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Shortest Path in Graph

BFS searching shortest path

Well done! Now, let's implement a method that helps us to find the length of shortest path between two vertices, i. e. minimum number of edges to reach end vertice from start.

You should store the length of the way from start to curr node, and you can do it by modifying visited array: visited[i] equals:

  • -1, if i not visited yet

  • 0, if i is visited as first node

  • 1, if i is a neighbor of node, that has mark 0

  • k, if i is a neighbor of node with mark k-1 etc.

This way, you'll store the distance between start and current node, like at the example:

So, the answer is a visited[end].

Compito

Swipe to start coding

bfs(start, end) returns a number of edges between start and end nodes. If there is no path, return -1.

Actually this method does traverse as BFT method, but until the end vertex is found. Copy & Paste your BFT algorithm, and add some changes.

Tutto è chiaro?

Come possiamo migliorarlo?

Grazie per i tuoi commenti!

Sezione 2. Capitolo 3
class Graph:
def __init__(self, vertices):
# init graph with its vertices
self.graph = {v : [] for v in vertices}
def addEdge(self, u, v):
self.graph[u].append(v)
def __str__(self):
out = ""
for vertex in self.graph:
out += vertex + ":"+self.graph[vertex]
return out
def bfs(self, start, end):
# do the bft()-like here, and modify the visited array
q = [] # queue

visited = [-1 for v in self.graph.keys()]
q.append(start)
visited[start] = 0

while len(q)>0:
node = q.pop(0)
for vertex in self.graph[node]: # iterate neighbors
if visited[vertex] < 0:
q.append(vertex)
visited[vertex] = visited[node]+1
if vertex == end:
break
return _________




g = Graph([0,1,2,3,4,5])
g.addEdge(0,1)
g.addEdge(2,1)
g.addEdge(1,0)
g.addEdge(3,1)
g.addEdge(0,2)
g.addEdge(0,3)
g.addEdge(4,1)
g.bfs(0, 4)
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