Shortest Path in DAG Using Topological Order in DSA C - Execution Trace

Concept Flow - Shortest Path in DAG Using Topological Order

Build graph with edges and weights

↓

Find topological order of nodes

↓

Initialize distances: start node=0, others=∞

↓

For each node in topological order

↓

Relax edges from current node

↓

Update shortest distances

↓

End: shortest paths found

We first build the graph, then find a topological order. We initialize distances and relax edges in that order to find shortest paths.

Execution Sample

DSA C

1. Create graph edges with weights
2. Get topological order
3. Set distance[start] = 0, others = infinity
4. For each node in order, relax edges
5. Print shortest distances

This code finds shortest paths from a start node in a DAG by processing nodes in topological order.

Execution Table

Step	Operation	Current Node	Distances Array	Relaxed Edge	Distance Update	Visual State
1	Initialize distances	N/A	[0, ∞, ∞, ∞, ∞]	N/A	Start node distance set to 0	0 -> ∞ -> ∞ -> ∞ -> ∞
2	Process node 0	0	[0, ∞, ∞, ∞, ∞]	0->1 (weight 3)	Distance[1] = min(∞, 0+3)=3	0 -> 3 -> ∞ -> ∞ -> ∞
3	Process node 0	0	[0, 3, ∞, ∞, ∞]	0->2 (weight 6)	Distance[2] = min(∞, 0+6)=6	0 -> 3 -> 6 -> ∞ -> ∞
4	Process node 1	1	[0, 3, 6, ∞, ∞]	1->2 (weight 4)	Distance[2] = min(6, 3+4)=6 (no change)	0 -> 3 -> 6 -> ∞ -> ∞
5	Process node 1	1	[0, 3, 6, ∞, ∞]	1->3 (weight 4)	Distance[3] = min(∞, 3+4)=7	0 -> 3 -> 6 -> 7 -> ∞
6	Process node 2	2	[0, 3, 6, 7, ∞]	2->4 (weight 2)	Distance[4] = min(∞, 6+2)=8	0 -> 3 -> 6 -> 7 -> 8
7	Process node 3	3	[0, 3, 6, 7, 8]	3->4 (weight 1)	Distance[4] = min(8, 7+1)=8 (no change)	0 -> 3 -> 6 -> 7 -> 8
8	Process node 4	4	[0, 3, 6, 7, 8]	No outgoing edges	No update	0 -> 3 -> 6 -> 7 -> 8
9	End	N/A	[0, 3, 6, 7, 8]	N/A	All nodes processed	0 -> 3 -> 6 -> 7 -> 8

💡 All nodes processed in topological order; shortest distances finalized.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 5	After Step 6	After Step 8	Final
Distances	[0, ∞, ∞, ∞, ∞]	[0, 3, ∞, ∞, ∞]	[0, 3, 6, ∞, ∞]	[0, 3, 6, 7, ∞]	[0, 3, 6, 7, 8]	[0, 3, 6, 7, 8]	[0, 3, 6, 7, 8]
Current Node	N/A	0	0	1	2	4	N/A

Key Moments - 3 Insights

Why do we initialize distances with infinity except the start node?

Because we don't know the shortest path to other nodes yet. Setting them to infinity means they are unreachable until updated. See execution_table step 1.

Why do we process nodes in topological order?

Topological order ensures all edges from earlier nodes to later nodes are relaxed in correct sequence, so shortest paths are found without revisiting nodes. See execution_table steps 2-8.

Why does relaxing an edge sometimes not update the distance?

Because the new calculated distance is not smaller than the current distance. For example, step 4 shows no update since 6 ≤ 7.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is the distance to node 3?

A 4

B ∞

C 7

D 3

💡 Hint

Check the Distances Array column at step 5 in execution_table.

At which step does the distance to node 4 first get updated?

A Step 7

B Step 6

C Step 4

D Step 3

💡 Hint

Look for the first Distance Update involving node 4 in execution_table.

If the edge 1->3 had weight 2 instead of 4, what would be the distance to node 3 after step 5?

A 5

B 7

C 3

D 6

💡 Hint

Calculate distance[3] = min(∞, distance[1] + new weight) using values from execution_table.

Concept Snapshot

Shortest Path in DAG Using Topological Order:
- Build graph with weighted edges
- Find topological order of nodes
- Initialize distances: start=0, others=∞
- Relax edges in topological order
- Update distances if shorter path found
- Result: shortest distances from start node

Full Transcript

This visualization shows how to find shortest paths in a Directed Acyclic Graph (DAG) using topological order. We start by building the graph and finding a topological order of nodes. We initialize distances with zero for the start node and infinity for others. Then, we process each node in topological order, relaxing edges to update distances if a shorter path is found. The execution table tracks each step, showing current node, distances array, edges relaxed, and updates made. Key moments clarify why distances start at infinity, why topological order is important, and why some edge relaxations do not update distances. The visual quiz tests understanding of distances at specific steps and effects of edge weight changes. The concept snapshot summarizes the method in a few lines.

Practice

1. What is the main advantage of using topological order to find the shortest path in a Directed Acyclic Graph (DAG)?

easy

A. It guarantees the graph has cycles.

B. It allows processing vertices in a linear order without revisiting nodes.

C. It sorts vertices by their weights.

D. It finds the longest path instead of the shortest.

Solution

1. Step 1: Understand topological order in DAG

   Topological order arranges vertices so that every directed edge goes from an earlier vertex to a later one, ensuring no cycles.

2. Step 2: Use topological order for shortest path

   Processing vertices in this order allows updating shortest distances without revisiting nodes, making the algorithm efficient.

3. Final Answer:

   It allows processing vertices in a linear order without revisiting nodes. -> Option B

4. Quick Check:

   Topological order = linear processing [OK]

Hint: Topological order processes nodes once in DAG [OK]

Common Mistakes:

• Thinking DAG has cycles
• Confusing topological order with sorting by weights
• Assuming it finds longest path by default

2. Which of the following C code snippets correctly initializes the distance array for shortest path calculation in a DAG with 5 vertices?

easy

A. int dist[5] = {0, 0, 0, 0, 0};

B. int dist[5] = {-1, -1, -1, -1, -1};

C. int dist[5]; for(int i=0; i<5; i++) dist[i] = INT_MAX;

D. int dist[5] = {INT_MIN, INT_MIN, INT_MIN, INT_MIN, INT_MIN};

Solution

1. Step 1: Understand initialization for shortest path

   Distances should start as very large values (infinity) to represent unknown paths.

2. Step 2: Identify correct initialization

   Using INT_MAX sets all distances to a large number, suitable for shortest path updates.

3. Final Answer:

   int dist[5]; for(int i=0; i<5; i++) dist[i] = INT_MAX; -> Option C

4. Quick Check:

   Initialize distances to INT_MAX [OK]

Hint: Initialize distances with INT_MAX for shortest path [OK]

Common Mistakes:

• Initializing distances to zero (wrong for shortest path)
• Using INT_MIN which is for max path, not min
• Using -1 which can cause incorrect comparisons

3. Consider the following DAG with edges and weights:
0 -> 1 (2), 0 -> 2 (4), 1 -> 2 (1), 1 -> 3 (7), 2 -> 3 (3)
If we run shortest path from vertex 0 using topological order, what is the shortest distance to vertex 3?

medium

A. 6

B. 7

C. 9

D. 10

Solution

1. Step 1: Find topological order and initialize distances

   Topological order can be [0,1,2,3]. Initialize dist[0]=0, others = infinity.

2. Step 2: Relax edges in topological order

   From 0: dist[1]=2, dist[2]=4.
   From 1: dist[2]=min(4,2+1)=3, dist[3]=2+7=9.
   From 2: dist[3]=min(9,3+3)=6.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Shortest path to 3 = 6 [OK]

Hint: Relax edges in topological order to update distances [OK]

Common Mistakes:

• Not updating dist[2] to 3 from 1->2 edge
• Ignoring better path from 2 to 3
• Using wrong topological order

4. Given this code snippet for shortest path in DAG, what is the error?

for (int i = 0; i < V; i++) {
  for (auto edge : adj[i]) {
    int v = edge.first;
    int weight = edge.second;
    if (dist[i] + weight < dist[v])
      dist[v] = dist[i] + weight;
  }
}

Assume dist is initialized and adj is adjacency list.

medium

A. The outer loop should follow topological order, not 0 to V-1.

B. The weight variable is not assigned correctly.

C. The condition should be dist[v] + weight < dist[i].

D. The adjacency list should be reversed.

Solution

1. Step 1: Check loop order for shortest path in DAG

   The outer loop must process vertices in topological order to ensure correct relaxation.

2. Step 2: Identify the error in loop order

   Looping from 0 to V-1 may not respect topological order, causing incorrect distance updates.

3. Final Answer:

   The outer loop should follow topological order, not 0 to V-1. -> Option A

4. Quick Check:

   Use topological order for outer loop [OK]

Hint: Always loop vertices in topological order for DAG shortest path [OK]

Common Mistakes:

• Ignoring topological order in loops
• Mixing up edge weight usage
• Reversing adjacency list unnecessarily

5. You have a DAG with 6 vertices and edges:
0->1(5), 0->2(3), 1->3(6), 1->2(2), 2->4(4), 2->5(2), 2->3(7), 3->4(-1), 4->5(-2)
Using shortest path with topological order from vertex 0, what is the shortest distance to vertex 5?

hard

A. 7

B. 8

C. 4

D. 5

Solution

1. Step 1: Determine topological order and initialize distances

   One topological order: [0,1,2,3,4,5]. Initialize dist[0]=0, others infinity.

2. Step 2: Relax edges in order

   From 0: dist[1]=5, dist[2]=3.
   From 1: dist[3]=min(inf,5+6=11)=11, dist[2]=min(3,5+2=7)=3 (unchanged).
   From 2: dist[4]=min(inf,3+4=7)=7, dist[5]=min(inf,3+2=5)=5, dist[3]=min(11,3+7=10)=10.
   From 3: dist[4]=min(7,10+(-1)=9)=7 (unchanged).
   From 4: dist[5]=min(5,7+(-2)=5)=5 (unchanged).

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Shortest distance to 5 = 5 [OK]

Hint: Relax edges in topological order, watch negative weights carefully [OK]

Common Mistakes:

• Ignoring negative weights in relaxation
• Not updating distances with better paths
• Wrong topological order causing errors