Introduction

This post provides a detailed proof that the Spanning Tree with Exact Leaves problem is NP-complete by reducing from Rudrata Path. The problem asks whether a graph has a spanning tree whose set of leaves is precisely a given set L.

Problem Definitions

Spanning Tree with Exact Leaves Problem

Input:

  • An undirected graph G = (V, E)
  • A set of nodes L ⊆ V

Output: YES if G has a spanning tree such that its set of leaves is precisely the set L, NO otherwise.

Note: The leaves of a tree are vertices of degree 1. The problem asks for a spanning tree where exactly the vertices in L are leaves, and all other vertices are not leaves.

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 L ≤ V Leaves: Solution Validation

Spanning Tree with Exact 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. Identify leaves (vertices of degree 1): O(n) time
  4. Check that the set of leaves equals L: O(n) time

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

Conclusion: Spanning Tree with Exact Leaves ∈ NP.

2. Reduce Rudrata Path to Spanning Tree with Exact Leaves

Key Insight: Given a Rudrata Path instance, we can reduce it to Spanning Tree with Exact Leaves by setting L = {s, t}. A Rudrata path has exactly 2 leaves (the endpoints s and t), and a spanning tree that is a path has exactly these two vertices as leaves.

Hint: A path has exactly 2 leaves (the endpoints). A spanning tree that is a path has exactly 2 leaves: the start and end vertices. Therefore, finding a Rudrata (s,t)-path is equivalent to finding a spanning tree with leaves exactly {s, t}.

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 Exact Leaves instance (G’, L).

Construction:

Step 1: Copy Graph

  • G’ = G (same graph, no changes)

Step 2: Set Leaf Set

  • L = {s, t} (we want a spanning tree with exactly s and t as leaves)

Step 3: Result

  • Spanning Tree with Exact Leaves instance: (G’, {s, t})

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)
  • L = {1, 4} (s = 1, t = 4)

Spanning Tree with Exact Leaves instance: (G’, {1, 4})

2.2 Output Conversion

Given: Spanning Tree with Exact Leaves solution (spanning tree T with leaves exactly L = {s, t})

Extract Rudrata Path:

  • Since T has exactly 2 leaves (s and t) and T is a tree on n vertices, T must be a path
  • A tree with exactly 2 leaves is a path
  • The path has endpoints s and t
  • The path visits all vertices exactly once
  • Therefore, T is a Rudrata (s,t)-path

Verify Satisfaction:

  • T is a spanning tree: ✓
  • T has leaves exactly {s, t}: ✓ (by assumption)
  • T is a path: ✓ (since it has exactly 2 leaves)
  • The path visits all vertices: ✓ (since T is spanning)
  • The path goes from s to t: ✓ (since s and t are the endpoints)
  • Therefore, T is a Rudrata (s,t)-path: ✓

3. Correctness Justification

3.1 If Rudrata Path has a solution, then Spanning Tree with Exact 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 leaves exactly {s, t} = L

Verify Satisfaction:

  • T spans all vertices: ✓ (since P visits all vertices)
  • T is a tree: ✓ (path is a tree)
  • T has leaves exactly {s, t}: ✓ (s and t are the endpoints, all other vertices have degree 2)
  • Therefore, the set of leaves equals L = {s, t}: ✓

Conclusion: Spanning Tree with Exact Leaves has a solution.

3.2a If Rudrata Path does not have a solution, then Spanning Tree with Exact 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 Exact Leaves instance (G’, {s, t}) has a solution (spanning tree T with leaves exactly {s, t}).

Extract Path:

  • Since T has exactly 2 leaves (s and t) and T is a tree, T must be a path
  • A tree with exactly 2 leaves is a path
  • T has endpoints s and t
  • T visits all n vertices exactly once
  • T is a path from s to t in G’ = G
  • Therefore, T is a Rudrata (s,t)-path

Contradiction:

  • T is a Rudrata (s,t)-path in G
  • This contradicts the assumption that G has no Rudrata (s,t)-path

Conclusion: Spanning Tree with Exact Leaves has no solution.

3.2b If Spanning Tree with Exact Leaves has a solution, then Rudrata Path has a solution

Given: Spanning Tree with Exact Leaves instance (G’, {s, t}) has solution (spanning tree T with leaves exactly {s, t}).

Extract Path:

  • Since T has exactly 2 leaves (s and t) and T is a tree, T must be a path
  • A tree with exactly 2 leaves is a path
  • T has endpoints s and t
  • T visits all n vertices exactly once
  • T is a path from s to t in G’ = G

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: ✓ (since s and t are the endpoints)
  • All edges in path exist in G: ✓ (since T is a subgraph of G)
  • Therefore, T is a Rudrata (s,t)-path

Key Insight:

  • A spanning tree with exactly 2 leaves is a path
  • If the leaves are exactly {s, t}, then the path goes from s to t
  • A path visiting all vertices from s to t is exactly a Rudrata (s,t)-path
  • Therefore, any solution to Spanning Tree with Exact Leaves gives us a Rudrata (s,t)-path

Conclusion: The spanning tree T 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 Exact Leaves: Graph G’ with n vertices and m edges, set L = {s, t}

Construction Time:

  • Copy graph G: O(n + m) time
  • Set L = {s, t}: O(1) time
  • Total: O(n + m) time

Conclusion: Reduction is polynomial-time.

Summary

We have shown:

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

Therefore, Spanning Tree with Exact 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. Exact Leaf Constraint: Requiring exactly {s, t} as leaves forces the path to go from s to t
  4. Preservation: The reduction preserves the path structure through the exact leaf set constraint

Practice Questions

  1. Modify the reduction to reduce Rudrata Cycle to Spanning Tree with Exact Leaves. What set L 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 Exact Leaves to Rudrata Path? How?
  5. Investigate special cases: For what sets L 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.