## Python, graph, dijkstra_algorithm.py

``````# Title: Dijkstra's Algorithm for finding single source shortest path from scratch
# Author: Shubham Malik
# References: https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm

import math
import sys

# For storing the vertex set to retrieve node with the lowest distance

class PriorityQueue:
# Based on Min Heap
def __init__(self):
self.cur_size = 0
self.array = []
self.pos = {}  # To store the pos of node in array

def isEmpty(self):
return self.cur_size == 0

def min_heapify(self, idx):
lc = self.left(idx)
rc = self.right(idx)
if lc < self.cur_size and self.array(lc)[0] < self.array(idx)[0]:
smallest = lc
else:
smallest = idx
if rc < self.cur_size and self.array(rc)[0] < self.array(smallest)[0]:
smallest = rc
if smallest != idx:
self.swap(idx, smallest)
self.min_heapify(smallest)

def insert(self, tup):
# Inserts a node into the Priority Queue
self.pos[tup[1]] = self.cur_size
self.cur_size += 1
self.array.append((sys.maxsize, tup[1]))
self.decrease_key((sys.maxsize, tup[1]), tup[0])

def extract_min(self):
# Removes and returns the min element at top of priority queue
min_node = self.array[0][1]
self.array[0] = self.array[self.cur_size - 1]
self.cur_size -= 1
self.min_heapify(1)
del self.pos[min_node]
return min_node

def left(self, i):
# returns the index of left child
return 2 * i + 1

def right(self, i):
# returns the index of right child
return 2 * i + 2

def par(self, i):
# returns the index of parent
return math.floor(i / 2)

def swap(self, i, j):
# swaps array elements at indices i and j
# update the pos{}
self.pos[self.array[i][1]] = j
self.pos[self.array[j][1]] = i
temp = self.array[i]
self.array[i] = self.array[j]
self.array[j] = temp

def decrease_key(self, tup, new_d):
idx = self.pos[tup[1]]
# assuming the new_d is atmost old_d
self.array[idx] = (new_d, tup[1])
while idx > 0 and self.array[self.par(idx)][0] > self.array[idx][0]:
self.swap(idx, self.par(idx))
idx = self.par(idx)

class Graph:
def __init__(self, num):
self.adjList = {}  # To store graph: u -> (v,w)
self.num_nodes = num  # Number of nodes in graph
# To store the distance from source vertex
self.dist = [0] * self.num_nodes
self.par = [-1] * self.num_nodes  # To store the path

#  Edge going from node u to v and v to u with weight w
# u (w)-> v, v (w) -> u
# Check if u already in graph
else:

# Assuming undirected graph
else:

def show_graph(self):
# u -> v(w)
print(u, "->", " -> ".join(str(f"{v}({w})") for v, w in self.adjList[u]))

def dijkstra(self, src):
# Flush old junk values in par[]
self.par = [-1] * self.num_nodes
# src is the source node
self.dist[src] = 0
Q = PriorityQueue()
Q.insert((0, src))  # (dist from src, node)
if u != src:
self.dist[u] = sys.maxsize  # Infinity
self.par[u] = -1

while not Q.isEmpty():
u = Q.extract_min()  # Returns node with the min dist from source
# Update the distance of all the neighbours of u and
# if their prev dist was INFINITY then push them in Q
new_dist = self.dist[u] + w
if self.dist[v] > new_dist:
if self.dist[v] == sys.maxsize:
Q.insert((new_dist, v))
else:
Q.decrease_key((self.dist[v], v), new_dist)
self.dist[v] = new_dist
self.par[v] = u

# Show the shortest distances from src
self.show_distances(src)

def show_distances(self, src):
print(f"Distance from node: {src}")
for u in range(self.num_nodes):
print(f"Node {u} has distance: {self.dist[u]}")

def show_path(self, src, dest):
# To show the shortest path from src to dest
# WARNING: Use it *after* calling dijkstra
path = []
cost = 0
temp = dest
# Backtracking from dest to src
while self.par[temp] != -1:
path.append(temp)
if temp != src:
if v == self.par[temp]:
cost += w
break
temp = self.par[temp]
path.append(src)
path.reverse()

print(f"----Path to reach {dest} from {src}----")
for u in path:
print(f"{u}", end=" ")
if u != dest:
print("-> ", end="")

print("\nTotal cost of path: ", cost)

if __name__ == "__main__":
graph = Graph(9)
graph.show_graph()
graph.dijkstra(0)
graph.show_path(0, 4)

# OUTPUT
# 0 -> 1(4) -> 7(8)
# 1 -> 0(4) -> 2(8) -> 7(11)
# 7 -> 0(8) -> 1(11) -> 6(1) -> 8(7)
# 2 -> 1(8) -> 3(7) -> 8(2) -> 5(4)
# 3 -> 2(7) -> 4(9) -> 5(14)
# 8 -> 2(2) -> 6(6) -> 7(7)
# 5 -> 2(4) -> 3(14) -> 4(10) -> 6(2)
# 4 -> 3(9) -> 5(10)
# 6 -> 5(2) -> 7(1) -> 8(6)
# Distance from node: 0
# Node 0 has distance: 0
# Node 1 has distance: 4
# Node 2 has distance: 12
# Node 3 has distance: 19
# Node 4 has distance: 21
# Node 5 has distance: 11
# Node 6 has distance: 9
# Node 7 has distance: 8
# Node 8 has distance: 14
# ----Path to reach 4 from 0----
# 0 -> 7 -> 6 -> 5 -> 4
# Total cost of path:  21
``````