Articulation Points in Graph in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the discovery time slice for all vertices.

DSA Go

discovery := make([]int, [1])
for i := range discovery {
    discovery[i] = -1
}

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

Alen(graph)

Bcap(graph)

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using capacity instead of length for slice initialization.

Initializing with zero length causing index errors.

We initialize the discovery slice with the length of the graph to store discovery times for each vertex.

2fill in blank

medium

Complete the code to update the low time of the current vertex after visiting a neighbor.

DSA Go

if discovery[neighbor] != -1 {
    low[current] = min(low[current], [1])
}

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

Alow[neighbor]

Bdiscovery[neighbor]

Cdiscovery[current]

Dlow[current]

Attempts:

3 left

💡 Hint

Common Mistakes

Using low[neighbor] instead of discovery[neighbor].

Using low[current] which does not update low time correctly.

We update low[current] with the discovery time of the neighbor if it was already visited.

3fill in blank

hard

Fix the error in the articulation point check condition for non-root vertices.

DSA Go

if parent != -1 && low[neighbor] >= [1] {
    articulation[current] = true
}

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

Adiscovery[neighbor]

Bparent

Clow[current]

Ddiscovery[current]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with discovery[neighbor] instead of discovery[current].

Using low[current] which is incorrect for this condition.

We compare low[neighbor] with discovery[current] to check if current is an articulation point.

4fill in blank

hard

Fill both blanks to correctly implement the DFS call and update the child count.

DSA Go

for _, neighbor := range graph[current] {
    if discovery[neighbor] == -1 {
        parent[neighbor] = current
        [1](neighbor)
        childCount++
    }
}

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

Adfs

Bbfs

Cvisit

Dsearch

Attempts:

3 left

💡 Hint

Common Mistakes

Using BFS or other function names instead of dfs.

Not incrementing childCount causing wrong articulation point detection.

We call dfs recursively on the neighbor and increment childCount for articulation point logic.

5fill in blank

hard

Fill all three blanks to correctly update low time and mark articulation points for root vertex.

DSA Go

if parent[current] == -1 && [1] > 1 {
    articulation[current] = [2]
}

low[current] = min(low[current], [3])

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

AchildCount

Btrue

Cdiscovery[neighbor]

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Marking root as articulation point incorrectly when childCount <= 1.

Using false instead of true to mark articulation point.

Updating low time with wrong value.

Root vertex is articulation point if it has more than one child; low time updated with discovery time of neighbor.

Practice

1. What is an articulation point in a graph?

easy

A. A vertex with the highest degree in the graph

B. A vertex that, when removed, increases the number of connected components

C. A vertex that has no edges connected to it

D. A vertex that forms a cycle with other vertices

Solution

1. Step 1: Understand the definition of articulation point

   An articulation point is a vertex that, if removed, breaks the graph into more pieces (connected components).

2. Step 2: Compare options with definition

   Only A vertex that, when removed, increases the number of connected components matches this definition exactly.

3. Final Answer:

   A vertex that, when removed, increases the number of connected components -> Option B

4. Quick Check:

   Articulation point = vertex increasing components when removed [OK]

Hint: Articulation point breaks graph connectivity when removed [OK]

Common Mistakes:

• Confusing articulation point with highest degree vertex
• Thinking articulation point is a vertex with no edges
• Mixing articulation point with cycle vertices

2. Which Go code snippet correctly declares a graph as an adjacency list for articulation point detection?

easy

A. graph := make(map[int][]int)

B. graph := make(map[int]int)

C. graph := make([]int, n)

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

Solution

1. Step 1: Identify correct data structure for adjacency list

   An adjacency list maps each vertex to a list of connected vertices, so map[int][]int is correct.

2. Step 2: Check other options

   graph := make([][]int, n) is a slice of slices, which can work but usually fixed size; graph := make(map[int]int) is map[int]int (single int), not list; graph := make([]int, n) is slice of ints, not adjacency list.

3. Final Answer:

   graph := make(map[int][]int) -> Option A

4. Quick Check:

   Adjacency list = map vertex to list of neighbors [OK]

Hint: Adjacency list uses map[int][]int in Go [OK]

Common Mistakes:

• Using map[int]int instead of map[int][]int
• Using slice of ints instead of slice of slices or map
• Confusing slice initialization with map initialization

3. What is the output of the following Go code snippet that finds articulation points in a graph with edges: 0-1, 1-2, 1-3?

package main
import "fmt"
func main() {
    n := 4
    graph := make(map[int][]int)
    graph[0] = []int{1}
    graph[1] = []int{0, 2, 3}
    graph[2] = []int{1}
    graph[3] = []int{1}
    articulationPoints := []int{}
    // Assume articulation point detection returns [1]
    articulationPoints = append(articulationPoints, 1)
    fmt.Println(articulationPoints)
}

medium

A. []

B. [0, 1]

C. [1]

D. [2, 3]

Solution

1. Step 1: Analyze the graph structure

   Vertices 0,1,2,3 with edges 0-1, 1-2, 1-3 form a star with center 1.

2. Step 2: Identify articulation points

   Removing vertex 1 disconnects the graph, so 1 is an articulation point; others are not.

3. Final Answer:

   [1] -> Option C

4. Quick Check:

   Center vertex in star graph is articulation point [OK]

Hint: Center vertex connecting others is articulation point [OK]

Common Mistakes:

• Including leaf nodes as articulation points
• Forgetting that removing center disconnects graph
• Assuming multiple articulation points without checking connectivity

4. Find the bug in this Go code snippet for articulation point detection:

func dfs(u, parent int, graph map[int][]int, visited []bool, disc []int, low []int, time *int, ap []bool) {
    visited[u] = true
    disc[u] = *time
    low[u] = *time
    *time = *time + 1
    children := 0
    for _, v := range graph[u] {
        if !visited[v] {
            children++
            dfs(v, u, graph, visited, disc, low, time, ap)
            low[u] = min(low[u], low[v])
            if parent != -1 && low[v] >= disc[u] {
                ap[u] = true
            }
        } else if v != parent {
            low[u] = min(low[u], disc[v])
        }
    }
    if parent == -1 && children > 1 {
        ap[u] = true
    }
}

What is the error?

medium

A. Wrong comparison operator in if condition

B. Missing initialization of visited array

C. Incorrect base case for DFS

D. Incorrect increment syntax for time pointer (*time++)

Solution

1. Step 1: Check pointer increment syntax

   The line '*time++' is invalid in Go; pointer dereference and increment must be done as '*time = *time + 1'.

2. Step 2: Verify other parts

   Visited array initialization, if conditions, and DFS base case are correct.

3. Final Answer:

   Incorrect increment syntax for time pointer (*time++) -> Option D

4. Quick Check:

   Pointer increment in Go uses '*time = *time + 1' [OK]

Hint: Use '*time = *time + 1' to increment pointer value in Go [OK]

Common Mistakes:

• Using '*time++' which is invalid in Go
• Not initializing visited slice before DFS
• Confusing parent check condition
• Missing DFS base case for recursion

5. Given a connected graph with vertices 0 to 5 and edges: 0-1, 1-2, 2-0, 1-3, 3-4, 4-5, 5-3, which vertices are articulation points?

hard

A. [1,3]

B. [1,4]

C. [3]

D. [0,2,4]

Solution

1. Step 1: Analyze the graph structure

   Vertices 0,1,2 form a triangle (cycle), so none are articulation points there. Vertices 3,4,5 form another cycle. Vertex 1 connects these two parts.

2. Step 2: Identify articulation points

   Removing vertex 1 disconnects {0,2} from {3,4,5}. Removing vertex 3 disconnects {0,1,2} from {4,5}. Removing vertex 4 does not disconnect since 5 connects back to 3.

3. Final Answer:

   [1,3] -> Option A

4. Quick Check:

   Vertices connecting cycles or bridges are articulation points [OK]

Hint: Articulation points connect cycles or separate components [OK]

Common Mistakes:

• Including vertices inside cycles as articulation points
• Missing vertex 3 as articulation point
• Assuming all cycle vertices are articulation points