/***********************************************************************
* Author: Isai Damier
* Title: Bellman-Ford's Single Source Shortest Path
* Project: geekviewpoint
* Package: algorithm.graph
*
* Statement:
* Given a source vertex and a directed, weighted graph; find the
* shortest path between the source vertex and each of the other
* vertices. Some of the edges may have negative weights. If the graph
* contains any negative weight cycle, throw an exception.
*
* Time-complexity: O(V*E)
* where V is the number of vertices and E is the number of edges.
*
* The Bellman-Ford Strategy:
*
* The Bellman-Ford argument is that the longest path in any graph
* can have at most V-1 edges, where V is the number of vertices.
* Furthermore, if we perform relaxation on the set of edges once, then
* we will at least have determined all the one-edged shortest paths;
* if we traverse the set of edges twice, we will have solved at least
* all the two-edged shortest paths; ergo, after the V-1 iteration thru
* the set of edges, we will have determined all the shortest paths
* in the graph.
*
* But then Bellman-Ford continues: If after V-1 iterations we are still
* able to relax any path, then the graph has a negative edge cycle.
*
* From that argument, even before looking at any code, we can see
* that the time complexity is (V-1)*E +1*E = V*E. The 1*E is the
* cycle-detection pass.
*
* The argument also implicitly tells us that this algorithm does not
* mind handling negatives edges -- only negative cycles are scary.
*
* Hence, while Bellman-Ford takes longer than all versions of
* Dijkstra's algorithm, it has the advantage of accepting all
* directed graphs.
*
* I use the term relaxation a few times. Here is the explanation.
* The computation starts by estimating that all vertices are infinitely
* far away from the source vertex. Then as we compute, we relax that
* outrageous assumption by checking if there is actually a shorter
* distance. For instance, say we have already computed the shortest
* distance from s to v as sv and now we are seeking the solution from
* s to x, where x is adjacent to v. Then we know sx <= sv + vx; which
* simply means, either the path from s to v to x is the shortest path
* or there is yet a shorter path than that. If however we find that
* our "outrageous" estimate so far is that sx > sv+vx, then we can
* set sx = sv+vx as our new better estimate. That's it. That's
* relaxation.
**********************************************************************/
package algorithm.graph;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class BellmanFord {
public Integer[][] singleSourceShortestPath(Integer[][] weight,
int source) throws Exception
{
//auxiliary constants
final int SIZE = weight.length;
final int EVE = -1;//to indicate no predecessor
final int INFINITY = Integer.MAX_VALUE;
//declare and initialize pred to EVE and minDist to INFINITY
Integer[] pred = new Integer[SIZE];
Integer[] minDist = new Integer[SIZE];
Arrays.fill(pred, EVE);
Arrays.fill(minDist, INFINITY);
//set minDist[source] = 0 because source is 0 distance from itself.
minDist[source] = 0;
//relax the edge set V-1 times to find all shortest paths
for (int i = 1; i < minDist.length - 1; i++) {
for (int v = 0; v < SIZE; v++) {
for (int x : adjacency(weight, v)) {
if (minDist[x] > minDist[v] + weight[v][x]) {
minDist[x] = minDist[v] + weight[v][x];
pred[x] = v;
}
}
}
}
//detect cycles if any
for (int v = 0; v < SIZE; v++) {
for (int x : adjacency(weight, v)) {
if (minDist[x] > minDist[v] + weight[v][x]) {
throw new Exception("Negative cycle found");
}
}
}
Integer[][] result = {pred, minDist};
return result;
}
/******************************************************************
* Retrieve all the neighbors of vertex v.
*****************************************************************/
private List<Integer> adjacency(Integer[][] G, int v) {
List<Integer> result = new ArrayList<Integer>();
for (int x = 0; x < G.length; x++) {
if (G[v][x] != null) {
result.add(x);
}
}
return result;
}
}
package algorithm.graph;
import org.junit.Test;
import static org.junit.Assert.*;
public class BellmanFordTest {
public BellmanFordTest() {
}
@Test
public void testBellmanFordWithPositiveEdges() throws Exception {
System.out.println("singleSourceShortestPath");
Integer[][] weight = {
{null, 10, null, null, 3},
{null, null, 2, null, 1},
{null, null, null, 7, null},
{null, null, 9, null, null},
{null, 4, 8, 2, null}
};
int source = 0;
BellmanFord instance = new BellmanFord();
Integer[][] expResult = {{-1, 4, 1, 4, 0}, {0, 7, 9, 5, 3}};
Integer[][] result = instance.singleSourceShortestPath(weight, source);
assertArrayEquals(expResult, result);
}
@Test
public void testBellmanFordWithNegativeEdges() throws Exception {
System.out.println("singleSourceShortestPath");
Integer[][] weight = {
{null, -1, 4, null, null},
{null, null, 3, 2, 2},
{null, null, null, null, null},
{null, 1, 5, null, null},
{null, null, null, -3, null}
};
int source = 0;
BellmanFord instance = new BellmanFord();
Integer[][] expResult = {{-1, 0, 1, 4, 1}, {0, -1, 2, -2, 1}};
Integer[][] result = instance.singleSourceShortestPath(weight, source);
assertArrayEquals(expResult, result);
}
@Test
public void testBellmanFordWithNegativeCycle() {
System.out.println("singleSourceShortestPath");
Integer[][] weight = {
{null, -1, 4, null, null},
{null, null, 3, 2, 2},
{null, -6, null, null, null},
{null, 1, 5, null, null},
{null, null, null, -3, null}
};
int source = 0;
BellmanFord instance = new BellmanFord();
try {
instance.singleSourceShortestPath(weight, source);
fail("Should have thrown an exception: Negative weight cycle");
} catch (Exception ex) {
assertTrue(true);
}
}
}