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, uint64_t maxIter) {
// beware M and C are stored in row major C style!!!
using namespace lemon;
uint64_t 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<uint64_t> 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, (int) (n + m), n * m, maxIter);
// Set supply and demand, don't account for 0 values (faster)
cur=0;
for (uint64_t 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 (uint64_t i=0; i<n2; i++) {
double val=*(Y+i);
if (val>0) {
weights2[ cur ] = -val;
indJ[cur++]=i;
}
}
net.supplyMap(&weights1[0], (int) n, &weights2[0], (int) m);
// Set the cost of each edge
int64_t idarc = 0;
for (uint64_t i=0; i<n; i++) {
for (uint64_t 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();
uint64_t 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, uint64_t maxIter, int numThreads) {
// beware M and C are stored in row major C style!!!
using namespace lemon_omp;
uint64_t 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<uint64_t> 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, (int) (n + m), n * m, maxIter, numThreads);
// Set supply and demand, don't account for 0 values (faster)
cur=0;
for (uint64_t 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 (uint64_t i=0; i<n2; i++) {
double val=*(Y+i);
if (val>0) {
weights2[ cur ] = -val;
indJ[cur++]=i;
}
}
net.supplyMap(&weights1[0], (int) n, &weights2[0], (int) m);
// Set the cost of each edge
int64_t idarc = 0;
for (uint64_t i=0; i<n; i++) {
for (uint64_t 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();
uint64_t 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;
}
|
