Additional #1: Yes, every bipartite Eulerian graph has even number of edges.
Proof 1. Consider one partition set X. The Euler’s theorem says that all the vertices in X must have even degrees. The number of edges, which equals to degree sum of X, is therefore even.
Proof 2. Assume the number of edges, or the length of the Euler tour is odd. We know that every odd closed walk contains an odd cycle. Hence the graph contains an odd cycle, so it is not bipartite.
Additional #2. Count the number of trees on n labeled vertices.
The size of trees could be 1, 2, … n. For a tree of size k, we first choose its vertex set and then apply Cayley’s formula.
The answer is ∑{n choose k} k^{k-2}, for k=1 to n, where {n choose k} is a binomial coefficient, n(n-1)..(n-k+1)/k!
2.1.38: if there is no edge between U-W and W-U.
2.2.7 Skipped.
2.2 22. Use the fact that G-e has at most one more component than G.
2.3: 21, 2.4: 7 Skipped.
3.1.14. Since G has e(G)+1 (odd number) vertices, if every degree is odd, then the degree sum is odd, contradicting with a Euler’s theorem.
3.1.25. Proof 1. Pick two non-adjacent vertices x, y. We need to find a (x,y)-path.
Then neighbor sets N(x) and N(y) are two subsets of V-x-y.
Since deg(x) + deg(y) ≥ 2δ(G) ≥ n-1, N(x) and N(y) have at least one vertex z in common. Hence the path x-z-y connects x to y.
Proof 2. Assume G is disconnected with components C1, C2… Ck. Since every vertex and its neighbors must be in the same component,
each Ci
has at least (n-1)/2+1 = (n+1)/2 vertices. Therefore the sum of C1 and C2 is
greater than n, contradiction!