Uniform_Cost_Search.py

# Python3 implementation of above approach

# returns the minimum cost in a vector( if
# there are multiple goal states)
def uniform_cost_search(goal, start):
	
	# minimum cost upto
	# goal state from starting
	global graph,cost
	answer = []

	# create a priority queue
	queue = []

	# set the answer vector to max value
	for i in range(len(goal)):
		answer.append(10**8)

	# insert the starting index
	queue.append([0, start])

	# map to store visited node
	visited = {}

	# count
	count = 0

	# while the queue is not empty
	while (len(queue) > 0):

		# get the top element of the
		queue = sorted(queue)
		p = queue[-1]

		# pop the element
		del queue[-1]

		# get the original value
		p[0] *= -1

		# check if the element is part of
		# the goal list
		if (p[1] in goal):

			# get the position
			index = goal.index(p[1])

			# if a new goal is reached
			if (answer[index] == 10**8):
				count += 1

			# if the cost is less
			if (answer[index] > p[0]):
				answer[index] = p[0]

			# pop the element
			del queue[-1]

			queue = sorted(queue)
			if (count == len(goal)):
				return answer

		# check for the non visited nodes
		# which are adjacent to present node
		if (p[1] not in visited):
			for i in range(len(graph[p[1]])):

				# value is multiplied by -1 so that
				# least priority is at the top
				queue.append( [(p[0] + cost[(p[1], graph[p[1]][i])])* -1, graph[p[1]][i]])

		# mark as visited
		visited[p[1]] = 1

	return answer

# main function
if __name__ == '__main__':
	
	# create the graph
	graph,cost = [[] for i in range(8)],{}

	# add edge
	graph[0].append(1)
	graph[0].append(3)
	graph[3].append(1)
	graph[3].append(6)
	graph[3].append(4)
	graph[1].append(6)
	graph[4].append(2)
	graph[4].append(5)
	graph[2].append(1)
	graph[5].append(2)
	graph[5].append(6)
	graph[6].append(4)

	# add the cost
	cost[(0, 1)] = 2
	cost[(0, 3)] = 5
	cost[(1, 6)] = 1
	cost[(3, 1)] = 5
	cost[(3, 6)] = 6
	cost[(3, 4)] = 2
	cost[(2, 1)] = 4
	cost[(4, 2)] = 4
	cost[(4, 5)] = 3
	cost[(5, 2)] = 6
	cost[(5, 6)] = 3
	cost[(6, 4)] = 7

	# goal state
	goal = []

	# set the goal
	# there can be multiple goal states
	goal.append(6)

	# get the answer
	answer = uniform_cost_search(goal, 0)

	# print the answer
	print("Minimum cost from 0 to 6 is = ",answer[0])