Introduction

This post provides a detailed proof that the Traveling Salesman Problem (TSP) is NP-complete by reducing from Rudrata Cycle. The reduction transforms the problem of finding a Hamiltonian cycle into finding a minimum-cost tour visiting all cities.

Problem Definitions

Traveling Salesman Problem (TSP)

Input:

  • A complete undirected graph G = (V, E) with n vertices (cities)
  • Edge weights w: E → ℤ⁺ (distances/costs)
  • Integer B ≥ 0 (budget)

Output: YES if there exists a tour (cycle visiting all vertices) with total cost ≤ B, NO otherwise.

Rudrata Cycle Problem

Input:

  • An undirected graph G = (V, E)

Output: YES if G contains a Rudrata cycle (Hamiltonian cycle visiting all vertices exactly once), NO otherwise.

1. NP-Completeness Proof of TSP: Solution Validation

TSP ∈ NP

Verification Algorithm: Given a candidate solution (tour π = v₁, v₂, …, vₙ, v₁):

  1. Check that π visits all n vertices exactly once: O(n) time
  2. Check that π forms a cycle: O(1) time (v₁ = vₙ₊₁)
  3. Compute total cost: ∑ᵢ₌₁ⁿ w(vᵢ, vᵢ₊₁): O(n) time
  4. Check if total cost ≤ B: O(1) time

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

Conclusion: TSP ∈ NP.

2. Reduce Rudrata Cycle to TSP

Key Insight: Given a Rudrata Cycle instance, we can reduce it to TSP by making the graph complete, setting edge weights to 1 for existing edges and 2 for missing edges, and setting B = n. Then G has a Rudrata cycle if and only if the TSP instance has a tour of cost n.

Hint: A Rudrata cycle uses exactly n edges, all of which must exist in the original graph. By setting existing edges to weight 1 and missing edges to weight 2, and B = n, we ensure that only tours using existing edges are feasible.

2.1 Input Conversion

Given a Rudrata Cycle instance G = (V, E) with n vertices, we construct a TSP instance (G’, w, B).

Construction:

Step 1: Create Complete Graph

  • G’ = (V, E’) where E’ contains all pairs of vertices (complete graph)
  • |E'| = n(n-1)/2

Step 2: Set Edge Weights

  • For each edge (u, v) ∈ E (existing edge): w(u, v) = 1
  • For each edge (u, v) ∉ E (missing edge): w(u, v) = 2

Step 3: Set Budget

  • B = n (we want a tour using exactly n edges, all of weight 1)

Step 4: Result

  • TSP instance: (G’, w, n)

Detailed Example:

Consider Rudrata Cycle instance:

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

Transformation:

  • G’ = complete graph on {1, 2, 3, 4}
  • Edge weights:
    • w(1,2) = 1, w(2,3) = 1, w(3,4) = 1, w(4,1) = 1, w(1,3) = 1
    • w(2,4) = 2 (missing edge)
  • B = 4

TSP instance: (G’, w, 4)

2.2 Output Conversion

Given: TSP solution (tour π with cost ≤ B = n)

Extract Rudrata Cycle:

  • The tour π visits all n vertices exactly once
  • Since cost ≤ n and each edge has weight ≥ 1, the tour uses exactly n edges
  • Since B = n and missing edges have weight 2, all edges in the tour must have weight 1
  • Therefore, all edges in the tour exist in the original graph G
  • π forms a Rudrata cycle in G

Verify Satisfaction:

  • π visits all n vertices exactly once: ✓
  • π forms a cycle: ✓
  • All edges in π exist in G: ✓ (since they have weight 1)
  • Therefore, π is a Rudrata cycle in G

3. Correctness Justification

3.1 If Rudrata Cycle has a solution, then TSP has a solution

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

Construct TSP Solution:

  • Use cycle C as the tour
  • C visits all n vertices exactly once
  • C uses exactly n edges, all of which exist in G
  • Therefore, all edges in C have weight 1
  • Total cost = n ≤ B = n

Verify Satisfaction:

  • Tour visits all vertices: ✓ (by definition of Rudrata cycle)
  • Tour forms a cycle: ✓ (by definition)
  • All edges have weight 1: ✓ (since they exist in G)
  • Total cost = n ≤ B: ✓

Conclusion: TSP has a solution.

3.2a If Rudrata Cycle does not have a solution, then TSP has no solution

Given: Rudrata Cycle instance G has no solution (no Hamiltonian cycle).

Proof:

Assume: TSP instance (G’, w, n) has a solution (tour π with cost ≤ n).

Extract Cycle:

  • π visits all n vertices exactly once
  • π forms a cycle
  • Since cost ≤ n and each edge has weight ≥ 1, π uses exactly n edges
  • Since missing edges have weight 2, all edges in π must have weight 1
  • Therefore, all edges in π exist in G
  • π is a Rudrata cycle in G

Contradiction:

  • π is a Rudrata cycle in G
  • This contradicts the assumption that G has no Rudrata cycle

Conclusion: TSP has no solution.

3.2b If TSP has a solution, then Rudrata Cycle has a solution

Given: TSP instance (G’, w, n) has solution (tour π with cost ≤ n).

Extract Cycle:

  • π visits all n vertices exactly once
  • π forms a cycle
  • Since cost ≤ n and each edge has weight ≥ 1, π uses exactly n edges
  • Since B = n and missing edges have weight 2, all edges in π must have weight 1
  • Therefore, all edges in π exist in the original graph G

Verify Satisfaction:

Rudrata Cycle Requirements:

  • π visits all n vertices exactly once: ✓ (by definition of tour)
  • π forms a cycle: ✓ (by definition)
  • All edges in π exist in G: ✓ (since they have weight 1)
  • Therefore, π is a Rudrata cycle in G

Key Insight:

  • A tour of cost n must use exactly n edges, each of weight 1
  • Edges of weight 1 correspond to edges in the original graph
  • Therefore, the tour is a Rudrata cycle in the original graph

Conclusion: The tour π forms a Rudrata cycle in G, so Rudrata Cycle has a solution.

Polynomial Time Analysis

Input Size:

  • Rudrata Cycle: Graph G with n vertices and m edges
  • TSP: Complete graph G’ with n vertices and n(n-1)/2 edges, weights w, integer B = n

Construction Time:

  • Create complete graph: O(n²) time
  • Set edge weights: O(n²) time (check each pair)
  • Set B = n: O(1) time
  • Total: O(n²) time

Conclusion: Reduction is polynomial-time.

Summary

We have shown:

  1. TSP ∈ NP: Solutions can be verified in polynomial time
  2. Rudrata Cycle ≤ₚ TSP: Polynomial-time reduction exists
  3. Correctness: Rudrata Cycle has solution ↔ TSP has solution

Therefore, TSP is NP-complete.

Key Insights

  1. Complete Graph: Making the graph complete allows any tour while controlling feasibility through weights
  2. Weight Encoding: Weight 1 for existing edges, weight 2 for missing edges ensures only existing edges are used
  3. Budget Constraint: Setting B = n forces the tour to use exactly n edges, all of weight 1
  4. Cycle Preservation: The reduction preserves the cycle structure through the weight assignment

Practice Questions

  1. Modify the reduction to reduce Rudrata Path to TSP. How would you handle the start and end vertices?
  2. Extend the reduction to handle weighted graphs. How would you set the weights?
  3. Consider metric TSP: What if we require the triangle inequality? Does the reduction still work?
  4. Prove the reverse reduction: Can we reduce TSP to Rudrata Cycle? How?
  5. Investigate approximation: How does this reduction relate to approximation algorithms for TSP?

This reduction demonstrates how edge weights can encode graph structure, enabling reductions between cycle-finding and optimization problems.