Shortest Path in DAG Using Topological Order in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a Directed Acyclic Graph (DAG)?

A DAG is a graph with directed edges and no cycles, meaning you cannot start at one node and follow a path that loops back to it.

intermediate

Why do we use topological order for shortest path in a DAG?

Topological order arranges nodes so all edges go from earlier to later nodes, allowing us to relax edges in one pass without revisiting nodes.

beginner

What does 'relaxing an edge' mean in shortest path algorithms?

Relaxing an edge means checking if the current known shortest distance to a node can be improved by going through another node connected by that edge.

beginner

In the shortest path algorithm for DAG, what initial distance values do we assign?

We set the distance to the source node as 0 and all other nodes as infinity (or a very large number) to indicate they are not yet reachable.

intermediate

What is the time complexity of finding shortest paths in a DAG using topological order?

The time complexity is O(V + E), where V is the number of vertices and E is the number of edges, because each node and edge is processed once.

What property must a graph have to apply shortest path using topological order?

A It must be a Directed Acyclic Graph (DAG)

B It must be undirected

C It must contain cycles

D It must be a tree

Only DAGs have a valid topological order, which is essential for this shortest path method.

What is the first step in finding shortest paths in a DAG using topological order?

A Find the topological order of the nodes

B Initialize all distances to zero

C Relax all edges randomly

D Run Dijkstra's algorithm

We first find the topological order to process nodes in a sequence that respects edge directions.

When relaxing edges in shortest path for DAG, what do we update?

A The graph structure

B Distance to the target node if a shorter path is found

C The topological order

D The source node

Relaxation updates the shortest known distance to a node if a better path is found.

What initial distance do we assign to nodes other than the source in shortest path for DAG?

A Negative one

B Zero

C Infinity or a very large number

D Random values

Nodes start with infinite distance to indicate they are not reachable yet.

What is the advantage of using topological order for shortest path in DAG over Dijkstra's algorithm?

A It finds longest paths

B It works with negative cycles

C It uses less memory

D It runs faster with O(V + E) time complexity

Topological order allows processing each node once, making it faster than Dijkstra's O(E log V) in DAGs.

Explain how topological order helps in finding the shortest path in a DAG.

Think about how ordering nodes prevents revisiting and simplifies edge relaxation.

Describe the steps to find shortest paths from a source node in a DAG using topological order.

Start from ordering nodes, then update distances step-by-step.

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