Bipartite Graph Check in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a bipartite graph?

A bipartite graph is a graph whose nodes can be divided into two groups so that no edge connects nodes within the same group.

beginner

How can you check if a graph is bipartite using colors?

You assign one color to a starting node, then assign the opposite color to all its neighbors. If you find a neighbor with the same color, the graph is not bipartite.

intermediate

What data structure is commonly used to perform the bipartite check?

A queue is used in a Breadth-First Search (BFS) to traverse nodes level by level while assigning colors.

intermediate

Why does a cycle of odd length in a graph mean it is not bipartite?

Because when coloring nodes alternately, an odd cycle forces a node to have the same color as its neighbor, breaking bipartite rules.

beginner

In Go, what type is typically used to represent a graph for bipartite checking?

A slice of slices of integers ([][]int) is used, where each inner slice holds neighbors of a node.

What does it mean if a graph is bipartite?

A Graph is fully connected

B Graph has no cycles

C Nodes can be split into two groups with no edges inside the same group

D All nodes have the same color

A bipartite graph means nodes can be divided into two groups with edges only between groups.

Which algorithm is commonly used to check bipartite graphs?

A Depth-First Search (DFS)

B Breadth-First Search (BFS)

C Dijkstra's Algorithm

D Kruskal's Algorithm

BFS is commonly used to assign colors level by level for bipartite checking.

What happens if you find two connected nodes with the same color during bipartite check?

A Graph is not bipartite

B Graph is bipartite

C Graph is disconnected

D Graph has no edges

Two connected nodes with the same color mean the graph is not bipartite.

What kind of cycle makes a graph non-bipartite?

A Odd length cycle

B Even length cycle

C No cycle

D Self-loop

Odd length cycles prevent proper two-coloring, so the graph is not bipartite.

In Go, how is the graph usually represented for bipartite check?

A map[int]int

B map[string]string

C []string

D [][]int

A slice of slices of ints ([][]int) represents adjacency lists for each node.

Explain how to check if a graph is bipartite using BFS and coloring.

Think about coloring nodes level by level.

Why does the presence of an odd cycle in a graph mean it cannot be bipartite?

Consider what happens when you alternate colors around a cycle.

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