Articulation Points in Graph in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is an articulation point in a graph?

An articulation point is a vertex in a graph that, if removed along with its edges, increases the number of connected components. It is also called a cut vertex.

beginner

Why are articulation points important in network design?

Articulation points represent critical nodes whose failure can disconnect the network. Identifying them helps improve network reliability and fault tolerance.

intermediate

What two arrays are commonly used in Tarjan's algorithm to find articulation points?

The two arrays are:
1. discovery time array (disc) - stores the time a node is first visited.
2. low value array (low) - stores the earliest visited vertex reachable from the subtree rooted at that node.

intermediate

In Tarjan's algorithm, what condition identifies a node as an articulation point when it is not the root?

If a child node v of u has low[v] >= disc[u], then u is an articulation point. This means v or its descendants cannot reach an ancestor of u without going through u.

intermediate

How is the root node of DFS tree treated differently in articulation point detection?

The root node is an articulation point if it has two or more children in the DFS tree. This means removing it disconnects the graph.

What happens if you remove an articulation point from a connected graph?

A Nothing changes

B The graph remains connected

C The graph becomes a tree

D The graph becomes disconnected

Removing an articulation point increases the number of connected components, disconnecting the graph.

Which of these arrays is NOT used in Tarjan's algorithm for articulation points?

A Distance array

B Low value array

C Parent array

D Discovery time array

Tarjan's algorithm uses discovery time, low value, and parent arrays, but not a distance array.

In Tarjan's algorithm, if low[v] >= disc[u], what does it imply about u?

A u and v are the same node

B v is an articulation point

C u is an articulation point

D No articulation point

This condition means u is an articulation point separating v's subtree from ancestors.

How many children must the root of DFS tree have to be an articulation point?

A At least 2

B At least 1

C Exactly 1

D None

The root is an articulation point only if it has two or more children in the DFS tree.

What is the time complexity of Tarjan's algorithm to find articulation points?

A O(V^2)

B O(V + E)

C O(E^2)

D O(log V)

Tarjan's algorithm runs in linear time relative to vertices and edges.

Explain how Tarjan's algorithm finds articulation points in a graph.

Think about how DFS helps track back edges and subtree connections.

Describe why articulation points are critical in network reliability.

Consider what happens if a key node fails in a communication network.

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