Topological Sort Using Kahn's Algorithm BFS in DSA C - Execution Trace

Concept Flow - Topological Sort Using Kahn's Algorithm BFS

Calculate in-degree of all nodes

↓

Add nodes with in-degree 0 to queue

↓

While queue not empty

↓

Pop node from queue

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Add node to topological order

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Decrease in-degree of neighbors

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If neighbor in-degree becomes 0, add to queue

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Repeat until queue empty

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Check if all nodes processed

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Output topological order or detect cycle

This flow shows how Kahn's algorithm uses in-degree counts and a queue to produce a topological order by repeatedly removing nodes with zero in-degree.

Execution Sample

DSA C

int indegree[V] = {0};
queue<int> q;
for (int u = 0; u < V; u++) {
  for (auto v : adj[u]) {
    indegree[v]++;
  }
}
for (int u = 0; u < V; u++) {
  if (indegree[u] == 0) q.push(u);
}

This code calculates in-degree for each node and initializes the queue with nodes having zero in-degree.

Execution Table

Step	Operation	Node Processed	Queue State	In-degree Array	Topological Order	Visual State
1	Calculate in-degree	-	[ ]	[A:0, B:1, C:1, D:2, E:1, F:0]	[]	Graph: A->B, A->C, B->D, C->D, D->E
2	Initialize queue with zero in-degree nodes	-	[A, F]	[A:0, B:1, C:1, D:2, E:1, F:0]	[]	Queue contains nodes with no incoming edges: A, F
3	Pop from queue	A	[F]	[A:0, B:1, C:1, D:2, E:1, F:0]	[A]	Remove A, process neighbors B and C
4	Decrease in-degree of B	-	[F]	[A:0, B:0, C:1, D:2, E:1, F:0]	[A]	B in-degree decreased from 1 to 0
5	Decrease in-degree of C	-	[F]	[A:0, B:0, C:0, D:2, E:1, F:0]	[A]	C in-degree decreased from 1 to 0
6	Add B and C to queue	-	[F, B, C]	[A:0, B:0, C:0, D:2, E:1, F:0]	[A]	B and C added to queue as their in-degree is zero
7	Pop from queue	F	[B, C]	[A:0, B:0, C:0, D:2, E:1, F:0]	[A, F]	Remove F, no neighbors to process
8	Pop from queue	B	[C]	[A:0, B:0, C:0, D:2, E:1, F:0]	[A, F, B]	Remove B, process neighbor D
9	Decrease in-degree of D	-	[C]	[A:0, B:0, C:0, D:1, E:1, F:0]	[A, F, B]	D in-degree decreased from 2 to 1
10	Pop from queue	C	[]	[A:0, B:0, C:0, D:1, E:1, F:0]	[A, F, B, C]	Remove C, process neighbor D
11	Decrease in-degree of D	-	[]	[A:0, B:0, C:0, D:0, E:1, F:0]	[A, F, B, C]	D in-degree decreased from 1 to 0
12	Add D to queue	-	[D]	[A:0, B:0, C:0, D:0, E:1, F:0]	[A, F, B, C]	D added to queue as in-degree is zero
13	Pop from queue	D	[]	[A:0, B:0, C:0, D:0, E:1, F:0]	[A, F, B, C, D]	Remove D, process neighbor E
14	Decrease in-degree of E	-	[]	[A:0, B:0, C:0, D:0, E:0, F:0]	[A, F, B, C, D]	E in-degree decreased from 1 to 0
15	Add E to queue	-	[E]	[A:0, B:0, C:0, D:0, E:0, F:0]	[A, F, B, C, D]	E added to queue as in-degree is zero
16	Pop from queue	E	[]	[A:0, B:0, C:0, D:0, E:0, F:0]	[A, F, B, C, D, E]	Remove E, no neighbors to process
17	Check if all nodes processed	-	[]	[A:0, B:0, C:0, D:0, E:0, F:0]	[A, F, B, C, D, E]	All nodes processed, topological order complete

💡 Queue empty and all nodes processed, topological order found

Variable Tracker

Variable	Start	After Step 2	After Step 6	After Step 11	After Step 15	Final
Queue	[]	[A, F]	[F, B, C]	[]	[E]	[]
In-degree	[A:0, B:1, C:1, D:2, E:1, F:0]	[A:0, B:1, C:1, D:2, E:1, F:0]	[A:0, B:0, C:0, D:2, E:1, F:0]	[A:0, B:0, C:0, D:0, E:1, F:0]	[A:0, B:0, C:0, D:0, E:0, F:0]	[A:0, B:0, C:0, D:0, E:0, F:0]
Topological Order	[]	[]	[A]	[A, F, B, C]	[A, F, B, C, D]	[A, F, B, C, D, E]

Key Moments - 3 Insights

Why do we add nodes with zero in-degree to the queue at the start?

Nodes with zero in-degree have no dependencies and can be processed first. This is shown in execution_table row 2 where nodes A and F are added to the queue initially.

What happens when we decrease the in-degree of neighbors?

Decreasing in-degree simulates removing the processed node's edges. When a neighbor's in-degree becomes zero, it means all its dependencies are processed, so it is added to the queue (see rows 4-6 and 11-12).

How do we know if the graph has a cycle?

If after processing, some nodes still have non-zero in-degree and the queue is empty, it means a cycle exists. Here, all nodes reach zero in-degree and are processed (row 17), so no cycle.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what nodes are in the queue?

A [F, B, C]

B [A, F]

C [B, C]

D [C]

💡 Hint

Check the 'Queue State' column at step 6 in execution_table

At which step does node D get added to the queue?

A Step 11

B Step 12

C Step 13

D Step 14

💡 Hint

Look for the step where 'Add D to queue' operation happens in execution_table

If node E had an extra incoming edge, how would the in-degree array change after step 14?

A E's in-degree would be 1 instead of 0

B E's in-degree would be 0 as before

C D's in-degree would increase

D F's in-degree would increase

💡 Hint

Check the 'In-degree Array' column at step 14 and consider how extra edges affect in-degree

Concept Snapshot

Topological Sort (Kahn's Algorithm):
- Calculate in-degree for all nodes
- Add nodes with zero in-degree to a queue
- While queue not empty:
  - Remove node, add to order
  - Decrease in-degree of neighbors
  - Add neighbors with zero in-degree to queue
- If all nodes processed, order is valid
- Detect cycle if nodes remain unprocessed

Full Transcript

Topological Sort using Kahn's Algorithm works by first calculating the in-degree of each node, which counts how many edges come into it. Nodes with zero in-degree have no dependencies and are added to a queue. Then, the algorithm repeatedly removes nodes from the queue, adds them to the topological order, and decreases the in-degree of their neighbors. If a neighbor's in-degree becomes zero, it is added to the queue. This process continues until the queue is empty. If all nodes are processed, the topological order is complete. If some nodes remain with non-zero in-degree, it means the graph has a cycle and no valid topological order exists. The execution table shows each step, including queue contents, in-degree array, and the growing topological order. Key moments include understanding why zero in-degree nodes start the process, how in-degree decreases simulate edge removal, and how cycle detection works by checking unprocessed nodes. The visual quiz tests understanding of queue states, step when nodes are added, and effects of extra edges on in-degree.