Bipartite Graph Check in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bipartite Check on a Simple Graph

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

DSA Python

def is_bipartite(graph):
    color = {}
    def dfs(node, c):
        if node in color:
            return color[node] == c
        color[node] = c
        return all(dfs(nei, 1 - c) for nei in graph[node])
    return all(dfs(node, 0) for node in graph if node not in color)

graph = {0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2]}
print(is_bipartite(graph))

A KeyError

B False

C True

D TypeError

💡 Hint

Think about whether the graph can be colored with two colors without conflicts.

The graph is a square cycle (0-1-2-3-0), which is bipartite. The code colors nodes alternatingly and returns True.

❓ Predict Output

intermediate

Output of Bipartite Check on a Triangle Graph

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

DSA Python

def is_bipartite(graph):
    color = {}
    def dfs(node, c):
        if node in color:
            return color[node] == c
        color[node] = c
        return all(dfs(nei, 1 - c) for nei in graph[node])
    return all(dfs(node, 0) for node in graph if node not in color)

graph = {0: [1, 2], 1: [0, 2], 2: [0, 1]}
print(is_bipartite(graph))

A False

B KeyError

C True

D RecursionError

💡 Hint

A triangle graph cannot be colored with two colors without conflict.

The graph is a triangle (3 nodes all connected), which is not bipartite. The code detects a conflict and returns False.

🔧 Debug

advanced

Identify the Error in Bipartite Check Code

What error will this code raise when checking bipartiteness of the graph?

DSA Python

def is_bipartite(graph):
    color = {}
    def dfs(node, c):
        if node in color:
            return color[node] == c
        color[node] = c
        for nei in graph[node]:
            if not dfs(nei, 1 - c):
                return False
        return True
    return all(dfs(node, 0) for node in graph if node not in color)

graph = {0: [1], 1: [0, 2], 2: [1, 3], 3: [2, 4], 4: [3, 5], 5: [4, 6], 6: [5, 7], 7: [6, 8], 8: [7, 9], 9: [8]}
print(is_bipartite(graph))

A True

B RecursionError

C False

D KeyError

💡 Hint

Check if the graph is connected and if the recursion depth is reasonable.

The graph is a simple chain of 10 nodes, which is bipartite. The code runs without error and returns True.

🚀 Application

advanced

Number of Connected Components That Are Bipartite

Given the graph below, how many connected components are bipartite?

DSA Python

graph = {
  0: [1],
  1: [0],
  2: [3, 4],
  3: [2, 4],
  4: [2, 3],
  5: []
}

# Use the bipartite check function from previous examples.

A 0

B 3

C 1

D 2

💡 Hint

Check each connected component separately for bipartiteness.

There are three components: {0,1} bipartite, {2,3,4} not bipartite (triangle), {5} single node bipartite. So 2 bipartite components.

🧠 Conceptual

expert

Why Odd Cycles Prevent Bipartiteness?

Why does the presence of an odd-length cycle in a graph mean it cannot be bipartite?

A Because odd cycles increase the graph's diameter beyond two colors.

B Because odd cycles force a node to have the same color as its neighbor, causing a conflict.

C Because odd cycles create disconnected components.

D Because odd cycles cause infinite recursion in bipartite check algorithms.

💡 Hint

Think about coloring nodes alternately along the cycle.

In an odd cycle, alternating colors will cause the first and last node to have the same color but are connected, causing a conflict.

Practice

1. What does it mean if a graph is bipartite?

easy

A. It has an equal number of vertices and edges

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

C. It contains at least one cycle

D. All vertices have the same degree

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph means its vertices can be split into two sets where no edge connects vertices within the same set.

2. Step 2: Compare options with definition

   Only Its vertices can be divided into two groups with no edges inside the same group matches this definition exactly. Other options describe different graph properties.

3. Final Answer:

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

4. Quick Check:

   Bipartite = Two groups, no internal edges [OK]

Hint: Bipartite means two groups with edges only between groups [OK]

Common Mistakes:

• Confusing bipartite with equal vertices and edges
• Thinking bipartite means no cycles
• Assuming all vertices have same degree

2. Which of the following Python code snippets correctly initializes a graph as an adjacency list for bipartite check?

easy

A. graph = {0: 1, 1: 2, 2: 3, 3: 0}

B. graph = [[0, 1], [1, 2], [2, 3], [3, 0]]

C. graph = {0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2]}

D. graph = [0, 1, 2, 3]

Solution

1. Step 1: Identify adjacency list format

   An adjacency list uses a dictionary mapping each vertex to a list of connected vertices.

2. Step 2: Check each option

   graph = {0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2]} uses a dictionary with lists, correctly representing adjacency. graph = [[0, 1], [1, 2], [2, 3], [3, 0]] is a list of edges, not adjacency list. graph = {0: 1, 1: 2, 2: 3, 3: 0} maps to single vertex, not list. graph = [0, 1, 2, 3] is just a list of vertices.

3. Final Answer:

   graph = {0: [1, 3], 1: [0, 2], 2: [1, 3], 3: [0, 2]} -> Option C

4. Quick Check:

   Adjacency list = dict with lists [OK]

Hint: Adjacency list uses dict with vertex keys and list of neighbors [OK]

Common Mistakes:

• Using list of edges instead of adjacency list
• Mapping vertex to single vertex instead of list
• Using just a list of vertices without edges

3. What is the output of this Python code that checks if the graph is bipartite?

def is_bipartite(graph):
    color = {}
    def dfs(node, c):
        if node in color:
            return color[node] == c
        color[node] = c
        return all(dfs(nei, 1 - c) for nei in graph[node])
    return all(dfs(node, 0) for node in graph if node not in color)

graph = {0: [1], 1: [0, 2], 2: [1, 3], 3: [2]}
print(is_bipartite(graph))

medium

A. True

B. False

C. None

D. Error

Solution

1. Step 1: Understand DFS coloring logic

   The function colors nodes 0 or 1 alternately. If a conflict occurs, it returns False.

2. Step 2: Trace the graph coloring

   Start at node 0 with color 0, node 1 gets color 1, node 2 color 0, node 3 color 1. No conflicts found.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Graph is bipartite, so output is True [OK]

Hint: DFS colors nodes alternately; no conflicts means bipartite [OK]

Common Mistakes:

• Assuming graph with chain edges is not bipartite
• Confusing color assignment logic
• Ignoring disconnected nodes

4. Find the error in this bipartite check code:

def is_bipartite(graph):
    color = {}
    def dfs(node, c):
        if node in color:
            return color[node] != c
        color[node] = c
        return all(dfs(nei, 1 - c) for nei in graph[node])
    return all(dfs(node, 0) for node in graph if node not in color)

medium

A. The graph parameter should be a list, not dict

B. The color dictionary should be a list

C. The dfs function should return True unconditionally

D. The condition should be color[node] == c, not != c

Solution

1. Step 1: Analyze the color check condition

   The code returns False if a node is already colored with the same color c, so it should check equality to detect conflict.

2. Step 2: Identify the incorrect operator

   The code uses != which returns True if colors differ, which is wrong. It should be == to detect conflict and return False.

3. Final Answer:

   The condition should be color[node] == c, not != c -> Option D

4. Quick Check:

   Conflict check needs equality, not inequality [OK]

Hint: Conflict if same color found, so use == not != [OK]

Common Mistakes:

• Using != instead of == in color conflict check
• Changing color dict to list unnecessarily
• Ignoring return values in dfs

5. Given a graph with edges [(0,1), (1,2), (2,3), (3,0), (2,4)], which statement is true about its bipartite property?

hard

A. The graph is bipartite because it has no odd cycles

B. The graph is not bipartite because it contains an odd cycle

C. The graph is bipartite because it has an even cycle

D. The graph is not bipartite because it is disconnected

Solution

1. Step 1: Identify cycles in the graph

   The edges form a cycle 0-1-2-3-0 which has 4 edges (even), so no odd cycle exists.

2. Step 2: Check for odd cycles

   Since the only cycle is even length, the graph is bipartite.

3. Final Answer:

   The graph is bipartite because it has no odd cycles -> Option A

4. Quick Check:

   Odd cycle means not bipartite; no odd cycle means bipartite [OK]

Hint: Odd cycle means not bipartite; even cycle is allowed [OK]

Common Mistakes:

• Confusing even and odd cycle lengths
• Assuming disconnected means not bipartite
• Ignoring extra edges when checking cycles