1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218
|
/* 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 "network_simplex_simple.h"
#include "network_simplex_simple_omp.h"
#include "EMD.h"
#include <cstdint>
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 are stored in row major C style!!!
using namespace lemon;
int n, m, cur;
typedef FullBipartiteDigraph Digraph;
DIGRAPH_TYPEDEFS(Digraph);
// 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, ((int64_t)n)*((int64_t)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
int64_t idarc = 0;
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(idarc), val);
++idarc;
}
}
// Solve the problem with the network simplex algorithm
int ret=net.run();
int i, j;
if (ret==(int)net.OPTIMAL || ret==(int)net.MAX_ITER_REACHED) {
*cost = 0;
Arc a; di.first(a);
for (; a != INVALID; di.next(a)) {
i = di.source(a);
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_omp(int n1, int n2, double *X, double *Y, double *D, double *G,
double* alpha, double* beta, double *cost, int maxIter, int numThreads) {
// beware M and C are stored in row major C style!!!
using namespace lemon_omp;
int n, m, cur;
typedef FullBipartiteDigraph Digraph;
DIGRAPH_TYPEDEFS(Digraph);
// 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, ((int64_t)n)*((int64_t)m), maxIter, numThreads);
// 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
int64_t idarc = 0;
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(idarc), val);
++idarc;
}
}
// Solve the problem with the network simplex algorithm
int ret=net.run();
int i, j;
if (ret==(int)net.OPTIMAL || ret==(int)net.MAX_ITER_REACHED) {
*cost = 0;
Arc a; di.first(a);
for (; a != INVALID; di.next(a)) {
i = di.source(a);
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;
}
|