Bridges in Graph Tarjan's Algorithm in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Tarjan's Algorithm on a Simple Graph

What is the output (list of bridges) of the following Tarjan's algorithm code on the given graph?

DSA C

/* Graph edges: 0-1, 1-2, 2-0, 1-3 */
#include <stdio.h>
#include <stdlib.h>
#define MAX 4

int time_counter = 0;
int disc[MAX], low[MAX];
int visited[MAX];
int parent[MAX];

int graph[MAX][MAX] = {
    {0,1,1,0},
    {1,0,1,1},
    {1,1,0,0},
    {0,1,0,0}
};

void dfs(int u) {
    visited[u] = 1;
    disc[u] = low[u] = ++time_counter;
    for (int v = 0; v < MAX; v++) {
        if (graph[u][v]) {
            if (!visited[v]) {
                parent[v] = u;
                dfs(v);
                if (low[v] > disc[u]) {
                    printf("Bridge found: %d - %d\n", u, v);
                }
                if (low[v] < low[u]) low[u] = low[v];
            } else if (v != parent[u]) {
                if (disc[v] < low[u]) low[u] = disc[v];
            }
        }
    }
}

int main() {
    for (int i = 0; i < MAX; i++) {
        visited[i] = 0;
        parent[i] = -1;
    }
    for (int i = 0; i < MAX; i++) {
        if (!visited[i]) dfs(i);
    }
    return 0;
}

A No bridges found

B Bridge found: 0 - 1\nBridge found: 1 - 3

C Bridge found: 2 - 0

D Bridge found: 1 - 3

💡 Hint

Focus on edges that, if removed, increase the number of connected components.

The graph has a cycle between nodes 0,1,2, so edges among them are not bridges. The edge 1-3 connects node 3 to the rest, so it's a bridge.

🧠 Conceptual

intermediate

Understanding Low and Discovery Times

In Tarjan's algorithm for finding bridges, what does the 'low' value of a node represent?

A The earliest discovery time reachable from the node including itself and its descendants

B The total number of edges connected to the node

C The maximum depth of the node in the DFS tree

D The number of connected components in the graph

💡 Hint

Think about what 'low' helps to detect in the graph.

The 'low' value is the smallest discovery time reachable from the node by following DFS tree edges and back edges.

🔧 Debug

advanced

Identify the Bug in Tarjan's Bridge Detection Code

What is the bug in this snippet of Tarjan's algorithm for bridges? void dfs(int u) { visited[u] = 1; disc[u] = low[u] = ++time_counter; for (int v = 0; v < MAX; v++) { if (graph[u][v]) { if (!visited[v]) { parent[v] = u; dfs(v); if (low[v] >= disc[u]) { printf("Bridge found: %d - %d\n", u, v); } if (low[v] < low[u]) low[u] = low[v]; } else if (v != parent[u]) { if (disc[v] < low[u]) low[u] = disc[v]; } } } }

A The graph adjacency matrix is not symmetric

B The parent array is not initialized before DFS

C The condition 'low[v] >= disc[u]' should be 'low[v] > disc[u]' to detect bridges correctly

D The visited array is not updated correctly

💡 Hint

Check the condition that identifies bridges carefully.

For bridges, the condition must be 'low[v] > disc[u]'. Using '>=' incorrectly marks edges that are not bridges.

🚀 Application

advanced

Number of Bridges in a Complex Graph

Given a graph with edges: 0-1, 1-2, 2-3, 3-4, 2-4, 4-5, 5-6, 6-4, how many bridges does Tarjan's algorithm find?

A 3

B 2

C 1

D 0

💡 Hint

Look for edges that, if removed, increase the number of connected components.

Edges 0-1 and 1-2 are bridges. The rest form cycles, so no other bridges.

❓ Predict Output

expert

Output of Tarjan's Algorithm on a Disconnected Graph

What is the output of the following code that runs Tarjan's algorithm on a disconnected graph with edges: 0-1, 1-2, 3-4? #include #include #define MAX 5 int time_counter = 0; int disc[MAX], low[MAX]; int visited[MAX]; int parent[MAX]; int graph[MAX][MAX] = { {0,1,0,0,0}, {1,0,1,0,0}, {0,1,0,0,0}, {0,0,0,0,1}, {0,0,0,1,0} }; void dfs(int u) { visited[u] = 1; disc[u] = low[u] = ++time_counter; for (int v = 0; v < MAX; v++) { if (graph[u][v]) { if (!visited[v]) { parent[v] = u; dfs(v); if (low[v] > disc[u]) { printf("Bridge found: %d - %d\n", u, v); } if (low[v] < low[u]) low[u] = low[v]; } else if (v != parent[u]) { if (disc[v] < low[u]) low[u] = disc[v]; } } } } int main() { for (int i = 0; i < MAX; i++) { visited[i] = 0; parent[i] = -1; } for (int i = 0; i < MAX; i++) { if (!visited[i]) dfs(i); } return 0; }

A Bridge found: 1 - 2\nBridge found: 0 - 1\nBridge found: 3 - 4

B Bridge found: 0 - 1\nBridge found: 1 - 2

C Bridge found: 3 - 4

D No bridges found

💡 Hint

Remember to run DFS on all disconnected components.

Edges 0-1 and 1-2 form a chain with no cycles, so both are bridges. Edge 3-4 connects two nodes with no cycle, so it is also a bridge.

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 edges that, if removed, increase the number of connected components

C. To find the shortest path between two nodes

D. To count the total number of edges in the graph

Solution

1. Step 1: Understand the definition of a bridge

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

2. Step 2: Relate Tarjan's algorithm to bridges

   Tarjan's algorithm efficiently finds such edges by using DFS and low-link values.

3. Final Answer:

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

4. Quick Check:

   Bridges = edges increasing components when removed [OK]

Hint: Bridges disconnect graph parts when removed [OK]

Common Mistakes:

• Confusing bridges with shortest paths
• Thinking bridges detect cycles
• Assuming bridges count edges

2. Which of the following is the correct way to declare the array that stores discovery times in C for Tarjan's algorithm?

easy

A. int disc = new int[MAX];

B. int disc[];

C. int disc[MAX];

D. int[] disc = new int[MAX];

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared with type and size like int arr[size];

2. Step 2: Match with discovery time array

   So, int disc[MAX]; is correct for fixed size MAX.

3. Final Answer:

   int disc[MAX]; -> Option C

4. Quick Check:

   C arrays use type name[size] [OK]

Hint: Use type name[size] for arrays in C [OK]

Common Mistakes:

• Using Java or C++ style array declarations in C
• Not specifying array size
• Using dynamic allocation syntax incorrectly

3. Consider the following code snippet for Tarjan's algorithm to find bridges. What will be printed?

void dfs(int u, int parent) {
  disc[u] = low[u] = ++time;
  for (int v : adj[u]) {
    if (disc[v] == -1) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] > disc[u])
        printf("Bridge: %d - %d\n", u, v);
    } else if (v != parent) {
      low[u] = min(low[u], disc[v]);
    }
  }
}

// Graph edges:
// 0-1, 1-2, 2-0, 1-3
// Start dfs(0, -1);

medium

A. Bridge: 1 - 3

B. Bridge: 0 - 1

C. Bridge: 2 - 0

D. No bridges

Solution

1. Step 1: Analyze the graph structure

   The graph has a triangle 0-1-2 and an edge 1-3. Triangle edges are not bridges because removing them doesn't disconnect the graph.

2. Step 2: Apply Tarjan's condition for bridges

   Edge 1-3 is a bridge because removing it disconnects node 3.

3. Final Answer:

   Bridge: 1 - 3 -> Option A

4. Quick Check:

   Only edge 1-3 disconnects graph [OK]

Hint: Bridges disconnect nodes; triangle edges are safe [OK]

Common Mistakes:

• Thinking all edges in a cycle are bridges
• Ignoring parent check in DFS
• Misinterpreting low and disc values

4. The following code snippet is intended to find bridges using Tarjan's algorithm. What is the bug?

void dfs(int u, int parent) {
  disc[u] = low[u] = ++time;
  for (int v : adj[u]) {
    if (disc[v] == -1) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] >= disc[u])
        printf("Bridge: %d - %d\n", u, v);
    } else if (v != parent) {
      low[u] = min(low[u], disc[v]);
    }
  }
}

medium

A. The condition should be low[v] > disc[u], not >=.

B. The disc array is not initialized.

C. The parent parameter is not used correctly.

D. The time variable is not incremented.

Solution

1. Step 1: Understand bridge condition in Tarjan's algorithm

   Bridge condition is low[v] > disc[u], meaning no back edge from v or its descendants to u or above.

2. Step 2: Identify the bug in condition

   Using >= incorrectly marks edges that are not bridges (like parent edges in cycles).

3. Final Answer:

   The condition should be low[v] > disc[u], not >=. -> Option A

4. Quick Check:

   Bridge if low[v] > disc[u] [OK]

Hint: Use strict greater than for bridge check [OK]

Common Mistakes:

• Using >= instead of > in bridge condition
• Not initializing discovery times
• Ignoring parent node in DFS

5. Given a connected undirected graph with 5 nodes and edges: 0-1, 1-2, 2-3, 3-4, 1-4, which edges are bridges according to Tarjan's algorithm?

hard

A. [0-1, 2-3]

B. [2-3, 3-4]

C. [1-4]

D. [0-1]

Solution

1. Step 1: Visualize the graph

   Edges: 0-1, 1-2, 2-3, 3-4, 1-4. The edge 1-4 creates a cycle with 1-2-3-4.

2. Step 2: Identify bridges

   Removing 0-1 disconnects node 0, so 0-1 is a bridge. Edges 2-3 and 3-4 are in cycle with 1-4, so not bridges.

3. Final Answer:

   [0-1] -> Option D

4. Quick Check:

   Only 0-1 disconnects graph [OK]

Hint: Edges in cycles are not bridges [OK]

Common Mistakes:

• Marking cycle edges as bridges
• Ignoring connectivity after edge removal
• Confusing bridge with articulation point