# Bellman-Ford's Shortest Pathby Isai Damier, Android Engineer @ Google

```/***********************************************************************
* 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) {
}
}
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);
}
}
}
```