Bridges in Graph Tarjan's Algorithm in DSA Go - Build from Scratch

Bridges in Graph Tarjan's Algorithm

📖 Scenario: You are working as a network engineer. You want to find critical connections in a network. These connections, called bridges, if removed, will split the network into disconnected parts. You will use Tarjan's algorithm to find these bridges in a graph.

🎯 Goal: Build a program that finds all bridges in an undirected graph using Tarjan's algorithm. You will create the graph, set up helper variables, implement the DFS logic to find bridges, and finally print the bridges found.

📋 What You'll Learn

Create a graph as an adjacency list with exact edges

Create helper slices for discovery time and low values

Implement Tarjan's DFS function with exact variable names

Print all bridges as pairs of nodes

💡 Why This Matters

🌍 Real World

Network engineers use bridge detection to find critical connections that, if broken, can split a network and cause failures.

💼 Career

Understanding Tarjan's algorithm and graph traversal is important for roles in network design, cybersecurity, and software engineering involving graph data.

1

Create the graph adjacency list

Create a variable called graph as a slice of slices of integers with length 5. Then add these edges exactly: 0-1, 1-2, 2-0, 1-3, 3-4. Use undirected edges by adding both directions.

DSA Go

graph := make([][]int, 5)
// Your code here

Hint

Remember to add edges in both directions for undirected graph.

2

Create helper slices for Tarjan's algorithm

Create three slices called disc, low, and visited each with length 5. Initialize disc and low with -1 for all elements. Initialize visited with false for all elements. Also create an integer variable time set to 0.

DSA Go

graph := make([][]int, 5)
// ... previous edges ...

// Your code here

Hint

Use a for loop to initialize disc and low with -1.

3

Implement Tarjan's DFS to find bridges

Write a function called dfs with parameters u int, parent int, and a pointer to a slice of pairs of integers called bridges. Inside, mark visited[u] true, set disc[u] and low[u] to time, then increment time. For each neighbor v in graph[u], if v is the parent, continue. If visited[v] is false, call dfs(v, u, bridges), then update low[u] to minimum of low[u] and low[v]. If low[v] > disc[u], append the pair {u, v} to *bridges. Else update low[u] to minimum of low[u] and disc[v].

DSA Go

graph := make([][]int, 5)
// ... previous code ...

// Your code here

Hint

Follow the exact steps of Tarjan's algorithm for bridges. Use the variable names given.

4

Run DFS and print all bridges

In main, create a variable bridges as a slice of pair. Then run a for loop from 0 to 4. If visited[i] is false, call dfs(i, -1, &bridges). Finally, print each bridge in bridges as "Bridge: u - v" where u and v are the nodes.

DSA Go

package main

// ... previous code ...

func main() {
    // ... previous code ...

    // Your code here
}

Hint

Run DFS from all unvisited nodes and print each bridge found.

Practice

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

easy

A. To identify edges that, if removed, increase the number of connected components

B. To find the shortest path between two nodes

C. To detect cycles in a graph

D. To count the total number of nodes 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: Purpose of Tarjan's algorithm for bridges

   Tarjan's algorithm efficiently finds all such edges (bridges) in a graph using DFS and low-link values.

3. Final Answer:

   To identify edges that, if removed, increase the number of connected components -> Option A

4. Quick Check:

   Bridges = edges increasing components when removed [OK]

Hint: Bridges split graph if removed, Tarjan finds them fast [OK]

Common Mistakes:

• Confusing bridges with cycles
• Thinking Tarjan finds shortest paths
• Mixing bridges with node counts

2. Which of the following Go code snippets correctly declares a slice to store adjacency lists for a graph with n nodes?

easy

A. adj := make([]int, n)

B. adj := make([][]int, n)

C. adj := [][]int{n}

D. adj := make([]int, 0, n)

Solution

1. Step 1: Understand adjacency list representation

   Adjacency list for n nodes is a slice of slices: each node has a slice of connected nodes.

2. Step 2: Correct Go syntax for adjacency list

   make([][]int, n) creates a slice of n slices of ints, perfect for adjacency lists.

3. Final Answer:

   adj := make([][]int, n) -> Option B

4. Quick Check:

   Adjacency list = slice of slices [OK]

Hint: Adjacency list needs slice of slices, not flat slice [OK]

Common Mistakes:

• Using make([]int, n) which is flat slice
• Incorrect slice literal syntax
• Confusing capacity with length

3. Consider the following Go code snippet implementing part of Tarjan's algorithm for bridges. What will be the output when run on the graph edges: (0-1), (1-2), (2-0), (1-3)?

package main
import "fmt"

var (
    time int
    adj [][]int
    disc []int
    low []int
    visited []bool
)

func dfs(u, parent int) {
    visited[u] = true
    disc[u] = time
    low[u] = time
    time++

    for _, v := range adj[u] {
        if v == parent {
            continue
        }
        if !visited[v] {
            dfs(v, u)
            if low[v] > disc[u] {
                fmt.Printf("Bridge found: %d - %d\n", u, v)
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

func main() {
    n := 4
    adj = make([][]int, n)
    adj[0] = []int{1, 2}
    adj[1] = []int{0, 2, 3}
    adj[2] = []int{0, 1}
    adj[3] = []int{1}

    disc = make([]int, n)
    low = make([]int, n)
    visited = make([]bool, n)

    dfs(0, -1)
}

medium

A. Bridge found: 2 - 0

B. Bridge found: 0 - 1
Bridge found: 1 - 3

C. No bridges found

D. Bridge found: 1 - 3

Solution

1. Step 1: Analyze the graph structure

   The graph has a triangle between nodes 0,1,2 and a single edge from 1 to 3. The triangle edges are not bridges because removing them doesn't disconnect the graph.

2. Step 2: Identify bridges using Tarjan's logic

   Edge 1-3 is a bridge because removing it disconnects node 3. The code prints "Bridge found: 1 - 3" only.

3. Final Answer:

   Bridge found: 1 - 3 -> Option D

4. Quick Check:

   Only edge 1-3 disconnects graph [OK]

Hint: Triangle edges are not bridges; check leaf edges [OK]

Common Mistakes:

• Marking triangle edges as bridges
• Ignoring parent check in DFS
• Confusing discovery and low values

4. The following Go code attempts to find bridges using Tarjan's algorithm but has a bug. What is the error?

func dfs(u, parent int) {
    visited[u] = true
    disc[u] = time
    low[u] = time
    time++

    for _, v := range adj[u] {
        if !visited[v] {
            dfs(v, u)
            if low[v] > disc[u] {
                fmt.Printf("Bridge: %d - %d\n", u, v)
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else if v != parent {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

medium

A. The time variable is not incremented

B. The else condition should check v == parent instead of v != parent

C. The parent check is missing before recursive DFS call

D. The low[u] update is done incorrectly after DFS call

Solution

1. Step 1: Check parent skipping logic

   The code does not skip the parent node before calling DFS recursively, which can cause revisiting the parent and incorrect low values.

2. Step 2: Identify missing parent check before DFS

   Correct code should have "if v == parent { continue }" before DFS call to avoid revisiting parent edge.

3. Final Answer:

   The parent check is missing before recursive DFS call -> Option C

4. Quick Check:

   Skip parent before DFS to avoid cycles [OK]

Hint: Always skip parent node before DFS recursion [OK]

Common Mistakes:

• Confusing parent check placement
• Not incrementing time variable
• Misusing else if condition

5. Given a connected undirected graph with n nodes and m edges, how can Tarjan's algorithm be adapted to find all bridges efficiently? Choose the best approach.

hard

A. Use DFS once from any node, track discovery and low times, and identify edges where low[v] > disc[u]

B. Remove edges one by one and check connectivity each time

C. Use BFS to find shortest paths and mark edges not in shortest paths as bridges

D. Run DFS from each node separately without tracking discovery times

Solution

1. Step 1: Understand Tarjan's algorithm approach

   Tarjan's algorithm uses a single DFS traversal from any node, tracking discovery and low times to find bridges efficiently.

2. Step 2: Identify condition for bridges

   An edge u-v is a bridge if low[v] > disc[u], meaning v cannot reach u or its ancestors without that edge.

3. Final Answer:

   Use DFS once from any node, track discovery and low times, and identify edges where low[v] > disc[u] -> Option A

4. Quick Check:

   Single DFS + low/disc times = bridges [OK]

Hint: Single DFS with low/disc times finds bridges fast [OK]

Common Mistakes:

• Running DFS multiple times unnecessarily
• Using BFS which doesn't track low times
• Brute force edge removal checking connectivity