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This numerical tour studies semi-discrete optimal transport, i.e. when one of the two measure is discrete.
The initial papers that proposed this approach are [Oliker89,Aurenhammer98]. We refer to [Mérigot11,Lévy15] for modern references and fast implementations.
This tour is not inteded to show efficient algorithm but only conveys the main underlying idea (c-transform, Laguerre cells, connexion to optimal quantization). In the Euclidean case, there exists efficient algorithm to compute Laguerre cells leveraging computational geometry algorithm for convex hulls [Aurenhammer87].
import numpy as np
import matplotlib.pyplot as plt
The primal Kantorovitch OT problem reads $$ W_c(\al,\be) = \umin{\pi} \enscond{\int_{\Xx \times \Yy} c(x,y) \text{d}\pi(x,y)}{ \pi_1=\al,\pi_2=\be }. $$ It dual is $$ W_c(\al,\be) = \umax{f,g} \enscond{ \int_\Xx f \text{d} \al + \int_\Yy g \text{d} \be }{ f(x)+g(y) \leq c(x,y) }. $$
We consider the case where $\al=\sum_i a_i \de_{x_i}$ is a discrete measure, so that the function $f(x)$ can be replaced by a vector $(f_i)_{i=1}^n \in \RR^n$. The optimal $g(y)$ function can the be replaced by the $c$-transform of $f$ $$ f^c(y) \eqdef \umin{i} c(x_i,y) - f_i. $$
The function to maximize is then $$ W_c(\al,\be) = \umax{f \in \RR^n} \Ee(f) \eqdef \sum_i f_i a_i + \int f^c(y) \text{d}\be(y). $$
We now implement a gradient ascent scheme for the maximization of $\Ee$. The evaluation of $\Ee$) can be computed via the introduction of the partition of the domain in Laguerre cells $$ \Yy = \bigcup_{i} L_i(f) \qwhereq L_i(f) \eqdef \enscond{y}{ \forall j, c(x_i,y) - f_i \leq c(x_j,y) - f_j }. $$ When $f=0$, this corrsponds to the partition in Voronoi cells.
One has that $\forall y \in L_i(f)$, $f^c(y) = c(x_i,y) - f_i$, i.e. $f^c$ is piecewise smooth according to this partition.
The grid for evaluation of the "continuous measure".
p = 300 # size of the image for sampling, m=p*p
t = np.linspace(0, 1, p)
[V, U] = np.meshgrid(t, t)
Y = np.concatenate((U.flatten()[None, :], V.flatten()[None, :]))
First measure, sums of Dirac masses $\al = \sum_{i=1}^n a_i \de_{x_i}$.
n = 30
X = .5+.5j + np.exp(1j*np.pi/4) * 1 * \
(.1*(np.random.rand(1, n)-.5)+1j*(np.random.rand(1, n)-.5))
X = np.concatenate((np.real(X), np.imag(X)))
a = np.ones(n)/n
Second measure $\be$, potentially a continuous one (i.e. with a density), mixture of Gaussians. Here we discretize $\beta = \sum_{j=1}^m b_j \de_{y_j}$ on a very fine grid.
def Gauss(mx, my, s): return np.exp((-(U-mx)**2-(V-my)**2)/(2*s**2))
Mx = [.6, .4] # means
My = [.9, .1]
S = [.07, .09] # variance
W = [.5, .5] # weights
b = W[0]*Gauss(Mx[0], My[0], S[0]) + W[1]*Gauss(Mx[1], My[1], S[1])
b = b/np.sum(b.flatten())
Display the two measures.
Col = np.random.rand(n, 3)
plt.imshow(-b[::-1, :], extent=[0, 1, 0, 1], cmap='gray')
plt.scatter(X[1, :], X[0, :], s=30, c=.8*Col);
Initial potentials.
f = np.zeros(n)
compute Laguerre cells and c-transform
def distmat(x, y): return np.sum(
x**2, 0)[:, None] + np.sum(y**2, 0)[None, :] - 2*x.transpose().dot(y)
D = distmat(Y, X) - f[:].transpose()
fC = np.min(D, axis=1)
I = np.reshape(np.argmin(D, axis=1), [p, p])
Dual value of the OT, $\dotp{f}{a}+\dotp{f^c}{\be}$.
OT = np.sum(f*a) + np.sum(fC*b.flatten())
print(OT)
0.10332540740714227
Display the Laguerre call partition (here this is equal to the Vornoi diagram since $f=0$).
plt.imshow(I[::-1, :], extent=[0, 1, 0, 1])
plt.scatter(X[1, :], X[0, :], s=20, c='k')
plt.contour(t, t, I, np.linspace(-.5, n-.5, n), colors='k')
plt.axis('off')
(0.0, 1.0, 0.0, 1.0)
Where $\be$ has a density with respect to Lebesgue measure, then $\Ee$ is smooth, and its gradient reads $$ \nabla \Ee(f)_i = a_i - \int_{L_i(f)} \text{d}\be(x). $$
sum area captured
Exercise 1
Implement a gradient ascent $$ f \leftarrow f + \tau \nabla \Ee(f). $$ Experiment on the impact of $\tau$, display the evolution of the OT value $\Ee$ and of the Laguerre cells.
tau = .02 # step size
niter = 200 # iteration for the descent
q = 6 # number of displays
ndisp = np.unique(np.round(1 + (niter/4-1)*np.linspace(0, 1, q)**2))
kdisp = 0
f = np.zeros(n)
E = np.zeros(niter)
for it in range(niter):
# compute Laguerre cells and c-transform
D = distmat(Y, X) - f[:].transpose()
fC = np.min(D, axis=1)
I = np.reshape(np.argmin(D, axis=1), [p, p])
E[it] = np.sum(f*a) + np.sum(fC*b.flatten())
# display
if (kdisp < len(ndisp)) and (ndisp[kdisp] == it):
plt.subplot(2, 3, kdisp+1)
plt.imshow(I[::-1, :], extent=[0, 1, 0, 1])
plt.scatter(X[1, :], X[0, :], s=20, c='k')
plt.contour(t, t, I, np.linspace(-.5, n-.5, n), colors='k')
plt.axis('off')
kdisp = kdisp+1
# gradient
R = (I[:, :, None] == np.arange(0, n)[None, None, :]) * b[:, :, None]
nablaE = a-np.sum(R, axis=(0, 1)).flatten()
f = f+tau*nablaE
Display the evolution of the estimated OT distance.
plt.plot(E, '-')
[<matplotlib.lines.Line2D at 0x7fc838ce8ac0>]
The function $\Ee$ to minimize can be written as an expectation over a random variable $Y \sim \be$ $$ \Ee(f)=\EE(E(f,Y)) \qwhereq E(f,y) = \dotp{f}{a} + f^c(y). $$
As proposed in [Genevay16], one can thus use a stochastic gradient ascent scheme to minimize this function, at iteration $\ell$ $$ f \leftarrow f + \tau_\ell \nabla E(f,y_\ell) $$ where $y_\ell \sim Y$ is a sample drawn according to $\be$ and the step size $\tau_\ell \sim 1/\ell$ should decay at a carefully chosen rate.
The gradient of the integrated functional reads $$ \nabla E(f,y)_i = a - 1_{L_i(f)}(y), $$ where $1_A$ is the binary indicator function of a set $A$.
Initialize the algorithm.
f = np.zeros(n)
Draw the sample.
k = np.int(np.random.rand(1) < W[1]) # select one of the two Gaussian
y = np.array((S[k] * np.random.randn(1) + Mx[k],
S[k] * np.random.randn(1) + My[k]))
Compute the randomized gradient: detect Laguerre cell where $y$ is.
R = np.sum(y**2) + np.sum(X**2, axis=0) - 2*y.transpose().dot(X) - f[:]
i = np.argmin(R)
Randomized gradient.
a = np.ones(n)/n
nablaEy = a.copy()
nablaEy[i] = nablaEy[i] - 1
Exercise 2
Implement the stochastic gradient descent. Test various step size selection rule.
niter = 300
q = 6
ndisp = np.unique(np.round(1 + (niter/2-1)*np.linspace(0, 1, q)**2))
kdisp = 0
E = np.zeros(niter)
for it in range(niter):
# sample
k = np.int(np.random.rand(1) < W[1]) # select one of the two Gaussian
y = np.array((S[k] * np.random.randn(1) + Mx[k],
S[k] * np.random.randn(1) + My[k]))
# detect Laguerre cell where y is
R = np.sum(y**2) + np.sum(X**2, axis=0) - 2*y.transpose().dot(X) - f[:]
i = np.argmin(R)
# gradient
nablaEy = a.copy()
nablaEy[i] = nablaEy[i] - 1
# gradient ascent
l0 = 10 # warmup phase.
tau = .1/(1 + it/l0)
f = f + tau*nablaEy
# compute Laguerre cells and c-transform
D = distmat(Y, X) - f[:].transpose()
fC = np.min(D, axis=1)
I = np.reshape(np.argmin(D, axis=1), [p, p])
E[it] = np.sum(f*a) + np.sum(fC*b.flatten())
# display
if (kdisp < len(ndisp)) and (ndisp[kdisp] == it):
plt.subplot(2, 3, kdisp+1)
plt.imshow(I[::-1, :], extent=[0, 1, 0, 1])
plt.scatter(X[1, :], X[0, :], s=20, c='k')
plt.contour(t, t, I, np.linspace(-.5, n-.5, n), colors='k')
plt.axis('off')
kdisp = kdisp+1
Display the evolution of the estimated OT distance (warning: recording this takes lot of time).
plt.plot(E)
[<matplotlib.lines.Line2D at 0x11cecb4a8>]
We consider the following optimal quantization problem [Gruber02] $$ \umin{ (a_i)_i,(x_i)_i } W_c\pa{ \sum_i a_i \de_{x_i},\be }. $$ This minimization is convex in $a$, and writing down the optimality condition, one has that the associated dual potential should be $f=0$, which means that the associated optimal Laguerre cells should be Voronoi cells $L_i(0)=V_i(x)$ associated to the sampling locations $$ V_i(x) = \enscond{y}{ \forall j, c(x_i,y) \leq c(x_j,y) }. $$
This problem is tightly connected to semi-discrete OT, and this connexion and its implications are studied in [Canas12].
The minimization is non-convex with respect to the positions $x=(x_i)_i$ and one needs to solve $$ \umin{x} \Ff(x) \eqdef \sum_{i=1}^n \int_{V_i(x)} c(x_i,y) \text{d} \be(y). $$ For the sake of simplicity, we consider the case where $c(x,y)=\frac{1}{2}\norm{x-y}^2$.
The gradient reads $$ \nabla \Ff(x)_i = x_i \int_{V_i(x)} \text{d}\be - \int_{V_i(x)} y \text{d}\be(y). $$ The usual algorithm to compute stationary point of this energy is Lloyd's algorithm [Lloyd82], which iterate the fixed point $$ x_i \leftarrow \frac{ \int_{V_i(x)} y \text{d}\be(y) }{ \int_{V_i(x)} \text{d}\be }, $$ i.e. one replaces the centroids by the barycenter of the cells.
Intialize the centroids positions.
X1 = X.copy()
Compute the Voronoi cells $V_i(x)$.
D = D = distmat(Y, X1)
fC = np.min(D, axis=1)
I = np.reshape(np.argmin(D, axis=1), [p, p])
Update the centroids to the barycenters.
A = (I[:, :, None] == np.arange(0, n)[None, None, :]) * b[:, :, None]
B = (I[:, :, None] == np.arange(0, n)[None, None, :]) * \
b[:, :, None] * (U[:, :, None] + 1j*V[:, :, None])
X1 = np.sum(B, axis=(0, 1)) / np.sum(A, axis=(0, 1))
X1 = np.concatenate((np.real(X1)[None, :], np.imag(X1)[None, :]))
Exercise 3
Implement Lloyd algortihm.
niter = 60
q = 6
ndisp = np.unique(np.round(1 + (niter/4-1)*np.linspace(0, 1, q)**2))
kdisp = 0
E = np.zeros(niter)
X1 = X.copy()
for it in range(niter):
# compute Voronoi cells
D = D = distmat(Y, X1)
fC = np.min(D, axis=1)
I = np.reshape(np.argmin(D, axis=1), [p, p])
E[it] = np.sum(fC*b.flatten())
# display
if (kdisp < len(ndisp)) and (ndisp[kdisp] == it):
plt.subplot(2, 3, kdisp+1)
plt.imshow(I[::-1, :], extent=[0, 1, 0, 1])
plt.scatter(X[1, :], X[0, :], s=20, c='k')
plt.contour(t, t, I, np.linspace(-.5, n-.5, n), colors='k')
plt.axis('off')
kdisp = kdisp+1
# update barycenter
A = (I[:, :, None] == np.arange(0, n)[None, None, :]) * b[:, :, None]
B = (I[:, :, None] == np.arange(0, n)[None, None, :]) * \
b[:, :, None] * (U[:, :, None] + 1j*V[:, :, None])
X1 = np.sum(B, axis=(0, 1)) / np.sum(A, axis=(0, 1))
X1 = np.concatenate((np.real(X1)[None, :], np.imag(X1)[None, :]))
Display the evolution of the estimated OT distance.
plt.plot(E[1:-1])
[<matplotlib.lines.Line2D at 0x119dddfd0>]