Bridges in Graph Tarjan's Algorithm in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the discovery time for the current node in Tarjan's algorithm.

DSA Go

disc[u] = [1]

Drag options to blanks, or click blank then click option'

A-1

B0

Ctime

Du

Attempts:

3 left

💡 Hint

Common Mistakes

Setting discovery time to 0 or -1 instead of current time.

In Tarjan's algorithm, we assign the current time as the discovery time for node u.

2fill in blank

medium

Complete the code to update the low-link value after visiting a neighbor in Tarjan's algorithm.

DSA Go

low[u] = min(low[u], [1])

Drag options to blanks, or click blank then click option'

Alow[v]

Bdisc[v]

Ctime

Du

Attempts:

3 left

💡 Hint

Common Mistakes

Using discovery time instead of low-link value for neighbor.

The low-link value of u is updated with the minimum of its current low-link and the low-link of neighbor v.

3fill in blank

hard

Fix the error in the condition that detects a bridge edge in Tarjan's algorithm.

DSA Go

if low[v] [1] disc[u] {
    // (u, v) is a bridge
}

Drag options to blanks, or click blank then click option'

A<

B<=

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= or < instead of > in the condition.

An edge (u, v) is a bridge if the low-link value of v is greater than the discovery time of u.

4fill in blank

hard

Fill both blanks to correctly skip the parent node during DFS and update low-link for back edges.

DSA Go

for _, v := range graph[u] {
    if v == [1] {
        continue
    }
    if disc[v] == -1 {
        dfs(v, u)
        low[u] = min(low[u], [2])
    } else {
        low[u] = min(low[u], disc[v])
    }
}

Drag options to blanks, or click blank then click option'

Aparent

Bu

Cdisc[v]

Dlow[v]

Attempts:

3 left

💡 Hint

Common Mistakes

Using u instead of parent to skip.

Updating low[u] with disc[v] instead of low[v].

We skip the parent node to avoid revisiting it. After DFS on v, update low[u] with low[v].

5fill in blank

hard

Fill all three blanks to correctly initialize and run Tarjan's algorithm to find bridges.

DSA Go

func findBridges(graph [][]int) [][]int {
    n := len(graph)
    disc := make([]int, n)
    low := make([]int, n)
    for i := 0; i < n; i++ {
        disc[i] = [1]
        low[i] = [2]
    }
    time := 0
    var bridges [][]int
    var dfs func(u, [3] int)
    dfs = func(u, parent int) {
        // DFS body omitted for brevity
    }
    for i := 0; i < n; i++ {
        if disc[i] == -1 {
            dfs(i, -1)
        }
    }
    return bridges
}

Drag options to blanks, or click blank then click option'

A-1

B0

Cparent

Dtime

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing disc or low with 0 instead of -1.

Using incorrect parameter name for DFS.

Discovery and low arrays are initialized to -1 to mark unvisited nodes. The DFS function takes parent as the second argument.

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