Bipartite Graph Check in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the color array with -1 for all vertices.

DSA Python

color = [-1] * [1]

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

Alen(graph)

Brange(len(graph))

Cgraph

Dlen(graph[0])

Attempts:

3 left

💡 Hint

Common Mistakes

Using graph instead of its length causes a TypeError.

Using len(graph[0]) assumes graph is a matrix, but it's an adjacency list.

We use len(graph) to get the number of vertices and initialize the color list with -1 for each vertex.

2fill in blank

medium

Complete the code to assign the starting vertex color as 0.

DSA Python

color[start] = [1]

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

A1

B-1

CNone

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning -1 again means uncolored, so no progress.

Assigning None causes errors in comparisons.

We assign color 0 to the starting vertex to begin the bipartite coloring.

3fill in blank

hard

Fix the error in the condition to check if the neighbor has the same color as the current vertex.

DSA Python

if color[neighbor] == [1]:

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

Acolor[neighbor]

Bcolor[node]

C-1

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing color[neighbor] to itself always true, causing wrong logic.

Comparing to -1 checks if uncolored, not conflict.

We check if the neighbor's color is the same as the current node's color to detect a conflict.

4fill in blank

hard

Fill both blanks to assign the opposite color to the neighbor and add it to the queue.

DSA Python

color[neighbor] = 1 [1] color[node]
queue.[2](neighbor)

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

A-

B+

Cappend

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using + instead of - gives wrong color assignment.

Using pop instead of append removes elements instead of adding.

We assign the opposite color by subtracting from 1 and add the neighbor to the queue using append.

5fill in blank

hard

Fill all three blanks to complete the BFS loop for bipartite check.

DSA Python

while queue:
    node = queue.[1]()
    for neighbor in graph[node]:
        if color[neighbor] == -1:
            color[neighbor] = 1 [2] color[node]
            queue.[3](neighbor)

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

Apop(0)

B-

Cappend

Dpop()

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes from the end, breaking BFS order.

Using + instead of - changes color logic incorrectly.

We use pop(0) to dequeue from the front, assign opposite color with -, and append neighbors to the queue.

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