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

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Tarjan's Bridge Finding on a Simple Graph

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

DSA Go

package main

import "fmt"

var (
    time int
    graph map[int][]int
    visited map[int]bool
    disc map[int]int
    low map[int]int
    bridges [][2]int
)

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

    for _, v := range graph[u] {
        if v == parent {
            continue
        }
        if !visited[v] {
            dfs(v, u)
            if low[v] > disc[u] {
                bridges = append(bridges, [2]int{u, v})
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

func main() {
    graph = map[int][]int{
        0: {1, 2},
        1: {0, 2},
        2: {0, 1, 3},
        3: {2, 4},
        4: {3},
    }
    visited = make(map[int]bool)
    disc = make(map[int]int)
    low = make(map[int]int)
    bridges = make([][2]int, 0)
    time = 0

    for u := range graph {
        if !visited[u] {
            dfs(u, -1)
        }
    }

    fmt.Println(bridges)
}

A [[1 2] [3 4]]

B [[0 1] [1 2]]

C [[0 2] [2 3]]

D [[2 3] [3 4]]

💡 Hint

Focus on edges where the low value of the adjacent node is greater than the discovery time of the current node.

The edges (2,3) and (3,4) are bridges because removing them increases the number of connected components. The other edges are part of cycles.

🧠 Conceptual

intermediate

Understanding Low and Discovery Times in Tarjan's Algorithm

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 maximum depth of the node in the DFS tree

C The total number of edges connected to the node

D The number of connected components in the graph

💡 Hint

Think about the earliest visited node reachable from the current node or its subtree.

The 'low' value is the smallest discovery time reachable from the node by following zero or more tree edges and possibly one back edge.

🔧 Debug

advanced

Identify the Bug in Tarjan's Bridge Detection Code

What error will occur when running this Go code snippet for Tarjan's algorithm?

DSA Go

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

    for _, v := range graph[u] {
        if v == parent {
            continue
        }
        if !visited[v] {
            dfs(v, u)
            if low[v] >= disc[u] {
                bridges = append(bridges, [2]int{u, v})
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

A It fails to compile due to syntax error

B It causes a runtime panic due to nil map access

C It incorrectly identifies some edges as bridges that are not bridges

D It causes infinite recursion due to missing visited check

💡 Hint

Check the condition that decides if an edge is a bridge.

The condition 'low[v] >= disc[u]' is incorrect for bridges; it should be 'low[v] > disc[u]'. Using '>=' causes edges that are part of cycles to be wrongly marked as bridges.

🚀 Application

advanced

Number of Bridges in a Complex Graph

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

A 2

B 1

C 3

D 0

💡 Hint

Look for edges whose removal increases connected components.

Edges (1-3) and (5-6) are bridges. The rest form cycles.

❓ Predict Output

expert

Output of Tarjan's Algorithm on a Disconnected Graph

What is the output of the following Go code that runs Tarjan's bridge-finding algorithm on a disconnected graph?

DSA Go

package main

import "fmt"

var (
    time int
    graph map[int][]int
    visited map[int]bool
    disc map[int]int
    low map[int]int
    bridges [][2]int
)

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

    for _, v := range graph[u] {
        if v == parent {
            continue
        }
        if !visited[v] {
            dfs(v, u)
            if low[v] > disc[u] {
                bridges = append(bridges, [2]int{u, v})
            }
            if low[v] < low[u] {
                low[u] = low[v]
            }
        } else {
            if disc[v] < low[u] {
                low[u] = disc[v]
            }
        }
    }
}

func main() {
    graph = map[int][]int{
        0: {1},
        1: {0},
        2: {3},
        3: {2, 4},
        4: {3},
    }
    visited = make(map[int]bool)
    disc = make(map[int]int)
    low = make(map[int]int)
    bridges = make([][2]int, 0)
    time = 0

    for u := range graph {
        if !visited[u] {
            dfs(u, -1)
        }
    }

    fmt.Println(bridges)
}

A [[0 1] [2 3]]

B [[0 1] [3 4] [2 3]]

C [[3 4]]

D []

💡 Hint

Remember to run DFS on all disconnected components.

The graph has three edges that are bridges: (0,1), (2,3), and (3,4). Each removal increases components.

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