Introduction

This post provides a detailed proof that the Reliable Network problem is NP-complete by reducing from Rudrata Cycle. Reliable Network asks for a graph that satisfies connectivity requirements within a budget, generalizing the Hamiltonian cycle problem.

Problem Definitions

Reliable Network Problem

Input:

  • Two n × n matrices:
    • Distance matrix d_ij (cost of edge between vertices i and j)
    • Connectivity requirement matrix r_ij (required number of vertex-disjoint paths between i and j)
  • A budget b ≥ 0

Output: YES if there exists a graph G = ({1, 2, …, n}, E) such that:

  1. The total cost of all edges is b or less: ∑_{(i,j) ∈ E} d_ij ≤ b
  2. Between any two distinct vertices i and j, there are r_ij vertex-disjoint paths

NO otherwise.

Note: Vertex-disjoint paths are paths that share no vertices except possibly their endpoints.

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 Reliable Network: Solution Validation

Reliable Network ∈ NP

Verification Algorithm: Given a candidate solution (graph G = ({1, 2, …, n}, E)):

  1. Check that total cost ≤ b: O(|E|) time
  2. For each pair (i, j) with i ≠ j:
    • Find r_ij vertex-disjoint paths: O(n²) time using max-flow algorithms
    • Check that at least r_ij paths exist: O(n²) time
  3. Check all pairs: O(n⁴) time

Total Time: O(n⁴), which is polynomial in input size.

Conclusion: Reliable Network ∈ NP.

2. Reduce Rudrata Cycle to Reliable Network

Key Insight: Given a Rudrata Cycle instance, we can reduce it to Reliable Network by setting all distances to 1 or 2, all connectivity requirements to 2, and budget to n. A graph with n vertices, n edges, cost ≤ n, and 2 vertex-disjoint paths between all pairs must be a cycle, which is exactly a Rudrata cycle.

Hint: If all d_ij’s are 1 or 2, b = n, and all r_ij’s are 2, then we need a graph with exactly n edges (each of cost 1) that is 2-connected. A cycle on n vertices has exactly n edges and provides 2 vertex-disjoint paths between any two vertices (going clockwise and counterclockwise).

2.1 Input Conversion

Given a Rudrata Cycle instance G = (V, E) with n vertices, we construct a Reliable Network instance (d, r, b).

Construction:

Step 1: Set Distance Matrix

  • For each pair (i, j) with i ≠ j:
    • If (i, j) ∈ E: d_ij = 1 (edge exists in original graph)
    • If (i, j) ∉ E: d_ij = 2 (edge doesn’t exist, higher cost)
  • d_ii = 0 (no self-loops)

Step 2: Set Connectivity Requirements

  • For all pairs (i, j) with i ≠ j: r_ij = 2
  • r_ii = 0 (no requirement for same vertex)

Step 3: Set Budget

  • b = n (we can afford exactly n edges of cost 1)

Step 4: Result

  • Reliable Network instance: (d, r, n) where:
    • d_ij ∈ {1, 2} for all i ≠ j
    • r_ij = 2 for all i ≠ j
    • b = n

Detailed Example:

Consider Rudrata Cycle instance:

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

Transformation:

  • Distance matrix d:
    • d₁₂ = d₂₃ = d₃₄ = d₄₁ = 1 (edges exist)
    • All other d_ij = 2 (edges don’t exist)
  • Connectivity matrix r:
    • r_ij = 2 for all i ≠ j
  • Budget: b = 4

Reliable Network instance: (d, r, 4)

Note: A cycle on 4 vertices has 4 edges (cost 4) and provides 2 vertex-disjoint paths between any pair.

2.2 Output Conversion

Given: Reliable Network solution (graph G’ = ({1, 2, …, n}, E’) with cost ≤ n and r_ij = 2 vertex-disjoint paths between all pairs)

Extract Rudrata Cycle:

  • Since b = n and all edges cost at least 1, G’ has exactly n edges, all of cost 1
  • Since r_ij = 2 for all pairs, G’ is 2-connected
  • A 2-connected graph with n vertices and n edges must be a cycle
  • Therefore, G’ is a cycle visiting all n vertices exactly once (a Rudrata cycle)

Verify Satisfaction:

  • G’ visits all n vertices: ✓ (by construction)
  • G’ forms a cycle: ✓ (2-connected with n vertices and n edges)
  • All edges in G’ exist in original graph G: ✓ (since d_ij = 1 only for edges in G)
  • Therefore, G’ is a Rudrata cycle in G

3. Correctness Justification

3.1 If Rudrata Cycle has a solution, then Reliable Network has a solution

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

Construct Reliable Network Solution:

  • Use cycle C as the graph G’
  • G’ has exactly n edges (one per vertex in the cycle)
  • All edges in C have cost 1 (since they exist in G, so d_ij = 1)
  • Total cost = n ≤ b = n ✓

Verify Connectivity:

  • For any two vertices i and j in the cycle:
    • Path 1: Follow the cycle in one direction from i to j
    • Path 2: Follow the cycle in the other direction from i to j
    • These two paths are vertex-disjoint (except at endpoints)
    • Therefore, there are 2 vertex-disjoint paths between i and j ✓
  • All connectivity requirements r_ij = 2 are satisfied

Conclusion: Reliable Network has a solution.

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

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

Proof:

Assume: Reliable Network instance (d, r, n) has a solution (graph G’ with cost ≤ n and 2 vertex-disjoint paths between all pairs).

Extract Cycle:

  • Since b = n and all edges cost at least 1, G’ has at most n edges
  • Since r_ij = 2 for all pairs, G’ is 2-connected
  • A 2-connected graph with n vertices needs at least n edges (by connectivity requirements)
  • Therefore, G’ has exactly n edges, all of cost 1
  • A 2-connected graph with n vertices and exactly n edges must be a cycle
  • Therefore, G’ is a cycle on n vertices
  • Since all edges have cost 1, all edges in G’ exist in the original graph G
  • Therefore, G’ is a Rudrata cycle in G

Contradiction:

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

Conclusion: Reliable Network has no solution.

3.2b If Reliable Network has a solution, then Rudrata Cycle has a solution

Given: Reliable Network instance (d, r, n) has solution (graph G’ with cost ≤ n and 2 vertex-disjoint paths between all pairs).

Extract Cycle:

  • Since b = n and all edges cost at least 1, G’ has at most n edges
  • Since r_ij = 2 for all pairs, G’ is 2-connected
  • A 2-connected graph with n vertices needs at least n edges
  • Therefore, G’ has exactly n edges, all of cost 1
  • A 2-connected graph with n vertices and exactly n edges must be a cycle
  • Since all edges have cost 1, they correspond to edges in the original graph G (where d_ij = 1)
  • Therefore, G’ is a cycle in G visiting all n vertices

Verify Satisfaction:

Rudrata Cycle Requirements:

  • G’ visits all n vertices: ✓ (by construction, G’ = ({1, 2, …, n}, E’))
  • G’ forms a cycle: ✓ (2-connected with n vertices and n edges)
  • All edges in G’ exist in G: ✓ (since d_ij = 1 only for edges in G)
  • Therefore, G’ is a Rudrata cycle in G

Key Insight:

  • With budget b = n and all edges costing at least 1, we can afford exactly n edges
  • Requiring r_ij = 2 vertex-disjoint paths between all pairs forces the graph to be 2-connected
  • A 2-connected graph with n vertices and n edges must be a cycle
  • A cycle provides exactly 2 vertex-disjoint paths between any two vertices (clockwise and counterclockwise)
  • Therefore, any solution to Reliable Network gives us a Rudrata cycle

Conclusion: The graph G’ 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
  • Reliable Network: Two n × n matrices (n² entries each), integer b = n

Construction Time:

  • Create distance matrix: O(n²) time (check each pair)
  • Create connectivity matrix: O(n²) time (set all r_ij = 2)
  • Set b = n: O(1) time
  • Total: O(n²) time

Conclusion: Reduction is polynomial-time.

Summary

We have shown:

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

Therefore, Reliable Network is NP-complete.

Key Insights

  1. Budget Constraint: Setting b = n with all edges costing at least 1 forces exactly n edges
  2. Connectivity Requirement: Requiring r_ij = 2 vertex-disjoint paths forces 2-connectivity
  3. Cycle Characterization: A 2-connected graph with n vertices and n edges must be a cycle
  4. Path Structure: A cycle provides exactly 2 vertex-disjoint paths between any two vertices
  5. Cost Encoding: Using d_ij = 1 for existing edges and d_ij = 2 for non-existing edges ensures only original edges are used

Practice Questions

  1. Modify the reduction to handle different connectivity requirements. What if r_ij varies?
  2. Extend the reduction to handle weighted graphs. How would you set the distance matrix?
  3. Consider the optimization version: How would you reduce the optimization version of Rudrata Cycle to Reliable Network?
  4. Prove the reverse reduction: Can we reduce Reliable Network to Rudrata Cycle? How?
  5. Investigate special cases: For what values of r_ij and b is Reliable Network polynomial-time solvable?

This reduction demonstrates how connectivity and budget constraints can encode Hamiltonian cycle requirements, enabling reductions between network design and cycle-finding problems.