Dijkstra's algorithm in python

Question

I am trying to implement Dijkstra\'s algorithm in python using arrays. This is my implementation.

def extract(Q, w):
    m=0
    minimum=w[0]
    for i in range(l         


        

Answer 1

import sys
import heapq

class Node:

     def __init__(self, name):
        self.name = name
        self.visited = False
        self.adjacenciesList = []
        self.predecessor = None
        self.mindistance = sys.maxsize    

    def __lt__(self, other):
        return self.mindistance < other.mindistance

class Edge:

    def __init__(self, weight, startvertex, endvertex):
        self.weight = weight
        self.startvertex = startvertex
        self.endvertex = endvertex

def calculateshortestpath(vertexlist, startvertex):
    q = []
    startvertex.mindistance = 0
    heapq.heappush(q, startvertex)

    while q:
        actualnode = heapq.heappop(q)
        for edge in actualnode.adjacenciesList:
            tempdist = edge.startvertex.mindistance + edge.weight
            if tempdist < edge.endvertex.mindistance:
                edge.endvertex.mindistance = tempdist
                edge.endvertex.predecessor = edge.startvertex
                heapq.heappush(q,edge.endvertex)
def getshortestpath(targetvertex):
    print("The value of it's minimum distance is: ",targetvertex.mindistance)
    node = targetvertex
    while node:
        print(node.name)
        node = node.predecessor

node1 = Node("A");
node2 = Node("B");
node3 = Node("C");
node4 = Node("D");
node5 = Node("E");
node6 = Node("F");
node7 = Node("G");
node8 = Node("H");

edge1 = Edge(5,node1,node2);
edge2 = Edge(8,node1,node8);
edge3 = Edge(9,node1,node5);
edge4 = Edge(15,node2,node4);
edge5 = Edge(12,node2,node3);
edge6 = Edge(4,node2,node8);
edge7 = Edge(7,node8,node3);
edge8 = Edge(6,node8,node6);
edge9 = Edge(5,node5,node8);
edge10 = Edge(4,node5,node6);
edge11 = Edge(20,node5,node7);
edge12 = Edge(1,node6,node3);
edge13 = Edge(13,node6,node7);
edge14 = Edge(3,node3,node4);
edge15 = Edge(11,node3,node7);
edge16 = Edge(9,node4,node7);

node1.adjacenciesList.append(edge1);
node1.adjacenciesList.append(edge2);
node1.adjacenciesList.append(edge3);
node2.adjacenciesList.append(edge4);
node2.adjacenciesList.append(edge5);
node2.adjacenciesList.append(edge6);
node8.adjacenciesList.append(edge7);
node8.adjacenciesList.append(edge8);
node5.adjacenciesList.append(edge9);
node5.adjacenciesList.append(edge10);
node5.adjacenciesList.append(edge11);
node6.adjacenciesList.append(edge12);
node6.adjacenciesList.append(edge13);
node3.adjacenciesList.append(edge14);
node3.adjacenciesList.append(edge15);
node4.adjacenciesList.append(edge16);

vertexlist = (node1,node2,node3,node4,node5,node6,node7,node8)

calculateshortestpath(vertexlist,node1)
getshortestpath(node7)

Answer 2

Here is my implementation of Dijkstra algorithm using min-priority-queue. Hope it will you.

from collections import defaultdict
from math import floor


class MinPQ:
    """
    each heap element is in form (key value, object handle), while heap
    operations works based on comparing key value and object handle points to
    the corresponding application object.
    """

    def __init__(self, array=[]):
        self._minheap = list(array)
        self._length = len(array)
        self._heapsize = 0
        self._build_min_heap()

    def _left(self, idx):
        return 2*idx+1

    def _right(self, idx):
        return 2*idx+2

    def _parent(self, idx):
        return floor((idx-1)/2)

    def _min_heapify(self, idx):
        left = self._left(idx)
        right = self._right(idx)
        min_idx = idx
        if left <= self._heapsize-1 and self._minheap[left] < self._minheap[min_idx]:
            min_idx = left
        if right <= self._heapsize-1 and self._minheap[right] < self._minheap[min_idx]:
            min_idx = right
        if min_idx != idx:
            self._minheap[idx], self._minheap[min_idx] = self._minheap[min_idx], self._minheap[idx]
            self._min_heapify(min_idx)

    def _build_min_heap(self):
        self._heapsize = self._length
        mid_id = int(self._heapsize-1)-1
        for i in range(mid_id, -1, -1):
            self._min_heapify(i)

    def decrease_key(self, idx, new_key):
        while idx > 0 and new_key < self._minheap[self._parent(idx)]:
            self._minheap[idx] = self._minheap[self._parent(idx)]
            idx = self._parent(idx)
        self._minheap[idx] = new_key

    def extract_min(self):
        minimum = self._minheap[0]
        self._minheap[0] = self._minheap[self._heapsize-1]
        self._heapsize = self._heapsize - 1
        self._min_heapify(0)
        return minimum

    def insert(self, item):
        self._minheap.append(item)
        self._heapsize = self._heapsize + 1
        self.decrease_key(self._heapsize-1, item)

    @property
    def minimum(self):
        return self._minheap[0]

    def is_empty(self):
        return self._heapsize == 0

    def __str__(self):
        return str(self._minheap)

    __repr__ = __str__

    def __len__(self):
        return self._heapsize


class DiGraph:
    def __init__(self, edges=None):
        self.adj_list = defaultdict(list)
        self.add_weighted_edges(edges)

    @property
    def nodes(self):
        nodes = set()
        nodes.update(self.adj_list.keys())
        for node in self.adj_list.keys():
            for neighbor, weight in self.adj_list[node]:
                nodes.add(neighbor)
        return list(nodes)

    def add_weighted_edges(self, edges):
        if edges is None:
            return None
        for edge in edges:
            self.add_weighted_edge(edge)

    def add_weighted_edge(self, edge):
        node1, node2, weight = edge
        self.adj_list[node1].append((node2, weight))

    def weight(self, tail, head):
        for node, weight in self.adj_list[tail]:
            if node == head:
                return weight
        return None


def relax(min_heapq, dist, graph, u, v):
    if dist[v] > dist[u] + graph.weight(u, v):
        dist[v] = dist[u] + graph.weight(u, v)
        min_heapq.insert((dist[v], v))


def dijkstra(graph, start):
    # initialize
    dist = dict.fromkeys(graph.nodes, float('inf'))
    dist[start] = 0
    min_heapq = MinPQ()
    min_heapq.insert((0, start))

    while not min_heapq.is_empty():
        distance, u = min_heapq.extract_min()
        # we may add a node multiple time in priority queue, but we process it
        # only once
        if distance > dist[u]:
            continue
        for neighbor, weight in graph.adj_list[u]:
            relax(min_heapq, dist, graph, u, neighbor)

    return dist

if __name__ == "__main__":
    edges = [('s', 't', 10), ('t', 'x', 1), ('s', 'y', 5), ('y', 't', 3), ('t', 'y', 2), 
             ('y', 'x', 9), ('y', 'z', 2), ('z', 's', 7), ('x', 'z', 4), ('z', 'x', 6)]
    digraph = DiGraph(edges)
    res = dijkstra(digraph, 's')
    print(res)

Answer 3

This is not my answer - my prof did it much more efficiently than my attempt. Here is his approach, obviously using helper-functions for the repetitive tasks

def dijkstra(graph, source):

    vertices, edges = graph
    dist = dict()
    previous = dict()

    for vertex in vertices:
        dist[vertex] = float("inf")
        previous[vertex] = None

    dist[source] = 0
    Q = set(vertices)

    while len(Q) > 0:
        u = minimum_distance(dist, Q)
        print('Currently considering', u, 'with a distance of', dist[u])
        Q.remove(u)

        if dist[u] == float('inf'):
            break

        n = get_neighbours(graph, u)
        for vertex in n:
            alt = dist[u] + dist_between(graph, u, vertex)
            if alt < dist[vertex]:
                dist[vertex] = alt
                previous[vertex] = u

    return previous

Given a graph

({'A', 'B', 'C', 'D'}, {('A', 'B', 5), ('B', 'A', 5), ('B', 'C', 10), ('B', 'D', 6), ('C', 'D', 2), ('D', 'C', 2)})

the command print(dijkstra(graph, 'A') yields

Currently considering A with a distance of 0

Currently considering B with a distance of 5

Currently considering D with a distance of 11

Currently considering C with a distance of 13

id est:

{'C': 'D', 'D': 'B', 'A': None, 'B': 'A'} => in random order

Answer 4

This implementation use only array and heap ds.

import heapq as hq
import math

def dijkstra(G, s):
    n = len(G)
    visited = [False]*n
    weights = [math.inf]*n
    path = [None]*n
    queue = []
    weights[s] = 0
    hq.heappush(queue, (0, s))
    while len(queue) > 0:
        g, u = hq.heappop(queue)
        visited[u] = True
        for v, w in G[u]:
            if not visited[v]:
                f = g + w
                if f < weights[v]:
                    weights[v] = f
                    path[v] = u
                    hq.heappush(queue, (f, v))
    return path, weights

G = [[(1, 6), (3, 7)],
     [(2, 5), (3, 8), (4, -4)],
     [(1, -2), (4, 7)],
     [(2, -3), (4, 9)],
     [(0, 2)]]

print(dijkstra(G, 0))

I hope this could help someone, it's a little bit late.

Answer 5

Set a breakpoint in extract. You will see that you delete entries from Q but never from w. Everything else is a dict, but Q/w are a paired array which you do not keep up to date. You have to keep those two in sync, or replace them with a dict. Special note: Eventually, after the algorithm is working, you may want to replace Q/w with a list of edges and re-code the "extract" function with a priority queue (heapq module).

Additionally, you will see that w always has weights of 0 for the source, and 'inf' for all other nodes. You completely skipped the critical step where you update the candidate distances.

So you are basically always taking the first path you encounter, rather than selecting the shortest path. You later compute the actual distance of that path, so the returned array of distances has actual values, but they were chosen arbitrarily, and you have no reason to expect them to be shortest.

After you (incorrectly) find the next node, you look at all of its edges. This should have been the critical step I mentioned above in the second paragraph, where you update the candidates for the next node. Instead you do something completely different: You appear to be looping through all the previous solutions (which are guaranteed correct and should be left alone, if you correctly implement dijkstra), and you look for a two-step solution from source->current->any. The correct intent for looking at those would have been to add the next candidate from previous paths to the next node, but because they never get added, you don't look at (previous shortest path) + (one step), you only look at literally two node solutions.

So basically, you are looping through all possible two-node paths from source in order to find the shortest paths. This is a complete error and has nothing to do with dijkstra. But it almost works on your tiny tiny graph where most of the correct shortest paths are two-step paths.

(ps: I agree with everyone about your variable names. You would have done much better if you use verbose names telling what those variables represent. I had to rename them before I got anywhere analyzing your code.)

Answer 6

I needed a solution which would also return the path so I put together a simple class using ideas from multiple questions/answers about Dijkstra:

class Dijkstra:

    def __init__(self, vertices, graph):
        self.vertices = vertices  # ("A", "B", "C" ...)
        self.graph = graph  # {"A": {"B": 1}, "B": {"A": 3, "C": 5} ...}

    def find_route(self, start, end):
        unvisited = {n: float("inf") for n in self.vertices}
        unvisited[start] = 0  # set start vertex to 0
        visited = {}  # list of all visited nodes
        parents = {}  # predecessors
        while unvisited:
            min_vertex = min(unvisited, key=unvisited.get)  # get smallest distance
            for neighbour, _ in self.graph.get(min_vertex, {}).items():
                if neighbour in visited:
                    continue
                new_distance = unvisited[min_vertex] + self.graph[min_vertex].get(neighbour, float("inf"))
                if new_distance < unvisited[neighbour]:
                    unvisited[neighbour] = new_distance
                    parents[neighbour] = min_vertex
            visited[min_vertex] = unvisited[min_vertex]
            unvisited.pop(min_vertex)
            if min_vertex == end:
                break
        return parents, visited

    @staticmethod
    def generate_path(parents, start, end):
        path = [end]
        while True:
            key = parents[path[0]]
            path.insert(0, key)
            if key == start:
                break
        return path

Example graph and usage (drawing is generated using this nifty tool):

input_vertices = ("A", "B", "C", "D", "E", "F", "G")
input_graph = {
    "A": {"B": 5, "D": 3, "E": 12, "F": 5},
    "B": {"A": 5, "D": 1, "G": 2},
    "C": {"E": 1, "F": 16, "G": 2},
    "D": {"A": 3, "B": 1, "E": 1, "G": 1},
    "E": {"A": 12, "C": 1, "D": 1, "F": 2},
    "F": {"A": 5, "C": 16, "E": 2},
    "G": {"B": 2, "C": 2, "D": 1}
}
start_vertex = "B"
end_vertex= "C"
dijkstra = Dijkstra(input_vertices, input_graph)
p, v = dijkstra.find_route(start_vertex, end_vertex)
print("Distance from %s to %s is: %.2f" % (start_vertex, end_vertex, v[end_vertex]))
se = dijkstra.generate_path(p, start_vertex, end_vertex)
print("Path from %s to %s is: %s" % (start_vertex, end_vertex, " -> ".join(se)))

Output

Distance from B to C is: 3.00
Path from B to C is: B -> D -> E -> C