Introduction

Rudrata Path (Hamiltonian Path) is a fundamental path problem proven NP-complete. This post provides detailed proofs following the standard template for reducing from Rudrata Path to prove other problems are NP-complete.

Problem Definition: Rudrata Path

Rudrata Path Problem:

  • Input: Graph G = (V, E)
  • Output: YES if G has a path visiting every vertex exactly once, NO otherwise

Hamiltonian Path: A path that visits each vertex exactly once.

Q1: How do you reduce Rudrata Path to Rudrata (s,t)-Path?

Answer: Add start and end vertices.

Key Insight:

  • Add universal start vertex s connected to all vertices
  • Add universal end vertex t connected from all vertices
  • Hamiltonian path can start/end anywhere ↔ Hamiltonian (s,t)-path exists

Hint: By connecting s to all vertices and all vertices to t, any Hamiltonian path can be extended to start at s and end at t. This removes the constraint on path endpoints.

1. NP-Completeness Proof of Rudrata (s,t)-Path: Solution Validation

Rudrata (s,t)-Path Problem:

  • Input: Graph G = (V, E) and vertices s, t
  • Output: YES if G has a path from s to t visiting every vertex exactly once, NO otherwise

Rudrata (s,t)-Path ∈ NP:

Verification Algorithm: Given a candidate solution (path P from s to t):

  1. Check that path starts at s and ends at t: O(1) time
  2. Check that all vertices appear exactly once: O(n) time
  3. Check that consecutive vertices are connected: O(n) time

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

Conclusion: Rudrata (s,t)-Path ∈ NP.

2. Reduce Rudrata Path to Rudrata (s,t)-Path

2.1 Input Conversion

Given a Rudrata Path instance: graph G = (V, E).

Construction:

  • Add two new vertices s and t
  • Connect s to all vertices in V
  • Connect all vertices in V to t
  • Return Rudrata (s,t)-Path instance: modified graph G’, vertices s and t

Key Property: Hamiltonian path ↔ Hamiltonian (s,t)-path

2.2 Output Conversion

Given: Rudrata (s,t)-Path solution (path P from s to t)

Extract Hamiltonian Path:

  • Remove s and t from path P
  • Remaining path visits all original vertices
  • Return as Hamiltonian path

3. Correctness Justification

3.1 If Rudrata Path has a solution, then Rudrata (s,t)-Path has a solution

Given: Rudrata Path instance has solution (Hamiltonian path P in G).

Construct (s,t)-Path:

  • Path P visits all vertices in V
  • Let u be first vertex and v be last vertex of P
  • Create path: s → u → … → v → t
  • This path visits all vertices exactly once

Conclusion: Rudrata (s,t)-Path has a solution.

3.2a If Rudrata Path does not have a solution, then Rudrata (s,t)-Path has no solution

Given: Rudrata Path instance has no Hamiltonian path.

Proof by Contradiction:

  • Assume Rudrata (s,t)-Path has solution P
  • Remove s and t from P
  • Remaining path is Hamiltonian path
  • Contradiction

Conclusion: Rudrata (s,t)-Path has no solution.

3.2b If Rudrata (s,t)-Path has a solution, then Rudrata Path has a solution

Given: Rudrata (s,t)-Path instance has solution (path P from s to t).

Extract Hamiltonian Path:

  • Path P visits all vertices including s and t
  • Remove s and t from P
  • Remaining path visits all original vertices exactly once
  • Therefore, Hamiltonian path exists

Conclusion: Rudrata Path has a solution.

Polynomial Time: O(n) to add vertices and edges.

Therefore, Rudrata (s,t)-Path is NP-complete.

Q2: How do you reduce Rudrata Path to Longest Path?

1. NP-Completeness Proof of Longest Path: Solution Validation

Longest Path Problem:

  • Input: Graph G = (V, E) and integer k
  • Output: YES if G has a path of length ≥ k, NO otherwise

Longest Path ∈ NP:

Verification Algorithm: Given a candidate solution (path P):

  1. Check that P is a valid path: O(n) time
  2. Check that length ≥ k: O(1) time

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

Conclusion: Longest Path ∈ NP.

2. Reduce Rudrata Path to Longest Path

2.1 Input Conversion

Given a Rudrata Path instance: graph G = (V, E).

Construction:

  • Return Longest Path instance: graph G, target length k = n - 1

Key Property: Hamiltonian path ↔ Path of length n - 1

2.2 Output Conversion

Given: Longest Path solution (path P of length ≥ n - 1)

Extract Hamiltonian Path:

  • Path P has length ≥ n - 1
  • Since graph has n vertices, path visits all vertices
  • P is Hamiltonian path

3. Correctness Justification

3.1 If Rudrata Path has a solution, then Longest Path has a solution

Given: Rudrata Path instance has solution (Hamiltonian path P).

Construct Longest Path:

  • Path P visits all n vertices
  • Length of P = n - 1
  • Therefore, path of length ≥ n - 1 exists

Conclusion: Longest Path has a solution.

3.2a If Rudrata Path does not have a solution, then Longest Path has no solution

Given: Rudrata Path instance has no Hamiltonian path.

Proof:

  • Maximum path length is n - 1 (visiting all vertices)
  • If no Hamiltonian path exists, maximum path length < n - 1
  • Therefore, no path of length ≥ n - 1

Conclusion: Longest Path has no solution.

3.2b If Longest Path has a solution, then Rudrata Path has a solution

Given: Longest Path instance has solution (path P of length ≥ n - 1).

Extract Hamiltonian Path:

  • Path P has length ≥ n - 1
  • Since graph has n vertices, P must visit all vertices
  • P is Hamiltonian path

Conclusion: Rudrata Path has a solution.

Polynomial Time: O(1) (trivial reduction).

Therefore, Longest Path is NP-complete.

Key Takeaways

  1. Adding Constraints: Rudrata Path → (s,t)-Path adds endpoints
  2. Optimization: Rudrata Path → Longest Path is special case
  3. Path Length: Maximum is n - 1
  4. Template Structure: All reductions follow rigorous format

Rudrata Path reductions demonstrate path constraints and optimization relationships.