Additional #1:
Part 1: Fix a vertex r as the root. Every vertex belongs to a (unique) path starting from r since T is a tree.
Vertices in a path must be colored alternatively when we color them by 2 colors. Therefore, once the color of r is known, we know the color of all other vertices. In other words, this 2-coloring of V(T) is unique.
Part 2: Fix a vertex in B as the root. Each non-leaf vertices in A must have at least one child in B; no two such vertices have the same child since T has no cycles. Therefore, the number of non-lead vertices in A is at most the |B|-1 (not counting the root from B). Hence the number of leaves in A is at least |A|-|B|+1.
Additional #2. shown in the review.
5.1.4: A 3-connected graph is 3-edge-connected (since κ ≤ κ’) and thus has no cut-edge (bridge).
5.1.17: you may use figure 5.1.1.
5.2 8: Use the definition of Harary graph; skipped.
6.1.26: Use Euler’s theorem: a graph is eulerian if and only if it is even (all degrees are even).
Consider an edge xy, which is a vertex in L(G). The degree of xy in L(G) equals to deg(x) + deg(y) -2, which is even since both deg(x) and deg(y) are even. Hence L(G) is even and thus eulerian.
6.2.6: shown in class.