Let $G$ be a weighted graph with edge weights greater than one and $G^{\prime}$ be the graph constructed by squaring the weights of edges in $G$. Let $T$ and $T^{\prime}$ be the minimum spanning trees of $G$ and $G^{\prime}$, respectively, with total weights $t$ and $t^{\prime}$. Which of the following statements is TRUE?

1. Preservation of Relative Ordering: The function $f(w) = w^2$ is strictly monotonically increasing for $w > 1$. Kruskal's and Prim's algorithms depend only on the relative order of edge weights. Therefore, squaring the edge weights preserves the relative order of the weights, meaning $T^{\prime} = T$ (assuming unique MSTs or corresponding choices of MSTs).

2. Total Weight Comparison: Let the edge weights of the MST $T$ be $w_1, w_2, \dots, w_{n-1}$, where $w_i > 1$.

   • Total weight of $T$: $t = \sum_{i=1}^{n-1} w_i$

   • Total weight of $T^{\prime}$: $t^{\prime} = \sum_{i=1}^{n-1} w_i^2$

   • By algebraic expansion:

     $$t^2 = \left(\sum_{i=1}^{n-1} w_i\right)^2 = \sum_{i=1}^{n-1} w_i^2 + 2 \sum_{1 \le i < j \le n-1} w_i w_j = t^{\prime} + 2 \sum_{i < j} w_i w_j$$

   • If the graph has $n \ge 3$ vertices, the cross terms are strictly positive, so $t^{\prime} < t^2$.

   • If $n = 2$, the tree has only 1 edge, so $t^{\prime} = t^2$.

Because of edge cases (such as graphs with multiple MSTs or a 2-vertex graph where $t^{\prime} = t^2$), the question was officially awarded Marks to All.