\(G = (V,E)\) is an undirected simple graph in which each edge has a distinct…

Answer: II only — the statement about the heaviest edge on a cycle is true; the statement about the lightest edge on a cycle is false.

Explanation:

• Statement I (If e is the lightest edge of some cycle, then every MST of G includes e): False. Being the lightest edge on a particular cycle does not imply that the edge is the unique minimum across any cut. The cut property is the correct criterion for forced inclusion: an edge belongs to every MST only if it is the unique minimum-weight edge crossing some cut. Therefore an edge can be light on a cycle yet still be avoidable in some MST.

• Statement II (If e is the heaviest edge of some cycle, then every MST of G excludes e): True. This follows from the cycle property. If an MST contained e (the heaviest edge of a cycle), removing e would split the tree into two components. The cycle gives an alternative path between those components consisting of lighter edges; adding any of those lighter edges and removing e yields a spanning tree of strictly smaller total weight, contradicting minimality. Hence no MST can contain e.

Thus the correct choice is the statement that every MST excludes an edge that is the heaviest on some cycle; the lightest-on-cycle statement is not guaranteed.