Cycle Detection in Undirected Graph in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the adjacency list for the graph.

DSA Python

graph = {i: [] for i in range([1])}

Drag options to blanks, or click blank then click option'

A0

Blen(edges)

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using the length of edges instead of number of nodes.

Starting range from 1 instead of 0.

We use n to create keys for all nodes from 0 to n-1 in the adjacency list.

2fill in blank

medium

Complete the code to add edges to the adjacency list for an undirected graph.

DSA Python

graph[u].append([1])
graph[v].append(u)

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Agraph

Bv

Cu

Dedges

Attempts:

3 left

💡 Hint

Common Mistakes

Appending the same node to itself.

Appending u to u's list.

For undirected graphs, add v to u's list and u to v's list.

3fill in blank

hard

Fix the error in the DFS function to correctly detect cycles.

DSA Python

def dfs(node, parent):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            if dfs(neighbor, [1]):
                return True
        elif neighbor != parent:
            return True
    return False

Drag options to blanks, or click blank then click option'

Anode

Bneighbor

Cparent

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the current neighbor as parent causing incorrect cycle detection.

Passing the wrong variable causing infinite recursion.

We pass node as the parent for the next DFS call to avoid false cycle detection.

4fill in blank

hard

Fill both blanks to complete the cycle detection loop over all nodes.

DSA Python

for i in range([1]):
    if i not in [2]:
        if dfs(i, -1):
            print("Cycle detected")
            break

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An

Bvisited

Cgraph

Dedges

Attempts:

3 left

💡 Hint

Common Mistakes

Looping over edges instead of nodes.

Checking visited incorrectly.

Loop over all nodes from 0 to n-1 and check if each node is visited before DFS.

5fill in blank

hard

Fill all three blanks to complete the function that returns True if a cycle exists.

DSA Python

def has_cycle(graph, [1]):
    visited = set()
    for node in graph:
        if node not in visited:
            if dfs(node, [2]):
                return [3]
    return False

Drag options to blanks, or click blank then click option'

An

B-1

CTrue

DFalse

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parent value in DFS call.

Returning False instead of True when cycle found.

The function takes n (number of nodes), starts DFS with parent -1, and returns True if a cycle is found.

Practice

1. What does cycle detection in an undirected graph help us find?

easy

A. The shortest path between two vertices

B. Whether there is a path that starts and ends at the same vertex without repeating edges

C. The total number of vertices in the graph

D. The maximum degree of any vertex

Solution

1. Step 1: Understand cycle meaning in undirected graph

   A cycle is a path that starts and ends at the same vertex without repeating edges or vertices (except start/end).

2. Step 2: Identify what cycle detection checks

   Cycle detection checks if such a path exists in the graph.

3. Final Answer:

   Whether there is a path that starts and ends at the same vertex without repeating edges -> Option B

4. Quick Check:

   Cycle detection = find loops [OK]

Hint: Cycle means a closed loop path in graph [OK]

Common Mistakes:

• Confusing cycle with shortest path
• Thinking cycle counts vertices only
• Mixing cycle with vertex degree

2. Which Python data structure is commonly used to keep track of visited nodes during cycle detection in an undirected graph?

easy

A. Set

B. List

C. Dictionary

D. Tuple

Solution

1. Step 1: Identify purpose of visited tracking

   We need to quickly check if a node was visited before to avoid revisiting and detect cycles.

2. Step 2: Choose data structure with fast membership check

   Set provides O(1) average time for membership checks, making it ideal for visited nodes.

3. Final Answer:

   Set -> Option A

4. Quick Check:

   Visited nodes stored in set [OK]

Hint: Use set for fast visited checks [OK]

Common Mistakes:

• Using list causes slow membership checks
• Using tuple which is immutable and not suitable
• Using dictionary unnecessarily when set suffices

3. What will be the output of the following code snippet detecting cycle in an undirected graph?

graph = {0: [1, 2], 1: [0, 2], 2: [0, 1]}
visited = set()
def dfs(node, parent):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            if dfs(neighbor, node):
                return True
        elif neighbor != parent:
            return True
    return False

print(dfs(0, -1))

medium

A. True

B. False

C. None

D. Error

Solution

1. Step 1: Analyze graph structure

   The graph has 3 nodes connected in a triangle: 0-1, 1-2, 2-0, which forms a cycle.

2. Step 2: Trace DFS cycle detection

   Starting at node 0, DFS visits 1, then 2. When 2 checks neighbors, it finds 0 visited and not parent, so returns True.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Triangle graph has cycle = True [OK]

Hint: Triangle graph always has cycle [OK]

Common Mistakes:

• Ignoring parent check causes false positives
• Assuming no cycle in small graphs
• Confusing return values in DFS

4. Find the bug in this cycle detection code for an undirected graph:

def dfs(node, parent, graph, visited):
    visited.add(node)
    for neighbor in graph[node]:
        if neighbor not in visited:
            dfs(neighbor, node, graph, visited)
        elif neighbor != parent:
            return True
    return False

medium

A. Graph should be directed

B. Visited set is not initialized

C. Parent parameter is not used

D. Missing return statement in recursive dfs call

Solution

1. Step 1: Check recursive call return usage

   The recursive call dfs(neighbor, node, graph, visited) is made but its return value is ignored.

2. Step 2: Understand impact on cycle detection

   If a cycle is found deeper in recursion, it won't propagate True back up, so function may miss cycles.

3. Final Answer:

   Missing return statement in recursive dfs call -> Option D

4. Quick Check:

   Return recursive call result to detect cycles [OK]

Hint: Always return recursive DFS results [OK]

Common Mistakes:

• Ignoring return value of recursive calls
• Not passing visited set properly
• Confusing parent usage

5. Given an undirected graph represented as adjacency list:

graph = {
  0: [1],
  1: [0, 2],
  2: [1, 3],
  3: [2, 4, 5],
  4: [3, 5],
  5: [3, 4]
}

Which nodes form a cycle in this graph?

hard

A. No cycles in this graph

B. Nodes 0, 1, 2 form a cycle

C. Nodes 3, 4, 5 form a cycle

D. All nodes form a cycle

Solution

1. Step 1: Analyze adjacency list for cycles

   Nodes 3, 4, and 5 connect in a triangle: 3-4, 4-5, 5-3, forming a cycle.

2. Step 2: Check other nodes for cycles

   Nodes 0, 1, 2 form a simple chain without loops, so no cycle there.

3. Final Answer:

   Nodes 3, 4, 5 form a cycle -> Option C

4. Quick Check:

   Triangle subgraph = cycle [OK]

Hint: Look for triangle connections to find cycles [OK]

Common Mistakes:

• Assuming chain is a cycle
• Ignoring cycle in subgraph
• Thinking all nodes must form cycle