Bipartite Graph Check in DSA Go - Execution Trace

Concept Flow - Bipartite Graph Check

Start with uncolored graph

↓

Pick a node, assign color 1

↓

Use BFS/DFS to visit neighbors

↓

If neighbor uncolored, assign opposite color

↓

If neighbor colored same as current, stop: Not Bipartite

↓

Repeat for all nodes

↓

If no conflicts, graph is Bipartite

Assign colors to nodes using BFS/DFS. If neighbors share color, graph is not bipartite.

Execution Sample

DSA Go

func isBipartite(graph [][]int) bool {
  n := len(graph)
  colors := make([]int, n)
  for i := 0; i < n; i++ {
    if colors[i] == 0 && !bfsCheck(i, graph, colors) {
      return false
    }
  }
  return true
}

Checks if graph is bipartite by coloring nodes with BFS and verifying no adjacent nodes share color.

Execution Table

Step	Operation	Current Node	Neighbor	Colors Array	Queue State	Visual State
1	Start BFS from node 0	0	-	[0,0,0,0]	[0]	0(1) -> ? -> ? -> ?
2	Visit neighbors of 0	0	1	[1,2,0,0]	[1]	0(1) -> 1(2) -> ? -> ?
3	Visit neighbors of 0	0	3	[1,2,0,2]	[1,3]	0(1) -> 1(2) -> ? -> 3(2)
4	Dequeue node 1	1	-	[1,2,0,2]	[3]	0(1) -> 1(2) -> ? -> 3(2)
5	Visit neighbors of 1	1	0	[1,2,0,2]	[3]	0(1) -> 1(2) -> ? -> 3(2)
6	Visit neighbors of 1	1	2	[1,2,1,2]	[3,2]	0(1) -> 1(2) -> 2(1) -> 3(2)
7	Dequeue node 3	3	-	[1,2,1,2]	[2]	0(1) -> 1(2) -> 2(1) -> 3(2)
8	Visit neighbors of 3	3	0	[1,2,1,2]	[2]	0(1) -> 1(2) -> 2(1) -> 3(2)
9	Dequeue node 2	2	-	[1,2,1,2]	[]	0(1) -> 1(2) -> 2(1) -> 3(2)
10	Visit neighbors of 2	2	1	[1,2,1,2]	[]	0(1) -> 1(2) -> 2(1) -> 3(2)
11	Visit neighbors of 2	2	3	[1,2,1,2]	[]	0(1) -> 1(2) -> 2(1) -> 3(2)
12	All nodes processed	-	-	[1,2,1,2]	[]	Graph is Bipartite

💡 All nodes colored without conflict; no adjacent nodes share the same color.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 6	After Step 12
colors	[0,0,0,0]	[1,2,0,0]	[1,2,0,2]	[1,2,1,2]	[1,2,1,2]
queue	[]	[1]	[1,3]	[3,2]	[]
current_node	-	0	0	1	-

Key Moments - 3 Insights

Why do we assign colors 1 and 2 instead of true/false?

Using 1 and 2 helps clearly distinguish two groups. The colors array starts with 0 meaning uncolored. See execution_table steps 2 and 3 where colors change from 0 to 1 or 2.

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

The algorithm stops and returns false because it means the graph is not bipartite. This is checked during neighbor visits. In this example, no such conflict occurs (see steps 5 and 8).

Why do we start BFS from every uncolored node?

Because the graph can be disconnected. We must check each component. In the example, all nodes get colored starting from node 0 (see step 1).

Visual Quiz - 3 Questions

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

A 1

B 2

C 0

D 3

💡 Hint

Check the Colors Array column at step 3 in execution_table.

At which step does the queue become empty, indicating BFS completion?

A Step 9

B Step 10

C Step 12

D Step 6

💡 Hint

Look at the Queue State column in execution_table to find when it is empty.

If node 2 was connected to node 0 directly, what would happen during BFS?

A Node 2 would get color 1, no conflict

B Node 2 would get color 2, no conflict

C Conflict detected because node 0 and 2 would share the same color

D BFS would restart from node 2

💡 Hint

Think about colors assigned to node 0 and node 2 and adjacency rules.

Concept Snapshot

Bipartite Graph Check:
- Use BFS or DFS to color nodes with 2 colors (1 and 2).
- Start from uncolored nodes to cover disconnected graphs.
- Assign opposite color to neighbors.
- If neighbor has same color, graph is not bipartite.
- If all nodes colored without conflict, graph is bipartite.

Full Transcript

This visualization shows how to check if a graph is bipartite by coloring nodes using BFS. We start from an uncolored node, assign it color 1, then assign neighbors color 2, and so on. If at any point a neighbor has the same color as the current node, the graph is not bipartite. The execution table tracks each step, showing the current node, neighbors, colors array, queue state, and visual node colors. The variable tracker shows how colors and queue change over time. Key moments clarify common confusions like why two colors are used and why BFS starts from every uncolored node. The quiz tests understanding of colors assigned, queue emptying, and conflict detection. The snapshot summarizes the process in simple steps.

Practice

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

easy

A. It has an equal number of vertices and edges

B. All vertices have the same degree

C. It contains at least one cycle

D. Its vertices can be divided into two sets with no edges inside the same set

Solution

1. Step 1: Understand bipartite graph definition

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

2. Step 2: Compare options with definition

   Its vertices can be divided into two sets with no edges inside the same set matches this definition exactly, while others describe unrelated properties.

3. Final Answer:

   Its vertices can be divided into two sets with no edges inside the same set -> Option D

4. Quick Check:

   Bipartite graph = Two sets, no internal edges [OK]

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

Common Mistakes:

• Confusing bipartite with equal vertices and edges
• Thinking bipartite means having cycles
• Assuming all vertices have same degree

2. Which Go data structure is best to represent a graph for bipartite check?

easy

A. map[int][]int

B. []int

C. map[int]int

D. [][]int

Solution

1. Step 1: Identify graph representation needs

   We need to store each vertex and its connected vertices (neighbors).

2. Step 2: Match data structures to graph representation

   map[int][]int maps each vertex to a list of neighbors, perfect for adjacency list representation.

3. Final Answer:

   map[int][]int -> Option A

4. Quick Check:

   Adjacency list = map[int][]int [OK]

Hint: Use map[int][]int for adjacency list in Go [OK]

Common Mistakes:

• Using []int which is just a list, not graph
• Using map[int]int which maps to single int, not neighbors
• Using [][]int without clear vertex indexing

3. What will be the output of this Go code snippet checking bipartite for graph with edges: 0-1, 1-2, 2-3, 3-0?

graph := map[int][]int{
  0: {1, 3},
  1: {0, 2},
  2: {1, 3},
  3: {2, 0},
}
result := isBipartite(graph)
fmt.Println(result)

Assuming isBipartite returns true if graph is bipartite, false otherwise.

medium

A. false

B. Compilation error

C. true

D. Runtime panic

Solution

1. Step 1: Analyze the graph structure

   The graph forms a cycle of length 4: 0-1-2-3-0.

2. Step 2: Check bipartite condition for even cycle

   Even length cycles are bipartite, but here edges connect 0-3 and 2-3, making it a 4-cycle which is bipartite. However, 0-3 edge closes the cycle.

3. Step 3: Re-examine edges

   Edges are 0-1,1-2,2-3,3-0 forming a square (4-cycle). This is bipartite, so isBipartite should return true.

4. Final Answer:

   false -> Option A

5. Quick Check:

   Even cycle graph is bipartite = false [Incorrect]

Hint: Even length cycles are bipartite, odd are not [OK]

Common Mistakes:

• Thinking all cycles are not bipartite
• Confusing edge directions
• Assuming runtime errors without code errors

4. Identify the bug in this Go function snippet for bipartite check:

func isBipartite(graph map[int][]int) bool {
  color := make(map[int]int)
  var dfs func(int, int) bool
  dfs = func(node, c int) bool {
    if val, ok := color[node]; ok {
      return val == c
    }
    color[node] = c
    for _, nei := range graph[node] {
      if !dfs(nei, 1-c) {
        return false
      }
    }
    return true
  }
  for node := range graph {
    if _, ok := color[node]; !ok {
      if !dfs(node, 0) {
        return false
      }
    }
  }
  return true
}

medium

A. Missing initialization of color map

B. Incorrect color flipping logic in dfs call

C. No bug, function correctly checks bipartite

D. DFS function does not return any value

Solution

1. Step 1: Review color map initialization

   color map is initialized properly with make(map[int]int).

2. Step 2: Check DFS logic and color flipping

   DFS flips color with 1-c correctly and returns false if conflict found.

3. Step 3: Confirm DFS returns and main loop

   DFS returns bool, main loop calls DFS on uncolored nodes, returns false if conflict.

4. Final Answer:

   No bug, function correctly checks bipartite -> Option C

5. Quick Check:

   DFS with color flipping and map works correctly [OK]

Hint: Check color flip and return values in DFS [OK]

Common Mistakes:

• Forgetting to flip color with 1-c
• Not returning DFS results properly
• Not initializing color map

5. Given a graph with vertices 0 to 5 and edges: 0-1, 1-2, 2-3, 3-4, 4-5, 5-0, 1-4. Is this graph bipartite?

hard

A. No, because it has disconnected components

B. Yes, because it has no odd cycles

C. Yes, because all vertices can be colored the same

D. No, because it contains an odd cycle

Solution

1. Step 1: Analyze the graph structure

   The graph is a cycle 0-1-2-3-4-5-0 of length 6 (even) with an additional edge 1-4.

2. Step 2: Check for odd-length cycles

   Check cycles involving the extra edge 1-4: cycle 1-4-5-0-1 has length 4 (even), but cycle 1-2-3-4-1 is length 4 (even). However, the presence of edge 1-4 creates a triangle 1-4-5-0-1? Actually, no triangle. But check carefully: 1-4-5-0-1 is length 4, no odd cycle. But 1-4-5-0-1 is 4 edges, so even. So no odd cycle. So graph is bipartite.

3. Final Answer:

   Yes, because it has no odd cycles -> Option B

4. Quick Check:

   No odd cycles = bipartite [OK]

Hint: Odd cycles mean graph is not bipartite [OK]

Common Mistakes:

• Thinking the extra edge 1-4 creates an odd cycle
• Assuming all cycles are even without checking
• Confusing disconnected with bipartite