Traveling Salesman Problem (TSP) Implementation in Python
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- Опубліковано 18 січ 2021
- Hey Guys,
In this video will learn how to solve the traveling salesman problem with the help of Python programming language.
The traveling salesman problem asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the origin city?
Source Code: github.com/pythonpool/python-...
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Related Questions:
what is the traveling salesman problem, how to solve traveling salesman problem, what is traveling salesman problem, how to solve the traveling salesman problem, traveling salesman problem where the cost depends on the order visited, how will you write a monte carlo type of solver for traveling salesman problem?, what did people discover while working on the traveling salesman problem, traveling salesman problem how to solve, now, consider a traveling salesman problem in which the salesman starts at city 0 a, for the traveling salesman problem applied to seven cities, how many distinct tours are possible?
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#TravelingSalesmanProblem #TSP #TSPPython
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This appears to be a brute force algorithm since all permutations are evaluated for cost and the minimum of the lowest-so-far cost and current cost is saved after each permutation. Thank you for sharing.
Very well explained and indeed this can be achieved in two ways either on Dynamic prog or through Brute force algorithm (first evaluate all possible tours and afterwards choose the one for which we get the shortest distance). I ran this prog in Spyder (5.3.3) but the output for me 95 instead of 80. I am still figuring it out why the output is 95. Thnaks for sharing this code. Cheers
Helped me..thank you☺️
sir which algorithm is been used ?
Can somebody tell me how to print the path you take through the nods ?
which algorithm did you use?
Thanks
Would this work, if any node has same distance to the any other two nodes?
Yes
what if its start from node 2 or 3 or 4??
It is not important is TSP because we can say that 1-5-3-2-4-1 is equal to the 2-4-1-5-3-2. Tsp creates a circle.
how can we print the path
You lost me when started defining the second function, could someone explain this part a bit? what is this function actually doing?
Can You use heurístic for tsp?
Is there any way to print the path which gives the min cost?
TRY THIS CODE!!!
from sys import maxsize
from itertools import permutations
V = 4
# implementation of traveling Salesman Problem
def travellingSalesmanProblem(graph, s):
# store all vertex apart from source vertex
vertex = []
for i in range(V):
if i != s:
vertex.append(i)
# store minimum weight Hamiltonian Cycle
min_path = maxsize
a = []
next_permutation = list(permutations(vertex))
for i in next_permutation:
# store current Path weight(cost)
current_pathweight = 0
# compute current path weight
k = s
for j in i:
current_pathweight += graph[k][j]
k = j
current_pathweight += graph[k][s]
# update minimum
a.append(current_pathweight)
min_path = min(min_path, current_pathweight)
for i in range(len(a)):
if min_path == a[i]:
l = list(next_permutation[i])
l.insert(0, s)
l.append(s)
print(l)
return min_path
# Driver Code
if _name_ == "__main__":
# matrix representation of graph
graph = [[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]]
s = 0
print(travellingSalesmanProblem(graph, s))
Can u tell how can we print the path also
TRY THIS CODE!!!
from sys import maxsize
from itertools import permutations
V = 4
# implementation of traveling Salesman Problem
def travellingSalesmanProblem(graph, s):
# store all vertex apart from source vertex
vertex = []
for i in range(V):
if i != s:
vertex.append(i)
# store minimum weight Hamiltonian Cycle
min_path = maxsize
a = []
next_permutation = list(permutations(vertex))
for i in next_permutation:
# store current Path weight(cost)
current_pathweight = 0
# compute current path weight
k = s
for j in i:
current_pathweight += graph[k][j]
k = j
current_pathweight += graph[k][s]
# update minimum
a.append(current_pathweight)
min_path = min(min_path, current_pathweight)
for i in range(len(a)):
if min_path == a[i]:
l = list(next_permutation[i])
l.insert(0, s)
l.append(s)
print(l)
return min_path
# Driver Code
if __name__ == "__main__":
# matrix representation of graph
graph = [[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]]
s = 0
print(travellingSalesmanProblem(graph, s))
Thanks for the code:) u need to really work on ur explanation. First, explain the logic and then the code.
Second function is wrongly implemented instead of comparing vertices index u should have used cost from graph matrix. How can you say by comparing the vertex numbers at it would be next min path without considering the cost comparison in next min path
Your code also does not check if there is valid edge present in between nodes in second function what if the 1-2 is 0
Please can you demo running against these cases
You should first explain your method before explaining it in code.. That's how followup become easy. BTW, thanks for the code
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Hindi mai hie pdha deta jada aacha se smjh aata