Shortest Path in DAG Using Topological Order in DSA Rust - 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 = infinity

↓

For each node in topological order

↓

Relax edges from current node

↓

Update distances if shorter path found

↓

Repeat until all nodes processed

↓

Result: shortest distances from start to all nodes

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

Execution Sample

DSA Rust

let graph = vec![
  (0, vec![(1, 2), (2, 4)]),
  (1, vec![(2, 1), (3, 7)]),
  (2, vec![(4, 3)]),
  (3, vec![]),
  (4, vec![(3, 2)])
];

This code sets up a DAG with weighted edges for shortest path calculation.

Execution Table

Step	Operation	Current Node	Distances Before	Edges Relaxed	Distances After	Visual State
1	Initialize distances	-	[0:0, 1:∞, 2:∞, 3:∞, 4:∞]	-	[0:0, 1:∞, 2:∞, 3:∞, 4:∞]	0(0) -> 1(∞) -> 2(∞) -> 3(∞) -> 4(∞)
2	Process node 0	0	[0:0, 1:∞, 2:∞, 3:∞, 4:∞]	Edges: 0->1(2), 0->2(4)	[0:0, 1:2, 2:4, 3:∞, 4:∞]	0(0) -> 1(2) -> 2(4) -> 3(∞) -> 4(∞)
3	Process node 1	1	[0:0, 1:2, 2:4, 3:∞, 4:∞]	Edges: 1->2(1), 1->3(7)	[0:0, 1:2, 2:3, 3:9, 4:∞]	0(0) -> 1(2) -> 2(3) -> 3(9) -> 4(∞)
4	Process node 2	2	[0:0, 1:2, 2:3, 3:9, 4:∞]	Edge: 2->4(3)	[0:0, 1:2, 2:3, 3:9, 4:6]	0(0) -> 1(2) -> 2(3) -> 3(9) -> 4(6)
5	Process node 4	4	[0:0, 1:2, 2:3, 3:9, 4:6]	Edge: 4->3(2)	[0:0, 1:2, 2:3, 3:8, 4:6]	0(0) -> 1(2) -> 2(3) -> 3(8) -> 4(6)
6	Process node 3	3	[0:0, 1:2, 2:3, 3:8, 4:6]	No edges	[0:0, 1:2, 2:3, 3:8, 4:6]	0(0) -> 1(2) -> 2(3) -> 3(8) -> 4(6)
7	End	-	[0:0, 1:2, 2:3, 3:8, 4:6]	-	-	Final shortest distances

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

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
distances	[0, ∞, ∞, ∞, ∞]	[0, 2, 4, ∞, ∞]	[0, 2, 3, 9, ∞]	[0, 2, 3, 9, 6]	[0, 2, 3, 8, 6]	[0, 2, 3, 8, 6]	[0, 2, 3, 8, 6]
current_node	-	0	1	2	4	3	-

Key Moments - 3 Insights

Why do we process nodes in topological order?

Processing nodes in topological order ensures that when we relax edges from a node, all shortest paths to that node are already computed, as shown in steps 2 to 6 in the execution_table.

Why do some distances update multiple times?

Distances update when a shorter path is found via a different route, for example, node 2's distance updates from 4 to 3 at step 3, reflecting a better path found through node 1.

What happens if a node has no outgoing edges?

If a node has no outgoing edges, like node 3 at step 6, no relaxation occurs and distances remain unchanged.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the distance to node 2 after relaxing edges from node 1?

A 4

B ∞

C 3

D 2

💡 Hint

Check the 'Distances After' column at step 3 in the execution_table.

At which step does the distance to node 3 first update to a finite value?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Distances After' column for node 3 in the execution_table rows.

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

A 9

B 8

C 11

D ∞

💡 Hint

Calculate distance to node 3 as distance to node 4 plus edge weight; check step 5 in 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 to all nodes

Full Transcript

This visualization shows how to find shortest paths in a Directed Acyclic Graph (DAG) using topological order. We first build the graph with weighted edges. Then we find a topological order of nodes, which means an order where all edges go from earlier to later nodes. We initialize distances with zero for the start node and infinity for others. Then, for each node in topological order, we relax its edges: we check if going through this node gives a shorter path to its neighbors. If yes, we update the distances. This process continues until all nodes are processed. The final distances array shows the shortest path lengths from the start node to every other node. The execution table tracks each step, showing how distances update and how the graph state changes. Key moments clarify why topological order is important and how updates happen. The quiz tests understanding of distance updates and edge relaxation.

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 finds the longest path instead of the shortest.

B. It guarantees the graph has cycles.

C. It sorts vertices by their weights.

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

Solution

1. Step 1: Understand topological order in DAG

   Topological order arranges vertices so all edges go from earlier to later vertices, ensuring no cycles.

2. Step 2: Use topological order for shortest path

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

3. Final Answer:

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

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

2. Which Rust code snippet correctly initializes a vector to store shortest distances with infinity for all vertices except the source (index 0)?

easy

A. let dist = vec![i32::MAX; n]; dist[0] = 0;

B. let dist = vec![0; n]; dist[0] = i32::MAX;

C. let dist = vec![None; n]; dist[0] = Some(0);

D. let dist = vec![0; n];

Solution

1. Step 1: Initialize distances with maximum value

   Using vec![i32::MAX; n] sets all distances to infinity (max int), representing unknown paths.

2. Step 2: Set source distance to zero

   dist[0] = 0 sets the source vertex distance to zero, as it is the start point.

3. Final Answer:

   let dist = vec![i32::MAX; n]; dist[0] = 0; -> Option A

4. Quick Check:

   Initialize with max, source zero [OK]

Hint: Use vec![i32::MAX; n] and set source to 0 [OK]

Common Mistakes:

• Setting source distance to max instead of zero
• Using zero for all distances initially
• Using Option types unnecessarily

3. Given the following Rust code snippet for shortest path in a DAG, what is the final distance vector printed?

let n = 4;
let edges = vec![
  (0, 1, 1),
  (0, 2, 4),
  (1, 2, 2),
  (1, 3, 6),
  (2, 3, 3),
];

let mut dist = vec![i32::MAX; n];
dist[0] = 0;

let topo_order = vec![0, 1, 2, 3];

for &u in &topo_order {
  for &(src, dst, w) in &edges {
    if src == u && dist[u] != i32::MAX && dist[u] + w < dist[dst] {
      dist[dst] = dist[u] + w;
    }
  }
}

println!("{:?}", dist);

medium

A. [0, 1, 4, 7]

B. [0, 1, 3, 6]

C. [0, 4, 6, 9]

D. [i32::MAX, i32::MAX, i32::MAX, i32::MAX]

Solution

1. Step 1: Initialize distances and process vertex 0

   dist = [0, MAX, MAX, MAX]. From 0: update dist[1] = 0+1=1, dist[2] = 0+4=4.

2. Step 2: Process vertex 1 and 2

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

3. Step 3: Process vertex 3

   No outgoing edges, distances remain.

4. Final Answer:

   [0, 1, 3, 6] -> Option B

5. Quick Check:

   Relax edges in topo order = [0,1,3,6] [OK]

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

Common Mistakes:

• Not updating dist[2] to 3 via vertex 1
• Ignoring better path to dist[3]
• Printing distances before updates

4. Identify the bug in this Rust code snippet for shortest path in a DAG using topological order:

let n = 3;
let edges = vec![(0, 1, 5), (1, 2, 10)];
let mut dist = vec![i32::MAX; n];
dist[0] = 0;
let topo_order = vec![0, 1, 2];

for u in topo_order {
  for &(src, dst, w) in &edges {
    if src == u && dist[dst] != i32::MAX && dist[u] + w < dist[dst] {
      dist[dst] = dist[u] + w;
    }
}
}
println!("{:?}", dist);

medium

A. The condition 'dist[dst] != i32::MAX' prevents updating distances correctly.

B. The topological order vector is incorrect.

C. The initial distance vector should be zeros.

D. Edges are not represented as tuples.

Solution

1. Step 1: Analyze the if condition in edge relaxation

   The condition 'dist[dst] != i32::MAX' wrongly blocks updating dist[dst] if it is still infinity, so no update happens initially.

2. Step 2: Correct the condition to allow updates

   Remove 'dist[dst] != i32::MAX' so that if dist[u] + w is less, dist[dst] updates even if dist[dst] was infinity.

3. Final Answer:

   The condition 'dist[dst] != i32::MAX' prevents updating distances correctly. -> Option A

4. Quick Check:

   Wrong condition blocks updates [OK]

Hint: Allow updates even if dist[dst] is infinity [OK]

Common Mistakes:

• Adding unnecessary condition on dist[dst]
• Incorrect topological order
• Initializing distances with zeros

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

hard

A. 11

B. 10

C. 9

D. 8

Solution

1. Step 1: Initialize distances and process vertex 0

   dist = [0, MAX, MAX, MAX, MAX]. From 0: dist[1] = 2, dist[2] = 3.

2. Step 2: Process vertex 1 and 2

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

3. Step 3: Process vertex 3 and 4

   From 3: dist[4] = 4 + 5 = 9. Vertex 4 has no outgoing edges.

4. Final Answer:

   9 -> Option C

5. Quick Check:

   Shortest path 0->2->3->4 = 9 [OK]

Hint: Relax edges in topo order, pick minimum sums [OK]

Common Mistakes:

• Choosing path 0->1->3->4 without checking shorter 0->2->3->4
• Adding weights incorrectly
• Ignoring topological order