{2} \norm{y-x}^2 + \la J_\epsilon(x)$$ where $J_\epsilon(x)$ is a smoothed total variation of the image. $$J_\epsilon(x) = \sum_i \norm{ (G x)_i }_{\epsilon}$$ where $(Gx)_i \in \RR^2$ is an approximation of the gradient of $x$ at pixel $i$ and for $u \in \RR^2$, we use the following smoothing of the $L^2$ norm in $\RR^2$ $$\norm{u}_\epsilon = \sqrt{ \epsilon^2 + \norm{u}^2 },$$ for a small value of $\epsilon>0$.

The gradient of the functional read $$\nabla f(x) = x-y + \lambda \nabla J_\epsilon(x)$$ where the gradient of the smoothed TV norm is $$\nabla J_\epsilon(x)_i = G^*( u ) \qwhereq u_i = \frac{ (G x)_i }{\norm{ (G x)_i }_\epsilon} $$ where $G^*$ is the adjoint operator of $G$ which corresponds to minus a discretized divergence

Value for $\lambda$.

la = 0.3 / 5

Value for $\epsilon$.

epsilon = 1e-3

TV norm.

norm_eps = lambda u: sqrt(epsilon**2 + sum(u * u, axis=2))
J = lambda x: sum(sum(norm_eps(grad(x))))

Function $f$ to minimize.

f = lambda x: 1 / 2 * norm(x - y) ** 2 + la * J(x)

Gradient of $J_\epsilon$. Note that |div| implement $-G^*$.

normalize_vec = lambda u: u / tile(norm_eps(u)[..., newaxis], (1, 1, 2))
gradTV = lambda x: -div(normalize_vec(grad(x)))

Gradient of the functional.

Gradf = lambda x: x - y + la * gradTV(x)

The step size should satisfy $$0 < \tau_k < \frac{2}{ 1 + 4 \lambda / \epsilon }.$$ Here we use a slightly larger step size, which still work in practice.

tau = 1.8 / (1 + la * 8 / epsilon)
tau = tau * 4

Exercice 3: Implement the gradient descent. Monitor the decay of $f$ through the iterations.

Display the resulting denoised image.

# imageplot(clamp(x))

Constrained Optimization Using Projected Gradient Descent

We consider a linear imaging operator $\Phi : x \mapsto \Phi(x)$ that maps high resolution images to low dimensional observations. Here we consider a pixel masking operator, that is diagonal over the spacial domain.

To emphasis the effect of the TV functional, we use a simple geometric image.

n = 64
radius = 0.6
t = linspace(-1, 1, n)
[Y, X] = meshgrid(t, t)
x0 = (maximum(abs(X), abs(Y)) < radius) * 1.0

We consider here the inpainting problem. This simply corresponds to a masking operator. Here we remove the central part of the image.

a = 4
Lambda = ones((n, n))
Lambda[n // 2 - a : n // 2 + a, :] = 0

Masking operator $\Phi$. Note that it is symmetric, i.e. $\Phi^*=\Phi$.

Phi = lambda x: x * Lambda
PhiS = Phi

Noiseless observations $y=\Phi x_0$.

y = Phi(x0)

Display.

imageplot(x0, "Original", (1, 2, 1))
imageplot(y, "Damaged", (1, 2, 2))
No description has been provided for this image

We want to solve the noiseless inverse problem $y=\Phi f$ using a total variation regularization: $$\umin{ y=\Phi x } J_\epsilon(x).$$ We use the following projected gradient descent $$x^{(k+1)} = \text{Proj}_{\Hh}( x^{(k)} - \tau_k \nabla J_{\epsilon}(x^{(k)}) )$$ where $\text{Proj}_{\Hh}$ is the orthogonal projection on the set of linear constraint $\Phi x = y$, and is easy to compute for inpainting

ProjH = lambda x: x + PhiS(y - Phi(x))

Exercice 4: Display the evolution of the inpainting process.

Exercice 5: Try with several values of $\epsilon$.

References and further reading