Minimum spanning tree (Prim's) in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the starting vertex in Prim's algorithm.

Data Structures Theory

visited = set()
start_vertex = [1]
visited.add(start_vertex)

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

ANone

B0

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from None causes errors because the vertex must be a valid node.

Using -1 or 1 may cause index errors if the graph is zero-indexed.

Prim's algorithm starts from a vertex, often vertex 0 in zero-indexed graphs.

2fill in blank

medium

Complete the code to select the edge with the minimum weight from the priority queue.

Data Structures Theory

while edges:
    weight, vertex = heapq.heappop([1])
    if vertex not in visited:
        break

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

Aqueue

Bvisited

Cgraph

Dedges

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'visited' instead of the priority queue causes errors.

Using 'graph' or 'queue' when those are not the priority queue variable.

Prim's algorithm uses a priority queue (often a heap) named 'edges' to pick the smallest edge.

3fill in blank

hard

Fix the error in the code that adds new edges to the priority queue.

Data Structures Theory

for neighbor, weight in graph[vertex]:
    if neighbor not in visited:
        heapq.heappush(edges, ([1], neighbor))

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

Aweight

Bneighbor

Cvertex

Dvisited

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'neighbor' or 'vertex' as the first tuple element causes incorrect sorting.

Using 'visited' is invalid because it's a set, not a weight.

Edges are pushed with their weight first to maintain the priority queue order by weight.

4fill in blank

hard

Fill both blanks to correctly update the total cost and add the new vertex to visited.

Data Structures Theory

total_cost [1]= weight
visited.[2](vertex)

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

A+

B-

Cadd

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using '-=' decreases total_cost incorrectly.

Using 'remove' on visited causes errors if vertex is not present.

We add the edge weight to total_cost and add the vertex to the visited set.

5fill in blank

hard

Fill all three blanks to complete the dictionary comprehension that maps vertices to their minimum edge weights.

Data Structures Theory

min_edges = {vertex: [1] for vertex, [2] in graph.items() if [3]

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

Amin(weight for _, weight in graph[vertex])

Bedges

Cvertex in visited

Dvertex not in visited

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'vertex in visited' includes already visited vertices.

Using 'edges' as the second variable in the loop is incorrect; it should be the adjacency list items.

This comprehension finds the minimum edge weight for each vertex not yet visited.

Practice

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

easy

A. To find the minimum spanning tree connecting all vertices with the least total edge weight

B. To find the shortest path between two vertices

C. To detect cycles in a graph

D. To sort vertices based on their degree

Solution

1. Step 1: Understand Prim's algorithm purpose

   Prim's algorithm builds a tree that connects all vertices with the minimum sum of edge weights.

2. Step 2: Compare with other graph problems

   Unlike shortest path or cycle detection, Prim's focuses on spanning all vertices with minimal total edge cost.

3. Final Answer:

   To find the minimum spanning tree connecting all vertices with the least total edge weight -> Option A

4. Quick Check:

   Prim's algorithm = Minimum spanning tree [OK]

Hint: Prim's always finds the cheapest tree connecting all points [OK]

Common Mistakes:

• Confusing minimum spanning tree with shortest path
• Thinking it detects cycles
• Assuming it sorts vertices

2. Which data structure is commonly used in Prim's algorithm to select the next edge with the smallest weight?

easy

A. Priority queue (min-heap)

B. Stack

C. Queue

D. Hash table

Solution

1. Step 1: Identify how Prim's picks edges

   Prim's algorithm always picks the edge with the smallest weight connecting the tree to a new vertex.

2. Step 2: Choose the right data structure

   A priority queue (min-heap) efficiently retrieves the smallest edge weight at each step.

3. Final Answer:

   Priority queue (min-heap) -> Option A

4. Quick Check:

   Smallest edge selection = Priority queue [OK]

Hint: Use a min-heap to always pick the cheapest edge next [OK]

Common Mistakes:

• Using stack or queue which don't sort by weight
• Confusing hash table with priority queue
• Ignoring edge weights in selection

3. Consider the following weighted graph edges:
A-B: 2, A-C: 3, B-C: 1, B-D: 4, C-D: 5
Using Prim's algorithm starting from vertex A, what is the total weight of the minimum spanning tree?

medium

A. 5

B. 8

C. 7

D. 6

Solution

1. Step 1: Start from vertex A and pick smallest edge

   Edges from A: A-B(2), A-C(3). Pick A-B with weight 2.

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

   From B and A, edges are B-C(1), B-D(4), A-C(3). Pick B-C with weight 1.

3. Step 3: Add next smallest edge to connect remaining vertex D

   Edges from B, C, A to D: B-D(4), C-D(5). Pick B-D with weight 4.

4. Step 4: Calculate total weight

   Total = 2 + 1 + 4 = 7.

5. Final Answer:

   7 -> Option C

6. Quick Check:

   Sum edges 2+1+4 = 7 [OK]

Hint: Add edges with smallest weight connecting new vertices [OK]

Common Mistakes:

• Choosing edges without checking if vertex is already included
• Adding edges that form cycles
• Forgetting to sum all chosen edges

4. In the following code snippet for Prim's algorithm, what is the error?

graph = { 'A': {'B': 2, 'C': 3}, 'B': {'A': 2, 'C': 1, 'D': 4}, 'C': {'A': 3, 'B': 1, 'D': 5}, 'D': {'B': 4, 'C': 5} }
visited = set()
keys = {v: float('inf') for v in graph}
keys['A'] = 0
while len(visited) < len(graph):
    u = min((k for k in keys if k not in visited), key=lambda x: keys[x])
    visited.add(u)
    for v, w in graph[u].items():
        if v not in visited and w < keys[v]:
            keys[v] = w

medium

A. The keys dictionary is not updated correctly

B. The algorithm does not track parent nodes to build the MST

C. The min function call is incorrect

D. The visited set is never updated

Solution

1. Step 1: Analyze the code's MST construction

   The code updates keys but does not store which vertex connects to which, so MST edges are unknown.

2. Step 2: Identify missing MST parent tracking

   Prim's algorithm requires tracking parent nodes to reconstruct the MST edges, which is missing here.

3. Final Answer:

   The algorithm does not track parent nodes to build the MST -> Option B

4. Quick Check:

   Missing parent tracking = MST incomplete [OK]

Hint: Track parents to know MST edges, not just weights [OK]

Common Mistakes:

• Assuming keys alone build MST
• Ignoring parent node tracking
• Thinking visited set is not updated

5. You have a graph with vertices representing cities and edges representing roads with costs. If some roads have zero cost, how does Prim's algorithm handle these edges when building the minimum spanning tree?

hard

A. It always includes zero-cost edges first, possibly creating multiple disconnected trees

B. It causes an error because zero-cost edges are invalid

C. It ignores zero-cost edges because they don't reduce total cost

D. It treats zero-cost edges as the cheapest and includes them first to connect vertices

Solution

1. Step 1: Understand edge weight role in Prim's

   Prim's algorithm always picks the edge with the smallest weight connecting new vertices.

2. Step 2: Consider zero-cost edges

   Zero-cost edges are the cheapest possible, so they will be chosen first to connect vertices without increasing total cost.

3. Step 3: Confirm no errors or ignoring

   Zero-cost edges are valid and included; they do not cause errors or get ignored.

4. Final Answer:

   It treats zero-cost edges as the cheapest and includes them first to connect vertices -> Option D

5. Quick Check:

   Zero-cost edges = chosen first [OK]

Hint: Zero-cost edges are cheapest, so Prim picks them first [OK]

Common Mistakes:

• Thinking zero-cost edges cause errors
• Assuming zero-cost edges are ignored
• Believing zero-cost edges create disconnected trees