## Shortest Path – Python

August 24, 2010 – 2:26 pmProblem: Write a python function to calculate the shortest path given a start node (or vertex), an end node and a graph. Use Dijkstra’s algorithm. Return the distance from the start node to the end node, as well as the path taken to get there.

The implementation below sticks pretty closely to the algorithm description in the wikipedia entry, which I turned into something a little more pseudo-code-ish to help me implement it:

## Initial state:

- give nodes two properties – node.visited and node.distance
- set node.distance = infinity for all nodes except the start node which is set to zero.
- set node.visited = false for all nodes
- set current node = start node.
## Current node loop:

- if current node = end node, finish and return current.distance & path
- for all unvisited neighbors, calc their tentative distance (current.distance + edge to neighbor).
- if tentative distance < neighbor's set distance, overwrite it.
- set current.isvisited = true.
- set current = remaining unvisited node with smallest node.distance

Here’s my implementation – it’s recursive, as suggested by the algorithm description:

import sys def shortestpath(graph,start,end,visited=[],distances={},predecessors={}): """Find the shortest path between start and end nodes in a graph""" # detect if it's the first time through, set current distance to zero if not visited: distances[start]=0 if start==end: # we've found our end node, now find the path to it, and return path=[] while end != None: path.append(end) end=predecessors.get(end,None) return distances[start], path[::-1] # process neighbors as per algorithm, keep track of predecessors for neighbor in graph[start]: if neighbor not in visited: neighbordist = distances.get(neighbor,sys.maxint) tentativedist = distances[start] + graph[start][neighbor] if tentativedist < neighbordist: distances[neighbor] = tentativedist predecessors[neighbor]=start # neighbors processed, now mark the current node as visited visited.append(start) # finds the closest unvisited node to the start unvisiteds = dict((k, distances.get(k,sys.maxint)) for k in graph if k not in visited) closestnode = min(unvisiteds, key=unvisiteds.get) # now we can take the closest node and recurse, making it current return shortestpath(graph,closestnode,end,visited,distances,predecessors) if __name__ == "__main__": graph = {'a': {'w': 14, 'x': 7, 'y': 9}, 'b': {'w': 9, 'z': 6}, 'w': {'a': 14, 'b': 9, 'y': 2}, 'x': {'a': 7, 'y': 10, 'z': 15}, 'y': {'a': 9, 'w': 2, 'x': 10, 'z': 11}, 'z': {'b': 6, 'x': 15, 'y': 11}} print shortestpath(graph,'a','a') print shortestpath(graph,'a','b') """ Result: (0, ['a']) (20, ['a', 'y', 'w', 'b']) """

You’ll see I turned the example in wikipedia into a graph for the test case in the code above.

I found this useful about how to represent and process graphs in python: http://www.python.org/doc/essays/graphs.html. I think this was written by Guido himself.

There’s a bunch of interesting implementations on the web, including two here. More elegant, but it took me a while to understand them.

## 8 Responses to “Shortest Path – Python”

Hey there,

I was wondering… This looks pretty cool and I’d like to know how to make it work with cities and distances like this:

NewYork Chicago 200

Chicago LA 100

Chicago SanFrancisco 1000

SanFrancisco Florida 1400

END

NewYork

Florida

This means, that the distance from NewYork to Chicago is 200, Chicago to LA is 100, etc. Note: Chicago to NewYork is unknown we only know NewYork to Chicago.

Then the user inputs ‘END’ and in each line the starting city and finishing city, respectively.

f = []

while True:

a = raw_input()

if a == ‘FIM’:

break

f.append(a)

city1 = raw_input()

city2 = raw_input()

This would work and create an array with all of them.

My question is how to “Convert” that array to a dictionary usable by this function?

Here’s an example of how the array would be:

NewYork Chicago 200

Chicago LA 100

Chicago SanFrancisco 1000

SanFrancisco Florida 1400

['NewYork Chicago 200', 'Chicago LA 100', 'Chicago SF 100', 'SF Florida 1400']

Thanks much for your time.

By

Omicron Alphaon Mar 3, 2011Can I get your email address, I need help to find out maximum capacity path from a source node to all other nodes.

By

Dipon Apr 28, 2011I think you have a bug:

If start == end at first iteration:

print shortestpath(graph,’a',’a')

Traceback (most recent call last):

File “./shortpath.py”, line 39, in

print shortestpath(graph,’a',’a')

File “./shortpath.py”, line 13, in shortestpath

return distances[start], path[::-1]

KeyError: ‘a’

You need to init distances[] before enter while() loop.

By

Int-0on May 31, 2011Thanks Int-0

Now fixed.

Dip

my email address is in the about page of the blog.

By

nolfonzoon Jun 18, 2011You should use float(“inf”) (aka python’s infinity) instead sys.maxint (a fixed value). Remember in python ints are automatically parsed to longs when required, so you can raise easily maxint on a graph without overflows.

By

Spayder26on Jun 28, 2011if i want all shortest path.

How to edit this?

By

naton Nov 30, 2011There is a bug in your code.

Using mutable objects as default values in your function definition may cause it to execute correctly only the first time it is called in a program, any subsequent calls that receive different input data may fail or return incorrect values.

This is because the function arguments that are mutable objects will carry over their contents to subsequent calls of this function if those arguments are omitted, they are not initialized to the default values in your function definition, as you are intending. The default values of mutable objects are only used when the function is first called in a program.

This behaviour is by design, see-

http://bugs.python.org/issue4181

and

http://bugs.python.org/issue4619

This behaviour can be avoided by not relying on the default argument values when calling the function, ie. always include all argument values that are mutable.

By

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