Reputation: 63
This is what I tried:
VISITED = 1
UNVISITED = -1
VISITING = 0
def dfs(g, start, num):
state = {}
for i in range(num):
state[i] = UNVISITED
stack = [start]
while(stack != []):
node = stack.pop()
if(state[node] == VISITED):
continue
state[node] = VISITING
if(node in g):
for i in g[node]:
stack.append(i)
if(state[i] == VISITED):
return True
state[node] = VISITED
def detect_cycle(n, edges):
g = {}
# adjacency list
for (x, y) in edges:
g[x] = g.get(x, []) + [y]
for i in range(n):
if(i in g):
if(dfs(g, i, n) == True):
return True
return False
print(detect_cycle(5, [[1,4],[2,4],[3,1],[3,2]])) # outputs True (should be false)
Image of the graph:
The above example where edges = [[1,4],[2,4],[3,1],[3,2]]
doesn't contain a cycle but it returns True
. So my algorithm does not work for that case.
I'm trying to use the coloring graph detect cycle algorithm but I am unsure how to do that without recursion.
The algorithm I'm trying to follow but iteratively: Detecting a cycle in a directed graph using DFS?
Upvotes: 1
Views: 608
Reputation: 63
A cycle is detected if a visiting node (gray) encounters an edge with another visiting node. The issue I had with initial code was there was no way to set a node VISITED (black) in a backtracking way. The new code now has ENTER = 0 and EXIT = 1. Enter = 0 means its the first time I visited that node and I set it to gray (visiting) the second time I visit the node exit will be 1 so i can then set it to black (fully visited).
WHITE = 0
GRAY = 1
BLACK = 2
ENTER = 0
EXIT = 1
def dfs(graph, start, n):
state = {}
for i in range(n):
state[i] = WHITE
stack = [(start, ENTER)]
while(stack != []):
node, pos = stack.pop()
if(pos == EXIT):
state[node] = BLACK
continue
state[node] = GRAY
stack.append((node, EXIT))
if(node in graph):
for v in graph[node]:
if state[v] == GRAY:
return True
elif(state[v] == WHITE):
stack.append((v, ENTER))
return False
def detect_cycle(n, edges):
g = {}
for (x, y) in edges:
g[x] = g.get(x, []) + [y]
for i in range(n):
if(i in g):
if(dfs(g, i, n) == True):
return True
return False
Upvotes: 1