Mean = 75.3
Median = 75
A: 88-100 (4)
B: 75-87 (8)
C: 68-74 (4)
D: 55-67 (8)
F: 0-54 (0)
Problem 1
One maximum match is {1-B, 2-C, 4-E}. This is maximum because {4, B, C} is a vertex cover and the size of any match is less than or equal to the size of any vertex cover.
I deducted 2 points if just the match and the cover were given, but no reason why the match is maximum. I allowed reference to Konig-Egervary, though that is technically not correct.
Problem 2
No 1-factor exists. Let S = {1, 3, 5, 9}. Then G - S consists of the isolated vertices {2, 4, 6, 7, 8, 10}, so o(G - S)=6 and |S|=4, violating Tutte's condition.
Problem 3
{e, f, h} is a (u, v) vertex cut and the paths u-a-e-i-v, u-b-f-g-j-v, and u-d-h-l-v are pairwise vertex disjoint, so, according to Menger's Theorem, kappa(u,v)=3.
{a-e, b-e, f-g, h-k, h-l} is a (u,v) edge cut and the paths u-a-e-i-v, u-b-e-j-v, u-b-f-g-j-v, u-c-h-k-v and u-d-h-l-v (with both u-b edges and both j-v edges being used) are pairwise edge disjoint. So, according to Menger's Theorem (edge version), kappa'(u,v)=5.
Some omitted the cut sets. Producing the disjoint paths only shows lambda(u,v)>=k, not lambda(u,v)=k. To get equality, you must produce a separating set of the same size as your set of paths. I deducted 4 points for this error.
Problem 4
(a) False. For example, the 4-cycle has kappa(G)=2, but it is not 3-connected.
(b) False. For example, the "bow tie" is 2-edge-connected, but it is not 2-connected since it has a cut vertex.
(c) False. An example is a 3-regular graph with a cut-vertex (see Exercise 3.3.8). Such a graph is 1-connected but not 2-connected or 3-connected.
(d) True. If the graph is 2-connected, it is 2-edge-connected. Therefore it has no cut edge, so this follows from Corollary 3.3.8.
Problem 5
rE, sC, tB, uA, vD.
Problem 6
Let S be a maximum independent set. For each vertex v in S, form the "ball" around v, B(v), consisting of v and all its neighbors and let T be the sum over all of S of |B(v)|. The union of B(v) over all vertices in S must be all of V(G), for if u is a vertex not in this union, then S+u would be a larger independent set. Thus, T >= n(G). But |B(v)| <= Delta(G)+1, so T <= alpha(G)(Delta(G)+1).
Problem 7
Let T be the block-cutpoint graph of G, which is a tree. Let m = the number of cut-vertices of G and p = the number of pairs (v, B) where v is a cut-vertex on block B. Then the number of vertices in T is m + k and the number of edges in T is p. Since T is a tree, p = m + k - 1.
The sum over all vertices of b(v) will count pairs (v, B) such that v is in block B. But this is also just the sum of the orders of the blocks, which will count non-cut-vertices exactly once and cut vertex v will be counted once for each block containing v. Thus, this sum is n(G) - m + p, which is n(G) + k - 1 from above.
This can also be done by induction on k, the number of blocks. It is clearly true if k = 1. Suppose it is true for all graphs with fewer than k blocks. Let G have k blocks. Pick a block which has only one cut vertex (why does such a block exist?). Removal of this block (except the cut-vertex) will leave a graph with one fewer blocks. Now apply induction to that graph and add the vertices in the removed block (less 1 for the cut-vertex) to give the formula for G.
Many argued that the (-1) in the sum compensated for the possible overcount in the b(v), but this was not made precise. You must use the tree structure of the block-cutpoint graph in some way. Depending on how precise the argument was, I typically gave 4-8 points for such arguments.
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