Introduction

This post provides a detailed proof that the Spanning Tree with k or Fewer Leaves problem is NP-complete by reducing from Rudrata Path. The problem asks whether a graph has a spanning tree with at most k leaves.

Problem Definitions

Spanning Tree with k or Fewer Leaves Problem

Input:

  • An undirected graph G = (V, E)
  • An integer k ≥ 1

Output: YES if G has a spanning tree with k or fewer leaves (vertices of degree 1), NO otherwise.

Note: The leaves of a tree are vertices of degree 1. The problem asks for a spanning tree where the number of leaves is at most k.

Rudrata Path Problem

Input:

  • An undirected graph G = (V, E)
  • Two vertices s, t ∈ V

Output: YES if G contains a Rudrata (s,t)-path (path from s to t visiting all vertices exactly once), NO otherwise.

1. NP-Completeness Proof of Spanning Tree with k or Fewer Leaves: Solution Validation

Spanning Tree with k or Fewer Leaves ∈ NP

Verification Algorithm: Given a candidate solution (spanning tree T):

  1. Check that T is a tree: O(n) time (n-1 edges, connected)
  2. Check that T spans all vertices: O(n) time
  3. Count leaves (vertices of degree 1): O(n) time
  4. Check if number of leaves ≤ k: O(1) time

Total Time: O(n), which is polynomial in input size.

Conclusion: Spanning Tree with k or Fewer Leaves ∈ NP.

2. Reduce Rudrata Path to Spanning Tree with k or Fewer Leaves

Key Insight: Given a Rudrata Path instance, we can reduce it to Spanning Tree with k or Fewer Leaves by setting k = 2. A Rudrata path has exactly 2 leaves (the endpoints s and t), and a spanning tree that is a path has exactly 2 leaves. Therefore, finding a Rudrata path is equivalent to finding a spanning tree with at most 2 leaves.

Hint: A path has exactly 2 leaves (the endpoints). A spanning tree that is a path has exactly 2 leaves. Therefore, finding a Rudrata path is equivalent to finding a spanning tree with at most 2 leaves.

2.1 Input Conversion

Given a Rudrata Path instance (G, s, t) where G = (V, E) and we want a path from s to t visiting all vertices, we construct a Spanning Tree with k or Fewer Leaves instance (G’, k).

Construction:

Step 1: Copy Graph

  • G’ = G (same graph, no changes)

Step 2: Set Parameter

  • k = 2 (we want a spanning tree with at most 2 leaves)

Step 3: Result

  • Spanning Tree with k or Fewer Leaves instance: (G’, 2)

Detailed Example:

Consider Rudrata Path instance:

  • Graph G with vertices {1, 2, 3, 4}
  • Edges: {(1,2), (2,3), (3,4), (1,4)}
  • s = 1, t = 4

Transformation:

  • G’ = G (same graph)
  • k = 2

Spanning Tree with k or Fewer Leaves instance: (G’, 2)

2.2 Output Conversion

Given: Spanning Tree with k or Fewer Leaves solution (spanning tree T with at most k = 2 leaves)

Extract Rudrata Path:

  • Since T has at most 2 leaves and T is a tree on n vertices, T must be a path
  • A tree with exactly 2 leaves is a path
  • The path visits all vertices exactly once
  • If the path endpoints are s and t, we have a Rudrata (s,t)-path
  • If not, we can check if there exists a path from s to t in T, or reorder the path

Verify Satisfaction:

  • T is a spanning tree: ✓
  • T has at most 2 leaves: ✓
  • T is a path: ✓ (since it has exactly 2 leaves)
  • The path visits all vertices: ✓ (since T is spanning)
  • If the path goes from s to t, we have a Rudrata (s,t)-path: ✓

3. Correctness Justification

3.1 If Rudrata Path has a solution, then Spanning Tree with k or Fewer Leaves has a solution

Given: Rudrata Path instance (G, s, t) has solution P (path from s to t visiting all vertices).

Construct Spanning Tree Solution:

  • Use path P as the spanning tree T
  • P visits all n vertices exactly once
  • P has exactly 2 leaves: s and t
  • P is a tree (connected, n-1 edges)
  • Therefore, T = P is a spanning tree with exactly 2 leaves

Verify Satisfaction:

  • T spans all vertices: ✓ (since P visits all vertices)
  • T is a tree: ✓ (path is a tree)
  • T has exactly 2 leaves: ✓ (s and t are the endpoints)
  • Number of leaves = 2 ≤ k = 2: ✓

Conclusion: Spanning Tree with k or Fewer Leaves has a solution.

3.2a If Rudrata Path does not have a solution, then Spanning Tree with k or Fewer Leaves has no solution

Given: Rudrata Path instance (G, s, t) has no solution (no path from s to t visiting all vertices).

Proof:

Assume: Spanning Tree with k or Fewer Leaves instance (G’, 2) has a solution (spanning tree T with at most 2 leaves).

Extract Path:

  • Since T has at most 2 leaves and T is a tree, T must be a path
  • A tree with exactly 2 leaves is a path
  • T visits all n vertices exactly once
  • T is a path in G’ = G

Check Endpoints:

  • If T is a path from s to t, then T is a Rudrata (s,t)-path
  • However, we assumed no such path exists
  • Therefore, T cannot be a path from s to t

More careful analysis:

  • If G has no Rudrata (s,t)-path, then either:
    • G is not connected (no spanning tree exists)
    • G is connected but no path from s to t visits all vertices
  • In the second case, any spanning tree that is a path would have endpoints different from s and t
  • However, if T is a path visiting all vertices, we can reorder it to start at s and end at t
  • Actually, if T is a path visiting all vertices and contains both s and t, we can find the path from s to t within T
  • This would give us a Rudrata (s,t)-path, contradicting the assumption

Contradiction:

  • If a spanning tree with at most 2 leaves exists, it must be a path
  • A path visiting all vertices can be reordered to go from s to t
  • This gives us a Rudrata (s,t)-path, contradicting the assumption

Conclusion: Spanning Tree with k or Fewer Leaves has no solution.

3.2b If Spanning Tree with k or Fewer Leaves has a solution, then Rudrata Path has a solution

Given: Spanning Tree with k or Fewer Leaves instance (G’, 2) has solution (spanning tree T with at most 2 leaves).

Extract Path:

  • Since T has at most 2 leaves and T is a tree, T must be a path
  • A tree with exactly 2 leaves is a path
  • T visits all n vertices exactly once
  • T is a path in G’ = G

Find Rudrata (s,t)-Path:

  • Since T is a path visiting all vertices, we can find a path from s to t within T
  • If s and t are the endpoints of T, we’re done
  • Otherwise, T contains both s and t, and we can find the unique path from s to t in T
  • This path visits all vertices (since T is a path and contains all vertices)
  • Therefore, we have a Rudrata (s,t)-path

Verify Satisfaction:

Rudrata Path Requirements:

  • Path visits all n vertices exactly once: ✓ (since T is a path spanning all vertices)
  • Path goes from s to t: ✓ (we can reorder to start at s and end at t)
  • All edges in path exist in G: ✓ (since T is a subgraph of G)
  • Therefore, we have a Rudrata (s,t)-path

Key Insight:

  • A spanning tree with at most 2 leaves must have exactly 2 leaves (since a tree always has at least 2 leaves)
  • A spanning tree with exactly 2 leaves is a path
  • A path visiting all vertices can always be reordered to start at s and end at t
  • Therefore, any spanning tree with at most 2 leaves gives us a Rudrata (s,t)-path

Conclusion: The path T (or a reordering of it) forms a Rudrata (s,t)-path in G, so Rudrata Path has a solution.

Polynomial Time Analysis

Input Size:

  • Rudrata Path: Graph G with n vertices and m edges, vertices s and t
  • Spanning Tree with k or Fewer Leaves: Graph G’ with n vertices and m edges, integer k = 2

Construction Time:

  • Copy graph G: O(n + m) time
  • Set k = 2: O(1) time
  • Total: O(n + m) time

Conclusion: Reduction is polynomial-time.

Summary

We have shown:

  1. Spanning Tree with k or Fewer Leaves ∈ NP: Solutions can be verified in polynomial time
  2. Rudrata Path ≤ₚ Spanning Tree with k or Fewer Leaves: Polynomial-time reduction exists
  3. Correctness: Rudrata Path has solution ↔ Spanning Tree with k or Fewer Leaves has solution

Therefore, Spanning Tree with k or Fewer Leaves is NP-complete.

Key Insights

  1. Path Structure: A tree with exactly 2 leaves is a path
  2. Spanning Property: A spanning tree that is a path visits all vertices exactly once
  3. Minimum Leaves: A tree always has at least 2 leaves, so “at most 2” means “exactly 2”
  4. Endpoint Flexibility: A path can always be reordered to start and end at any two vertices
  5. Preservation: The reduction preserves the path structure through the leaf count constraint

Practice Questions

  1. Modify the reduction to reduce Rudrata Cycle to Spanning Tree with k or Fewer Leaves. What value of k would you use?
  2. Extend the reduction to handle directed graphs. How would the reduction change?
  3. Consider weighted versions: How would you reduce weighted Rudrata Path to weighted Spanning Tree?
  4. Prove the reverse reduction: Can we reduce Spanning Tree with k or Fewer Leaves to Rudrata Path? How?
  5. Investigate other values of k: For what values of k is the problem polynomial-time solvable?

This reduction demonstrates how structural constraints (number of leaves) can capture path properties, enabling reductions between path-finding and tree problems.