/*********************************************************************
* Author: Isai Damier
* Title: Shortest Path with Constraint
* Project: geekviewpoint
* Package: algorithm.graph
*
* Statement:
* In http://www.geekviewpoint.com/java/graph/dijkstra_shortest_path
* we talked about finding the shortest path from my local airport in
* Silicon Valley to my parents' airport in Boston, MA. In the
* excitement over going to see my parents, I forgot a very important
* detail: the landing fee each airport charges. You see, not all
* airports charge the same toll. So I can't just blindly pick the
* shortest path from Silicon Valley to Boston. I must budget for tolls.
* So here is the real problem I actually needed to solve:
*
* In addition to knowing the distance between all pairs of airports
* that my small airplane could reach in one nonstop fight, I also
* know how much each airport would charge for toll. So given that
* I only have c dollars for toll, what is the shortest path from
* my airport to my parents' 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.
*
* Dijkstra Constrained:
*
* It is highly recommended that you read
* http://www.geekviewpoint.com/java/graph/dijkstra_shortest_path before
* proceeding.
*
* ... waiting ... waiting ... waiting ... waiting ... waiting ...
*
* Good. So you got a taste of how life can be when money is not an issue.
* Now to my problem: How do I factor in my financial limitations?
*
* - Instead of using a vector (i.e. 1D array) to track minDist, we must
* now use a matrix (i.e. 2D array), where the rows will represent the
* distance from s to the present airport and the columns will
* represent money left for the journey.
*
* Actual Algorithm:
*
* Note: we are still using s as my local airport and e as my parents'
* airport in Boston; v can be any airport.
*
* 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][90]=100 would mean the distance between s and v is
* 100 miles and I have $90 left in my budget after paying v.
*
* 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 and that we have not yet spent any money.
* Therefore set minDist[s][c]=0.
*
* 3] Create a priorityQueue containing all vertices v, 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, if x would not take us over-budget, recalculate the
* distance from s to x. To know whether x would take us over-budget,
* get the money left after v, say m, and subtract the cost of x
* from m: m - cost[x] >= 0. You may want to manually run the
* algorithm for a small graph to see how it works.
*
* 6] Re-prioritize the queue based on the recalculated values of
* minDist and repeat thru the while-loop.
*
* 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. I only care about the path so
* I extract the path from pred. But you can do so much more as you wish.
**********************************************************************/
package algorithm.graph;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.PriorityQueue;
public class ConstrainedShortestPath {
public List<Integer> constrainedShortestPath(
Integer[][] weight,
int[] cost,
int money,
int source,
int finalDestination)
{
//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][money + 1];
Arrays.fill(pred, EVE);
for (Integer[] m : minDist) {
Arrays.fill(m, INFINITY);
}
//set minDist[source][money]=0, as source is 0 distance from itself.
minDist[source][money] = 0;
PriorityQueue<Integer[]> pq = createPriorityQueue(money, minDist);
while (!pq.isEmpty()) {
updatePriorityQueue(pq);
int v = pq.remove()[0];
for (Integer nodeData[] : adjacency(weight, pq, v)) {
Integer x = nodeData[0];
if (null == x) {
continue;
}
//find how much landing at v actually cost.
//If you can't see this, run a small graph manually.
//Recall that the first v is actuall s with all the money intact.
int m = -1;
while (m < money && minDist[v][++m] == Integer.MAX_VALUE);
//triangle inequality
int c = m - cost[x];
if (c >= 0 && minDist[x][c] >(long) minDist[v][m]+weight[v][x]){
minDist[x][c] = minDist[v][m] + weight[v][x];
nodeData[1] = minDist[x][c];
pred[x] = v;
}
}
}
List<Integer> result = new ArrayList<Integer>();
int p = finalDestination;
while (-1 != pred[p]) {
result.add(p);
p = pred[p];
}
result.add(p);
Collections.reverse(result);
return result;
}//
private PriorityQueue<Integer[]> createPriorityQueue(
int money, Integer[][] minDist)
{
PriorityQueue<Integer[]> pq = new PriorityQueue<Integer[]>(11,
new Comparator<Integer[]>() {
public int compare(Integer[] A, Integer[] B) {
return A[0] < B[0] ? -1 : 1;
}
});
for (int v = 0; v < minDist.length; v++) {
pq.add(new Integer[]{v, minDist[v][money]});
}
return pq;
}// createPriorityQueue
/******************************************************************
* 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;
}// pqNeighbors
/*****************************************************************
* 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());
}
}// ConstrainedShortestPath
package algorithm.graph;
import static org.junit.Assert.*;
import java.util.Arrays;
import java.util.List;
import org.junit.After;
import org.junit.Before;
import org.junit.Test;
public class ConstrainedShortestPathTest {
ConstrainedShortestPath dijkstra;
@Before
public void setUp() throws Exception {
dijkstra = new ConstrainedShortestPath();
}
@After
public void tearDown() throws Exception {
dijkstra = null;
}
@Test
public void testConstrainedShortestPath() {
Integer[][] D = {
{null, 2, 2, null, null, null, null, null, null, null},
{null, null, null, 4, null, 16, null, null, null, null},
{null, null, null, null, 3, null, null, 8, null, null},
{null, null, null, null, null, null, 4, 4, null, null},
{null, null, null, null, null, null, null, 4, null, null},
{null, null, null, null, null, null, null, null, 13, null},
{null, null, null, null, null, null, null, null, null, 2},
{null, null, null, null, null, null, null, null, null, 6},
{null, null, null, null, null, null, null, null, null, 4},
{null, null, null, null, null, null, null, null, 5, null}};
int[] S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int money = 199;
List<Integer> expected = Arrays.asList(0, 1, 5, 8);
List<Integer> result =
dijkstra.constrainedShortestPath(D, S, money, 0, 8);
assertEquals(expected,result);
}
}