# Java : Dijkstra's Algorithm

#### Dijkstra’s Shortest Path Algorithm

• Dijkstra’s algorithm finds the shortest path in a weighted graph from a single source.
• The weighted graph for Dijkstra’s algorithm contains only positive edge weights.
• It uses a priority queue to select a node (vertex) nearest to the source that has not been edge relaxed.
The edge relaxation for a node is done as below
If ( distance [ adjacent-node ] > length-of-path-to-adjacent-node-from-current-source + distance [ current-source-node ] ) then
Note :
a) The distance [ ] array stores the shortest distance of every node from the source-node.
b) Dijkstra’s algorithm begins by initializing distance [ source ] = 0 i.e the distance from source node to itself is 0 and distance [ all_other_nodes ] = ∞.
• Since Dijkstra’s algorithm cannot handle negative edge weights, Bellman-Ford algorithm is used for finding the shortest path in a graph containing negative edge weights.
• In a graph with only positive edge weights, Dijkstra’s algorithm with a priority queue implementation runs faster in O ((E+V) log V) than Bellman-Ford O (E.V).
E represents the edges and V represents the vertices in the graph.

Algorithm : Dijkstra’s Shortest Path [ Java ]

1.     Initialize the distance from the source node S to all other nodes as infinite (999999999999) and to itself as 0.
2.    Insert an object of < node, distance > for source i.e < S, 0 > in a priority Queue
where the priority of the elements in the queue is based on the length of the distance.
3.    While the Priority Queue is not empty do
4.       object_at_top = PriorityQueue. peek();
Remove the object from the top of the Priority Queue.
current_source_node = object_at_top . node.
5.       For every adjacent_node to current_source_node do
6.            If ( distance [ adjacent_node ] > length_of_path_to_adjacent_node_from_source + distance [ current_source_node ] ) then
7.                 distance [ adjacent_node ] = length_of_path_to_adjacent_node_from_source + distance [ current_source_node ]
8.                 Update the Priority Queue with the new object < adjacent_node, distance [ adjacent_node ] >

Below data structure is used for storing the graph before applying Dijkstra’s algorithm
Adjacency List : List of pairs of adjacent nodes and their corresponding weights in the graph. Dijkstra’s algorithm step-by-step

This example of Dijkstra’s algorithm finds the shortest distance of all the nodes in the graph from a single / original source node 0.

Step 1 : Initialize the distance of the source node to itself as 0, and ∞ (a very large number) for the rest of the nodes.
Insert the object < distance_from_original_source, node > in the priority queue.
i.e Insert < 0, 0 > in the priority queue as the distance from the original source (0) to itself is 0. Step 2 : Remove the object < 0, 0 > from the front of the priority queue and relax the edge going towards every adjacent node(s) from the original source node ( 0 ).

Current
source node
node
from the original source ( 0 )
Edge relaxation
0 1 distance [ 1 ] = ∞ distance [ 1 ] > distance_between [ 0 - 1 ] + distance [ 0 ]
i.e Since ∞ > 5 + 0
Update distance, i.e distance [ 1 ] = 5 and insert object < 1, 5 > in the priority queue.
0 2 distance [ 2 ] = ∞ distance [ 2 ] > distance_between [ 0 - 2 ] + distance [ 0 ]
i.e Since ∞ > 1 + 0
Update distance, i.e distance [ 2 ] = 1 and insert object < 2, 1 > in the priority queue.
0 3 distance [ 3 ] = ∞ distance [ 3 ] > distance_between [ 0 - 3 ] + distance [ 0 ]
i.e Since ∞ > 4 + 0
Update distance, i.e distance [ 3 ] = 4 and insert object < 3, 4 > in the priority queue. Step 3 : Remove the object < 2, 1 > from the front of the priority queue and relax the edge going towards every adjacent node(s) from the current source ( 2 ).

Current
source node
node
from the original source ( 0 )
Edge relaxation
2 0 distance [ 0 ] = 0 distance [ 0 ] < distance_between [ 2 - 0 ] + distance [ 2 ]
i.e Since 0 < 1 + 1
No edge relaxation needed.
2 1 distance [ 1 ] = 5 distance [ 1 ] > distance_between [ 2 - 1 ] + distance [ 2 ]
i.e Since 5 > 3 + 1
Update distance, i.e distance [ 1 ] = 4 and insert object < 1, 4 > in the priority queue.
2 3 distance [ 3 ] = 4 distance [ 3 ] > distance_between [ 2 - 3 ] + distance [ 2 ]
i.e Since 4 > 2 + 1
Update distance, i.e distance [ 3 ] = 3 and insert object < 3, 3 > in the priority queue.
2 4 distance [ 4 ] = ∞ distance [ 4 ] > distance_between [ 2 - 4 ] + distance [ 2 ]
i.e Since ∞ > 1 + 1
Update distance, i.e distance [ 4 ] = 2 and insert object < 4, 2 > in the priority queue. Step 4 : Remove the object < 4, 2 > from the from of the priority queue and relax the edge going towards every adjacent node(s) from the current source ( 4 ).

Current
source node
node
from the original source ( 0 )
Edge relaxation
4 1 distance [ 1 ] = 4 distance [ 1 ] < distance_between [ 4 - 1 ] + distance [ 4 ]
i.e Since 4 < 8 + 2
No edge relaxation needed.
4 2 distance [ 2 ] = 1 distance [ 2 ] < distance_between [ 4 - 2 ] + distance [ 4 ]
i.e Since 1 < 1 + 2
No edge relaxation needed.
4 3 distance [ 3 ] = 3 distance [ 3 ] < distance_between [ 4 - 3 ] + distance [ 4 ]
i.e Since 3 < 2 + 2
No edge relaxation needed.
4 5 distance [ 5 ] = ∞ distance [ 5 ] > distance_between [ 4 - 5 ] + distance [ 4 ]
i.e Since ∞ > 3 + 2
Update distance, i.e distance [ 5 ] = 5 and insert object < 5, 5 > in the priority queue. Step 5 : Remove the object < 3, 3 > from the front of the priority queue and relax the edge going towards every adjacent node(s) from the current source ( 3 ).

Current
source node
node
from the original source ( 0 )
Edge relaxation
3 0 distance [ 0 ] = 0 distance [ 0 ] < distance_between [ 3 - 0 ] + distance [ 3 ]
i.e Since 0 < 4 + 3
No edge relaxation needed.
3 2 distance [ 2 ] = 1 distance [ 2 ] < distance_between [ 3 - 2 ] + distance [ 3 ]
i.e Since 1 < 2 + 3
No edge relaxation needed.
3 4 distance [ 4 ] = 2 distance [ 4 ] < distance_between [ 3 - 4 ] + distance [ 3 ]
i.e Since 2 < 2 + 3
No edge relaxation needed.
3 5 distance [ 5 ] = 5 distance [ 5 ] > distance_between [ 3 - 5 ] + distance [ 3 ]
i.e Since 5 > 1 + 3
Update distance, i.e distance [ 5 ] = 4 and insert object < 5, 4 > in the priority queue. Step 6 : Remove the object < 4, 1 > from the front of the priority queue and relax the edge going towards every adjacent node(s) from the current source ( 1 ).

Current
source node
node
from the original source ( 0 )
Edge relaxation
1 0 distance [ 0 ] = 0 distance [ 0 ] < distance_between [ 1 - 0 ] + distance [ 1 ]
i.e Since 0 < 5 + 4
No edge relaxation needed.
1 2 distance [ 2 ] = 1 distance [ 2 ] < distance_between [ 1 - 2 ] + distance [ 1 ]
i.e Since 1 < 3 + 4
No edge relaxation needed.
1 4 distance [ 4 ] = 2 distance [ 4 ] < distance_between [ 1 - 4 ] + distance [ 1 ]
i.e Since 2 < 8 + 4
No edge relaxation needed. Step 7 : Remove the object < 4, 5 > from the front of the priority queue and relax the edge going towards every adjacent node(s) from the current source ( 5 ).

Current
source node
node
from the original source ( 0 )
Edge relaxation
5 3 distance [ 3 ] = 3 distance [ 3 ] < distance_between [ 5 - 3 ] + distance [ 5 ]
i.e Since 3 < 1 + 4
No edge relaxation needed.
5 4 distance [ 4 ] = 2 distance [ 4 ] < distance_between [ 5 - 4 ] + distance [ 5 ]
i.e Since 2 < 3 + 4
No edge relaxation needed.

The algorithm terminates here as the priority queue is empty and we have calculated the shortest path to all the nodes from the original source 0 as shown below. Data structure used for running Dijkstra’s shortest path : Distance based priority queue for choosing the vertex nearest to the source.
Graph type: Designed for weighted (directed / un-directed) graph containing positve edge weights.
Time complexity of Dijkstra’s algorithm : O ( (E+V) Log(V) ) for an adjacency list implementation of a graph. V is the number of vertices and E is the number of edges in a graph.

``````import java.util.ArrayList;
import java.util.List;
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.Comparator;

class NodeDist {

long dist; // Distance to adjacent node

NodeDist (int node, long dist) {
this.node = node;
this.dist = dist;
}
}

class Dijkstras {

void ShortestPath (int source_node, int node_count, List<List<NodeDist>> graph) {

// Assume that the distance from the source_node to other nodes is infinite
// in the beginnging, i.e initialize the distance list to a max value
Long INF = (long) 999999999;
List<Long> dist = new ArrayList<Long>(Collections.nCopies(node_count, INF));

// Distance from the source vertex to itself is 0
dist.set(source_node, (long) 0); // (Node, Dist)

// Comparator lambda function that enables the priority queue to store the nodes
// based on the distance in the ascending order.
Comparator<NodeDist> NodeDistComparator = (obj1, obj2) -> {
if (obj1.dist < obj2.dist)
return 1;
if (obj1.dist > obj2.dist)
return -1;
return 0;
};

// Priority queue stores the object node-distance into the queue with
// the smallest distance node at the top.
PriorityQueue<NodeDist> pq = new PriorityQueue<>(NodeDistComparator);

while (!pq.isEmpty()) {

NodeDist obj = pq.peek();
pq.remove();

int current_source = obj.node;

for (NodeDist obj_node_dist : graph.get(current_source)) {

// Edge relaxation

// If the distance to the adjacent node is not INF, means the object <node, dist> is in the priority queue.
// Remove the object before updating it in the priority queue.
}
}
}
}

for (int i=0; i<node_count; i++)
System.out.println("Source Node(" + source_node + ") -> Destination Node(" + i + ") : " + dist.get(i));
}

public static void main(String args[]) {

int node_count = 6;
List<List<NodeDist>> graph = new ArrayList<>(node_count);

for(int i=0; i<node_count; i++) {
}

// Node 0: <1,5> <2,1> <3,4>
Collections.addAll(graph.get(0), new NodeDist(1, 5), new NodeDist(2, 1), new NodeDist(3, 4));

// Node 1: <0,5> <2,3> <4,8>
Collections.addAll(graph.get(1), new NodeDist(0, 5), new NodeDist(2, 3), new NodeDist(4, 8));

// Node 2: <0,1> <1,3> <3,2> <4,1>
Collections.addAll(graph.get(2), new NodeDist(0, 1), new NodeDist(1, 3), new NodeDist(3, 2), new NodeDist(4, 1));

// Node 3: <0,4> <2,2> <4,2> <5,1>
Collections.addAll(graph.get(3), new NodeDist(0, 4), new NodeDist(2, 2), new NodeDist(4, 2), new NodeDist(5, 1));

// Node 4: <1,8> <2,1> <3,2> <5,3>
Collections.addAll(graph.get(4), new NodeDist(1, 8), new NodeDist(2, 1), new NodeDist(3, 2), new NodeDist(5, 3));

// Node 5: <3,1> <4,3>
Collections.addAll(graph.get(5), new NodeDist(3, 1), new NodeDist(4, 3));

int source_node = 0;
Dijkstras d = new Dijkstras();
d.ShortestPath(source_node, node_count, graph);

System.out.println();
source_node = 5;
d.ShortestPath(source_node, node_count, graph);
}
}
``````

Output

``````Source Node(0) -> Destination Node(0) : 0
Source Node(0) -> Destination Node(1) : 4
Source Node(0) -> Destination Node(2) : 1
Source Node(0) -> Destination Node(3) : 3
Source Node(0) -> Destination Node(4) : 2
Source Node(0) -> Destination Node(5) : 4

Source Node(5) -> Destination Node(0) : 4
Source Node(5) -> Destination Node(1) : 6
Source Node(5) -> Destination Node(2) : 3
Source Node(5) -> Destination Node(3) : 1
Source Node(5) -> Destination Node(4) : 3
Source Node(5) -> Destination Node(5) : 0
``````