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/* This file is a c++ wrapper function for computing the transportation cost
* between two vectors given a cost matrix.
*
* It was written by Antoine Rolet (2014) and mainly consists of a wrapper
* of the code written by Nicolas Bonneel available on this page
* http://people.seas.harvard.edu/~nbonneel/FastTransport/
*
* It was then modified to make it more amenable to python inline calling
*
* Please give relevant credit to the original author (Nicolas Bonneel) if
* you use this code for a publication.
*
*/
#include "EMD.h"
int EMD_wrap(int n1, int n2, double *X, double *Y, double *D, double *G,
double* alpha, double* beta, double *cost, int maxIter) {
// beware M and C anre strored in row major C style!!!
int n, m, i, cur;
typedef FullBipartiteDigraph Digraph;
DIGRAPH_TYPEDEFS(FullBipartiteDigraph);
// Get the number of non zero coordinates for r and c
n=0;
for (int i=0; i<n1; i++) {
double val=*(X+i);
if (val>0) {
n++;
}else if(val<0){
return INFEASIBLE;
}
}
m=0;
for (int i=0; i<n2; i++) {
double val=*(Y+i);
if (val>0) {
m++;
}else if(val<0){
return INFEASIBLE;
}
}
// Define the graph
std::vector<int> indI(n), indJ(m);
std::vector<double> weights1(n), weights2(m);
Digraph di(n, m);
NetworkSimplexSimple<Digraph,double,double, node_id_type> net(di, true, n+m, n*m, maxIter);
// Set supply and demand, don't account for 0 values (faster)
cur=0;
for (int i=0; i<n1; i++) {
double val=*(X+i);
if (val>0) {
weights1[ cur ] = val;
indI[cur++]=i;
}
}
// Demand is actually negative supply...
cur=0;
for (int i=0; i<n2; i++) {
double val=*(Y+i);
if (val>0) {
weights2[ cur ] = -val;
indJ[cur++]=i;
}
}
net.supplyMap(&weights1[0], n, &weights2[0], m);
// Set the cost of each edge
for (int i=0; i<n; i++) {
for (int j=0; j<m; j++) {
double val=*(D+indI[i]*n2+indJ[j]);
net.setCost(di.arcFromId(i*m+j), val);
}
}
// Solve the problem with the network simplex algorithm
int ret=net.run();
if (ret==(int)net.OPTIMAL || ret==(int)net.MAX_ITER_REACHED) {
*cost = 0;
Arc a; di.first(a);
for (; a != INVALID; di.next(a)) {
int i = di.source(a);
int j = di.target(a);
double flow = net.flow(a);
*cost += flow * (*(D+indI[i]*n2+indJ[j-n]));
*(G+indI[i]*n2+indJ[j-n]) = flow;
*(alpha + indI[i]) = -net.potential(i);
*(beta + indJ[j-n]) = net.potential(j);
}
}
return ret;
}
int EMD_wrap_return_sparse(int n1, int n2, double *X, double *Y, double *D,
long *iG, long *jG, double *G, long * nG,
double* alpha, double* beta, double *cost, int maxIter) {
// beware M and C anre strored in row major C style!!!
// Get the number of non zero coordinates for r and c and vectors
int n, m, i, cur;
typedef FullBipartiteDigraph Digraph;
DIGRAPH_TYPEDEFS(FullBipartiteDigraph);
// Get the number of non zero coordinates for r and c
n=0;
for (int i=0; i<n1; i++) {
double val=*(X+i);
if (val>0) {
n++;
}else if(val<0){
return INFEASIBLE;
}
}
m=0;
for (int i=0; i<n2; i++) {
double val=*(Y+i);
if (val>0) {
m++;
}else if(val<0){
return INFEASIBLE;
}
}
// Define the graph
std::vector<int> indI(n), indJ(m);
std::vector<double> weights1(n), weights2(m);
Digraph di(n, m);
NetworkSimplexSimple<Digraph,double,double, node_id_type> net(di, true, n+m, n*m, maxIter);
// Set supply and demand, don't account for 0 values (faster)
cur=0;
for (int i=0; i<n1; i++) {
double val=*(X+i);
if (val>0) {
weights1[ cur ] = val;
indI[cur++]=i;
}
}
// Demand is actually negative supply...
cur=0;
for (int i=0; i<n2; i++) {
double val=*(Y+i);
if (val>0) {
weights2[ cur ] = -val;
indJ[cur++]=i;
}
}
// Define the graph
net.supplyMap(&weights1[0], n, &weights2[0], m);
// Set the cost of each edge
for (int i=0; i<n; i++) {
for (int j=0; j<m; j++) {
double val=*(D+indI[i]*n2+indJ[j]);
net.setCost(di.arcFromId(i*m+j), val);
}
}
// Solve the problem with the network simplex algorithm
int ret=net.run();
if (ret==(int)net.OPTIMAL || ret==(int)net.MAX_ITER_REACHED) {
*cost = 0;
Arc a; di.first(a);
cur=0;
for (; a != INVALID; di.next(a)) {
int i = di.source(a);
int j = di.target(a);
double flow = net.flow(a);
if (flow>0)
{
*cost += flow * (*(D+indI[i]*n2+indJ[j-n]));
*(G+cur) = flow;
*(iG+cur) = indI[i];
*(jG+cur) = indJ[j-n];
*(alpha + indI[i]) = -net.potential(i);
*(beta + indJ[j-n]) = net.potential(j);
cur++;
}
}
*nG=cur; // nb of value +1 for numpy indexing
}
return ret;
}
int EMD_wrap_all_sparse(int n1, int n2, double *X, double *Y,
long *iD, long *jD, double *D, long nD,
long *iG, long *jG, double *G, long * nG,
double* alpha, double* beta, double *cost, int maxIter) {
// beware M and C anre strored in row major C style!!!
// Get the number of non zero coordinates for r and c and vectors
int n, m, cur;
typedef FullBipartiteDigraph Digraph;
DIGRAPH_TYPEDEFS(FullBipartiteDigraph);
n=n1;
m=n2;
// Define the graph
std::vector<double> weights2(m);
Digraph di(n, m);
NetworkSimplexSimple<Digraph,double,double, node_id_type> net(di, true, n+m, n*m, maxIter);
// Set supply and demand, don't account for 0 values (faster)
// Demand is actually negative supply...
cur=0;
for (int i=0; i<n2; i++) {
double val=*(Y+i);
if (val>0) {
weights2[ cur ] = -val;
}
}
// Define the graph
net.supplyMap(X, n, &weights2[0], m);
// Set the cost of each edge
for (int k=0; k<nD; k++) {
int i = iD[k];
int j = jD[k];
net.setCost(di.arcFromId(i*m+j), D[k]);
}
// Solve the problem with the network simplex algorithm
int ret=net.run();
if (ret==(int)net.OPTIMAL || ret==(int)net.MAX_ITER_REACHED) {
*cost = net.totalCost();
Arc a; di.first(a);
cur=0;
for (; a != INVALID; di.next(a)) {
int i = di.source(a);
int j = di.target(a);
double flow = net.flow(a);
if (flow>0)
{
*(G+cur) = flow;
*(iG+cur) = i;
*(jG+cur) = j-n;
*(alpha + i) = -net.potential(i);
*(beta + j-n) = net.potential(j);
cur++;
}
}
*nG=cur; // nb of value +1 for numpy indexing
}
return ret;
}
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