Dijkstra shortest path algorithm based on python heapq heap implementation

dijkstra.py

from collections import defaultdict
from heapq import *

def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l,r,c in edges:
        g[l].append((c,r))

    q, seen, mins = [(0,f,())], set(), {f: 0}
    while q:
        (cost,v1,path) = heappop(q)
        if v1 not in seen:
            seen.add(v1)
            path = (v1, path)
            if v1 == t: return (cost, path)

            for c, v2 in g.get(v1, ()):
                if v2 in seen: continue
                prev = mins.get(v2, None)
                next = cost + c
                if prev is None or next < prev:
                    mins[v2] = next
                    heappush(q, (next, v2, path))

    return float("inf"), None

if __name__ == "__main__":
    edges = [
        ("A", "B", 7),
        ("A", "D", 5),
        ("B", "C", 8),
        ("B", "D", 9),
        ("B", "E", 7),
        ("C", "E", 5),
        ("D", "E", 15),
        ("D", "F", 6),
        ("E", "F", 8),
        ("E", "G", 9),
        ("F", "G", 11)
    ]

    print "=== Dijkstra ==="
    print edges
    print "A -> E:"
    print dijkstra(edges, "A", "E")
    print "F -> G:"
    print dijkstra(edges, "F", "G")

Very good algorithm, it helped me a lot in a task. I made a translation commenting on the Spanish code for a better understanding.

@hhu94 line 12 is not redundant either. For an existing node in q, heappush will keep adding different costs for that node, so without line 12, that node will be visited again and update with a higher cost later.

Can it be possible to optimise more? As I am getting run-time error(NZEC) in codechef.
I am working on this https://www.codechef.com/INOIPRAC/

@licensed @andreppires @romech

It may very simple by change line 14 into path += (v1, ), this will make output more clear and reverse the path in the meanwhile.

A -> E:
(14, ('A', 'B', 'E'))
F -> G:
(11, ('F', 'G'))

@waylonflinn

I think you are right. The key problem here is when node v2 is already in the heap, you should not put v2 into heap again, instead you need to heap.remove(v) and then head.insert(v2) if new cost of v2 is better then original cost of v2 recorded in the heap. But indeed remove node in heap is just O(n), so that will not be any better then original implementation of Dijkstra using distance array.

I change the code by taking the distance array into consideration which will record the min value of each node already put into the heap. So we will only put the v2 into the heap just on new value is better then the original one which I think in most case it will improve the performance of this algorithm. But just as @tjwudi mentioned, in worst case, it still will be O(V^2 logV) :)

Revised version:

def dijkstra_revised(edges, f, t):
    g = defaultdict(list)
    for l,r,c in edges:
        g[l].append((c,r))

    # dist records the min value of each node in heap.
    q, seen, dist = [(0,f,())], set(), {f: 0}
    while q:
        (cost,v1,path) = heappop(q)
        if v1 in seen: continue

        seen.add(v1)
        path += (v1,)
        if v1 == t: return (cost, path)

        for c, v2 in g.get(v1, ()):
            if v2 in seen: continue

            # Not every edge will be calculated. The edge which can improve the value of node in heap will be useful.
            if v2 not in dist or cost+c < dist[v2]:
                dist[v2] = cost+c
                heappush(q, (cost+c, v2, path))

    return float("inf")

@whiledoing Thanks! I've made an adjustment to the initial gist (slightly changed to avoid checking the same key from dist twice).

I dunno, @whiledoing's tuple output doesn't look as defunct as your latest change @kachayev.... I started getting some weird nested tuples with your version.

this is brilliant. Thanks @kachayev

seen is redundant

def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l,r,c in edges:
        g[l].append((c,r))
  
    q, dist = [(0,f,())], {f: 0}
    while q:
        (cost, node, path) = heappop(q) 
        if cost > dist[node]: continue
        path += (node, )
        if node == t: 
            return (cost, path)
        for w, n in g.get(node, ()):
            oldc = dist.get(n, float("inf"))
            newc = cost + w
            if newc < oldc:
                dist[n] = newc # relax
                heappush(q, (newc, n, path))

    return float("inf")

or you can just use seen, ignore mins/dist

from collections import defaultdict
from heapq import *

def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l,r,c in edges:
        g[l].append((c,r))

    q, seen = [(0,f,())], set()
    while q:
        (cost,v1,path) = heappop(q)
        if v1 not in seen:
            seen.add(v1)
            path = (v1, path)
            if v1 == t: return (cost, path)

            for c, v2 in g.get(v1, ()):
                if v2 in seen: continue
                next = cost + c
                heappush(q, (next, v2, path))

    return float("inf")

Thanks for your code very much. But I want to make some expansion on this basis. The situation is that our map is a matrix, and there are more than one shortest path to reach the destination, if I want to find all the road not just the one, how to modify the code to achieve this？ Thanks again.

This is a slightly simpler approach, following the wikipedia definition closely:
http://rebrained.com/?p=392

import sys
def shortestpath(graph,start,end,visited=[],distances={},predecessors={}):
"""Find the shortest path btw start & end nodes in a graph"""

# detect if first time through, set current distance to zero
if not visited: distances[start]=0

# if we've found our end node, find the path to it, and return
if start==end:
    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 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')

"""
Expected Result:
    (0, ['a']) 
    (20, ['a', 'y', 'w', 'b'])
    """

Flatten path version:

from collections import defaultdict
from heapq import *


def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l, r, c in edges:
        g[l].append((c, r))

    q, seen, mins = [(0, f, [])], set(), {f: 0}
    while q:
        (cost, v1, path) = heappop(q)
        if v1 not in seen:
            seen.add(v1)
            path = [v1] + path
            if v1 == t:
                return (cost, path)

            for c, v2 in g.get(v1, ()):
                if v2 in seen:
                    continue
                prev = mins.get(v2, None)
                next = cost + c
                if prev is None or next < prev:
                    mins[v2] = next
                    heappush(q, (next, v2, path))

    return (float("inf"), [])


if __name__ == "__main__":
    edges = [
        ("A", "B", 7),
        ("A", "D", 5),
        ("B", "C", 8),
        ("B", "D", 9),
        ("B", "E", 7),
        ("C", "E", 5),
        ("D", "E", 15),
        ("D", "F", 6),
        ("E", "F", 8),
        ("E", "G", 9),
        ("F", "G", 11)
    ]

    print("=== Dijkstra ===")
    print(edges)
    print("A -> E: ", end="")
    print(dijkstra(edges, "A", "E"))
    print("F -> G: ", end="")
    print(dijkstra(edges, "F", "G"))

=== Dijkstra ===
[('A', 'B', 7), ('A', 'D', 5), ('B', 'C', 8), ('B', 'D', 9), ('B', 'E', 7), ('C', 'E', 5), ('D', 'E', 15), ('D', 'F', 6), ('E', 'F', 8), ('E', 'G', 9), ('F', 'G', 11)]
A -> E: (14, ['E', 'B', 'A'])
F -> G: (11, ['G', 'F'])

Hi, I think I made a bit cleaner (subjectively :)) implementation in Python that uses RBTree as a priority queue with tests there

https://github.com/ehborisov/algorithms/blob/master/8.Graphs/dijkstra.py

Unless I am missing something here, this is a BFS with a min-heap, not a Dijkstra's algorithm.

@JixinSiND Dijkstra's algorithm is essentially a weighted version of BFS.

Just leaving a comment to let the author know that his code has been inappropriately taken and re-used as material for teaching at a University master in London. The authorship has been modified to report the lecturer's one instead.
https://www.dcs.bbk.ac.uk/~ale/pwd/2019-20/pwd-8/src/pwd-ex-dijkstra+heap.py

@alelom Thanks a lot for letting me know, such a kind of you! This is not the first time this code was copy-pasted into lecture materials and/or projects codebases. Honestly, if it helped students to learn - I would be glad and proud. I care less about authorship or any sort of attribution. On the one hand, I wouldn't want to encourage disrespectful actions, on the other hand, I don't have reliable way to prevent this from happening. So, choosing between spread of knowledge or nurturing morality, I would always vote for the former. Thanks again for letting me know!

I just care for what is right. If I were the lecturer, I'd quote the real author and the source – an action that does not diminish the teaching potential, and encourages sharing of good code lawfully.

Thank you so much for this gift, very clean and clever solution 😄

If anyone just wonders how to easily receive as output only the value of the solution remove the cost from the return at line 15:

if v1 == t: return cost
instead of
if v1 == t: return (cost, path)

Nice and clean

Thank you very much for this beautiful algorithm.

pretty sure this is not Dijkstra; you're doing heappush(q, (next, v2, path)) at the very end, but in True dijkstra it would need a call to "decrease_key", which in python is heap._siftdown

@xdavidliu I was confused by this until I saw https://stackoverflow.com/a/31123108. I think Dijkstra's algorithm is a higher level concept, so either implementation is valid.

More concise with path reconstruction. The node IDs are represented as integers while the edge weights as floats

from typing import *
from heapq import *


class Dijkstra:
    def __init__(self,
                 graph: Dict[int, Dict[int, float]],
                 origin: int):
        self.graph = graph
        self.edge_to: Dict[int, int] = {}
        self.distances: Dict[int, float] = {vertex: float('inf') for vertex in self.graph}
        self.origin = origin
        self._find(self.origin)

    def _find(self, node: int):

        self.distances[node] = 0

        # Priority queue which stores tuples from distance to node id
        # Distance is the first in the tuple order since it needs to have
        # priority when entries are inserted into the priority queue
        priority_queue: List[(float, int)] = [(node, 0)]

        while priority_queue:
            current_node, current_distance = heappop(priority_queue)

            # If the distance currently recorded at the distances dict is
            # bigger than the one pushed to the pq then we do not need to
            # process this entry
            if current_distance > self.distances[current_node]:
                continue

            for n, weight in self.graph[current_node].items():
                updated_distance = current_distance + weight

                if updated_distance < self.distances[n]:
                    self.distances[n] = updated_distance
                    self.edge_to[n] = current_node
                    heappush(priority_queue, (n, updated_distance))

    def reconstruct_path(self, destination: int) -> List[int]:
        node: int = destination
        path: List[int] = []

        while node != self.origin:
            path.append(node)
            node = self.edge_to[node]

        path.reverse()
        return path


if __name__ == '__main__':
    graph = {
        1: {2: 1, 3: 4},
        2: {1: 1, 3: 2, 4: 5},
        3: {1: 4, 2: 2, 4: 1},
        4: {2: 5, 3: 1}
    }

    dijkstra: Dijkstra = Dijkstra(graph, 1)

    print(dijkstra.distances)
    print(dijkstra.reconstruct_path(4))
    print(dijkstra.reconstruct_path(2))
    print(dijkstra.reconstruct_path(3))

friends don't let friends import * 😄

to quote pep8

Wildcard imports (from module import *) should be avoided, as they make it unclear which names are present in the namespace, confusing both readers and many automated tools. There is one defensible use case for a wildcard import, which is to republish an internal interface as part of a public API (for example, overwriting a pure Python implementation of an interface with the definitions from an optional accelerator module and exactly which definitions will be overwritten isn’t known in advance).

When republishing names this way, the guidelines below regarding public and internal interfaces still apply.