If a graph has 'n' vertices, what is the maximum size of the matrix required to represent its transitive closure using the Warshall algorithm?

To determine the maximum size of the matrix required to represent the transitive closure of a graph with $n$ vertices using Warshall's algorithm, we would follow these steps:

1. Break Down the Problem

We need to identify the size of the adjacency matrix for a graph with $n$ vertices and understand how Warshall's algorithm operates on it.

2. Relevant Concepts

• The adjacency matrix of a graph with $n$ vertices is an $n \times n$ matrix, where each entry $a_{ij}$ indicates whether there is an edge from vertex $i$ to vertex $j$.
• Warshall's algorithm modifies this matrix to compute the transitive closure, effectively meaning whether there is a path (direct or indirect) from vertex $i$ to vertex $j$.

3. Analysis and Detail

1. For a graph with $n$ vertices, the adjacency matrix has $n^2$ entries, as each pair of vertices can either have an edge or not.
2. Warshall's algorithm works by iterating over each vertex as a potential intermediate vertex and updating the matrix, which does not change the matrix size but may change its entries.
3. Thus, the size of the matrix remains $n \times n$ throughout the execution of the algorithm.

4. Verify and Summarize

The maximum size remains consistent regardless of the alterations made during the execution of Warshall's algorithm. Hence, we conclude:

Final Answer

The maximum size of the matrix required to represent the transitive closure of a graph with $n$ vertices using the Warshall algorithm is $n^2$.