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]:
using PyPlot
using NtToolBox

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]:
Cols  = (n0,n1) -> sparse(vec(repeat(1:n1, outer=(1,n0))'), vec(reshape(collect(1:n0*n1),n0,n1) ), Int.(ones(n0*n1)) );

Rows  = (n0,n1) -> sparse( vec(repeat(1:n0, outer=(1,n1))'), vec(reshape(collect(1:n0*n1),n0,n1)' ), Int.(ones(n0*n1) ));
Sigma = (n0,n1) -> [Rows(n0,n1); Cols(n0,n1)];

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

In [3]:
maxit = 1e4
tol = 1e-9
otransp = (C,p0,p1) -> reshape(perform_linprog(Array(Sigma(length(p0),length(p1))),[vec(p0);vec(p1)], C, maxit, tol), length(p0), length(p1));

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]:
gauss = (q,a,c) -> a*randn(2, q) + repeat(c', outer=(1,q))
X0 = randn(2,n0)*.3
X1 = [gauss(Base.div(n1,2),.5, [0 1.6]) gauss(Base.div(n1,4),.3, [-1 -1]) gauss(Base.div(n1,4),.3, [1 -1])];

Density weights $p_0, p_1$.

In [6]:
normalize = a-> a/sum(a)
p0 = normalize(rand(n0, 1))
p1 = normalize(rand(n1, 1));

Shortcut for display.

In [7]:
myplot = (x,y,ms,col) -> 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]:
figure(figsize = (10,7))
axis("off")

for i in 1:length(p0)
    myplot(X0[1,i], X0[2,i], p0[i]*length(p0)*10, "b")
end

for i in 1:length(p1)
    myplot(X1[1,i], X1[2,i], p1[i]*length(p1)*10, "r")
    
xlim(minimum(X1[1,:])-.1,maximum(X1[1,:])+.1)
ylim(minimum(X1[2,:])-.1,maximum(X1[2,:])+.1)
end

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

In [9]:
C = repeat( sum(X0.^2,1)', outer=(1, n1) ) + repeat( sum(X1.^2,1), outer=(n0,1) ) - 2*X0'*X1;
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]:
println("Number of non-zero: $(length(Gamma[Gamma.>0])) (n0 + n1-1 = $(n0 + n1-1))" )
Number of non-zero: 139 (n0 + n1-1 = 139)

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

In [12]:
println("Constraints deviation (should be 0): $(norm(sum(Gamma,2)-vec(p0))), $(norm(sum(Gamma, 1)'-vec(p1)))")
Constraints deviation (should be 0): 4.9279582884301896e-17, 7.359808400080195e-18

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,Gammaij = findnz(Gamma)
Out[13]:
([32,13,45,56,26,10,32,8,18,25  …  53,1,52,35,55,31,60,1,41,58],[1,2,2,3,4,5,6,7,7,7  …  76,77,77,78,78,79,79,80,80,80],[0.0153754,0.00758918,0.00179225,0.00931836,0.00540233,0.000804079,0.00391476,0.00715802,0.00813156,0.000680036  …  0.0196654,0.0141461,0.00265769,0.00844133,0.00146386,0.00760516,0.00800825,0.0124461,0.00522099,0.00752341])

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

In [14]:
length(Gammaij)
Out[14]:
139
In [15]:
figure(figsize =(15,10))
tlist = collect(linspace(0, 1, 6))

for i in 1:length(tlist)
    t = tlist[i]
    Xt = (1-t)*X0[:,I] + t*X1[:,J]
    subplot(2,3,i)
    axis("off")
    for j in 1:length(Gammaij)
        myplot(Xt[1,j],Xt[2,j],Gammaij[j]*length(Gammaij)*6,[t,0,1-t])
    end
    title("t = $t") 
    xlim(minimum(X1[1,:])-.1,maximum(X1[1,:])+.1)
    ylim(minimum(X1[2,:])-.1,maximum(X1[2,:])+.1)
    
end

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 [16]:
n0 = 40
n1 = n0;

Compute points clouds.

In [17]:
X0 = randn(2,n0)*.3
X1 = [gauss(Base.div(n1,2),.5, [0 1.6]) gauss(Base.div(n1,4),.3, [-1 -1]) gauss(Base.div(n1,4),.3, [1 -1])];

Constant distributions.

In [18]:
p0 = ones(n0)/n0
p1 = ones(n1)/n1;

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

In [19]:
C = repeat( sum(X0.^2,1)', outer=(1,n1) ) + repeat( sum(X1.^2,1), outer=(n0,1) ) - 2*X0'*X1;

Display the coulds.

In [20]:
figure(figsize = (10,7))
axis("off")

myplot(X0[1,:],X0[2,:],10,"b")
myplot(X1[1,:],X1[2,:],10,"r")
    
xlim(minimum(X1[1,:])-.1,maximum(X1[1,:])+.1)
ylim(minimum(X1[2,:])-.1,maximum(X1[2,:])+.1);

Solve the optimal transport.

In [21]:
Gamma = otransp(C, p0, p1);

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

In [22]:
figure(figsize = (5,5))
imageplot(Gamma)

Display the optimal assignement.

In [23]:
I,J = findn(Gamma)

figure(figsize = (10,7))
axis("off")

for k in 1:length(I)
    h = plot([X0[1,I[k]]; X1[1,J[k]]],[X0[2,I[k]]; X1[2,J[k]]],"k", lw = 2)
end
    
myplot(X0[1,:],X0[2,:],10,"b")
myplot(X1[1,:],X1[2,:],10,"r")
    
xlim(minimum(X1[1,:])-.1,maximum(X1[1,:])+.1)
ylim(minimum(X1[2,:])-.1,maximum(X1[2,:])+.1);