Bridges in Graph Tarjan's Algorithm in DSA C# - 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 DFS tree edges and back edges.

intermediate

Why do we compare low[v] > disc[u] to find a bridge in Tarjan's algorithm?

If low[v] > disc[u], it means vertex v cannot reach any ancestor of u through back edges, so the edge (u, v) is 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 performs a single DFS traversal.

What does a bridge in a graph do?

A Is part of a cycle

B Connects two disconnected components

C If removed, increases the number of connected components

D Is always a leaf edge

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

In Tarjan's algorithm, what does the discovery time represent?

A The order in which vertices are visited in DFS

B The number of edges connected to a vertex

C The lowest reachable vertex from a vertex

D The total number of vertices

Discovery time is the order in which vertices are first visited during DFS.

Which condition identifies a bridge edge (u, v) in Tarjan's algorithm?

A low[u] > disc[v]

B low[v] > disc[u]

C low[v] < disc[u]

D disc[u] > disc[v]

If low[v] > disc[u], edge (u, v) is a bridge.

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

A Priority Queue

B Stack

C Queue

D Depth-First Search (DFS) recursion

Tarjan's algorithm uses DFS recursion to explore the graph.

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

A O(V + E)

B O(E log V)

C O(V^2)

D O(log V)

Tarjan's algorithm runs in O(V + E) time due to a single DFS traversal.

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 values tell us about connectivity.

Practice

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

easy

A. To sort nodes in topological order

B. To find the shortest path between two nodes

C. To detect cycles in a graph

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

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 depth-first search and tracking discovery and low times.

3. Final Answer:

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

4. Quick Check:

   Bridges = edges increasing components when removed [OK]

Hint: Bridges disconnect graph if removed [OK]

Common Mistakes:

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

2. Which of the following is the correct way to declare the arrays used for discovery and low times in C# for Tarjan's algorithm?

easy

A. int disc = new int[n]; int low = new int[n];

B. int[] disc = new int[n]; int[] low = new int[n];

C. int disc[] = new int[n]; int low[] = new int[n];

D. int[] disc, low = new int[n];

Solution

1. Step 1: Recall C# array declaration syntax

   In C#, arrays are declared as type[] name = new type[size];

2. Step 2: Check each option

   int[] disc = new int[n]; int[] low = new int[n]; uses correct syntax for both arrays. int disc[] = new int[n]; int low[] = new int[n]; uses Java-style syntax, invalid in C#. int disc = new int[n]; int low = new int[n]; misses [] for arrays. int[] disc, low = new int[n]; declares only low as array, disc is just a variable.

3. Final Answer:

   int[] disc = new int[n]; int[] low = new int[n]; -> Option B

4. Quick Check:

   C# arrays = type[] name = new type[size] [OK]

Hint: Use type[] name = new type[size] in C# [OK]

Common Mistakes:

• Using Java array syntax in C#
• Forgetting [] in declaration
• Declaring multiple arrays incorrectly

3. Given the following graph edges and Tarjan's algorithm code snippet, what is the output list of bridges?

Edges: (0-1), (1-2), (2-0), (1-3)

// Assume graph is adjacency list
// Tarjan's algorithm finds bridges

Output bridges:

medium

A. [(0, 1)]

B. [(0, 1), (1, 3)]

C. [(1, 3)]

D. []

Solution

1. Step 1: Analyze the graph structure

   The edges (0-1), (1-2), and (2-0) form a cycle. The edge (1-3) connects node 3 to the cycle.

2. Step 2: Identify bridges

   Edges in a cycle are not bridges because removing them does not disconnect the graph. The edge (1-3) is a bridge because removing it disconnects node 3.

3. Final Answer:

   [(1, 3)] -> Option C

4. Quick Check:

   Only edge (1-3) disconnects graph [OK]

Hint: Edges in cycles are not bridges [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring leaf edges as bridges
• Confusing nodes and edges

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

void DFS(int u, int parent) {
  disc[u] = time;
  low[u] = time;
  time++;
  foreach (int v in graph[u]) {
    if (disc[v] == -1) {
      DFS(v, u);
      low[u] = Math.Min(low[u], low[v]);
      if (low[v] > disc[u])
        bridges.Add((u, v));
    } else {
      low[u] = Math.Min(low[u], disc[v]);
    }
  }
}

medium

A. The check 'if (v == parent) continue;' is missing

B. The 'disc[u]' and 'low[u]' are not initialized before DFS

C. The 'time' variable is not incremented correctly

D. The condition 'if (low[v] > disc[u])' should be 'if (low[v] >= disc[u])'

Solution

1. Step 1: Examine the DFS traversal logic

   The code lacks 'if (v == parent) continue;' after foreach. This check prevents processing the parent edge as a back edge, which would incorrectly lower low[u] and miss bridges.

2. Step 2: Rule out other options

   The condition 'if (low[v] > disc[u])' should be 'if (low[v] >= disc[u])': 'low[v] > disc[u]' is correct for bridges (strict >). The 'disc[u]' and 'low[u]' are not initialized before DFS: disc[u] and low[u] are set inside DFS. The 'time' variable is not incremented correctly: time++ follows assignment correctly.

3. Final Answer:

   The check 'if (v == parent) continue;' is missing -> Option A

4. Quick Check:

   Parent check avoids revisiting parent [OK]

Hint: Always skip parent node in DFS to avoid false bridges [OK]

Common Mistakes:

• Forgetting to skip parent node
• Using wrong comparison operator for bridges
• Not initializing discovery times

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

hard

A. [(1, 3)]

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

C. [(1, 3), (4, 5)]

D. []

Solution

1. Step 1: Analyze graph cycles and connections

   Edges (0-1), (1-2), (2-0) form a cycle. Edges (3-4), (4-5), (5-3) form another cycle. Edge (1-3) connects these two cycles.

2. Step 2: Identify bridges

   Only edge (1-3) is a bridge; removing it disconnects {0,1,2} from {3,4,5}. Cycle edges like (3-4) are not bridges (path 3-5-4 remains), similarly for others.

3. Final Answer:

   [(1, 3)] -> Option A

4. Quick Check:

   Only edge (1-3) disconnects graph [OK]

Hint: Bridges connect cycles or leaf nodes [OK]

Common Mistakes:

• Missing bridges inside connected components
• Confusing cycle edges as bridges
• Ignoring multi-cycle connections