Dijkstra's Shortest Pathby Isai Damier, Android Engineer @ Google

```/*********************************************************************
* Author: Isai Damier
* Title:  Dijkstra's Single Source Shortest Path
* Project: geekviewpoint
* Package: algorithm.graph
*
* Statement:
*   I live in the Bay Area (a.k.a. Silicon Valley). But my parents live
*   in Boston. Suppose I own a small airplane and know the distance
*   between all pairs of airports that my plane could reach in one
*   nonstop flight. How do I find the shortest path from my airport
*   to my parent's airport?
*
* Time-complexity: O((V+E) log V)
*   where V is the number of airports (i.e. vertices/nodes) and E is
*   the number of paths connecting all of the airports.
*
* Greed as a Virtue:
*
*   Oftentimes, the worst vices sublimate into the greatest virtues.
*   As shown by the late Professor Edsger Wybe Dijkstra, the best known
*   solution to this problem is a greedy algorithm. If we call my
*   starting airport s and my ending airport e, then the intuition
*   governing Dijkstra's "Single Source Shortest Path" algorithm
*   goes like this:
*
*   0] To begin with, my probable itinerary T only contains one airport,
*       s, since that's where I am starting from: T={s}
*   1] Then among all the airports that I could reach from s thru a
*      nonstop flight, I pick the nearest one (the greedy choice), say
*      v2. So now T={s,v2}. I also write down some notes next to v2:
*      the distance from s to v2 and that I went directly from s to v2.
*   2] Then among all the airports that I could fly directly into from
*      either s or v2, I pick the one nearest to s, say w3. Again, I
*      take some notes next to w3: the distance from s to w3 and whether
*      I reached w3 from v2 or from s. So now T={s,v2,w3}
*   3] Then I repeat the process until my probable itinerary contains e
*      (i.e. Logan Airport in Boston). As soon as I reach e I can stop
*      computing. Say T={s,v2,w3,w4,x5,y5,y6,z6,e}.
*   4] At this point, to trace out the actual path I am to take, I
*      backtrack from e to s. Recall that I always took note of how I
*      reached an airport. So if I reached e from z6, z6 from y5, y5
*      from w4, w4 from w3, w3 from s, then to go from s to e I will
*     travel s -> w3 -> w4 -> y5 -> z6 -> e as my shortest path.
*
*   You probably noticed that v2 and x5 are not on the final path.
*   That's okay. That's the price for being greedy. But on the bright
*   side (if you can see it), I know the shortest distance from s to all
*   the airports that's in T. Hence, not only do I now know the shortest
*   distance from s to e, I also know the shortest distance from s to z6,
*   from s to y5, from s to x5, etc.
*
*   Hence in solving for the shortest distance from s to e,
*   I inadvertently may have solved for the shortest distance to all
*   the airports in America without any additional effort. I say "may"
*   because you may choose to halt the search after reaching e. But in
*   reality Dijkstra's algorithm tend to not supply e as a parameter;
*   consequently, you normally solve for the distance from s to all
*   the other vertices. Then you use the result as you wish; say to
*   find the shortest path to e. Again, there is no additional cost:
*   Either way, time complexity is O((V+E) log V).
*
*
* Actual Algorithm:
*
*  0] If I am at airport v, then I need to know how I got there and how
*     far v is from s. Therefore, declare array pred{} to track the
*     predecessor of v and array minDist to track the distance of v
*     from s; so that pred[v]=u would mean I got to v from airport u,
*     and minDist[v]=100 would mean the distance between s and v is 100.
*
*  1] Since we have not done any computation yet, we will fill pred with
*     -1s and we will fill minDist with infinities. Both meaning
*     undefined.
*
*  2] Since we are starting at s, we can say that the distance from
*     where we are to s is zero. Therefore set minDist[s]=0.
*
*  3] Create a priorityQueue containing all vertices, sorted by
*     minDist; so that vertex s will head the queue since it is the
*     only vertex with a minDist of zero while all others are infinity.
*
*  4] While queue is not empty, remove the head vertex as v:
*
*  5] For each vertex x that is adjacent to v and that is still on the
*     queue, recalculate the distance from s to x. Note that as long as
*     a vertex is still on the queue it is effectively not in T
*    (the probable itinerary), and that by removing a vertex v from the
*    queue we are essentially adding it to T.
*
*  6] Re-prioritize the queue based on the recalculated values of minDist.
*
*  After step 6 the computation is done. We are left with two arrays:
*  pred and minDist. pred tells us the path from s to e, and minDist
*  tells us the distance along the path.
*
**********************************************************************/
package algorithm.graph;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.PriorityQueue;

public class DijkstraSingleSourceShortestPath {

public Integer[][] singleSourceShortestPath(Integer[][] weight,int source){
//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;

PriorityQueue<Integer[]> pq = createPriorityQueue(minDist);

while (!pq.isEmpty()) {
updatePriorityQueue(pq);
int v = pq.remove()[0];
for (Integer[] XD : adjacency(weight, pq, v)) {
Integer x = XD[0];
//triangle inequality
if (null != x && minDist[x] > minDist[v] + weight[v][x]) {
minDist[x] = minDist[v] + weight[v][x];
pred[XD[0]] = v;
XD[1] = minDist[x];//update pq.
}
}
}
Integer[][] result = {pred, minDist};
return result;
}

/*********************************************************************
* Create a priority queue and load it with the vertices sorted by
* minDist.
********************************************************************/
private PriorityQueue<Integer[]> createPriorityQueue(Integer[] dist) {
PriorityQueue<Integer[]> pq = new PriorityQueue<Integer[]>(11,
new Comparator<Integer[]>() {
public int compare(Integer[] A, Integer[] B) {
return A[1] < B[1] ? -1 : 1;
}
});
for (int v = 0; v < dist.length; v++) {
}
return pq;
}

/******************************************************************
*  Retrieve all the neighbors of vertex v that are
*  in the priority queue pq.
*****************************************************************/
private List<Integer[]> adjacency(Integer[][] G,
PriorityQueue<Integer[]> pq, int v) {
List<Integer[]> result = new ArrayList<Integer[]>();
for (Integer[] ent : pq) {// {u,key[u]}
int u = ent[0];
if (G[v][u] != null) {
}
}
return result;
}

/*****************************************************************
* Re-prioritize the queue based on changes in the
*    minDist array.
*
* Technical Details: Dijktra's algorithm requires a priority queue
*    that changes continuously to reflect changes in minDist.
*    For Java it does not suffice to simply pass new values to
*    the array objects that constitute the queue. The
*    PriorityQueue data structure in Java only checks its structure
*    when it is adding or removing elements. It is unaware of any
*    direct changes to the objects it comprises. Therefore to force
*    the queue to re-prioritize, an element is removed and then
*    immediately added back.
*
*****************************************************************/
private void updatePriorityQueue(PriorityQueue<Integer[]> pq) {
}
}
```
```package algorithm.graph;

import org.junit.Test;
import static org.junit.Assert.*;

public class DijkstraSingleSourceShortestPathTest {

/**
* Test of singleSourceShortestPath method,
* of class DijkstraSingleSourceShortestPath.
*/
@Test
public void testSingleSourceShortestPath() {
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;
DijkstraSingleSourceShortestPath dijkstra =
new DijkstraSingleSourceShortestPath();

Integer[][] expResult = {{-1, 4, 1, 4, 0}, {0, 7, 9, 5, 3}};
Integer[][] result = dijkstra.singleSourceShortestPath(weight, source);
assertArrayEquals(expResult, result);

weight = new Integer[][]{
{null, 6, 8, 18, null, null},
{null, null, null, null, 11, null},
{null, null, null, 9, null, null},
{null, null, null, null, null, null},
{null, null, null, null, null, 3},
{null, null, 7, 4, null, null},};

expResult = new Integer[][]{{-1, 0, 0, 2, 1, 4}, {0, 6, 8, 17, 17, 20}};
result = dijkstra.singleSourceShortestPath(weight, source);
assertArrayEquals(expResult, result);
}
}
```