Dependency cycles

Understand the idea

If Kahn emits fewer than V vertices, a directed cycle blocks the remaining work. The blocked set may include descendants of a cycle, so it is not itself an exact list of cycle vertices.

Work through an example

0→1→0 leaves both indegrees positive. Adding 1→2 also blocks 2 even though 2 does not lie on the cycle.

Time and space complexity

O(V+E) detection; DFS active-state tracking can reconstruct a particular cycle.

Why it works

A nonempty remaining graph with no zero-indegree vertex contains a directed cycle.

Common mistakes

• Do not report all blocked nodes as cycle members.
• A disconnected cycle still prevents a complete order.

Chapter lab · Python implementation

Dependency order and connectivity answer different questions. Kahn topological sorting removes zero-indegree vertices from a directed graph. Disjoint-set union merges undirected connectivity groups without maintaining paths. The lab implements both; choose the structure whose contract matches the question.

This shared lab connects the chapter's topics. Follow the current lesson's worked example first, then trace the complete implementation. Additional algorithms described in topic exercises are variations you can implement independently.

from collections import deque

def topological(n,edges):
    graph=[[] for _ in range(n)]; indegree=[0]*n
    for a,b in edges:
        graph[a].append(b); indegree[b]+=1
    queue=deque(i for i in range(n) if indegree[i]==0)
    order=[]
    while queue:
        a=queue.popleft(); order.append(a)
        for b in graph[a]:
            indegree[b]-=1
            if indegree[b]==0: queue.append(b)
    return order if len(order)==n else None

class DSU:
    def __init__(self,n):
        self.parent=list(range(n)); self.size=[1]*n
    def find(self,x):
        while x!=self.parent[x]:
            self.parent[x]=self.parent[self.parent[x]]
            x=self.parent[x]
        return x
    def union(self,a,b):
        a,b=self.find(a),self.find(b)
        if a==b: return False
        if self.size[a]<self.size[b]: a,b=b,a
        self.parent[b]=a; self.size[a]+=self.size[b]
        return True

Run and test the lab

Copy the implementation and these assertions into a Python 3 file and run it. An assertion failure means the actual result differs from the expectation. Add the current topic's exercise and boundary tests to deepen your understanding.

assert topological(4,[(0,1),(0,2),(1,3),(2,3)])==[0,1,2,3]
assert topological(2,[(0,1),(1,0)]) is None
d=DSU(4); assert d.union(0,1) and d.union(1,2)
assert d.find(0)==d.find(2) and not d.union(0,2)
assert d.find(3)!=d.find(0)