Introduction

The Rudrata Cycle Problem (also known as the Hamiltonian Cycle Problem) is one of the most fundamental graph problems in computer science. It asks whether a graph contains a cycle that visits each vertex exactly once. This problem is closely related to the Traveling Salesman Problem (TSP) and serves as a cornerstone for understanding NP-completeness in graph theory. The problem is named after Sir William Rowan Hamilton, who studied it in the context of the Icosian game.

What is a Rudrata Cycle?

A Rudrata cycle (also called a Hamiltonian cycle) in an undirected graph is a cycle that visits each vertex exactly once and returns to the starting vertex. Unlike an Eulerian cycle (which visits each edge exactly once), a Hamiltonian cycle visits each vertex exactly once.

Problem Definition

Rudrata Cycle Decision Problem:

Input:

  • An undirected graph G = (V, E)

Output: YES if G contains a Rudrata cycle (a cycle visiting every vertex exactly once), NO otherwise

Rudrata Cycle Optimization Problem (TSP):

Input:

  • A complete undirected graph G = (V, E) with edge weights w: E → ℝ⁺

Output: The minimum weight Rudrata cycle (this is the Traveling Salesman Problem)

Example

Consider the following graph:

    1---2---3
    |   |   |
    4---5---6

Rudrata Cycle:

  • Cycle: 1 to 2 to 5 to 4 to 1 ✗ (doesn’t visit all vertices)
  • Cycle: 1 to 2 to 3 to 6 to 5 to 4 to 1 ✓ (visits all 6 vertices exactly once)
  • Cycle: 1 to 4 to 5 to 2 to 1 ✗ (doesn’t visit vertices 3 and 6)

Visual Example

A graph with a Rudrata cycle highlighted:

    1---2
    |\ /|
    | X |
    |/ \|
    3---4
  • Rudrata cycle: 1 to 2 to 4 to 3 to 1 (visits all 4 vertices and returns to start)
  • This is a complete graph K_4, so it has many Hamiltonian cycles

Why Rudrata Cycle is in NP

To show that Rudrata Cycle is NP-complete, we first need to show it’s in NP.

Rudrata Cycle ∈ NP:

Given a candidate solution (a sequence of vertices representing a cycle), we can verify in polynomial time:

  1. Check that the cycle has exactly n vertices: O(n) time
  2. Check that each vertex appears exactly once: O(n) time (use a set or array)
  3. Check that consecutive vertices in the cycle are adjacent: O(n) time (check n edges, including the wrap-around edge from last to first)
  4. Check that the cycle returns to start: O(1) time

Total verification time: O(n), which is polynomial in the input size. Therefore, Rudrata Cycle is in NP.

NP-Completeness: Reduction from 3-SAT

The standard proof that Rudrata Cycle is NP-complete reduces directly from 3-SAT.

Construction

For a 3-SAT formula φ = C_1 ∧ C_2 ∧ … ∧ C_m with variables x₁, x₂, …, x_n:

Key Idea: Create a graph where a Rudrata cycle corresponds to a satisfying assignment.

  1. For each variable x_i:
    • Create a “variable gadget”: a row of vertices representing the variable
    • Typically: vertices arranged horizontally, where going “left” means x_i = TRUE and going “right” means x_i = FALSE
    • Example: For variable x_i, create vertices v_{i,1}, v_{i,2}, …, v_{i,k} where the cycle can traverse left-to-right (TRUE) or right-to-left (FALSE)
  2. For each clause C_j:
    • Create a “clause gadget”: vertices that can be visited if the clause is satisfied
    • Connect clause vertices to variable vertices corresponding to literals in the clause
    • The cycle must visit clause vertices, which is only possible if at least one literal is true
  3. Connect gadgets:
    • Chain variable gadgets together in sequence
    • Connect clause gadgets to appropriate variable positions
    • Ensure a Rudrata cycle visits all vertices exactly once

Simplified Construction Example

A common construction uses:

Variable Gadget:

  • For each variable x_i, create a horizontal chain of vertices
  • The cycle can traverse this chain in two ways (encoding TRUE/FALSE)

Clause Gadget:

  • For each clause C_j = (l_1 ∨ l_2 ∨ l_3), create vertices connected to the variable gadgets
  • If the cycle takes the path corresponding to a true literal, it can visit the clause vertex

Why This Works:

Forward Direction (3-SAT satisfiable → Rudrata Cycle exists):

  • If φ is satisfiable, construct cycle through variable gadgets based on assignment
  • Visit clause vertices for satisfied clauses
  • This gives a valid Rudrata cycle

Reverse Direction (Rudrata Cycle exists → 3-SAT satisfiable):

  • Extract truth assignment from cycle’s path through variable gadgets
  • Since all clause vertices are visited, each clause has at least one true literal
  • This gives a satisfying assignment

Relationship to Other Problems

The Rudrata Cycle Problem is closely related to several important problems:

Rudrata Path

As we saw in the previous post:

  • Rudrata Cycle ≤ₚ Rudrata Path: Break cycle into path
  • Rudrata Path ≤ₚ Rudrata Cycle: Connect path ends to form cycle
  • They are polynomially equivalent

Traveling Salesman Problem (TSP)

TSP Decision Problem:

  • Given complete graph with edge weights and bound B
  • Does there exist a Rudrata cycle with total weight ≤ B?

Relationship:

  • TSP is a weighted version of Rudrata Cycle
  • Rudrata Cycle reduces to TSP: set all edge weights to 1, ask if cycle of weight n exists
  • TSP reduces to Rudrata Cycle: use unweighted version
  • They are essentially the same problem

Eulerian Cycle

Eulerian Cycle:

  • Visits each edge exactly once (vs. Hamiltonian cycle visits each vertex exactly once)

Key Difference:

  • Eulerian cycle: polynomial-time solvable (check degrees, use Fleury’s or Hierholzer’s algorithm)
  • Hamiltonian cycle: NP-complete
  • This demonstrates how a small change in problem statement can dramatically affect complexity

Practical Implications

Why Rudrata Cycle is Hard

The Rudrata Cycle Problem is NP-complete, which means:

  1. No Known Polynomial-Time Algorithm: Best known algorithms have exponential time complexity
  2. Brute Force: Try all (n-1)!/2 possible cycles (for undirected graphs) - factorial time
  3. Dynamic Programming: Can solve in O(2^n · n^2) time using bitmask DP (similar to TSP)
  4. Backtracking: Practical for small instances, but still exponential worst-case

Dynamic Programming Solution

Subproblem: dp[mask][v] = true if there exists a path visiting all vertices in mask ending at vertex v, and this path can be extended to a cycle.

Recurrence:

  • Base case: dp[2^i][i] = true for all i (path of length 1)
  • Recurrence: dp[mask][v] = bigvee_{u in mask, (u,v) in E} dp[mask setminus {v}][u]
  • Final check: For each v, check if dp[2^n-1][v] = true and (v, start) is an edge

Algorithm:

Algorithm: RudrataCycleDP(G)
1. n = n
2. Let dp[0..2^n-1][0..n-1] be a boolean array
3. for i = 0 to n-1:
4.     dp[2^i][i] = true
5. for mask = 1 to 2^n - 1:
6.     for v = 0 to n-1:
7.         if v in mask:
8.             for each neighbor u of v:
9.                 if u in mask:
10.                    dp[mask][v] = dp[mask][v] OR dp[mask - 2^v][u]
11. for v = 0 to n-1:
12.     if dp[2^n - 1][v] AND (v, start) is an edge:
13.         return true
14. return false

Time Complexity: O(2^n · n^2) Space Complexity: O(2^n · n)

Runtime Analysis

Brute Force Approach

Algorithm: Try all (n-1)!/2 possible cycles (for undirected graphs)

  • Time Complexity: O((n-1)! · n) = O(n!)
  • Space Complexity: O(n) for storing current cycle
  • Analysis: For each permutation, verify it forms a valid cycle (O(n) checks including wrap-around)

Dynamic Programming (Held-Karp Algorithm)

Algorithm: Bitmask DP as described above

  • Time Complexity: O(2^n · n^2)
  • Space Complexity: O(2^n · n)
  • Subproblem: dp[mask][v] = true if path exists visiting all vertices in mask ending at v
  • Final Check: Verify edge exists from last vertex to start vertex

Backtracking

Algorithm: Systematic search with pruning

  • Time Complexity: Exponential worst-case, improved with pruning
  • Space Complexity: O(n) for recursion stack
  • Pruning: Stop if current path can’t form a cycle visiting all vertices

Special Cases

Complete Graphs:

  • Time Complexity: O(1) - trivial, any cycle works

Trees:

  • Time Complexity: O(n) - check if tree has exactly n-1 edges (then no cycle possible unless n=2)

Verification Complexity

Given a candidate cycle:

  • Time Complexity: O(n) - verify all n edges exist (including wrap-around)
  • Space Complexity: O(1) additional space
  • This polynomial-time verifiability shows Rudrata Cycle is in NP

Real-World Applications

Rudrata Cycle has numerous applications:

  1. Route Planning: Finding routes that visit all locations and return to start (TSP)
  2. Circuit Design: Designing circuits that visit all components
  3. DNA Sequencing: Finding cycles in overlap graphs
  4. Network Analysis: Analyzing connectivity and designing network topologies
  5. Game Theory: Solving puzzles and games
  6. Scheduling: Sequencing tasks with return constraints
  7. Logistics: Vehicle routing, delivery optimization

Special Cases

Some restricted versions of Rudrata Cycle are tractable:

  • Complete Graphs: Always has a Rudrata cycle (any permutation works)
  • Dirac’s Theorem: If deg(v) ≥ n/2 for all vertices, then Hamiltonian cycle exists (but finding it is still hard)
  • Ore’s Theorem: If deg(u) + deg(v) ≥ n for all non-adjacent u,v, then Hamiltonian cycle exists
  • Grid Graphs: Can be solved efficiently for certain grid structures
  • Bounded Treewidth: Can be solved efficiently using tree decomposition

Key Takeaways

  1. Rudrata Cycle is NP-Complete: Proven by reduction from 3-SAT
  2. TSP Connection: Traveling Salesman Problem is essentially weighted Rudrata Cycle
  3. Path vs Cycle: Rudrata Cycle and Rudrata Path are polynomially equivalent
  4. Dynamic Programming: O(2^n · n^2) time solution using bitmask DP works for small graphs
  5. Practical Algorithms: Despite NP-completeness, DP and heuristics work well for many practical instances

Reduction Summary

3-SAT ≤ₚ Rudrata Cycle:

  • Construct graph with variable and clause gadgets
  • Satisfying assignment ↔ Rudrata cycle
  • The construction ensures cycle visits all vertices exactly once

Rudrata Cycle ≤ₚ TSP:

  • Given Rudrata Cycle instance: graph G
  • Create complete graph G’ with edge weights: 1 if edge exists in G, ∈fty otherwise
  • Rudrata cycle exists ↔ TSP has solution of weight n

TSP ≤ₚ Rudrata Cycle:

  • Given TSP instance, ask if unweighted version has Hamiltonian cycle
  • They are essentially equivalent

All reductions are polynomial-time, establishing Rudrata Cycle as NP-complete.

Further Reading

  • Garey & Johnson: “Computers and Intractability” - Standard reference for NP-completeness proofs
  • Hamilton’s Icosian Game: Historical context of the problem
  • TSP: Understanding the relationship to Traveling Salesman Problem
  • Dynamic Programming: CLRS covers bitmask DP techniques for TSP
  • Graph Theory: Dirac’s and Ore’s theorems provide sufficient conditions

Practice Problems

  1. Find all Rudrata cycles: For the complete graph K_4 with vertices {1,2,3,4}, list all distinct Rudrata cycles. How many are there? (Consider cycles equivalent if they’re rotations or reversals)

  2. Prove the reduction: Show that Rudrata Cycle reduces to TSP. What edge weights should you use?

  3. DP implementation: Implement the dynamic programming algorithm for Rudrata Cycle. Test it on small graphs and analyze its performance.

  4. Path to Cycle: Show that Rudrata Path reduces to Rudrata Cycle. How do you modify the graph?

  5. Time complexity analysis: For the DP algorithm, verify the O(2^n · n^2) time complexity. Can you optimize the space complexity?

  6. Special cases: Research Dirac’s theorem. If a graph satisfies the conditions, does that mean we can find a Hamiltonian cycle in polynomial time?

  7. Eulerian vs Hamiltonian: Explain the key difference between Eulerian cycles and Hamiltonian cycles. Why is one polynomial-time and the other NP-complete?

  8. Extension: Research approximation algorithms for TSP. What approximation ratios can be achieved? What about metric TSP?

Understanding the Rudrata Cycle Problem provides crucial insight into cycle problems in graph theory and their connection to optimization problems like TSP. The relationship to 3-SAT demonstrates how logical constraints can be encoded as graph structures.