Topological Sort Using Kahn's Algorithm BFS in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the in-degree array to zero for all vertices.

DSA C

for (int i = 0; i < V; i++) {
    in_degree[i] = [1];
}

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

A0

BV

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting in-degree to 1 initially

Using negative values for in-degree

In-degree counts the number of incoming edges. Initially, it should be zero for all vertices.

2fill in blank

medium

Complete the code to enqueue vertices with zero in-degree into the queue.

DSA C

for (int i = 0; i < V; i++) {
    if (in_degree[i] == [1]) {
        enqueue(&q, i);
    }
}

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

AV

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Enqueuing vertices with in-degree 1

Forgetting to check in-degree

Vertices with zero in-degree have no dependencies and can be processed first.

3fill in blank

hard

Fix the error in the code that decreases the in-degree of adjacent vertices after removing a vertex from the queue.

DSA C

while (!is_empty(&q)) {
    int u = dequeue(&q);
    for (int i = 0; i < adj[u].size; i++) {
        int v = adj[u].array[i];
        in_degree[v] [1] 1;
        if (in_degree[v] == 0) {
            enqueue(&q, v);
        }
    }
}

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

A*=

B-=

C+=

D/=

Attempts:

3 left

💡 Hint

Common Mistakes

Increasing in-degree instead of decreasing

Using multiplication or division operators

When a vertex is removed, the in-degree of its adjacent vertices decreases by one.

4fill in blank

hard

Fill both blanks to correctly update the topological order array and its index.

DSA C

topo_order[[1]] = u;
[2]++;

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

Aindex

Bcount

Ci

Dpos

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for assignment and increment

Forgetting to increment the index

The variable 'index' tracks the position in the topological order array and should be incremented after assignment.

5fill in blank

hard

Fill all three blanks to complete the condition that checks if a cycle exists in the graph.

DSA C

if ([1] != [2]) {
    printf("Cycle detected, topological sort not possible.\n");
    return [3];
}

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

Acount

BV

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of -1

Comparing wrong variables

If the number of vertices processed is not equal to total vertices, a cycle exists. Returning -1 indicates failure.

Practice

1. What is the main purpose of Kahn's Algorithm in graph theory?

easy

A. To calculate the minimum spanning tree

B. To find a topological ordering of a Directed Acyclic Graph (DAG)

C. To detect cycles in an undirected graph

D. To find the shortest path between two nodes

Solution

1. Step 1: Understand Kahn's Algorithm goal

   Kahn's Algorithm is designed to produce a linear ordering of vertices in a Directed Acyclic Graph (DAG) such that for every directed edge u -> v, u comes before v in the ordering.

2. Step 2: Compare with other graph algorithms

   Other options like shortest path, cycle detection, and minimum spanning tree solve different problems unrelated to topological sorting.

3. Final Answer:

   To find a topological ordering of a Directed Acyclic Graph (DAG) -> Option B

4. Quick Check:

   Topological sort = Kahn's Algorithm purpose [OK]

Hint: Kahn's Algorithm always finds order in DAGs [OK]

Common Mistakes:

• Confusing topological sort with shortest path
• Thinking it detects cycles in undirected graphs
• Mixing it up with minimum spanning tree algorithms

2. Which data structure is primarily used in Kahn's Algorithm to keep track of nodes with zero incoming edges?

easy

A. Priority Queue

B. Stack

C. Queue

D. Hash Map

Solution

1. Step 1: Identify the data structure for zero in-degree nodes

   Kahn's Algorithm uses a queue to process nodes with zero incoming edges in a first-in-first-out manner, ensuring correct topological order.

2. Step 2: Why not other data structures?

   A stack would reverse the order, a priority queue is not necessary for basic Kahn's Algorithm, and a hash map stores key-value pairs, not order.

3. Final Answer:

   Queue -> Option C

4. Quick Check:

   Zero in-degree nodes stored in queue [OK]

Hint: Use queue for BFS order in Kahn's Algorithm [OK]

Common Mistakes:

• Using stack instead of queue
• Confusing priority queue with normal queue
• Thinking hash map manages order

3. Consider the directed graph with edges: 5->2, 5->0, 4->0, 4->1, 2->3, 3->1. What is the topological order produced by Kahn's Algorithm?

medium

A. [4, 5, 2, 0, 3, 1]

B. [5, 4, 2, 3, 1, 0]

C. [4, 5, 0, 2, 3, 1]

D. [5, 4, 0, 2, 3, 1]

Solution

1. Step 1: Calculate in-degree of each node

   Nodes: 0(in-degree 2),1(2),2(1),3(1),4(0),5(0). Start with nodes 4 and 5 in queue (pushed in ascending order: queue [4,5]).

2. Step 2: Process nodes in queue and update in-degree

   Pop 4: output [4], reduce 0->1, 1->1. Queue: [5]
   Pop 5: output [4,5], reduce 2->0 (push 2), 0->0 (push 0). Queue: [2,0]
   Pop 2: output [4,5,2], reduce 3->0 (push 3). Queue: [0,3]
   Pop 0: output [4,5,2,0], no neighbors. Queue: [3]
   Pop 3: output [4,5,2,0,3], reduce 1->0 (push 1). Queue: [1]
   Pop 1: output [4,5,2,0,3,1]. All nodes processed.

3. Final Answer:

   [4, 5, 2, 0, 3, 1] -> Option A

4. Quick Check:

   Topological order = [4, 5, 2, 0, 3, 1] [OK]

Hint: Start with zero in-degree nodes, update neighbors' in-degree [OK]

Common Mistakes:

• Not updating in-degree after removing nodes
• Wrong initial zero in-degree nodes
• Mixing order of nodes in queue

4. Given the following code snippet for Kahn's Algorithm, what is the error?

int indegree[V] = {0};
for (int i = 0; i < V; i++) {
  for (int j = 0; j < adj[i].size(); j++) {
    indegree[adj[i][j]]++;
  }
}
queue<int> q;
for (int i = 0; i < V; i++) {
  if (indegree[i] == 0) q.push(i);
}
while (!q.empty()) {
  int u = q.front(); q.pop();
  cout << u << " ";
  for (int v : adj[u]) {
    indegree[v]--;
    if (indegree[v] == 0) q.push(v);
  }
}

medium

A. No error, the code is correct

B. The adjacency list is not declared properly

C. The indegree array is not initialized correctly

D. The loop to push zero indegree nodes is missing

Solution

1. Step 1: Check indegree initialization

   Indegree array is initialized to zero for all vertices correctly using {0} syntax.

2. Step 2: Verify adjacency list and loops

   Loops correctly calculate indegree, push zero indegree nodes, and process queue with BFS logic.

3. Final Answer:

   No error, the code is correct -> Option A

4. Quick Check:

   Code matches Kahn's Algorithm steps [OK]

Hint: Check initialization and loops carefully for correctness [OK]

Common Mistakes:

• Forgetting to initialize indegree array
• Incorrect adjacency list declaration
• Missing zero indegree node insertion

5. You have a graph with edges: 1->2, 1->3, 3->4, 2->4, 4->5, and 5->3. What will Kahn's Algorithm detect when run on this graph?

hard

A. A valid topological order exists

B. The algorithm will run infinitely without stopping

C. The graph is disconnected but topological order exists

D. The graph contains a cycle, so no topological order

Solution

1. Step 1: Identify cycle in the graph

   The edges 3->4, 4->5, and 5->3 form a cycle (3->4->5->3), so the graph is not a DAG.

2. Step 2: Understand Kahn's Algorithm behavior on cycles

   Kahn's Algorithm fails to process all nodes if a cycle exists because some nodes never reach zero in-degree, indicating no valid topological order.

3. Final Answer:

   The graph contains a cycle, so no topological order -> Option D

4. Quick Check:

   Cycle detected = no topological order [OK]

Hint: Cycle means no topological order in Kahn's Algorithm [OK]

Common Mistakes:

• Assuming topological order exists despite cycle
• Thinking disconnected graph always causes failure
• Believing algorithm runs infinitely on cycle