Bipartite Graph Check in DSA Python - Build from Scratch

Bipartite Graph Check

📖 Scenario: Imagine you are organizing teams for a friendly sports event. You want to split players into two groups so that no two players who dislike each other end up in the same group. This is like checking if the relationship graph is bipartite.

🎯 Goal: Build a program that checks if a given graph (players and their dislikes) is bipartite. The program will use colors to separate players into two groups without conflicts.

📋 What You'll Learn

Create a graph using a dictionary where keys are player names and values are lists of disliked players.

Create a dictionary called colors to store the color (group) assigned to each player.

Write a function is_bipartite that uses BFS to check if the graph can be colored with two colors without conflicts.

Print True if the graph is bipartite, otherwise False.

💡 Why This Matters

🌍 Real World

Bipartite graph checks are useful in team assignments, scheduling, and matching problems where two groups must be separated without conflicts.

💼 Career

Understanding bipartite graphs helps in roles involving graph algorithms, network analysis, and optimization problems in software engineering and data science.

1

Create the graph dictionary

Create a dictionary called graph with these exact entries: 'Alice': ['Bob', 'Charlie'], 'Bob': ['Alice', 'David'], 'Charlie': ['Alice', 'David'], 'David': ['Bob', 'Charlie']

DSA Python

# Create the graph dictionary with players and their disliked players
# Your code here

Hint

Use a dictionary where each key is a player and the value is a list of players they dislike.

2

Create the colors dictionary

Create an empty dictionary called colors to store the color assigned to each player.

DSA Python

graph = {
    'Alice': ['Bob', 'Charlie'],
    'Bob': ['Alice', 'David'],
    'Charlie': ['Alice', 'David'],
    'David': ['Bob', 'Charlie']
}

# Create an empty dictionary to store colors
# Your code here

Hint

The colors dictionary will hold player names as keys and their assigned color (0 or 1) as values.

3

Write the is_bipartite function

Write a function called is_bipartite that takes graph as input and returns True if the graph is bipartite, otherwise False. Use a BFS approach with a queue, assign colors 0 and 1 to players, and check for conflicts.

DSA Python

graph = {
    'Alice': ['Bob', 'Charlie'],
    'Bob': ['Alice', 'David'],
    'Charlie': ['Alice', 'David'],
    'David': ['Bob', 'Charlie']
}

colors = {}

# Write the is_bipartite function below
# Your code here

Hint

Use a queue to explore nodes level by level. Assign colors 0 and 1 alternately. If a neighbor has the same color, return False.

4

Print the bipartite check result

Call the function is_bipartite with graph and print the result.

DSA Python

graph = {
    'Alice': ['Bob', 'Charlie'],
    'Bob': ['Alice', 'David'],
    'Charlie': ['Alice', 'David'],
    'David': ['Bob', 'Charlie']
}

colors = {}

from collections import deque

def is_bipartite(graph):
    for node in graph:
        if node not in colors:
            colors[node] = 0
            queue = deque([node])
            while queue:
                current = queue.popleft()
                for neighbor in graph[current]:
                    if neighbor not in colors:
                        colors[neighbor] = 1 - colors[current]
                        queue.append(neighbor)
                    elif colors[neighbor] == colors[current]:
                        return False
    return True

# Print the result of the bipartite check
# Your code here

Hint

Simply call print(is_bipartite(graph)) to show if the graph is bipartite.

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