# Optimal Transport with Linear Programming¶

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This numerical tours details how to solve the discrete optimal transport problem (in the case of measures that are sums of Diracs) using linear programming.

In [1]:
from __future__ import division

import numpy as np
import scipy as scp
import pylab as pyl
import matplotlib.pyplot as plt

from nt_toolbox.general import *
from nt_toolbox.signal import *

import warnings
warnings.filterwarnings('ignore')

%matplotlib inline


## Optimal Transport of Discrete Distribution¶

We consider two dicretes distributions $$\forall k=0,1, \quad \mu_k = \sum_{i=1}^{n_k} p_{k,i} \de_{x_{k,i}}$$ where $n_0,n_1$ are the number of points, $\de_x$ is the Dirac at location $x \in \RR^d$, and $X_k = ( x_{k,i} )_{i=1}^{n_k} \subset \RR^d$ for $k=0,1$ are two point clouds.

We define the set of couplings between $\mu_0,\mu_1$ as

$$\Pp = \enscond{ (\ga_{i,j})_{i,j} \in (\RR^+)^{n_0 \times n_1} }{ \forall i, \sum_j \ga_{i,j} = p_{0,i}, \: \forall j, \sum_i \ga_{i,j} = p_{1,j} }$$

The Kantorovitch formulation of the optimal transport reads

$$\ga^\star \in \uargmin{\ga \in \Pp} \sum_{i,j} \ga_{i,j} C_{i,j}$$

where $C_{i,j} \geq 0$ is the cost of moving some mass from $x_{0,i}$ to $x_{1,j}$.

The optimal coupling $\ga^\star$ can be shown to be a sparse matrix with less than $n_0+n_1-1$ non zero entries. An entry $\ga_{i,j}^\star \neq 0$ should be understood as a link between $x_{0,i}$ and $x_{1,j}$ where an amount of mass equal to $\ga_{i,j}^\star$ is transfered.

In the following, we concentrate on the $L^2$ Wasserstein distance. $$C_{i,j}=\norm{x_{0,i}-x_{1,j}}^2.$$

The $L^2$ Wasserstein distance is then defined as $$W_2(\mu_0,\mu_1)^2 = \sum_{i,j} \ga_{i,j}^\star C_{i,j}.$$

The coupling constraint $$\forall i, \sum_j \ga_{i,j} = p_{0,i}, \: \forall j, \sum_i \ga_{i,j} = p_{1,j}$$ can be expressed in matrix form as $$\Sigma(n_0,n_1) \ga = [p_0;p_1]$$ where $\Sigma(n_0,n_1) \in \RR^{ (n_0+n_1) \times (n_0 n_1) }$.

In [2]:
from scipy import sparse

Rows  = lambda n0,n1: sparse.coo_matrix((np.ones(n0*n1),
(np.ravel(np.tile(np.arange(0,n0),(n1,1)),order='C'),
np.arange(0,n0*n1))))
Cols  = lambda n0,n1: sparse.coo_matrix((np.ones(n0*n1),
(np.ravel(np.tile(np.arange(0,n1),(n0,1)),order='F'),
np.arange(0,n0*n1))))
Sigma = lambda n0,n1: sparse.vstack((Rows(n0,n1),Cols(n0,n1)))


We use a simplex algorithm to compute the optimal transport coupling $\ga^\star$.

In [3]:
from nt_toolbox.perform_linprog import *

maxit = 1e4
tol = 1e-9
otransp = lambda C,p0,p1: np.reshape(perform_linprog(Sigma(len(p0),len(p1)).toarray(),np.vstack((p0,p1)),C,maxit,tol),(n0,n1),order="F")


Dimensions $n_0, n_1$ of the clouds.

In [4]:
n0 = 60
n1 = 80


Compute a first point cloud $X_0$ that is Gaussian, and a second point cloud $X_1$ that is Gaussian mixture.

In [5]:
from numpy import random

gauss = lambda q,a,c: a*random.randn(2, q) + np.transpose(np.tile(c, (q,1)))
X0 = random.randn(2,n0)*.3
X1 = np.hstack((gauss(n1/2,.5,[0,1.6]),np.hstack((gauss(n1/4,.3,[-1,-1]),gauss(n1/4,.3,[1,-1])))))


Density weights $p_0, p_1$.

In [6]:
normalize = lambda a: a/np.sum(a)
p0 = normalize(random.rand(n0, 1))
p1 = normalize(random.rand(n1, 1))


Shortcut for display.

In [7]:
myplot = lambda x,y,ms,col: plt.scatter(x,y, s=ms*20, edgecolors="k", c=col, linewidths=2)


Display the point clouds. The size of each dot is proportional to its probability density weight.

In [8]:
plt.figure(figsize = (10,7))
plt.axis("off")

for i in range(len(p0)):
myplot(X0[0,i], X0[1,i], p0[i]*len(p0)*10, 'b')

for i in range(len(p1)):
myplot(X1[0,i], X1[1,i], p1[i]*len(p1)*10, 'r')

plt.xlim(np.min(X1[0,:])-.1,np.max(X1[0,:])+.1)
plt.ylim(np.min(X1[1,:])-.1,np.max(X1[1,:])+.1)

plt.show()


Compute the weight matrix $(C_{i,j})_{i,j}.$

In [9]:
C = np.transpose(np.tile(np.transpose(np.sum(X0**2,0)),(n1,1))) + np.tile(np.sum(X1**2,0),(n0,1)) - 2*np.dot(np.transpose(X0),X1)


Compute the optimal transport plan.

In [10]:
gamma = otransp(C, p0, p1)


Check that the number of non-zero entries in $\ga^\star$ is $n_0+n_1-1$.

In [11]:
print("Number of non-zero: %d (n0 + n1-1 = %d)" %(len(gamma[gamma>0]), n0 + n1-1))

Number of non-zero: 139 (n0 + n1-1 = 139)


Check that the solution satifies the constraints $\ga \in \Cc$.

In [12]:
from numpy import linalg

print("Constraints deviation (should be 0): %.2f, %.2f" %(linalg.norm(np.sum(gamma,1)-np.ravel(p0)),linalg.norm(np.transpose(np.sum(gamma, 0))-np.ravel(p1))))

Constraints deviation (should be 0): 0.00, 0.00


## Displacement Interpolation¶

For any $t \in [0,1]$, one can define a distribution $\mu_t$ such that $t \mapsto \mu_t$ defines a geodesic for the Wasserstein metric.

Since the $W_2$ distance is a geodesic distance, this geodesic path solves the following variational problem

$$\mu_t = \uargmin{\mu} (1-t)W_2(\mu_0,\mu)^2 + t W_2(\mu_1,\mu)^2.$$

This can be understood as a generalization of the usual Euclidean barycenter to barycenter of distribution. Indeed, in the case that $\mu_k = \de_{x_k}$, one has $\mu_t=\de_{x_t}$ where $x_t = (1-t)x_0+t x_1$.

Once the optimal coupling $\ga^\star$ has been computed, the interpolated distribution is obtained as

$$\mu_t = \sum_{i,j} \ga^\star_{i,j} \de_{(1-t)x_{0,i} + t x_{1,j}}.$$

Find the $i,j$ with non-zero $\ga_{i,j}^\star$.

In [13]:
I,J = np.nonzero(gamma)
gammaij = gamma[I,J]


Display the evolution of $\mu_t$ for a varying value of $t \in [0,1]$.

In [14]:
plt.figure(figsize =(15,10))
tlist = np.linspace(0, 1, 6)

for i in range(len(tlist)):
t = tlist[i]
Xt = (1-t)*X0[:,I] + t*X1[:,J]
plt.subplot(2,3,i+1)
plt.axis("off")
for j in range(len(gammaij)):
myplot(Xt[0,j],Xt[1,j],gammaij[j]*len(gammaij)*6,[t,0,1-t])

plt.title("t = %.1f" %t)
plt.xlim(np.min(X1[0,:])-.1,np.max(X1[0,:])+.1)
plt.ylim(np.min(X1[1,:])-.1,np.max(X1[1,:])+.1)

plt.show()


## Optimal Assignement¶

In the case where the weights $p_{0,i}=1/n, p_{1,i}=1/n$ (where $n_0=n_1=n$) are constants, one can show that the optimal transport coupling is actually a permutation matrix. This properties comes from the fact that the extremal point of the polytope $\Cc$ are permutation matrices.

This means that there exists an optimal permutation $\si^\star \in \Sigma_n$ such that

$$\ga^\star_{i,j} = \choice{ 1 \qifq j=\si^\star(i), \\ 0 \quad\text{otherwise}. }$$

where $\Si_n$ is the set of permutation (bijections) of $\{1,\ldots,n\}$.

This permutation thus solves the so-called optimal assignement problem

$$\si^\star \in \uargmin{\si \in \Sigma_n} \sum_{i} C_{i,\si(j)}.$$

Same number of points.

In [15]:
n0 = 40
n1 = n0


Compute points clouds.

In [16]:
X0 = random.randn(2,n0)*.3
X1 = np.hstack((gauss(n1/2,.5,[0,1.6]),np.hstack((gauss(n1/4,.3,[-1,-1]),gauss(n1/4,.3,[1,-1])))))


Constant distributions.

In [17]:
p0 = np.ones([n0,1])/n0
p1 = np.ones([n1,1])/n1


Compute the weight matrix $(C_{i,j})_{i,j}.$

In [18]:
C = np.transpose(np.tile(np.transpose(np.sum(X0**2,0)),(n1,1))) + np.tile(np.sum(X1**2,0),(n0,1)) - 2*np.dot(np.transpose(X0),X1)


Display the coulds.

In [19]:
plt.figure(figsize = (10,7))
plt.axis('off')

myplot(X0[0,:],X0[1,:],10,'b')
myplot(X1[0,:],X1[1,:],10,'r')

plt.xlim(np.min(X1[0,:])-.1,np.max(X1[0,:])+.1)
plt.ylim(np.min(X1[1,:])-.1,np.max(X1[1,:])+.1)
plt.show()


Solve the optimal transport.

In [20]:
gamma = otransp(C, p0, p1)


Show that $\ga$ is a binary permutation matrix.

In [21]:
plt.figure(figsize = (5,5))
imageplot(gamma)


Display the optimal assignement.

In [22]:
I,J = np.nonzero(gamma)

plt.figure(figsize = (10,7))
plt.axis('off')

for k in range(len(I)):
h = plt.plot(np.hstack((X0[0,I[k]],X1[0,J[k]])),np.hstack(([X0[1,I[k]], X1[1,J[k]]])),'k', lw = 2)

myplot(X0[0,:], X0[1,:], 10, 'b')
myplot(X1[0,:], X1[1,:], 10, 'r')

plt.xlim(np.min(X1[0,:])-.1,np.max(X1[0,:])+.1)
plt.ylim(np.min(X1[1,:])-.1,np.max(X1[1,:])+.1)
plt.show()