# Detect Cycles

Eugene Low Nov 04, 2020 · 5 mins read

LC 207
LC 210

### Detect Cycles in Directed Graph 1

Steps:

1. Keep two lists visited and completed same length of number of nodes and initialised to false
2. define a function(haveCycle) to solve for node:
1. visited[node] = true
2. for each children of node:
1. if visited[children] , but not completed[children]:
1. HAVE CYCLE
2. else:
1. if children haveCycle :
1. HAVE CYCLE
3. completed[node] = true
3. for each node:
1. if not visited and haveCycle(node):
1. HAVE CYCLE

### Detect Cycles in Directed Graph 2

Steps:

1. Keep a lists indegree initialised to number of child point into that node.
2. Keep a queue to queue that contains the nodes that do not have child
3. while queue is not empty:
1. node = queue.pop()
2. for each parent of node:
1. detected dependency and we remove this dependency indegree[parent] -= 1
2. if indegree[parent] is 0: there is no child anymore
1. push parent into queue
4. if all indegree is not 0: there is still some dependencies
1. HAVE CYCLE

How this works? If the node is not a leaf, we the dependencies of parent by reducing the number of dependencies of parent. If the parent have no more dependencies the parent become the leaf and we can do this continuously. If the parent still have dependencies and its child dependent on something its depend on the child will never be pushed into the list, because child was never resolved. Therefore parent will never be solved too (indegree more than 1).

### Flatten a tree with dependencies

Resolve the tree by solving the leaf first then the parent. This algorithm is the similar to the one above.

Steps:

1. Keep two lists visited and completed same length of number of nodes and initialised to false
2. A list of solution = [];
3. define a function solve to solve for node:
1. visited[node] = true
2. for each children of node:
1. if visited[children] , but not completed[children]:
1. HAVE CYCLE
2. else:
1. if children haveCycle :
1. HAVE CYCLE
3. completed[node] = true
4. push node to solution
4. for each node:
1. if not visited and solve for node:
1. HAVE CYCLE
5. print solution

### Code

Method 1

def haveCycle(node, graph, visited, completed):
visited[node] = True

for out_node in graph[node]:
if not visited[out_node] and \
haveCycle(out_node, graph, visited, completed):
return True
else if not completed[out_node]:
return True

completed[node] = True
return False

numNodes = 3
graph = [
0: [1,2],  # 0 to 1 and 0 to 2
1: [0],
2: [2]
]
visited = [False] * numNodes
completed = [False] * numNodes

for node in range(numNodes):
if not visited[node] and haveCycle(node, graph, visited, completed):
print('Cycle found')


Method 2

from collections import deque

numNodes = 3
graph = [
0: [1,2],  # 0 to 1 and 0 to 2
1: [0],
2: [2]
]

indegree = [0] * numNodes

queue = deque()
for node in graph:
for out_node in graph[node]:
indegree[out_node] += 1

for node, indeg in enumerate(indegree):
if indeg == 0:
queue.append(node)

while queue:
node = queue.popleft()
for out_node in graph[node]:
indegree[out_node] -= 1
if indegree[out_node] == 0:  # no more indegree all of them are solved
queue.append(out_node)

if sum(indegree) != 0:
print('Cycle found')

Hi, I am Eugene!