Bridges in Graph Tarjan's Algorithm in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a bridge in a graph?

A bridge is an edge in a graph that, if removed, increases the number of connected components. It means the edge is critical for keeping the graph connected.

intermediate

What is the main idea behind Tarjan's algorithm for finding bridges?

Tarjan's algorithm uses a depth-first search (DFS) to assign discovery times and low values to each vertex. An edge is a bridge if the low value of the connected vertex is greater than the discovery time of the current vertex.

intermediate

In Tarjan's algorithm, what does the 'low' value represent?

The 'low' value of a vertex is the earliest discovery time reachable from that vertex by following zero or more edges, including back edges.

advanced

Why do we compare low[v] > disc[u] to find a bridge between u and v?

If low[v] > disc[u], it means vertex v and its descendants cannot reach any ancestor of u without using the edge (u, v). So, removing (u, v) disconnects the graph, making it a bridge.

beginner

What is the time complexity of Tarjan's algorithm for finding bridges?

The time complexity is O(V + E), where V is the number of vertices and E is the number of edges, because it uses a single DFS traversal.

What does a bridge in a graph do?

A Disconnects the graph if removed

B Connects two vertices with the shortest path

C Is part of a cycle

D Is a vertex with highest degree

A bridge is an edge that, when removed, increases the number of connected components, thus disconnecting the graph.

In Tarjan's algorithm, what is the purpose of the discovery time array?

A To mark visited vertices only

B To store the order vertices are visited in DFS

C To count the number of edges

D To store the parent of each vertex

Discovery time records when each vertex is first visited during DFS, helping to identify bridges.

What does it mean if low[v] <= disc[u] for an edge (u, v)?

A Edge (u, v) is not a bridge

B Edge (u, v) is a bridge

C Vertex v is not reachable

D Graph is disconnected

If low[v] <= disc[u], vertex v or its descendants can reach an ancestor of u, so (u, v) is not a bridge.

Which data structure is primarily used in Tarjan's algorithm for bridges?

A Queue

B Stack

C Depth-First Search recursion

D Priority Queue

Tarjan's algorithm uses DFS recursion to explore the graph and compute discovery and low values.

What is the output of Tarjan's algorithm for bridges?

A Minimum spanning tree

B List of all cycles

C Shortest path tree

D List of all bridges (critical edges)

Tarjan's algorithm outputs all edges that are bridges in the graph.

Explain how Tarjan's algorithm finds bridges in a graph step-by-step.

Think about how DFS explores and back edges affect low values.

Describe the significance of the 'low' value in Tarjan's algorithm and how it helps detect bridges.

Focus on what low value tells about connectivity.

Practice

1. What is the main purpose of Tarjan's algorithm when finding bridges in a graph?

easy

A. To detect cycles in a graph

B. To find the shortest path between two nodes

C. To find edges that, if removed, increase the number of connected components

D. To sort nodes in topological order

Solution

1. Step 1: Understand what a bridge is in a graph

   A bridge is an edge that, when removed, increases the number of connected components in the graph.

2. Step 2: Identify Tarjan's algorithm goal

   Tarjan's algorithm finds such edges efficiently by using DFS and tracking discovery and low times.

3. Final Answer:

   To find edges that, if removed, increase the number of connected components -> Option C

4. Quick Check:

   Bridges = edges increasing components [OK]

Hint: Bridges break connectivity when removed [OK]

Common Mistakes:

• Confusing bridges with cycles
• Thinking it finds shortest paths
• Mixing bridges with articulation points

2. Which of the following Rust code snippets correctly declares a vector to store adjacency lists for a graph with 5 nodes?

easy

A. let graph: Vec = vec![vec![0]; 5];

B. let graph: Vec<usize> = vec![0; 5];

C. let graph: Vec = vec![vec![]; 4];

D. let graph: Vec = vec![vec![]; 5];

Solution

1. Step 1: Understand adjacency list representation

   Each node has a vector of neighbors, so graph is a vector of vectors.

2. Step 2: Check vector initialization for 5 nodes

   vec![vec![]; 5] creates 5 empty vectors, one per node.

3. Final Answer:

   let graph: Vec = vec![vec![]; 5]; -> Option D

4. Quick Check:

   Adjacency list for 5 nodes = vec![vec![]; 5] [OK]

Hint: Use vec![vec![]; n] for adjacency lists [OK]

Common Mistakes:

• Using wrong vector type (Vec<usize> instead of Vec)
• Initializing with wrong size
• Filling inner vectors with zeros instead of empty

3. Given the following Rust code snippet implementing part of Tarjan's algorithm, what will be printed as bridges for the graph edges?

let edges = vec![(0,1), (1,2), (2,0), (1,3)];
// Graph: triangle 0-1-2 and a bridge edge 1-3
// After running Tarjan's algorithm, which edges are bridges?
println!("Bridges: {:?}", vec![(1,3)]);

medium

A. Bridges: [(1, 3)]

B. Bridges: [(0, 1), (1, 2), (2, 0)]

C. Bridges: []

D. Bridges: [(0, 3)]

Solution

1. Step 1: Analyze the graph structure

   The edges (0,1), (1,2), (2,0) form a cycle (triangle), so none are bridges.

2. Step 2: Identify the bridge edge

   The edge (1,3) connects node 3 to the triangle and is a bridge because removing it disconnects node 3.

3. Final Answer:

   Bridges: [(1, 3)] -> Option A

4. Quick Check:

   Only edge (1,3) breaks connectivity [OK]

Hint: Edges in cycles are not bridges [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring leaf node connections
• Confusing bridge edges with articulation points

4. Identify the bug in this Rust snippet of Tarjan's algorithm for finding bridges:

fn dfs(u: usize, parent: usize, graph: &Vec, disc: &mut Vec<usize>, low: &mut Vec<usize>, time: &mut usize, bridges: &mut Vec<(usize, usize)>) {
    disc[u] = *time;
    low[u] = *time;
    *time += 1;
    for &v in &graph[u] {
        if v == parent {
            continue;
        }
        if disc[v] == 0 {
            dfs(v, u, graph, disc, low, time, bridges);
            low[u] = low[u].min(low[v]);
            if low[v] > disc[u] {
                bridges.push((u, v));
            }
        } else {
            low[u] = low[u].min(disc[v]);
        }
    }
}

medium

A. disc and low arrays are not initialized with zeros

B. The parent check should be if v != parent instead of v == parent

C. The time variable should be passed by value, not mutable reference

D. The condition low[v] > disc[u] should be low[v] < disc[u]

Solution

1. Step 1: Check initialization of disc and low arrays

   disc and low must be initialized with zeros to detect unvisited nodes (disc[v] == 0).

2. Step 2: Verify other logic correctness

   Parent check and time passing are correct; condition low[v] > disc[u] is correct for bridges.

3. Final Answer:

   disc and low arrays are not initialized with zeros -> Option A

4. Quick Check:

   Uninitialized disc causes wrong visited checks [OK]

Hint: Initialize disc and low with zeros before DFS [OK]

Common Mistakes:

• Not initializing disc and low arrays
• Wrong parent check condition
• Passing time by value instead of mutable reference

5. You have a graph with 6 nodes and edges: (0,1), (1,2), (2,0), (1,3), (3,4), (4,5), (5,3). Using Tarjan's algorithm, which edges are bridges in this graph?

hard

A. [(3, 4), (4, 5), (5, 3)]

B. [(1, 3)]

C. [(0, 1), (1, 2), (2, 0)]

D. [(1, 3), (3, 4)]

Solution

1. Step 1: Analyze graph components

   Nodes 0,1,2 form a cycle (triangle), so edges among them are not bridges. Nodes 3,4,5 form a cycle, so edges among them are not bridges.

2. Step 2: Identify bridge edges

   The edge (1,3) connects the two cycles. Removing (1,3) disconnects the graph, so it is a bridge.

3. Final Answer:

   [(1, 3)] -> Option B

4. Quick Check:

   Only edge connecting components is bridge [OK]

Hint: Bridges connect separate cycles or components [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring connectivity between components
• Confusing bridges with articulation points