Bipartite Graph Check in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

color := make([]int, len(graph))
for i := range color {
	color[i] = [1]
}

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

A-1

B1

C0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 as initial color causes confusion with actual colors.

We use -1 to mark vertices as uncolored initially.

2fill in blank

medium

Complete the code to assign the opposite color to the neighbor vertex.

DSA Go

color[neighbor] = [1]

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

Acolor[current]

Bcolor[current] - 1

C1 - color[current]

Dcolor[current] + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Adding or subtracting 1 can cause invalid colors like 2 or -1.

We flip color 0 to 1 and 1 to 0 using 1 - current color.

3fill in blank

hard

Fix the error in the condition that checks if the neighbor has the same color as current.

DSA Go

if color[neighbor] [1] color[current] {
	return false
}

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

A!=

B==

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == causes wrong bipartite detection.

If neighbor has the same color as current, graph is not bipartite.

4fill in blank

hard

Fill both blanks to complete the BFS loop that colors neighbors.

DSA Go

for len(queue) > 0 {
	current := queue[0]
	queue = queue[[1]:]
	for _, neighbor := range graph[current] {
		if color[neighbor] == -1 {
			color[neighbor] = [2]
			queue = append(queue, neighbor)
		}
	}
}

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

A1

B0

C1 - color[current]

Dcolor[current]

Attempts:

3 left

💡 Hint

Common Mistakes

Using queue[0:] does not remove the first element.

Assigning same color to neighbor causes wrong bipartite check.

We remove the first element with queue[1:] and assign opposite color to neighbor.

5fill in blank

hard

Fill all three blanks to complete the bipartite check function using BFS.

DSA Go

func isBipartite(graph [][]int) bool {
	color := make([]int, len(graph))
	for i := range color {
		color[i] = [1]
	}
	for i := 0; i < len(graph); i++ {
		if color[i] == -1 {
			queue := []int{i}
			color[i] = [2]
			for len(queue) > 0 {
				current := queue[0]
				queue = queue[[3]:]
				for _, neighbor := range graph[current] {
					if color[neighbor] == -1 {
						color[neighbor] = 1 - color[current]
						queue = append(queue, neighbor)
					} else if color[neighbor] == color[current] {
						return false
					}
				}
			}
		}
	}
	return true
}

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

A-1

B0

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing colors with 0 causes confusion with actual colors.

Removing queue element incorrectly causes infinite loops.

Initialize colors with -1, start coloring with 0, and remove first queue element with queue[1:].

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