Bipartite Graph Check in DSA Go - Build from Scratch

Bipartite Graph Check

📖 Scenario: You are working on a social network app. You want to check if the users can be split into two groups so that no two friends are in the same group. This means the friendship graph is bipartite.

🎯 Goal: Build a Go program that checks if a given graph is bipartite using a coloring method.

📋 What You'll Learn

Create a graph using an adjacency list representation

Use a slice to keep track of colors for each node

Implement a BFS to assign colors and check bipartiteness

Print true if the graph is bipartite, otherwise false

💡 Why This Matters

🌍 Real World

Social networks, job assignment problems, and scheduling tasks often require checking if a graph is bipartite to separate entities into two compatible groups.

💼 Career

Understanding bipartite graphs is important for software engineers working on graph algorithms, network analysis, and optimization problems.

1

Create the graph adjacency list

Create a variable called graph of type [][]int with these exact edges: node 0 connected to 1 and 3, node 1 connected to 0 and 2, node 2 connected to 1 and 3, node 3 connected to 0 and 2.

DSA Go

graph := [][]int{
    // Your code here
}

Hint

Use a slice of slices of int to represent the graph. Each inner slice lists neighbors of that node.

2

Create a color slice to track node colors

Create a variable called color of type []int with length equal to the number of nodes in graph. Initialize all values to -1 to indicate uncolored nodes.

DSA Go

package main

import "fmt"

func main() {
    graph := [][]int{
        {1, 3},
        {0, 2},
        {1, 3},
        {0, 2},
    }
    color := make([]int, len(graph))
    // Your code here
}

Hint

Use make to create a slice of ints. Use a loop to set all values to -1.

3

Implement BFS to check bipartite property

Write a loop to iterate over each node index i in graph. If color[i] is -1, assign color[i] = 0 and use a queue to perform BFS. For each neighbor v of the current node u, if color[v] is -1, assign color[v] = 1 - color[u] and enqueue v. If color[v] equals color[u], print false and return immediately.

DSA Go

package main

import "fmt"

func main() {
    graph := [][]int{
        {1, 3},
        {0, 2},
        {1, 3},
        {0, 2},
    }
    color := make([]int, len(graph))
    for i := range color {
        color[i] = -1
    }
    // Your code here
}

Hint

Use a slice as a queue for BFS. Assign colors alternating between 0 and 1. If neighbors share the same color, the graph is not bipartite.

4

Print true if graph is bipartite

After the BFS loop completes without conflicts, print true to indicate the graph is bipartite.

DSA Go

package main

import "fmt"

func main() {
    graph := [][]int{
        {1, 3},
        {0, 2},
        {1, 3},
        {0, 2},
    }
    color := make([]int, len(graph))
    for i := range color {
        color[i] = -1
    }

    for i := 0; i  0 {
                u := queue[0]
                queue = queue[1:]
                for _, v := range graph[u] {
                    if color[v] == -1 {
                        color[v] = 1 - color[u]
                        queue = append(queue, v)
                    } else if color[v] == color[u] {
                        fmt.Println(false)
                        return
                    }
                }
            }
        }
    }
    // Your code here
}

Hint

Print true only if no conflicts were found during BFS.

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