Mean = 86.2
Median = 87
A: 90-100 (12)
B: 80-89 (7)
C: 70-79 (4)
D: 60-69 (2)
F: 0-59 (0)
Problem 1
(a) (1), (2), (3), (4), (7)
(b) (1), (2), (3), (5), (7)
(c) (1)
(d) (3), (4), (7)
(e) (3)
(f) (3) (7)
(g) (1), (2), (4)
Problem 2
(a) True. If there is no u-v path, then u and v must be in different components of G. The subgraph of G consisting of the component containing u (say) has all even vertices except u, which is impossible, since there must be an even number of odd vertices.
Some misinterpreted this problem to read that G had only 2 vertices, not 2 vertices of odd degree. I deducted 2 points.
(b) True. The handshake condition is satisfied (the sum is 18, which is 2*10-2). So either construct a tree by forming a Prufer code with 3 1's, 2 2's, and 1 each of 3, 4, and 5, or appeal to Ex. 2.1.27.
Showing that a simple graph has this degree sequence is not enough. Either construct the graph or refer to the exercise.
(c) False. Let G be the union of two 3-cycles.
(d) True. Make two copies of G and connect corresponding vertices. Each vertex in the resulting graph will have degree k+1, and each copy of G is a subgraph.
It is incorrect to add a vertex and connect it to some (or even worse, all) of the vertices of G. Also, the problem asks to construct a larger graph from G which has G as a subgraph, not start with an arbitrary k+1 regular graph and produce a k regular subgraph.
Problem 3
(a) Impossible. A has a non-zero diagonal entry.
(b) Possible. Any nonconnected graph will suffice.
(c) Possible. Apply Theorem 1.3.31 repeatedly.
Problem 4
The edges are 14, 23, 37, 39, 47, 57, 68, 69.
Problem 5
The first and third are bipartite, 3-regular, so must both be K_{3,3}. The second is also 3-regular, but is not bipartite. So the first and third are isomorphic, while the second is not isomorphic to the other two.
You must include an argument for the isomorphism (explicit bijection, both K_{3,3}, etc.) It is not enough to say they are both bipartite. You must also include an argument for the non-isomorphism (one has 3-cycle, one is not bipartite, etc.)
Problem 6
Let t be the number of leaves. Let d_i be the degree of vertex v_i and suppose d_1 >=k. By the handshake theorem, 2n-2 = d_1 + d_2 + ... + d_n >=k + t + 2(n-1-t), since the remaining n-1-t vertices have degree >=2. Solving this inequality gives t >= k, as required.
Alternatively, the Prufer code of a tree with a vertex of degree at least k can have at most (n-2-(k-1)+1) = n-k different values appearing. That leaves at least k values not appearing, which are leaves.
Most people use some version of the following reasoning. If v has degree at least k, then it has k branches, each of which must have at least one leaf. However, it requires some careful wording to make this correct, and I usually deducted 1 or 2 points, not because the basic idea is incorrect, but because the exposition was poor.
One correct version of this would go as follows. Suppose v has degree m (at least k). The vertex v must be a cut vertex and removing it breaks the tree into m components, each of which is a tree containing exactly one neighbor of v. This is because if two neighbors were in the same component, there would be two paths between them, one through v and one within the component. Now either the component containing the neighbor w consists of w (in which case w is a leaf) or it is a tree with 2 or more vertices, in which case it has at least 2 leaves. One might be w, but at least one other will be a leaf of the original tree. In either case, each component contributes at least one leaf to the tree.
Problem 7
Since T is a tournament, either u beats v or v beats u. WLOG, assume u beats v. Then d(u,v)=1. If d(v,u)=1, then v beats u, which cannot happen in a tournament.
Please note that this proof is one line, not 3 pages.
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