Bipartite Graph Check in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the color array for the graph nodes.

DSA C#

int[] color = new int[[1]];

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

Agraph.Count

Bgraph.Length

Cgraph.Size

Dgraph.Nodes

Attempts:

3 left

💡 Hint

Common Mistakes

Using graph.Count which is for lists, not arrays.

Using graph.Size or graph.Nodes which are invalid properties.

The color array size should match the number of nodes, which is graph.Length in C# arrays.

2fill in blank

medium

Complete the code to initialize all colors to -1 (uncolored).

DSA C#

for (int i = 0; i < color.Length; i++) {
    color[i] = [1];
}

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

A-1

Bint.MinValue

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 1 which are valid colors, not uncolored.

Using int.MinValue which is unnecessary and confusing.

We use -1 to mark nodes as uncolored before starting the bipartite check.

3fill in blank

hard

Fix the error in the DFS method to assign the opposite color to neighbors.

DSA C#

color[neighbor] = color[node] [1] 1;

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

A^

B-

C+

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using + or - which can produce invalid colors.

Using * which does not toggle colors.

Using XOR (^) with 1 flips the color between 0 and 1 correctly.

4fill in blank

hard

Fill both blanks to check if a neighbor has the same color and return false.

DSA C#

if (color[neighbor] != [1] && color[neighbor] == color[[2]]) {
    return false;
}

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

A-1

Bnode

Cneighbor

Dcolor[node]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with neighbor instead of node.

Checking for color[node] instead of node index.

We check if neighbor is colored (not -1) and if its color equals the current node's color.

5fill in blank

hard

Fill all three blanks to complete the main bipartite check loop.

DSA C#

for (int [1] = 0; [2] < graph.Length; [3]++) {
    if (color[i] == -1 && !DFS(graph, i, color)) {
        return false;
    }
}

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

Ai

Bj

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables in loop parts causing errors.

Using variables not declared in the loop.

The loop variable is i, used consistently in initialization, condition, and increment.

Practice

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

easy

A. All vertices have the same degree

B. It has an equal number of vertices and edges

C. It contains at least one cycle

D. Its vertices can be divided into two groups with no edges inside the same group

Solution

1. Step 1: Understand bipartite graph definition

   A bipartite graph is one where vertices can be split into two sets with edges only between sets, not inside a set.

2. Step 2: Compare options with definition

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

3. Final Answer:

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

4. Quick Check:

   Bipartite graph = two groups, no internal edges [OK]

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

Common Mistakes:

• Confusing bipartite with equal vertices and edges
• Thinking bipartite means no cycles
• Assuming all vertices have same degree

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

easy

A. List adjacency list

B. Dictionary adjacency list

C. int[,] adjacency matrix

D. HashSet<int> for vertices only

Solution

1. Step 1: Identify graph representation needs

   We need to store edges efficiently and access neighbors quickly for bipartite check.

2. Step 2: Evaluate options

   Dictionary allows flexible vertex keys and fast neighbor lookup, better than fixed-size arrays or sets.

3. Final Answer:

   Dictionary adjacency list -> Option B

4. Quick Check:

   Dictionary adjacency list = best for flexible graphs [OK]

Hint: Use Dictionary for flexible graph edges [OK]

Common Mistakes:

• Using HashSet only stores vertices, no edges
• Adjacency matrix wastes space for sparse graphs
• List lacks vertex key mapping

3. What is the output of this C# code snippet checking bipartite for a graph with edges: 0-1, 1-2, 2-3, 3-0?

var graph = new Dictionary {
  {0, new List<int>{1,3}},
  {1, new List<int>{0,2}},
  {2, new List<int>{1,3}},
  {3, new List<int>{2,0}}
};
bool IsBipartite(Dictionary g) {
  var color = new Dictionary();
  foreach(var node in g.Keys) {
    if(!color.ContainsKey(node)) {
      var queue = new Queue<int>();
      queue.Enqueue(node);
      color[node] = 0;
      while(queue.Count > 0) {
        var u = queue.Dequeue();
        foreach(var v in g[u]) {
          if(!color.ContainsKey(v)) {
            color[v] = 1 - color[u];
            queue.Enqueue(v);
          } else if(color[v] == color[u]) {
            return false;
          }
        }
      }
    }
  }
  return true;
}
Console.WriteLine(IsBipartite(graph));

medium

A. True

B. False

C. Compilation error

D. Runtime exception

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 on even cycle

   Even cycles are bipartite, but here the cycle is length 4 (even), so it should be bipartite. However, the code returns false if adjacent nodes have same color.

3. Step 3: Trace coloring

   Coloring alternates 0 and 1 around the cycle, no conflict occurs, so function returns true.

4. Final Answer:

   True -> Option A

5. Quick Check:

   Even cycle graph is bipartite = True [OK]

Hint: Even cycles are bipartite; odd cycles are not [OK]

Common Mistakes:

• Assuming all cycles are not bipartite
• Confusing color assignment logic
• Ignoring queue BFS traversal

4. Identify the bug in this bipartite check code snippet:

bool CheckBipartite(Dictionary graph) {
  var color = new Dictionary();
  foreach(var node in graph.Keys) {
    if(!color.ContainsKey(node)) {
      color[node] = 0;
      var stack = new Stack<int>();
      stack.Push(node);
      while(stack.Count > 0) {
        var u = stack.Pop();
        foreach(var v in graph[u]) {
          if(!color.ContainsKey(v)) {
            color[v] = color[u];
            stack.Push(v);
          } else if(color[v] == color[u]) {
            return false;
          }
        }
      }
    }
  }
  return true;
}

medium

A. Colors assigned to neighbors are same as current node, should be opposite

B. Stack should be replaced with queue for BFS

C. Missing initialization of color dictionary

D. No return statement for true case

Solution

1. Step 1: Review color assignment logic

   Neighbors must have opposite colors to current node for bipartite check.

2. Step 2: Identify incorrect assignment

   Code assigns neighbor color same as current node, causing false negatives.

3. Final Answer:

   Colors assigned to neighbors are same as current node, should be opposite -> Option A

4. Quick Check:

   Neighbor colors must differ [OK]

Hint: Neighbors must have opposite colors, not same [OK]

Common Mistakes:

• Assigning same color to neighbors
• Confusing stack vs queue (both work but BFS preferred)
• Forgetting to initialize colors

5. Given a graph with vertices 0 to 5 and edges: 0-1, 1-2, 2-3, 3-4, 4-5, 5-0, and 1-4, is the graph bipartite? Explain why.

hard

A. Yes, because all cycles are even length

B. No, because it has a self-loop

C. No, because it contains an odd cycle including edges 1-2-3-4-1

D. Yes, because it has no cycles

Solution

1. Step 1: Identify cycles in the graph

   The graph has a large cycle 0-1-2-3-4-5-0 (length 6, even), and with edge 1-4, cycles 1-2-3-4-1 (length 4, even) and 0-1-4-5-0 (length 4, even).

2. Step 2: Check for odd-length cycles

   No odd-length cycles exist. A graph is bipartite if and only if it has no odd-length cycles.

3. Final Answer:

   No, because it contains an odd cycle including edges 1-2-3-4-1 -> Option C

4. Quick Check:

   Odd cycle means not bipartite [OK]

Hint: Check for odd cycles to decide bipartite [OK]

Common Mistakes:

• Miscounting cycle lengths
• Assuming any cycle means not bipartite
• Ignoring extra edges creating odd cycles