Minimum Spanning Tree Kruskal's Algorithm in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the parent array for Union-Find.

DSA C#

int[] parent = new int[n];
for (int i = 0; i < n; i++) {
    parent[i] = [1];
}

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

A0

Bi

C-1

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Setting all parents to 0 causes incorrect union-find behavior.

Using -1 or n as parent values breaks the find operation.

Each node is initially its own parent, so we assign parent[i] = i.

2fill in blank

medium

Complete the code to find the root parent of a node with path compression.

DSA C#

int Find(int[] parent, int x) {
    if (parent[x] != x) {
        parent[x] = [1];
    }
    return parent[x];
}

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

Aparent[x]

Bx

CFind(parent, parent[x])

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning parent[x] to x causes infinite recursion.

Not using recursion misses path compression benefits.

Path compression sets parent[x] to the root found by recursively calling Find.

3fill in blank

hard

Fix the error in the union operation to merge two sets correctly.

DSA C#

void Union(int[] parent, int x, int y) {
    int rootX = Find(parent, x);
    int rootY = Find(parent, y);
    if (rootX != rootY) {
        parent[[1]] = rootY;
    }
}

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

Ax

By

CrootY

DrootX

Attempts:

3 left

💡 Hint

Common Mistakes

Using x or y instead of root parents breaks union correctness.

Assigning parent[rootY] = rootX merges sets in wrong direction.

We set the parent of rootX to rootY to merge the sets.

4fill in blank

hard

Fill both blanks to check if adding an edge creates a cycle and to add the edge weight.

DSA C#

foreach (var edge in edges) {
    if (Find(parent, edge.u) [1] Find(parent, edge.v)) {
        Union(parent, edge.u, edge.v);
        totalWeight [2]= edge.weight;
    }
}

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

A!=

B+

C==

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' causes cycles to be added.

Using '-=' subtracts weights incorrectly.

We add the edge only if the roots are different (no cycle), and add the edge weight to totalWeight.

5fill in blank

hard

Fill all three blanks to sort edges by weight and initialize variables for Kruskal's algorithm.

DSA C#

edges.Sort((a, b) => a.[1].CompareTo(b.[2]));
int n = [3];
int[] parent = new int[n];

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

Aweight

CnodesCount

DedgesCount

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by wrong property causes incorrect MST.

Using edgesCount instead of nodesCount for parent array size.

We sort edges by their weight property, and initialize n with the number of nodes.

Practice

1. What is the main goal of Kruskal's Algorithm in graph theory?

easy

A. To sort the vertices of a graph

B. To find the shortest path between two nodes

C. To find the minimum spanning tree of a connected weighted graph

D. To detect cycles in a graph

Solution

1. Step 1: Understand the purpose of Kruskal's Algorithm

   Kruskal's Algorithm is designed to find a subset of edges that connects all vertices with the minimum total edge weight and no cycles.

2. Step 2: Compare with other graph algorithms

   Unlike shortest path algorithms or cycle detection, Kruskal's focuses on building a minimum spanning tree.

3. Final Answer:

   To find the minimum spanning tree of a connected weighted graph -> Option C

4. Quick Check:

   Minimum Spanning Tree = To find the minimum spanning tree of a connected weighted graph [OK]

Hint: Kruskal connects all nodes with least total edge weight [OK]

Common Mistakes:

• Confusing MST with shortest path
• Thinking Kruskal detects cycles only
• Assuming it sorts vertices

2. Which of the following C# code snippets correctly initializes a list of edges for Kruskal's Algorithm?

easy

A. var edges = new List<(int, int, int)> { (0, 1, 4), (1, 2, 3) };

B. var edges = new List { [0, 1, 4], [1, 2, 3] };

C. var edges = new List<(int, int, int)> { [0, 1, 4], [1, 2, 3] };

D. var edges = new List<(int, int)> { (0, 1), (1, 2) };

Solution

1. Step 1: Check tuple syntax for edges

   In C#, tuples are declared with parentheses and commas, e.g., (int, int, int). var edges = new List<(int, int, int)> { (0, 1, 4), (1, 2, 3) }; uses this correctly.

2. Step 2: Identify incorrect list initializations

   var edges = new List { [0, 1, 4], [1, 2, 3] }; uses square brackets inside List initialization which is invalid. var edges = new List<(int, int, int)> { [0, 1, 4], [1, 2, 3] }; mixes tuple and array syntax incorrectly. var edges = new List<(int, int)> { (0, 1), (1, 2) }; misses the weight part of edges.

3. Final Answer:

   var edges = new List<(int, int, int)> { (0, 1, 4), (1, 2, 3) }; -> Option A

4. Quick Check:

   Correct tuple list syntax = var edges = new List<(int, int, int)> { (0, 1, 4), (1, 2, 3) }; [OK]

Hint: Use parentheses for tuples, curly braces for list initialization [OK]

Common Mistakes:

• Using square brackets for tuples
• Omitting edge weights
• Mixing array and tuple syntax

3. What is the output MST edges after running Kruskal's Algorithm on this graph?

var edges = new List<(int, int, int)>{ (0,1,1), (1,2,2), (0,2,3), (2,3,4) };
// Vertices: 0,1,2,3
// Edges: (0-1:1), (1-2:2), (0-2:3), (2-3:4)

medium

A. [(0,1,1), (0,2,3), (2,3,4)]

B. [(0,1,1), (1,2,2), (2,3,4)]

C. [(1,2,2), (0,2,3), (2,3,4)]

D. [(0,1,1), (1,2,2)]

Solution

1. Step 1: Sort edges by weight

   Edges sorted: (0-1:1), (1-2:2), (0-2:3), (2-3:4)

2. Step 2: Pick edges without forming cycles

   Pick (0-1:1), then (1-2:2). Adding (0-2:3) creates a cycle, so skip. Add (2-3:4).

3. Final Answer:

   [(0,1,1), (1,2,2), (2,3,4)] -> Option B

4. Quick Check:

   Edges chosen without cycle = [(0,1,1), (1,2,2), (2,3,4)] [OK]

Hint: Sort edges by weight, skip edges forming cycles [OK]

Common Mistakes:

• Including edges that form cycles
• Not sorting edges by weight
• Missing edges to connect all vertices

4. Identify the error in this Kruskal's Algorithm snippet:

var edges = new List<(int, int, int)> { (0,1,1), (1,2,2), (0,2,3) };
var parent = new int[3];
for(int i=0; i<3; i++) parent[i] = i;

int Find(int x) {
  if(parent[x] == x) return x;
  return Find(parent[x]);
}

void Union(int a, int b) {
  int rootA = Find(a);
  int rootB = Find(b);
  parent[rootA] = rootB;
}

foreach(var edge in edges) {
  int u = edge.Item1, v = edge.Item2;
  if(Find(u) != Find(v)) {
    Union(u, v);
    Console.WriteLine($"Edge: {u}-{v}");
  }
}

medium

A. Union function does not use union by rank or size, causing inefficiency

B. Find function lacks path compression, causing stack overflow

C. Parent array size is incorrect for the number of vertices

D. Edges list is missing weights, so MST cannot be formed

Solution

1. Step 1: Check parent array size

   There are 3 vertices (0,1,2), so parent array size 3 is correct for these vertices.

2. Step 2: Verify if all vertices are covered

   Edges use vertices 0,1,2 only, so parent array size matches vertices count. No error here.

3. Step 3: Eliminate incorrect options

   Parent array size [3] is correct for vertices 0-2 (C wrong). Edges include weights (D wrong). Path compression absent but no stack overflow for small graph (B wrong). Thus, A is correct: lacking union by rank or size causes inefficiency.

4. Final Answer:

   Union function does not use union by rank or size, causing inefficiency -> Option A

5. Quick Check:

   Missing union by rank causes inefficiency = Union function does not use union by rank or size, causing inefficiency [OK]

Hint: Check array sizes and union-find optimizations carefully [OK]

Common Mistakes:

• Confusing inefficiency with error
• Assuming missing weights when present
• Ignoring union-find optimization importance

5. Given a graph with vertices 0 to 4 and edges:

var edges = new List<(int, int, int)> {
  (0,1,10), (0,2,6), (0,3,5), (1,3,15), (2,3,4), (3,4,2)
};

Apply Kruskal's Algorithm to find the total weight of the minimum spanning tree.

hard

A. 17

B. 27

C. 37

D. 21

Solution

1. Step 1: Sort edges by weight

   Sorted edges: (3-4:2), (2-3:4), (0-3:5), (0-2:6), (0-1:10), (1-3:15)

2. Step 2: Pick edges without cycles to connect all vertices

   Pick (3-4:2), (2-3:4), (0-3:5), (0-1:10). Adding (0-2:6) or (1-3:15) creates cycles or is unnecessary.

3. Step 3: Calculate total weight

   Total = 2 + 4 + 5 + 10 = 21

4. Final Answer:

   21 -> Option D

5. Quick Check:

   Sum MST edges weights = 21 [OK]

Hint: Sort edges, pick smallest without cycle until all connected [OK]

Common Mistakes:

• Adding edges that form cycles
• Forgetting to connect all vertices
• Misordering edges by weight