Introduction

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

Problem Definition: Rudrata (s,t)-Path

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

Hamiltonian (s,t)-Path: A path from s to t that visits each vertex exactly once.

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

Answer: Remove endpoint constraints (if graph allows).

Key Insight:

  • Hamiltonian (s,t)-path is a special case of Hamiltonian path
  • If Hamiltonian path exists, may be able to reorder to start at s and end at t
  • Reduction works if endpoints can be adjusted

Hint: This reduction is conditional - it works if the graph structure allows reordering. For unconditional reduction, better to reduce from Rudrata Cycle by removing an edge.

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

Rudrata Path Problem:

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

Rudrata Path ∈ NP:

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

  1. Check that all vertices appear exactly once: O(n) time
  2. Check that consecutive vertices are connected: O(n) time

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

Conclusion: Rudrata Path ∈ NP.

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

2.1 Input Conversion

Given a Rudrata (s,t)-Path instance: graph G = (V, E), vertices s, t.

Construction:

  • Return Rudrata Path instance: graph G

Key Property: Hamiltonian (s,t)-path → Hamiltonian path (if edge exists)

Note: This reduction works if we can verify that a Hamiltonian path can be reordered to start at s and end at t. For general reduction, may need to add/remove edges.

2.2 Output Conversion

Given: Rudrata Path solution (path P)

Extract (s,t)-Path:

  • If path P starts at s and ends at t, return P
  • Otherwise, check if path can be reordered (if graph allows)

3. Correctness Justification

3.1 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).

Construct Hamiltonian Path:

  • Path P visits all vertices
  • P is Hamiltonian path (may need reordering if endpoints don’t matter)

Conclusion: Rudrata Path has a solution.

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

Given: Rudrata (s,t)-Path instance has no Hamiltonian (s,t)-path.

Proof:

  • If Hamiltonian path exists, can check if it can start at s and end at t
  • But if no (s,t)-path exists, may still have Hamiltonian path
  • Need more careful construction

Refined Approach:

  • If (s,t)-path doesn’t exist, Hamiltonian path may still exist
  • This reduction shows (s,t)-path is at least as hard, but not necessarily equivalent

Conclusion: Rudrata Path may or may not have solution (reduction is not straightforward).

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

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

Extract (s,t)-Path:

  • If P starts at s and ends at t, return P
  • Otherwise, need to check if reordering is possible

Conclusion: Rudrata (s,t)-Path may have solution (depends on graph structure).

Note: This reduction is not straightforward. Better to reduce from Rudrata Cycle.

Polynomial Time: O(1) (trivial, but correctness is conditional).

Therefore, Rudrata Path is NP-complete (via other reductions).

Q2: How do you reduce Rudrata (s,t)-Path to Rudrata Cycle?

1. NP-Completeness Proof of Rudrata Cycle: Solution Validation

Rudrata Cycle Problem:

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

Rudrata Cycle ∈ NP:

Verification Algorithm: Given a candidate solution (cycle C):

  1. Check that all vertices appear exactly once: O(n) time
  2. Check that cycle is closed: O(1) time

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

Conclusion: Rudrata Cycle ∈ NP.

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

2.1 Input Conversion

Given a Rudrata (s,t)-Path instance: graph G = (V, E), vertices s, t.

Construction:

  • Add edge (t, s) to G if it doesn’t exist
  • Return Rudrata Cycle instance: modified graph G’

Key Property: Hamiltonian (s,t)-path ↔ Hamiltonian cycle (with added edge)

2.2 Output Conversion

Given: Rudrata Cycle solution (cycle C)

Extract (s,t)-Path:

  • Remove edge (t, s) from cycle C
  • Remaining path is Hamiltonian (s,t)-path

3. Correctness Justification

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

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

Construct Hamiltonian Cycle:

  • Path P visits all vertices from s to t
  • Add edge (t, s) to form cycle
  • Cycle visits all vertices exactly once

Conclusion: Rudrata Cycle has a solution.

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

Given: Rudrata (s,t)-Path instance has no Hamiltonian (s,t)-path.

Proof:

  • If Hamiltonian cycle exists in G’:
    • Remove edge (t, s) from cycle
    • Remaining path is Hamiltonian (s,t)-path
    • Contradiction

Conclusion: Rudrata Cycle has no solution.

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

Given: Rudrata Cycle instance has solution (cycle C).

Extract (s,t)-Path:

  • Cycle C visits all vertices
  • Remove edge (t, s) from C
  • Remaining path is Hamiltonian (s,t)-path from s to t

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

Polynomial Time: O(1) to add edge.

Therefore, Rudrata Cycle is NP-complete.

Key Takeaways

  1. Adding Edges: (s,t)-Path → Cycle adds return edge
  2. Removing Constraints: Cycle → Path removes edge
  3. Path-Cycle Relationship: Natural transformation
  4. Template Structure: All reductions follow rigorous format

Rudrata (s,t)-Path reductions demonstrate path-cycle relationships and constraint manipulation.