Bipartite Graph Check in DSA Python - Execution Trace

Concept Flow - Bipartite Graph Check

Start with graph

↓

Pick uncolored node

↓

Assign color 0

↓

Use BFS/DFS to visit neighbors

↓

For each neighbor

↓

If uncolored: assign opposite color

↓

Add neighbor to queue

↓

If colored and same color: Not Bipartite

↓

Repeat until all nodes processed

↓

If no conflicts found

↓

Graph is Bipartite

We start from an uncolored node, assign a color, then use BFS or DFS to color neighbors with opposite colors. If a conflict arises, graph is not bipartite.

Execution Sample

DSA Python

graph = {0: [1,3], 1: [0,2], 2: [1,3], 3: [0,2]}
colors = {}
queue = [0]
colors[0] = 0
while queue:
  node = queue.pop(0)
  for nei in graph[node]:
    if nei not in colors:
      colors[nei] = 1 - colors[node]
      queue.append(nei)
    elif colors[nei] == colors[node]:
      print('Not Bipartite')
      break

This code uses BFS to assign colors 0 and 1 to nodes and checks for conflicts to determine bipartiteness.

Execution Table

Step	Operation	Current Node	Neighbor	Colors Map	Queue	Visual State
1	Start BFS	None	None	{}	[]	Graph: 0--1, 0--3, 1--2, 2--3; No colors assigned
2	Assign color to start node	None	None	{0:0}	[0]	Node 0 colored 0; Queue=[0]
3	Pop node from queue	0	None	{0:0}	[]	Processing node 0
4	Check neighbor	0	1	{0:0}	[]	Neighbor 1 uncolored
5	Assign opposite color	0	1	{0:0,1:1}	[1]	Node 1 colored 1; Queue=[1]
6	Check neighbor	0	3	{0:0,1:1}	[1]	Neighbor 3 uncolored
7	Assign opposite color	0	3	{0:0,1:1,3:1}	[1,3]	Node 3 colored 1; Queue=[1,3]
8	Pop node from queue	1	None	{0:0,1:1,3:1}	[3]	Processing node 1
9	Check neighbor	1	0	{0:0,1:1,3:1}	[3]	Neighbor 0 colored 0, different from 1's color 1, OK
10	Check neighbor	1	2	{0:0,1:1,3:1}	[3]	Neighbor 2 uncolored
11	Assign opposite color	1	2	{0:0,1:1,3:1,2:0}	[3,2]	Node 2 colored 0; Queue=[3,2]
12	Pop node from queue	3	None	{0:0,1:1,3:1,2:0}	[2]	Processing node 3
13	Check neighbor	3	0	{0:0,1:1,3:1,2:0}	[2]	Neighbor 0 colored 0, different from 3's color 1, OK
14	Check neighbor	3	2	{0:0,1:1,3:1,2:0}	[2]	Neighbor 2 colored 0, different from 3's color 1, OK
15	Pop node from queue	2	None	{0:0,1:1,3:1,2:0}	[]	Processing node 2
16	Check neighbor	2	1	{0:0,1:1,3:1,2:0}	[]	Neighbor 1 colored 1, different from 2's color 0, OK
17	Check neighbor	2	3	{0:0,1:1,3:1,2:0}	[]	Neighbor 3 colored 1, different from 2's color 0, OK
18	Queue empty	None	None	{0:0,1:1,3:1,2:0}	[]	All nodes processed, no conflicts
19	Result	None	None	{0:0,1:1,3:1,2:0}	[]	Graph is Bipartite

💡 Queue is empty, all nodes colored without conflict

Variable Tracker

Variable	Start	After Step 2	After Step 5	After Step 7	After Step 11	After Step 15	Final
colors	{}	{0:0}	{0:0,1:1}	{0:0,1:1,3:1}	{0:0,1:1,3:1,2:0}	{0:0,1:1,3:1,2:0}	{0:0,1:1,3:1,2:0}
queue	[]	[0]	[1]	[1,3]	[3,2]	[]	[]

Key Moments - 3 Insights

Why do we assign opposite colors to neighbors?

Assigning opposite colors ensures no two connected nodes share the same color, which is the definition of a bipartite graph. See execution_table rows 5,7,11 where neighbors get opposite colors.

What happens if a neighbor already has the same color as current node?

This means the graph is not bipartite because two connected nodes share the same color. In the execution_table, if such a case occurred, we would detect it and stop. Here, no such conflict appears.

Why do we use a queue in BFS for this check?

The queue helps process nodes level by level, assigning colors systematically. This avoids missing any node and ensures all connected nodes are checked. See variable_tracker 'queue' changes at steps 2,5,7,11.

Visual Quiz - 3 Questions

Look at the execution_table at Step 7, what is the color assigned to node 3?

A Not colored yet

B 0

C 1

D Same as node 0

💡 Hint

Check the 'Colors Map' column at Step 7 in execution_table

At which step does the queue become empty, indicating all nodes are processed?

A Step 18

B Step 15

C Step 11

D Step 3

💡 Hint

Look at the 'Queue' column in execution_table rows to find when it becomes []

If node 2 was assigned color 1 instead of 0 at Step 11, what would happen?

A Graph would still be bipartite

B Conflict detected at neighbor 3

C Queue would become empty earlier

D No change in colors

💡 Hint

Refer to colors of node 3 and node 2 in execution_table and how conflicts are detected

Concept Snapshot

Bipartite Graph Check:
- Assign colors 0 or 1 to nodes using BFS/DFS
- Start from uncolored node, color it 0
- For each neighbor, assign opposite color
- If neighbor has same color, graph is not bipartite
- If all nodes colored without conflict, graph is bipartite

Full Transcript

To check if a graph is bipartite, we color nodes with two colors so that no two connected nodes share the same color. We start from an uncolored node, assign color 0, then use BFS to visit neighbors. Each neighbor gets the opposite color. If a neighbor already has the same color as the current node, the graph is not bipartite. We continue until all nodes are processed. If no conflicts arise, the graph is bipartite. The execution table shows step-by-step how nodes are colored and how the queue changes. The variable tracker shows colors and queue states after key steps. Key moments clarify why opposite colors are assigned, what conflicts mean, and why BFS queue is used. The visual quiz tests understanding of colors assigned, queue emptying, and conflict detection.

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