forking code for mfas from 1dsfm
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#include "mfas.h"
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#include <map>
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#include <set>
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#include <vector>
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using std::map;
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using std::pair;
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using std::set;
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using std::vector;
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void reindex_problem(vector<Edge> &edges, map<int, int> &reindexing_key) {
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// get the unique set of notes
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set<int> nodes;
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for (int i = 0; i < edges.size(); ++i) {
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nodes.insert(edges[i].first);
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nodes.insert(edges[i].second);
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}
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// iterator through them and assign a new name to each vertex
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std::set<int>::const_iterator it;
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reindexing_key.clear();
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int i = 0;
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for (it = nodes.begin(); it != nodes.end(); ++it) {
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reindexing_key[*it] = i;
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++i;
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}
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// now renumber the edges
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for (int i = 0; i < edges.size(); ++i) {
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edges[i].first = reindexing_key[edges[i].first];
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edges[i].second = reindexing_key[edges[i].second];
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}
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}
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void flip_neg_edges(vector<Edge> &edges, vector<double> &weights) {
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// now renumber the edges
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for (int i = 0; i < edges.size(); ++i) {
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if (weights[i] < 0.0) {
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double tmp = edges[i].second;
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edges[i].second = edges[i].first;
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edges[i].first = tmp;
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weights[i] *= -1;
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}
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}
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}
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void mfas_ratio(const std::vector<Edge> &edges,
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const std::vector<double> &weights, std::vector<int> &order) {
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// find the number of nodes in this problem
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int n = -1;
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int m = edges.size();
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for (int i = 0; i < m; ++i) {
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n = (edges[i].first > n) ? edges[i].first : n;
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n = (edges[i].second > n) ? edges[i].second : n;
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}
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n += 1; // 0 indexed
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// initialize data structures
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vector<double> win_deg(n, 0.0);
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vector<double> wout_deg(n, 0.0);
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vector<bool> unchosen(n, 1);
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vector<vector<pair<int, double> > > inbrs(n);
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vector<vector<pair<int, double> > > onbrs(n);
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// stuff data structures
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for (int ii = 0; ii < m; ++ii) {
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int i = edges[ii].first;
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int j = edges[ii].second;
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double w = weights[ii];
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win_deg[j] += w;
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wout_deg[i] += w;
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inbrs[j].push_back(pair<int, double>(i, w));
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onbrs[i].push_back(pair<int, double>(j, w));
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}
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while (order.size() < n) {
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// choose an unchosen node
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int choice = -1;
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double max_score = 0.0;
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for (int i = 0; i < n; ++i) {
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if (unchosen[i]) {
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// is this a source
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if (win_deg[i] < 1e-8) {
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choice = i;
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break;
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} else {
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double score = (wout_deg[i] + 1) / (win_deg[i] + 1);
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if (score > max_score) {
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max_score = score;
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choice = i;
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}
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}
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}
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}
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// find its inbrs, adjust their wout_deg
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vector<pair<int, double> >::iterator it;
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for (it = inbrs[choice].begin(); it != inbrs[choice].end(); ++it)
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wout_deg[it->first] -= it->second;
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// find its onbrs, adjust their win_deg
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for (it = onbrs[choice].begin(); it != onbrs[choice].end(); ++it)
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win_deg[it->first] -= it->second;
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order.push_back(choice);
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unchosen[choice] = 0;
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}
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}
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void broken_weight(const std::vector<Edge> &edges,
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const std::vector<double> &weight,
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const std::vector<int> &order, std::vector<double> &broken) {
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// clear the output vector
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int m = edges.size();
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broken.resize(m);
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broken.assign(broken.size(), 0.0);
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// find the number of nodes in this problem
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int n = -1;
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for (int i = 0; i < m; ++i) {
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n = (edges[i].first > n) ? edges[i].first : n;
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n = (edges[i].second > n) ? edges[i].second : n;
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}
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n += 1; // 0 indexed
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// invert the permutation
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std::vector<int> inv_perm(n, 0.0);
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for (int i = 0; i < n; ++i) inv_perm[order[i]] = i;
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// find the broken edges
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for (int i = 0; i < m; ++i) {
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int x0 = inv_perm[edges[i].first];
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int x1 = inv_perm[edges[i].second];
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if ((x1 - x0) * weight[i] < 0)
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broken[i] += weight[i] > 0 ? weight[i] : -weight[i];
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}
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}
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@ -0,0 +1,25 @@
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/*
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This file contains the code to solve a Minimum feedback arc set (MFAS) problem
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Copyright (c) 2014, Kyle Wilson
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All rights reserved.
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*/
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#ifndef __MFAS_H__
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#define __MFAS_H__
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#include <map>
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#include <vector>
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typedef std::pair<int, int> Edge;
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void mfas_ratio(const std::vector<Edge> &edges,
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const std::vector<double> &weight, std::vector<int> &order);
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void reindex_problem(std::vector<Edge> &edges,
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std::map<int, int> &reindexing_key);
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void flip_neg_edges(std::vector<Edge> &edges, std::vector<double> &weights);
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void broken_weight(const std::vector<Edge> &edges,
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const std::vector<double> &weight,
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const std::vector<int> &order, std::vector<double> &broken);
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#endif // __MFAS_H__
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