Question

Write a polynomial-time verification algorithm for the Hamiltonian Circuits Decision Problem.

Short answer

A polynomial-time verification algorithm for the Hamiltonian Circuits Decision Problem checks if a given sequence of vertices forms a Hamiltonian cycle in a given graph. The algorithm confirms two conditions for each pair of consecutive vertices (vi, vi+1): first, if there is an edge between these vertices in the graph, and second, if each vertex is visited only once. At the end, it confirms that the first and the last vertices are connected, closing the cycle. This algorithm runs with a time complexity of O(n), which is considered polynomial.

Step-by-step solution

Verify Input

Before beginning the problem, verify that the provided input is indeed a graph. A graph is a collection of vertices (nodes) and edges connecting these vertices.

Explanation of Algorithm: Polynomial Verification

As a decision problem, it is enough to show an algorithm that will confirm if a valid solution (Hamiltonian Circuit) of the problem indeed exists. The algorithm expects a graph G and a sequence of vertices {v1, v2, ..., vn} as input. It then checks if this sequence forms a Hamiltonian Circuit in G. The order of the process is O(n), hence it's polynomial. The algorithm processes the sequence of vertices and checks two conditions for each pair (vi, vi+1): if there is an edge between them in the graph G and each vertex is visited only once. At the end, it checks if v1 and vn are connected, to begin and end the path at the same vertex.

Formulation of Algorithm

A broad pseudo code for this algorithm could be formed as follows: pick a starting node on the graph, create a cycle by moving to adjacent vertices while avoiding places visited before. If you run out of new places to visit, check if there's a direct path back to the start and you've visited every vertex exactly once. If you have, it's a Hamiltonian Cycle. The algorithm runs with time complexity O(n), as iterating through each edge from every vertex takes at most n steps per vertex, thus giving us n*n or n^2 steps for the entire graph.