/*********************************************************************
* 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++) {
pq.add(new Integer[]{v, dist[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) {
result.add(ent);
}
}
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) {
pq.add(pq.remove());
}
}
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);
}
}
