Probabilistic shortest path routing algorithm for optical networks

Data transfer operations is a crucial aspect in case of networking and routing. So efficient data transfer operations is a must need, with minimum hardware cost (Optical Cables, WDM Network components, Decoders, Multiplexers) and also in the minimum time possible. Thus, the need is to propose an algorithm that finds the shortest path between two nodes (source node and destination node).

Let’s see a completely new algorithm unlike Dijkstra’s Shortest Path or any other algorithm for finding Shortest Path.

Given a graph and two nodes (source node and destination node), find the shortest path between them.

Let’s Calculate the distance ratio for each link :

Distance of link AB [denoted by d(AB)] = 10
Distance of link AC [denoted by d(AC)] = 12

For link AB, Distance Ratio of AB = d(AB) / (d(AB) + d(AC))
For link AC, Distance Ratio of AC = d(AC) / (d(AB) + d(AC))

Algorithm :

Given a graph and two nodes -

1. Find all the paths connecting the two nodes.
2. For each path calculate probability = (Distance Ratio). 
3. After looping over all such paths, find the path for
   which the probability turns out to be minimum.

Examples :

Input : 

Output : Shortest Path is [A -> B]


Explanation :
All possible paths are
P1 = [A->B]
P2 = [A->C->B]
P3 = [A->D->B] 

total distance D = d(P1) + d(P2) + d(P3) 
                 = (3) + (2 + 5) + (4 + 3)
                 = 17

distance ratio for P1 = d(P1) / D = 3/17
distance ratio for P2 = d(P2) / D = 7/17
distance ratio for P3 = d(P3) / D = 7/17

So the shortest path is  P1 = [A->B]




Input : 

Output : Shortest Path is [A -> B]

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Let’s illustrate the algorithm with a 7-node network and find out the Probabilistic shortest path between node 1 and node 5.

Below is the implementation :

# Python program to find Probabilistic 
# shortest path routing algorithm for 
# optical networks 
  
# importing random module 
import random 
  
# Number of nodes 
NODES = 7 
  
# very small invalid  
# when no link exists 
INVALID = 0.001  
  
  
distance_links = [[INVALID for i in range(NODES)] 
                           for j in range(NODES)] 
# distance of each link                            
distance_links[0][1] = 7
distance_links[1][0] = 7
distance_links[1][2] = 8
distance_links[2][1] = 8
distance_links[0][2] = 9
distance_links[2][0] = 9
distance_links[3][0] = 9
distance_links[0][3] = 9
distance_links[4][3] = 4
distance_links[3][4] = 4
distance_links[5][4] = 6
distance_links[4][5] = 6
distance_links[5][2] = 4
distance_links[2][5] = 4
distance_links[4][6] = 8
distance_links[6][4] = 8
distance_links[0][6] = 5
distance_links[6][0] = 5
  
  
  
# Finds next node from current node 
def next_node(s):  
    nxt = [] 
      
    for i in range(NODES): 
        if(distance_links[s][i] != INVALID): 
            nxt.append(i) 
    return nxt 
  
# Find simple paths for each 
def find_simple_paths(start, end): 
    visited = set() 
    visited.add(start) 
  
    nodestack = list() 
    indexstack = list() 
    current = start 
    i = 0
  
    while True: 
          
        # get a list of the neighbors 
        # of the current node 
        neighbors = next_node(current) 
  
        # Find the next unvisited neighbor 
        # of this node, if any 
        while i < len(neighbors) and neighbors[i] in visited:  
            i += 1
  
        if i >= len(neighbors): 
            visited.remove(current) 
              
            if len(nodestack) < 1:  
                break
                  
            current = nodestack.pop() 
            i = indexstack.pop() 
              
        elif neighbors[i] == end: 
            yield nodestack + [current, end] 
            i += 1
              
        else: 
            nodestack.append(current) 
            indexstack.append(i + 1) 
            visited.add(neighbors[i]) 
            current = neighbors[i] 
            i = 0
  
# Find the shortest path 
def solution(sour, dest): 
  
    block = 0
    l = [] 
    for path in find_simple_paths(sour, dest):  
        l.append(path)      
  
    k = 0
    for i in range(len(l)): 
        su = 0
        for j in range(1, len(l[i])): 
            su += (distance_links[l[i][j-1]] 
                                 [l[i][j]]) 
        k += su 
          
    # print k  
    dist_prob = [] 
    probability = [] 
  
    for i in range(len(l)): 
        s, su = 0, 0
      
        for j in range(1, len(l[i])):      
                  
            su += (distance_links[l[i][j-1]] 
                                 [l[i][j]]) 
                                   
        dist_prob.append(su/(1.0 * k)) 
          
      
    for m in range(len(dist_prob)): 
        z = (dist_prob[m]) 
        probability.append(z) 
  
    for i in range(len(probability)): 
        if(probability[i] == min(probability)): 
  
            z = l[i] 
            print("Shortest Path is", end = " ") 
            print(z) 
              
# Driver Code 
if __name__ == '__main__' : 
    source, dest = 1, 5
      
    # Calling the solution function 
    solution(source, dest)  

Output :

Shortest Path is [1, 2, 5]

Advantage over common Shortest Path Algorithms :
Most of the shortest path algorithms are greedy algorithms. So its based on the fact that an optimal solution leads to a globally optimal solution. In most of the cases, due to greedy property, it may not always lead to an optimal solution. But using this algorithm, one can always guarantee an optimal solution and hence the accuracy is 100%.

Probabilistic shortest path routing algorithm for optical networks

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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