Minimum Spanning Tree Prim's Algorithm in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the key values for all vertices in Prim's algorithm.

DSA C++

for (int i = 0; i < V; i++) {
    key[i] = [1];
    mstSet[i] = false;
}

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

A0

BINT_MAX

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing key values to 0 causes incorrect minimum edge selection.

Using negative values for key leads to wrong comparisons.

In Prim's algorithm, key values are initialized to a very large number (infinity) to ensure any edge weight will be smaller.

2fill in blank

medium

Complete the code to find the vertex with the minimum key value not yet included in MST.

DSA C++

int minKey(int key[], bool mstSet[]) {
    int min = INT_MAX, min_index;
    for (int v = 0; v < V; v++) {
        if (mstSet[v] == false && key[v] < [1]) {
            min = key[v];
            min_index = v;
        }
    }
    return min_index;
}

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

Av

BmstSet[v]

CINT_MAX

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing key[v] with v instead of min causes wrong minimum selection.

Using mstSet[v] in comparison leads to logical errors.

We compare key[v] with the current minimum value stored in 'min' to find the smallest key.

3fill in blank

hard

Fix the error in updating key and parent arrays inside the main loop of Prim's algorithm.

DSA C++

for (int v = 0; v < V; v++) {
    if (graph[u][v] && mstSet[v] == false && graph[u][v] < [1]) {
        parent[v] = u;
        key[v] = graph[u][v];
    }
}

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Agraph[u][v]

Bparent[v]

Ckey[v]

DmstSet[v]

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing with parent[v] instead of key[v] causes wrong updates.

Using graph[u][v] < graph[u][v] is always false.

We update the key[v] only if the new edge weight graph[u][v] is less than the current key[v].

4fill in blank

hard

Fill both blanks to complete the loop that constructs the MST result output.

DSA C++

for (int i = 1; i < V; i++) {
    cout << parent[i] << " - " << [1] << " \t" << [2] << endl;
}

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Ai

Bgraph[parent[i]][i]

Cparent[i]

Dkey[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Printing parent[i] twice instead of i.

Using key[i] instead of graph[parent[i]][i] for weight.

We print the edge from parent[i] to i and its weight graph[parent[i]][i].

5fill in blank

hard

Fill all three blanks to complete the main Prim's algorithm function header and initialization.

DSA C++

void primMST([1] graph[[2]][[2]]) {
    int parent[[2]];
    int key[[2]];
    bool mstSet[[2]];

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Aint

BV

Cconst int

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'const int' in function parameter causes syntax issues here.

Using 'size' instead of V leads to undefined variable errors.

The function takes an int 2D array graph of size VxV; arrays parent, key, mstSet are sized by V.

Practice

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

easy

A. To find the shortest path between two nodes

B. To detect cycles in a graph

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

D. To sort the vertices of a graph

Solution

1. Step 1: Understand Prim's Algorithm purpose

   Prim's Algorithm is designed to find a tree that connects all vertices with the minimum total edge weight.

2. Step 2: Differentiate from other graph algorithms

   It is not used for shortest path, cycle detection, or sorting vertices, which are different problems.

3. Final Answer:

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

4. Quick Check:

   Prim's Algorithm = Minimum Spanning Tree [OK]

Hint: Prim's always builds a tree with minimum total edge weight [OK]

Common Mistakes:

• Confusing Prim's with shortest path algorithms
• Thinking it detects cycles
• Assuming it sorts nodes

2. Which of the following is the correct way to initialize the key array in Prim's Algorithm in C++?

easy

A. int key[V]; for(int i=0; i

B. int key[V] = {0};

C. int key[V]; for(int i=0; i

D. int key[V]; for(int i=0; i

Solution

1. Step 1: Understand key array role

   The key array stores minimum edge weights to include each vertex; initially, all are set to a very large value (infinity).

2. Step 2: Check initialization options

   Setting key[i] = INT_MAX correctly represents infinity; other options set to 0 or -1 are incorrect for this purpose.

3. Final Answer:

   int key[V]; for(int i=0; i -> Option A

4. Quick Check:

   Initialize keys with INT_MAX for minimum edge tracking [OK]

Hint: Initialize keys with INT_MAX to represent infinity [OK]

Common Mistakes:

• Initializing keys with 0 instead of INT_MAX
• Using negative values incorrectly
• Not initializing keys at all

3. Consider the following graph with 4 vertices and edges with weights:

0 - 1 : 10
0 - 2 : 6
0 - 3 : 5
1 - 3 : 15
2 - 3 : 4

What is the total weight of the minimum spanning tree found by Prim's Algorithm starting from vertex 0?

medium

A. 21

B. 19

C. 20

D. 25

Solution

1. Step 1: Start from vertex 0 and pick minimum edges

   Edges from 0: (0-3)=5, (0-2)=6, (0-1)=10. Pick (0-3)=5 first.

2. Step 2: Add next minimum edge connecting new vertices

   From vertices {0,3}, edges are (3-2)=4, (3-1)=15, (0-2)=6, (0-1)=10. Pick (3-2)=4 next.

3. Step 3: Add last vertex with minimum edge

   From {0,3,2}, edges to 1 are (0-1)=10 and (3-1)=15. Pick (0-1)=10.

4. Step 4: Calculate total weight

   Total = 5 + 4 + 10 = 19.

5. Final Answer:

   19 -> Option B

6. Quick Check:

   Sum of chosen edges = 19 [OK]

Hint: Pick smallest edge connecting new vertex each step [OK]

Common Mistakes:

• Choosing edges with larger weights first
• Not updating the set of included vertices
• Adding edges that form cycles

4. The following C++ code snippet is intended to implement Prim's Algorithm but has a bug. What is the error?

const int V = 4;
int key[V] = {INT_MAX, INT_MAX, INT_MAX, INT_MAX};
bool mstSet[V] = {false, false, false, false};
key[0] = 0;
for (int count = 0; count < V - 1; count++) {
    int u = -1;
    for (int v = 0; v < V; v++) {
        if (!mstSet[v] && (u == -1 || key[v] < key[u]))
            u = v;
    }
    mstSet[u] = true;
    for (int v = 0; v < V; v++) {
        if (graph[u][v] && !mstSet[v] && graph[u][v] < key[v])
            key[v] = graph[u][v];
    }
}

medium

A. The condition 'graph[u][v]' should check for non-zero edges

B. The loop should run from 0 to V instead of V - 1

C. The key array is not initialized with INT_MAX

D. The variable 'u' is initialized incorrectly inside the loop

Solution

1. Step 1: Analyze the neighbor update condition

   The condition if(graph[u][v] && !mstSet[v] && graph[u][v] < key[v]) uses graph[u][v] as a boolean check (true if non-zero).

2. Step 2: Identify the bug

   This implicit check assumes positive weights and treats 0 as no edge. Standard implementations explicitly use graph[u][v] != 0 to clearly indicate checking for edge existence (non-zero weight).

3. Step 3: Rule out other options

   A: 'u' is correctly initialized to -1 inside each iteration. B: V-1 iterations build the tree correctly. C: key array is initialized with INT_MAX.

4. Final Answer:

   The condition 'graph[u][v]' should check for non-zero edges -> Option A

5. Quick Check:

   Edge check: graph[u][v] != 0 [OK]

Hint: Check edge existence condition carefully for zero weights [OK]

Common Mistakes:

• Misinitializing 'u' outside the loop
• Incorrect loop range
• Not initializing key array properly

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

0-1: 3, 0-3: 1, 1-2: 1, 1-3: 3, 1-4: 10, 2-4: 5, 3-4: 6, 4-5: 2

If Prim's Algorithm starts from vertex 0, which edge will NOT be included in the minimum spanning tree?

hard

A. Edge 1-2 with weight 1

B. Edge 0-1 with weight 3

C. Edge 4-5 with weight 2

D. Edge 3-4 with weight 6

Solution

1. Step 1: Build MST step-by-step from vertex 0

   Start at 0: edges (0-3)=1, (0-1)=3. Pick (0-3)=1 first.

2. Step 2: Add next minimum edges connecting new vertices

   From {0,3}, edges: (3-1)=3, (3-4)=6, (0-1)=3. Pick (0-1)=3 or (3-1)=3 (same weight). Pick (0-1)=3.

3. Step 3: Continue adding minimum edges

   From {0,1,3}, edges: (1-2)=1, (1-4)=10, (3-4)=6. Pick (1-2)=1.

4. Step 4: Add edge to vertex 4

   From {0,1,2,3}, edges: (2-4)=5, (3-4)=6, (1-4)=10. Pick (2-4)=5.

5. Step 5: Add edge to vertex 5

   From {0,1,2,3,4}, edge: (4-5)=2.

6. Step 6: Identify excluded edge

   Edge (3-4)=6 is not chosen because (2-4)=5 is cheaper and connects 4 earlier.

7. Final Answer:

   Edge 3-4 with weight 6 -> Option D

8. Quick Check:

   Exclude heavier edge when cheaper path exists [OK]

Hint: Always pick smallest edge connecting new vertex, exclude heavier alternatives [OK]

Common Mistakes:

• Including heavier edges when cheaper ones exist
• Not updating candidate edges after each addition
• Assuming all edges connected to MST are included