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

Concept Flow - Shortest Path in DAG Using Topological Order

Build graph with edges and weights

↓

Find topological order of nodes

↓

Initialize distances: source=0, others=∞

↓

For each node in topological order

↓

Relax edges from current node

↓

Update distances if shorter path found

↓

Repeat until all nodes processed

↓

Final shortest distances from source

We first order nodes so all edges go forward. Then we update shortest distances step-by-step following this order.

Execution Sample

DSA Go

graph := map[int][]Edge{
  0: {{1, 2}, {2, 4}},
  1: {{2, 1}, {3, 7}},
  2: {{4, 3}},
  3: {{5, 1}},
  4: {{3, 2}, {5, 5}},
}
source := 0
// Compute shortest paths

This code finds shortest paths from node 0 in a directed acyclic graph with weighted edges.

Execution Table

Step	Operation	Current Node	Distances Before	Edges Relaxed	Distances After	Visual State
1	Topological order found	-	-	-	-	Order: 0 -> 1 -> 2 -> 4 -> 3 -> 5
2	Initialize distances	-	-	-	0:0,1:∞,2:∞,3:∞,4:∞,5:∞	Distances: [0,∞,∞,∞,∞,∞]
3	Relax edges from node 0	0	0:0,1:∞,2:∞,3:∞,4:∞,5:∞	0->1(2), 0->2(4)	0:0,1:2,2:4,3:∞,4:∞,5:∞	Distances: [0,2,4,∞,∞,∞]
4	Relax edges from node 1	1	0:0,1:2,2:4,3:∞,4:∞,5:∞	1->2(1), 1->3(7)	0:0,1:2,2:3,3:9,4:∞,5:∞	Distances: [0,2,3,9,∞,∞]
5	Relax edges from node 2	2	0:0,1:2,2:3,3:9,4:∞,5:∞	2->4(3)	0:0,1:2,2:3,3:9,4:6,5:∞	Distances: [0,2,3,9,6,∞]
6	Relax edges from node 4	4	0:0,1:2,2:3,3:9,4:6,5:∞	4->3(2), 4->5(5)	0:0,1:2,2:3,3:8,4:6,5:11	Distances: [0,2,3,8,6,11]
7	Relax edges from node 3	3	0:0,1:2,2:3,3:8,4:6,5:11	3->5(1)	0:0,1:2,2:3,3:8,4:6,5:9	Distances: [0,2,3,8,6,9]
8	Relax edges from node 5	5	0:0,1:2,2:3,3:8,4:6,5:9	-	0:0,1:2,2:3,3:8,4:6,5:9	Distances unchanged
9	End	-	-	-	-	All nodes processed, shortest distances finalized

💡 All nodes processed in topological order, no more edges to relax

Variable Tracker

Variable	Start	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	Final
Distances	[0,∞,∞,∞,∞,∞]	[0,2,4,∞,∞,∞]	[0,2,3,9,∞,∞]	[0,2,3,9,6,∞]	[0,2,3,8,6,11]	[0,2,3,8,6,9]	[0,2,3,8,6,9]	[0,2,3,8,6,9]
Current Node	-	0	1	2	4	3	5	-

Key Moments - 3 Insights

Why do we process nodes in topological order?

Processing nodes in topological order ensures all edges from earlier nodes are relaxed before visiting a node, so distances are updated correctly as shown in steps 3 to 8.

Why do distances start as infinity except the source?

Distances start as infinity to represent unknown paths. Only the source node distance is zero because it's the start point, as initialized in step 2.

Why does relaxing edges update distances only if shorter?

Relaxing edges updates distances only if the new path is shorter, preventing longer paths from overwriting shorter ones. See step 4 where distance to node 2 updates from 4 to 3.

Visual Quiz - 3 Questions

Look at the execution table at step 4, what is the distance to node 3 after relaxing edges from node 1?

A 9

B 7

C ∞

D 2

💡 Hint

Check 'Distances After' column in step 4 of execution_table

At which step does the distance to node 5 first get updated to 9?

A Step 6

B Step 5

C Step 7

D Step 8

💡 Hint

Look at 'Distances After' column for node 5 in steps 6, 7, and 8

If the edge 4->5 had weight 1 instead of 5, how would the distance to node 5 change after step 6?

A It would remain 11

B It would be 7

C It would be 6

D It would be 9

💡 Hint

Calculate distance to node 5 as distance to node 4 plus new edge weight

Concept Snapshot

Shortest Path in DAG Using Topological Order:
- Find topological order of nodes
- Initialize distances: source=0, others=∞
- For each node in order, relax all outgoing edges
- Update distances if shorter path found
- Result: shortest distances from source to all nodes

Full Transcript

This visualization shows how to find shortest paths in a Directed Acyclic Graph (DAG) using topological order. We first find an order of nodes so that all edges go forward. Then we set the source node distance to zero and others to infinity. We process each node in this order, relaxing edges to update distances if a shorter path is found. The execution table tracks each step, showing current node, distances before and after relaxing edges, and the visual state of distances. Key moments clarify why topological order is needed, why distances start as infinity, and why we only update distances when shorter paths are found. The quiz tests understanding of distance updates at specific steps and effects of changing edge weights.

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 that 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 means each vertex is processed after all its predecessors, so shortest paths can be updated without revisiting.

3. Final Answer:

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

4. Quick Check:

   Topological order = linear processing without revisits [OK]

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

Common Mistakes:

• Thinking topological order applies to graphs with cycles
• Confusing topological order with sorting by weights
• Assuming it finds longest path by default

2. Which Go code snippet correctly initializes the distance array for shortest path calculation in a DAG with 5 vertices, starting from vertex 0?

easy

A. dist := make([]int, 5); for i := range dist { dist[i] = 1<<31 - 1 }; dist[0] = 0

B. dist := make([]int, 5); for i := range dist { dist[i] = 0 }; dist[0] = 1<<31 - 1

C. dist := make([]int, 5); for i := range dist { dist[i] = -1 }; dist[0] = 0

D. dist := make([]int, 5); dist[0] = 0

Solution

1. Step 1: Initialize distances with max int

   We set all distances to a large number (infinity) to represent unknown paths, here 1<<31 - 1 for max int.

2. Step 2: Set source distance to zero

   The starting vertex distance is zero because the shortest path from source to itself is zero.

3. Final Answer:

   dist := make([]int, 5); for i := range dist { dist[i] = 1<<31 - 1 }; dist[0] = 0 -> Option A

4. Quick Check:

   Initialize with max int, source zero [OK]

Hint: Set all distances max int, source distance zero [OK]

Common Mistakes:

• Setting all distances to zero initially
• Not setting source distance to zero
• Using -1 as initial distance

3. Given the following DAG edges with weights: 0->1 (2), 0->2 (4), 1->2 (1), 1->3 (7), 2->3 (3). What is the shortest distance array from source vertex 0 after running shortest path using topological order?

dist := []int{0, 2147483647, 2147483647, 2147483647}
// After processing edges in topological order

medium

A. [0, 4, 5, 8]

B. [0, 2, 4, 7]

C. [0, 2, 3, 6]

D. [0, 3, 4, 7]

Solution

1. Step 1: Relax edges from source 0

   dist[1] = min(inf, 0+2) = 2; dist[2] = min(inf, 0+4) = 4

2. Step 2: Relax edges from vertex 1 and 2

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

3. Final Answer:

   [0, 2, 3, 6] -> Option C

4. Quick Check:

   Relax edges in topological order = [0,2,3,6] [OK]

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

Common Mistakes:

• Not updating dist[2] to 3 after relaxing from vertex 1
• Ignoring relaxation from vertex 2 to 3
• Using initial distances without updates

4. Consider this Go code snippet for shortest path in DAG:

dist := make([]int, 4)
for i := range dist {
    dist[i] = 1<<31 - 1
}
dist[0] = 0
for _, u := range topoOrder {
    for _, edge := range graph[u] {
        if dist[u]+edge.weight < dist[edge.to] {
            dist[edge.to] = dist[u] + edge.weight
        }
    }
}

If the graph contains a cycle, what problem will occur?

medium

A. The code will produce correct shortest paths regardless.

B. The code will skip some vertices but finish normally.

C. The code will run infinitely because topological order is invalid.

D. The code will panic due to nil pointer dereference.

Solution

1. Step 1: Understand topological order requirement

   Topological order exists only if the graph has no cycles. If a cycle exists, topoOrder is invalid or partial.

2. Step 2: Effect of cycle on code execution

   Standard topological sort algorithms produce a partial order (not all vertices included), so the loop will skip some vertices but finish normally since topoOrder is a finite slice.

3. Final Answer:

   The code will skip some vertices but finish normally. -> Option B

4. Quick Check:

   Cycle causes partial topoOrder, skips vertices [OK]

Hint: DAG must be acyclic; cycles break topological order [OK]

Common Mistakes:

• Assuming code works fine with cycles
• Expecting panic instead of infinite loop
• Ignoring the need for cycle detection

5. You have a DAG with 6 vertices and weighted 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. 5

B. 4

C. 3

D. 7

Solution

1. Step 1: Initialize distances

   dist = [0, inf, inf, inf, inf, inf]

2. Step 2: Relax edges in topological 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
   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
   From 4: dist[5]=min(5,7+(-2)=5)=5

3. Step 3: Check final distance to vertex 5

   dist[5] = 5

4. Final Answer:

   5 -> Option A

5. Quick Check:

   Relax edges carefully to get dist[5] = 5 [OK]

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

Common Mistakes:

• Ignoring negative edge weights
• Not updating distances after relaxation
• Assuming negative weights cause cycles