问题
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(len(w)):
if w[i]<minimum:
minimum=w[i]
m=i
return m, Q[m]
def dijkstra(G, s, t='B'):
Q=[s]
p={s:None}
w=[0]
d={}
for i in G:
d[i]=float('inf')
Q.append(i)
w.append(d[i])
d[s]=0
S=[]
n=len(Q)
while Q:
u=extract(Q,w)[1]
S.append(u)
#w.remove(extract(Q, d, w)[0])
Q.remove(u)
for v in G[u]:
if d[v]>=d[u]+G[u][v]:
d[v]=d[u]+G[u][v]
p[v]=u
return d, p
B='B'
A='A'
D='D'
G='G'
E='E'
C='C'
F='F'
G={B:{A:5, D:1, G:2}, A:{B:5, D:3, E:12, F:5}, D:{B:1, G:1, E:1, A:3}, G:{B:2, D:1, C:2}, C:{G:2, E:1, F:16}, E:{A:12, D:1, C:1, F:2}, F:{A:5, E:2, C:16}}
print "Assuming the start vertex to be B:"
print "Shortest distances", dijkstra(G, B)[0]
print "Parents", dijkstra(G, B)[1]
I expect the answer to be:
Assuming the start vertex to be B:
Shortest distances {'A': 4, 'C': 4, 'B': 0, 'E': 2, 'D': 1, 'G': 2, 'F': 4}
Parents {'A': 'D', 'C': 'G', 'B': None, 'E': 'D', 'D': 'B', 'G': 'D', 'F': 'E'}
However, the answer that I get is this:
Assuming the start vertex to be B:
Shortest distances {'A': 4, 'C': 4, 'B': 0, 'E': 2, 'D': 1, 'G': 2, 'F': 10}
Parents {'A': 'D', 'C': 'G', 'B': None, 'E': 'D', 'D': 'B', 'G': 'D', 'F': 'A'}.
For the node F, the program gives the incorrect answer. Can someone please tell me why?
回答1:
As others have pointed out, due to not using understandable variable names, it is almost impossible to debug your code.
Following the wiki article about Dijkstra's algorithm, one can implement it along these lines (and in a million other manners):
nodes = ('A', 'B', 'C', 'D', 'E', 'F', 'G')
distances = {
'B': {'A': 5, 'D': 1, 'G': 2},
'A': {'B': 5, 'D': 3, 'E': 12, 'F' :5},
'D': {'B': 1, 'G': 1, 'E': 1, 'A': 3},
'G': {'B': 2, 'D': 1, 'C': 2},
'C': {'G': 2, 'E': 1, 'F': 16},
'E': {'A': 12, 'D': 1, 'C': 1, 'F': 2},
'F': {'A': 5, 'E': 2, 'C': 16}}
unvisited = {node: None for node in nodes} #using None as +inf
visited = {}
current = 'B'
currentDistance = 0
unvisited[current] = currentDistance
while True:
for neighbour, distance in distances[current].items():
if neighbour not in unvisited: continue
newDistance = currentDistance + distance
if unvisited[neighbour] is None or unvisited[neighbour] > newDistance:
unvisited[neighbour] = newDistance
visited[current] = currentDistance
del unvisited[current]
if not unvisited: break
candidates = [node for node in unvisited.items() if node[1]]
current, currentDistance = sorted(candidates, key = lambda x: x[1])[0]
print(visited)
This code is more verbous than necessary and I hope comparing your code with mine you might spot some differences.
The result is:
{'E': 2, 'D': 1, 'G': 2, 'F': 4, 'A': 4, 'C': 3, 'B': 0}
回答2:
I wrote it in a more verbose form to make it clearer for a novice reader:
from collections import defaultdict
def get_shortest_path(weighted_graph, start, end):
"""
Calculate the shortest path for a directed weighted graph.
Node can be virtually any hashable datatype.
:param start: starting node
:param end: ending node
:param weighted_graph: {"node1": {"node2": "weight", ...}, ...}
:return: ["START", ... nodes between ..., "END"] or None, if there is no
path
"""
# We always need to visit the start
nodes_to_visit = {start}
visited_nodes = set()
distance_from_start = defaultdict(lambda: float("inf"))
# Distance from start to start is 0
distance_from_start[start] = 0
tentative_parents = {}
while nodes_to_visit:
# The next node should be the one with the smallest weight
current = min(
[(distance_from_start[node], node) for node in nodes_to_visit]
)[1]
# The end was reached
if current == end:
break
nodes_to_visit.discard(current)
visited_nodes.add(current)
for neighbour, distance in weighted_graph[current]:
if neighbour in visited_nodes:
continue
neighbour_distance = distance_from_start[current] + distance
if neighbour_distance < distance_from_start[neighbour]:
distance_from_start[neighbour] = neighbour_distance
tentative_parents[neighbour] = current
nodes_to_visit.add(neighbour)
return _deconstruct_path(tentative_parents, end)
def _deconstruct_path(tentative_parents, end):
if end not in tentative_parents:
return None
cursor = end
path = []
while cursor:
path.append(cursor)
cursor = tentative_parents.get(cursor)
return list(reversed(path))
回答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
回答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.
回答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.)
回答6:
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)
回答7:
I broke down the wikipedia description into the following pseudo-code on my blog rebrained.com:
Initial state:
- give nodes two properties - node.visited and node.distance
- set node.distance = infinity for all nodes except set start node 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
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'])
"""
回答8:
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)
回答9:
I implement Dijkstra using priority-queue. Apart from that, I also implement min-heap myself. Hope this will help you.
from collections import defaultdict
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 int((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)/2)-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):
if self._heapsize < 1:
raise IndexError
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
回答10:
Implementation based on CLRS 2nd Ed. Chapter 24.3
d is deltas, p is predecessors
import heapq
def dijkstra(g, s, t):
q = []
d = {k: sys.maxint for k in g.keys()}
p = {}
d[s] = 0
heapq.heappush(q, (0, s))
while q:
last_w, curr_v = heapq.heappop(q)
for n, n_w in g[curr_v]:
cand_w = last_w + n_w # equivalent to d[curr_v] + n_w
# print d # uncomment to see how deltas are updated
if cand_w < d[n]:
d[n] = cand_w
p[n] = curr_v
heapq.heappush(q, (cand_w, n))
print "predecessors: ", p
print "delta: ", d
return d[t]
def test():
og = {}
og["s"] = [("t", 10), ("y", 5)]
og["t"] = [("y", 2), ("x", 1)]
og["y"] = [("t", 3), ("x", 9), ("z", 2)]
og["z"] = [("x", 6), ("s", 7)]
og["x"] = [("z", 4)]
assert dijkstra(og, "s", "x") == 9
if __name__ == "__main__":
test()
Implementation assumes all nodes are represented as keys. If say node(e.g "x" in the example above) was not defined as a key in the og, deltas d would be missing that key and check if cand_w < d[n] wouldn't work correctly.
来源：https://stackoverflow.com/questions/22897209/dijkstras-algorithm-in-python