In this article, we are going to talk about how Dijkstras algorithm finds the shortest path between nodes in a network and write a Python script to illustrate the same.
Dijkstra's Shortest Path Algorithm in Network routing using Python
We will use an example of network routing.
A basic understanding of Python and its OOP concepts is needed.
We will first talk about some basic graph concepts because we are going to use them in this article.
A basic introduction to Graphs
Graphs are pictorial representations of connections between pairs of elements. The graphs in our case represent a network topology.
The connections are referred to as edges while the elements are called nodes.
We have three types of graphs:
- Undirected: You can move using the edges towards any direction.
- Directed: The direction you can move is specified and shown using arrows.
- Weighted: The edges of weighted graphs denote a certain metric like distance, time taken to move using the edges, etc.
Dijkstra's shortest path algorithm
This algorithm is used to calculate and find the shortest path between nodes using the weights given in a graph. (In a network, the weights are given by link-state packets and contain information such as the health of the routers, traffic costs, etc.).
Summary of the working
It starts with the source node and finds the rest of the distances from the source node. Dijkstra's algorithm keeps track of the currently known distance from the source node to the rest of the nodes and dynamically updates these values if a shorter path is found.
A node is then marked as visited and added to the path if the distance between it and the source node is the shortest. This continues until all the nodes have been added to the path, and finally, we get the shortest path from the source node to all other nodes, which packets in a network can follow to their destination.
- We need positive weights because they have to be added to the computations to achieve our goal. Negative weights would make the algorithm not give the desired results.
An example illustrating the working of the algorithm
The source node here is node 0. We assume the weights show the distances.
Initially, we have this list of distances. We mark the initial distances as INF (infinity) because we have not yet determined the actual distance except for node 0. After all, the distance from the node 0 to itself is 0.
NODE | DISTANCE |
---|---|
0 | 0 |
1 | INF |
2 | INF |
3 | INF |
4 | INF |
5 | INF |
6 | INF |
We also have a list to keep track of only the visited nodes, and since we have started with node 0, we add it to the list (we denote a visited node by adding an asterisk beside it in the table and a red border around it on the graph).
{0}
We check the distances 0 -> 1
and 0 -> 2
, which are 2 and 6, respectively. We first update the distances from nodes 1 and 2 in the table.
NODE | DISTANCE |
---|---|
0 | 0 |
1 | 2 |
2 | 6 |
3 | INF |
4 | INF |
5 | INF |
6 | INF |
We then choose the shortest one, which is 0 -> 1
and mark node 1 as visited and add it to the visited path list.
NODE | DISTANCE |
---|---|
0 | 0 |
1 | 2* |
2 | 6 |
3 | INF |
4 | INF |
5 | INF |
6 | INF |
{0,1}
Next, we check the nodes adjacent to the nodes added to the path(Nodes 2 and 3). We then update our distance table with the distance from the source node to the new adjacent node, node 3 (2 + 5 = 7).
To choose what to add to the path, we select the node with the shortest currently known distance to the source node, which is 0 -> 2
with distance 6.
NODE | DISTANCE |
---|---|
0 | 0 |
1 | 2* |
2 | 6* |
3 | 7 |
4 | INF |
5 | INF |
6 | INF |
{0,1,2}
Next we have the distances 0 -> 1 -> 3
(2 + 5 = 7) and 0 -> 2 -> 3
(6 + 8 = 14) in which 7 is clearly the shorter distance, so we add node 3 to the path and mark it as visited.
NODE | DISTANCE |
---|---|
0 | 0 |
1 | 2* |
2 | 6* |
3 | 7* |
4 | INF |
5 | INF |
6 | INF |
{0,1,2,3}
We then check the next adjacent nodes (node 4 and 5) in which we have 0 -> 1 -> 3 -> 4
(7 + 10 = 17) for node 4 and 0 -> 1 -> 3 -> 5
(7 + 15 = 22) for node 5. We add node 4.
NODE | DISTANCE |
---|---|
0 | 0 |
1 | 2* |
2 | 6* |
3 | 7* |
4 | 17* |
5 | 22 |
6 | INF |
{0,1,2,3,4}
In the same way, we check the adjacent nodes(nodes 5 and 6).
Node 5:
- Option 1:
0 -> 1 -> 3 -> 5
(7 + 15 = 22) - Option 2:
0 -> 1 -> 3 -> 4 -> 5
(17 + 6 = 23) - Option 3:
0 -> 1 -> 3 -> 4 -> 6 -> 5
(17 + 2 + 6 = 25) We choose 22.
Node 6
0 -> 1 -> 3 -> 4 -> 6
(17 + 2 = 19)
NODE | DISTANCE |
---|---|
0 | 0 |
1 | 2* |
2 | 6* |
3 | 7* |
4 | 17* |
5 | 22* |
6 | 19* |
{0,1,2,3,4,5,6}
Python code and explanation
We have the Python code below to illustrate the process above:
class Graph():
# A constructor to iniltialize the values
def __init__(self, nodes):
#distance array initialization
self.distArray = [0 for i in range(nodes)]
#visited nodes initialization
self.vistSet = [0 for i in range(nodes)]
#initializing the number of nodes
self.V = nodes
#initializing the infinity value
self.INF = 1000000
#initializing the graph matrix
self.graph = [[0 for column in range(nodes)]
for row in range(nodes)]
def dijkstra(self, srcNode):
for i in range(self.V):
#initialise the distances to infinity first
self.distArray[i] = self.INF
#set the visited nodes set to false for each node
self.vistSet[i] = False
#initialise the first distance to 0
self.distArray[srcNode] = 0
for i in range(self.V):
# Pick the minimum distance node from
# the set of nodes not yet processed.
# u is always equal to srcNode in first iteration
u = self.minDistance(self.distArray, self.vistSet)
# Put the minimum distance node in the
# visited nodes set
self.vistSet[u] = True
# Update dist[v] only if is not in vistSet, there is an edge from
# u to v, and total weight of path from src to v through u is
# smaller than current value of dist[v]
for v in range(self.V):
if self.graph[u][v] > 0 and self.vistSet[v] == False and self.distArray[v] > self.distArray[u] + self.graph[u][v]:
self.distArray[v] = self.distArray[u] + self.graph[u][v]
self.printSolution(self.distArray)
#A utility function to find the node with minimum distance value, from
# the set of nodes not yet included in shortest path tree
def minDistance(self, distArray, vistSet):
# Initilaize minimum distance for next node
min = self.INF
# Search not nearest node not in the
# unvisited nodes
for v in range(self.V):
if distArray[v] < min and vistSet[v] == False:
min = distArray[v]
min_index = v
return min_index
def printSolution(self, distArray):
print ("Node \tDistance from 0")
for i in range(self.V):
print (i, "\t", distArray[i])
#Display our table
ourGraph = Graph(7)
ourGraph.graph = [[0, 2, 6, 0, 0, 0, 0],
[2, 0, 0, 5, 0, 0, 0],
[6, 6, 0, 8, 0, 0, 0],
[0, 0, 8, 0, 10, 15, 0],
[0, 0, 0, 10, 0, 6, 2],
[0, 0, 0, 15, 6, 0, 6],
[0, 0, 0, 0, 2, 6, 0],
];
ourGraph.dijkstra(0)
Explanation
We have a constructor for giving initial _init_
values and three user-defined functions:
printSolution()
minDistance()
dijkstra()
The constructor takes the parameter nodes
, which is the number of nodes to analyze.
def __init__(self, nodes):
#distance array initialization
self.distArray = [0 for i in range(nodes)]
#visited nodes initialization
self.vistSet = [0 for i in range(nodes)]
#initializing the number of nodes
self.V = nodes
#initializing the infinity value
self.INF = 1000000
#initializing the graph matrix
self.graph = [[0 for column in range(nodes)]
for row in range(nodes)]
dijkstra()
takes a parameter, the source node (srcNode
). It then first initializes each distance to infinity and visited status to false to show the node is unvisited using a for loop and the initial distance from the source node to 0.
In the next loop, it first picks the node with the minimum distance from the set of nodes not yet processed.u
is always equal to srcNode
in the first iteration.
It then adds the node with the minimum distance in the visited nodes set by setting the value to True
. In the last loop, which is in the second loop, the code updates the distance of the node from node 0.
dist[v]
only if it is not in visited list array, vistSet[],
and if there is an edge from u
to v,
and the total distance of path from srcNode
to v
through u
is less than the current value of dist[v].
It then calls the printSolution()
to display the table after passing the distance array to the function.
def dijkstra(self, srcNode):
for i in range(self.V):
self.distArray[i] = self.INF
self.vistSet[i] = False
self.distArray[srcNode] = 0
for i in range(self.V):
u = self.minDistance(self.distArray, self.vistSet)
self.vistSet[u] = True
for v in range(self.V):
if self.graph[u][v] > 0 and self.vistSet[v] == False and self.distArray[v] > self.distArray[u] + self.graph[u][v]:
self.distArray[v] = self.distArray[u] + self.graph[u][v]
self.printSolution(self.distArray)
minDistance()
checks for the nearest node in the distArray
not included in the unvisited nodes in the array vistSet[v]
. It then returns the node's index. It takes two arrays as parameters distArray
and vistSet[v]
.
def minDistance(self, distArray, vistSet):
min = self.INF
for v in range(self.V):
if distArray[v] < min and vistSet[v] == False:
min = distArray[v]
min_index = v
return min_index
printSolution()
is used to display the final results, which are the nodes and their respective tables stored in an array distArray
, that it takes as a parameter.
def printSolution(self, distArray):
print ("Node \tDistance from 0")
for i in range(self.V):
print (i, "\t", distArray[i])
We then create an object ourGraph
from our Graph()
class and pass to it the number of nodes.
ourGraph = Graph(7)
Next, create the matrix to store the distances.
ourGraph.graph = [[0, 2, 6, 0, 0, 0, 0],
[2, 0, 0, 5, 0, 0, 0],
[6, 6, 0, 8, 0, 0, 0],
[0, 0, 8, 0, 10, 15, 0],
[0, 0, 0, 10, 0, 6, 2],
[0, 0, 0, 15, 6, 0, 6],
[0, 0, 0, 0, 2, 6, 0],
];
The matrix is the same as the table shown below:
0 | 1 | 2 | 3 | 4 | 5 | 6 | |
---|---|---|---|---|---|---|---|
0 | 0 | 2 | 6 | 0 | 0 | 0 | 0 |
1 | 2 | 0 | 0 | 5 | 0 | 0 | 0 |
2 | 6 | 6 | 0 | 8 | 0 | 0 | 0 |
3 | 0 | 0 | 8 | 0 | 10 | 15 | 0 |
4 | 0 | 0 | 0 | 10 | 0 | 6 | 2 |
5 | 0 | 0 | 0 | 15 | 6 | 0 | 6 |
6 | 0 | 0 | 0 | 0 | 2 | 6 | 0 |
The topmost row and most left column represent the nodes. We read a node from the left column and check its distance with the topmost row. The intersection shows the distance. The distance is 0 if the nodes are not adjacent.
For example:
The distance of 0 from 0 is 0.
0 | 1 | 2 | 3 | 4 | 5 | 6 | |
---|---|---|---|---|---|---|---|
0 | 0 | 2 | 6 | 0 | 0 | 0 | 0 |
1 | 2 | 0 | 0 | 5 | 0 | 0 | 0 |
2 | 6 | 6 | 0 | 8 | 0 | 0 | 0 |
3 | 0 | 0 | 8 | 0 | 10 | 15 | 0 |
4 | 0 | 0 | 0 | 10 | 0 | 6 | 2 |
5 | 0 | 0 | 0 | 15 | 6 | 0 | 6 |
6 | 0 | 0 | 0 | 0 | 2 | 6 | 0 |
The distance of 5 from 3 is 15.
0 | 1 | 2 | 3 | 4 | 5 | 6 | |
---|---|---|---|---|---|---|---|
0 | 0 | 2 | 6 | 0 | 0 | 0 | 0 |
1 | 2 | 0 | 0 | 5 | 0 | 0 | 0 |
2 | 6 | 6 | 0 | 8 | 0 | 0 | 0 |
3 | 0 | 0 | 8 | 0 | 10 | 15 | 0 |
4 | 0 | 0 | 0 | 10 | 0 | 6 | 2 |
5 | 0 | 0 | 0 | 15 | 6 | 0 | 6 |
6 | 0 | 0 | 0 | 0 | 2 | 6 | 0 |
Finally, we display our results.
ourGraph.dijkstra(0)
The output will be:
Node Distance from 0
0 0
1 2
2 6
3 7
4 17
5 22
6 19
That's all for now. We now have a better idea on how Dijkstra's Algorithm works. I hope you can work with different graphs and language of your own.
Have a good one.
The images used were sourced from Free Code Camp.
Happy coding.
Top comments (2)
Brilliant explanation. Thank you!
Thank you @b3pr0