An undirected, unweighted, simple graph G(V,E) is said to be 2-colorable if there exists a function c : V → {0,1} such that for every (u,v) ∈ E,c(u) ̸ = c(v). Which of the following statements about 2-colorable graphs is/are true?

The correct statements about 2-colorable graphs are:
1. If G is 2-colorable, then G may contain cycles of even length
2. An optimal algorithm for testing whether G is 2-colorable runs in time $\Theta \left(\right|V|+|E\left|\right)$, if G is represented as an adjacency list

Let's break down the logic and explanation for each statement:

1. Understanding 2-Colorability and Cycles:
A graph is 2-colorable if and only if it is bipartite. A fundamental theorem in graph theory states that a graph is bipartite (and thus 2-colorable) if and only if it contains no odd cycles.
Since it cannot contain any cycle of odd length, the statement "If G is 2-colorable, then G may contain cycles of odd length" is false.
However, 2-colorable graphs can definitely contain cycles of even length. For example, a simple cycle graph on 4 vertices (C₄) is 2-colorable (we can color alternating vertices with 0 and 1). Thus, the statement "If G is 2-colorable, then G may contain cycles of even length" is true.

2. Testing 2-Colorability Algorithm Complexity:
We can test if a graph G(V, E) is 2-colorable (bipartite) by using a graph traversal algorithm such as Breadth-First Search (BFS) or Depth-First Search (DFS).
During the traversal, we assign colors (0 or 1) to vertices level-by-level. If we encounter an edge between two vertices of the same color, the graph is not 2-colorable. Otherwise, it is 2-colorable.
For a graph represented as an adjacency list, both BFS and DFS visit every vertex and edge of the graph. The time complexity of this traversal is:

Since we must examine the entire graph in the worst case to determine colorability, this linear time complexity is optimal. Therefore, the statement "An optimal algorithm for testing whether G is 2-colorable runs in time $\Theta \left(\right|V|+|E\left|\right)$, if G is represented as an adjacency list" is true.