Witam. Mam następujący kod, za pomocą, którego użytkownik otrzymuje informację czy graf ma drogę, cykl Eulera lub nie jest Eulerowski. Niestety, natknąłem się na pewien problem. W niektórych przypadkach program pokazuje mi błędną wartość? Bardzo proszę o dopieszczenie kodu
'''
Python program to check if a given graph is Eulerian or not
Complexity : O(V+E)
from collections import defaultdict
This class represents a undirected graph using adjacency list
representation
class Graph:
def __init__(self, vertices):
self.V = vertices # No. of vertices
self.graph = defaultdict(list) # default dictionary to store graph
# function to add an edge to graph
def addEdge(self, u, v):
self.graph[u].append(v)
self.graph[v].append(u)
# A function used by isConnected
def DFSUtil(self, v, visited):
# Mark the current node as visited
visited[v] = True
# Recur for all the vertices adjacent to this vertex
for i in self.graph[v+1]:
if visited[i] == False:
self.DFSUtil(i, visited)
'''Method to check if all non-zero degree vertices are
connected. It mainly does DFS traversal starting from
node with non-zero degree'''
def isConnected(self):
# Mark all the vertices as not visited
visited = [False] * (self.V)
# Find a vertex with non-zero degree
for i in range(self.V):
if len(self.graph[i]) > 1:
break
# If there are no edges in the graph, return true
if i == self.V - 1:
return True
# Start DFS traversal from a vertex with non-zero degree
self.DFSUtil(i, visited)
# Check if all non-zero degree vertices are visited
for i in range(self.V):
if visited[i] == False and len(self.graph[i]) > 0:
return False
return True
'''The function returns one of the following values
0 --> If grpah is not Eulerian
1 --> If graph has an Euler path (Semi-Eulerian)
2 --> If graph has an Euler Circuit (Eulerian) '''
def isEulerian(self):
# Check if all non-zero degree vertices are connected
if self.isConnected() == False:
return 0
else:
# Count vertices with odd degree
odd = 0
for i in range(self.V):
if len(self.graph[i]) % 2 != 0:
odd += 1
'''If odd count is 2, then semi-eulerian.
If odd count is 0, then eulerian
If count is more than 2, then graph is not Eulerian
Note that odd count can never be 1 for undirected graph'''
if odd == 0:
return 2
elif odd == 2:
return 1
elif odd > 2:
return 0
# Function to run test cases
def test(self):
res = self.isEulerian()
if res == 0:
print "graph is not Eulerian"
elif res == 1:
print "graph has a Euler path"
else:
print "graph has a Euler cycle"
ln1 = [1, 2, 1, 6, 2, 3, 3, 4, 4, 5, 5, 6]
#ln1 = [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 1, 2, 1, 4, 1, 5, 2, 3, 2, 4, 3, 5, 4, 5] #euler path
#ln1 = [1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 2, 3, 2, 5, 2, 6, 3, 4, 3, 5, 4, 6, 5, 6] #euler path
g1 = Graph(len(ln1)/2)
i = 0
j = 1
while i + 2 <= len(ln1) and j + 1 <= len(ln1):
g1.addEdge(ln1[i], ln1[j])
i += 2
j += 2
g1.test()
'''
Python 