Bipartite Graph Check in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bipartite Check on Simple Graph

What is the output of this Rust code that checks if the given graph is bipartite?

DSA Rust

use std::collections::VecDeque;

fn is_bipartite(graph: &Vec<Vec<usize>>) -> bool {
    let n = graph.len();
    let mut colors = vec![-1; n];

    for start in 0..n {
        if colors[start] == -1 {
            let mut queue = VecDeque::new();
            queue.push_back(start);
            colors[start] = 0;

            while let Some(u) = queue.pop_front() {
                for &v in &graph[u] {
                    if colors[v] == -1 {
                        colors[v] = 1 - colors[u];
                        queue.push_back(v);
                    } else if colors[v] == colors[u] {
                        return false;
                    }
                }
            }
        }
    }
    true
}

fn main() {
    let graph = vec![
        vec![1, 3],
        vec![0, 2],
        vec![1, 3],
        vec![0, 2],
    ];
    println!("{}", is_bipartite(&graph));
}

A Compilation error

B false

C panic at runtime

D true

💡 Hint

Think about whether the graph can be colored with two colors without adjacent nodes sharing the same color.

The graph is a square cycle (0-1-2-3-0), which is bipartite. The code colors nodes alternatingly and finds no conflict, so it returns true.

❓ Predict Output

intermediate

Output of Bipartite Check on Non-Bipartite Graph

What is the output of this Rust code that checks if the given graph is bipartite?

DSA Rust

use std::collections::VecDeque;

fn is_bipartite(graph: &Vec<Vec<usize>>) -> bool {
    let n = graph.len();
    let mut colors = vec![-1; n];

    for start in 0..n {
        if colors[start] == -1 {
            let mut queue = VecDeque::new();
            queue.push_back(start);
            colors[start] = 0;

            while let Some(u) = queue.pop_front() {
                for &v in &graph[u] {
                    if colors[v] == -1 {
                        colors[v] = 1 - colors[u];
                        queue.push_back(v);
                    } else if colors[v] == colors[u] {
                        return false;
                    }
                }
            }
        }
    }
    true
}

fn main() {
    let graph = vec![
        vec![1, 2],
        vec![0, 2],
        vec![0, 1],
    ];
    println!("{}", is_bipartite(&graph));
}

A false

B true

C Compilation error

D panic at runtime

💡 Hint

Check if the triangle graph can be colored with two colors without conflict.

The graph is a triangle (3 nodes all connected), which is not bipartite. The code detects adjacent nodes with the same color and returns false.

🔧 Debug

advanced

Identify the Error in Bipartite Check Implementation

What error will this Rust code produce when checking if the graph is bipartite?

DSA Rust

use std::collections::VecDeque;

fn is_bipartite(graph: &Vec<Vec<usize>>) -> bool {
    let n = graph.len();
    let mut colors = vec![-1; n];

    for start in 0..n {
        if colors[start] == -1 {
            let mut queue = VecDeque::new();
            queue.push_back(start);
            colors[start] = 0;

            while let Some(u) = queue.pop_front() {
                for &v in &graph[u] {
                    if colors[v] == -1 {
                        colors[v] = 1 - colors[u];
                        queue.push_back(v);
                    } else if colors[v] == colors[u] {
                        return false;
                    }
                }
            }
        }
    }
    true
}

fn main() {
    let graph = vec![
        vec![1],
        vec![0],
    ];
    println!("{}", is_bipartite(&graph));
}

A Runtime panic due to index out of bounds

B Returns false incorrectly

C SyntaxError due to missing semicolon

D Returns true correctly

💡 Hint

Look carefully at the line where colors[start] is assigned.

The line 'colors[start] = 0' is missing a semicolon, causing a syntax error.

❓ Predict Output

advanced

Output of Bipartite Check on Disconnected Graph

What is the output of this Rust code that checks if the given disconnected graph is bipartite?

DSA Rust

use std::collections::VecDeque;

fn is_bipartite(graph: &Vec<Vec<usize>>) -> bool {
    let n = graph.len();
    let mut colors = vec![-1; n];

    for start in 0..n {
        if colors[start] == -1 {
            let mut queue = VecDeque::new();
            queue.push_back(start);
            colors[start] = 0;

            while let Some(u) = queue.pop_front() {
                for &v in &graph[u] {
                    if colors[v] == -1 {
                        colors[v] = 1 - colors[u];
                        queue.push_back(v);
                    } else if colors[v] == colors[u] {
                        return false;
                    }
                }
            }
        }
    }
    true
}

fn main() {
    let graph = vec![
        vec![1],
        vec![0],
        vec![3],
        vec![2, 4],
        vec![3],
    ];
    println!("{}", is_bipartite(&graph));
}

A false

B true

C Compilation error

D panic at runtime

💡 Hint

Consider each disconnected component separately for bipartite check.

Both disconnected components are bipartite, so the function returns true.

❓ Predict Output

expert

Output of Bipartite Check on Complex Graph with Odd Cycle

What is the output of this Rust code that checks if the given complex graph with an odd cycle is bipartite?

DSA Rust

use std::collections::VecDeque;

fn is_bipartite(graph: &Vec<Vec<usize>>) -> bool {
    let n = graph.len();
    let mut colors = vec![-1; n];

    for start in 0..n {
        if colors[start] == -1 {
            let mut queue = VecDeque::new();
            queue.push_back(start);
            colors[start] = 0;

            while let Some(u) = queue.pop_front() {
                for &v in &graph[u] {
                    if colors[v] == -1 {
                        colors[v] = 1 - colors[u];
                        queue.push_back(v);
                    } else if colors[v] == colors[u] {
                        return false;
                    }
                }
            }
        }
    }
    true
}

fn main() {
    let graph = vec![
        vec![1, 4],
        vec![0, 2],
        vec![1, 3, 4],
        vec![2, 4],
        vec![0, 2, 3],
    ];
    println!("{}", is_bipartite(&graph));
}

A false

B true

C Compilation error

D panic at runtime

💡 Hint

Look for an odd cycle in the graph that prevents bipartite coloring.

The graph contains an odd cycle (2-3-4-2), so it is not bipartite. The code detects color conflict and returns false.

Practice

1. What does it mean for a graph to be bipartite?

easy

A. Its vertices can be divided into two groups with no edges inside the same group

B. It has an equal number of vertices and edges

C. It contains at least one cycle of odd length

D. All vertices have the same degree

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph is one where vertices can be split into two sets so that no edge connects vertices within the same set.

2. Step 2: Check options against definition

   Its vertices can be divided into two groups with no edges inside the same group matches this definition exactly. Options A, B, and D describe unrelated properties.

3. Final Answer:

   Its vertices can be divided into two groups with no edges inside the same group -> Option A

4. Quick Check:

   Bipartite graph = two groups, no internal edges [OK]

Hint: Bipartite means two groups with no edges inside groups [OK]

Common Mistakes:

• Confusing bipartite with equal vertices and edges
• Thinking bipartite graphs must have cycles
• Assuming all vertices have same degree

2. Which Rust code snippet correctly declares an adjacency list for an undirected graph to check bipartiteness?

easy

A. let graph: Vec<(usize, usize)> = Vec::new();

B. let graph: Vec<usize> = vec![0; n];

C. let graph: HashMap = HashMap::new();

D. let graph: Vec = vec![vec![]; n];

Solution

1. Step 1: Identify adjacency list type

   An adjacency list for a graph is usually a vector of vectors, where each inner vector holds neighbors of a vertex.

2. Step 2: Match code snippet to adjacency list

   let graph: Vec = vec![vec![]; n]; declares a vector of vectors of usize, initialized with empty vectors for each vertex, which is correct. Options A, B, and C do not represent adjacency lists properly.

3. Final Answer:

   let graph: Vec = vec![vec![]; n]; -> Option D

4. Quick Check:

   Adjacency list = Vec of Vecs [OK]

Hint: Adjacency list is Vec in Rust [OK]

Common Mistakes:

• Using Vec<usize> instead of Vec
• Using HashMap with wrong types
• Using tuple vector which is edge list, not adjacency list

3. Given the following Rust code snippet to check bipartiteness, what is the output?

let graph = vec![
    vec![1, 3],
    vec![0, 2],
    vec![1, 3],
    vec![0, 2],
];

fn is_bipartite(graph: &Vec) -> bool {
    let n = graph.len();
    let mut colors = vec![-1; n];
    for i in 0..n {
        if colors[i] == -1 {
            let mut stack = vec![i];
            colors[i] = 0;
            while let Some(u) = stack.pop() {
                for &v in &graph[u] {
                    if colors[v] == -1 {
                        colors[v] = 1 - colors[u];
                        stack.push(v);
                    } else if colors[v] == colors[u] {
                        return false;
                    }
                }
            }
        }
    }
    true
}

println!("{}", is_bipartite(&graph));

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Analyze graph structure

   The graph has 4 vertices with edges forming a square (0-1-2-3-0). This is an even cycle, which is bipartite.

2. Step 2: Trace DFS coloring

   The code colors vertex 0 as 0, then neighbors 1 and 3 as 1, then neighbors 2 as 0. No conflicts occur, so it returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Even cycle graph is bipartite = true [OK]

Hint: Even cycles are bipartite, so output is true [OK]

Common Mistakes:

• Thinking all cycles are non-bipartite
• Confusing stack usage with queue
• Missing color conflict check

4. The following Rust code attempts to check if a graph is bipartite but has a bug. What is the bug?

fn is_bipartite(graph: &Vec) -> bool {
    let n = graph.len();
    let mut colors = vec![-1; n];
    for i in 0..n {
        if colors[i] == -1 {
            let mut stack = vec![i];
            colors[i] = 0;
            while let Some(u) = stack.pop() {
                for &v in &graph[u] {
                    if colors[v] == -1 {
                        colors[v] = colors[u]; // Bug here
                        stack.push(v);
                    } else if colors[v] == colors[u] {
                        return false;
                    }
                }
            }
        }
    }
    true
}

medium

A. Stack is used instead of queue, causing infinite loop

B. Colors assigned to neighbors are same as current vertex, causing wrong bipartite check

C. colors vector is initialized with zeros instead of -1

D. Graph parameter should be mutable

Solution

1. Step 1: Identify color assignment bug

   The code assigns neighbors the same color as current vertex, which breaks bipartite property requiring opposite colors.

2. Step 2: Understand impact on bipartite check

   This causes false negatives or positives because neighbors must have opposite colors, so the check fails.

3. Final Answer:

   Colors assigned to neighbors are same as current vertex, causing wrong bipartite check -> Option B

4. Quick Check:

   Neighbors must have opposite colors [OK]

Hint: Neighbors must get opposite color, not same [OK]

Common Mistakes:

• Assigning same color to neighbors
• Confusing stack and queue usage
• Initializing colors incorrectly

5. You have a graph with vertices 0 to 5 and edges: (0-1), (1-2), (2-3), (3-0), (3-4), (4-5). Which of the following statements is true about its bipartiteness?

hard

A. The graph is not bipartite because it contains an odd cycle (3-4-5)

B. The graph is not bipartite because it contains an even cycle (0-1-2-3)

C. The graph is bipartite because it contains no odd cycles

D. The graph is bipartite because all vertices have degree 2 or less

Solution

1. Step 1: Identify cycles in the graph

   The graph has a cycle 0-1-2-3-0 which is length 4 (even). The path 3-4-5 is a chain, not a cycle.

2. Step 2: Check bipartite condition

   Graphs with no odd cycles are bipartite. Since the only cycle is even, the graph is bipartite.

3. Final Answer:

   The graph is bipartite because it contains no odd cycles -> Option C

4. Quick Check:

   No odd cycles means bipartite [OK]

Hint: Check for odd cycles; none means bipartite [OK]

Common Mistakes:

• Mistaking chains for cycles
• Thinking even cycles break bipartiteness
• Assuming degree affects bipartiteness