COMPSCI 575/MATH 513: Combinatorics and Graph Theory

Solutions to First Midterm Exam, Fall 2016

David Mix Barrington

22 October 2016

Directions:

• Answer the problems on the exam pages.
• There are eight problems (some with multiple parts) for 100 total points. Actual scale was A = 93, C = 63.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.

  Q1: 10 points
  Q2: 10 points
  Q3: 10 points
  Q4: 10 points
  Q5: 15 points
  Q6: 15 points
  Q7: 15 points
  Q8: 15 points
 Total: 100 points

Question text is in black, solutions in blue.

For Questions 1-4, the graph H has node set {a, b, c, d, e, f} and twelve edges: (a, b), (a, d), (a, e), (a, f), (b, c), (b, d), (b, e), (c, d), (c, e), (c, f), (d, f), and (e, f).

• Question 1 (10): Are all graphs with six nodes and twelve edges isomorphic to H? Prove your answer.

  They are not. H has degree sequence {4, 4, 4, 4, 4, 4). Let H' be the graph with the same six vertices and all edges except (a, b), (a, c), and (b, c). Since it has degree sequence (3, 3, 3, 5, 5, 5), and isomorphic graphs have identical degree sequences, H and H' are not isomorphic even though they each have six nodes and twelve edges.

• Question 2 (10): Is H a planar graph? Prove your answer. What is the number of regions in a planar embedding of H, given by Euler's Formula?

  H is planar. If we draw the Hamilton cycle a-b-d-c-f-e-a as a circle in the plane, we can place the edges (a, d), (d, f), and (f, a) outside the circle and the edges (b, c), (c, e), and (e, b) inside it. Euler's Formula r = e - v + 2 gives 12 - 6 + 2 = 8 regions. In the planar embedding above there are four regions inside the circle and four outside it, the latter including the region outside the entire drawing. (We could also put this graph on a sphere, with the nodes and edges of a regular octohedron.)

• Question 3 (10): Does H have an Euler path and/or Euler circuit? Does H have a Hamilton path and/or Hamilton circuit? Justify your answers.

  Since all vertices have degree 4, H has an Euler circuit but no Euler path (an Euler path is defined to not be an Euler circuit). There are several Hamilton circuits such as the one referred to in the solution to Question 2 above. If we delete any one edge from any of these circuits, we get a Hamilton path.

• Question 4 (10): What is the chromatic number of H? (This is the "vertex-chromatic" rather than the "edge-chromatic" number.) Prove your answer.

  H's chromatic number is 3. We get a 3-coloring by making a and c red, b and f blue, and d and e green. There cannot be a 2-coloring because H contains a number of triangles, including a-b-d.

• Question 5 (15): Here are two questions about "graph-theoretic" trees, by which we mean connected acyclic (undirected) graphs.
  • (a, 5) Using the average degree of a node, argue that every tree has at least one leaf. (This should have said "every tree with n ≥ 2 nodes".)

    We know that a tree of n nodes has n-1 edges, and so the average degree of its nodes is 2(n-1)/n = 2 - 2/n < 2. Thus the tree must contain at least one node of degree less than 2. If the tree has more than one node, no node has degree 0 because the tree is connected. So the node of degree less than 2 must have degree 1 and be a leaf. (I should have ruled out the case of a 1-node tree, but there we can consider the single node of degree 0 to be a leaf.)

  • (b, 10) Prove by induction that for all integers n with n ≥ 2, every tree with n nodes has at least two leaves.

    We use induction on n, starting with n = 2. In the base case, both nodes have degree 1 and are thus leaves. Assume that every tree of n nodes has at least two leaves. Let G be an arbitrary tree of n+1 nodes. By part (a), G has a leaf, so let G' be the graph obtained by deleting this leaf (and its edge) from G. G' is still connected and cannot have a cycle, so it is a tree and by the IH has at least two leaves. At most one of these two leaves can be adjacent to the node we removed, so the other one is still a leaf and, with the leaf we removed, gives us at least two leaves in G.

    Note: Many people assumed that the tree of n+1 nodes had to be obtained by adding a node (or even adding a leaf) to a tree of n nodes. This is not immediate -- we justify it above only by appealing to part (a) to say that an arbitrary tree of n+1 nodes must be obtainable in this way.

• Question 6 (15): Let J be the following 16-node weighted (undirected) graph. The nodes are pairs (i, j) where 0 ≤ i, j ≤ 3. There is an edge from (i, j) to (i', j') if and only if i = i' and |j=j'| = 1 or |i-i'| = 1 and j = j'. (The nodes thus form a Manhattan-style grid enclosing nine blocks.) The weight of the edge between (i, j) and (i', j'), where i > i' or j *gt; j', is 1 if i+j is odd and 2 if i+j is even.
  • (a, 10) Determine the weight of a minimum spanning tree of J.

    If we run Kruskal's Algorithm, it will take all 12 weight-1 edges, since there is no cycle formed by these edges alone. This leaves us with four connected components, and the algorithm will select three more edges to complete an MST of weight 18.

    An alternate argument is as follows: Any spanning tree must contain exactly fifteen edges, so the minimum possible weight would be all 12 1-edges and 3 2-edges for total weight 18. Thus if we exhibit any spanning tree of this weight, we know it is minimal. I did insist that you justify the claim that the tree you found was minimal. If you said that you were using either the Kruskal or Prim algorithms, this was fine, because it was proved in lecture and in the text that these algorithms always find an MST.

  • (b, 5) Generalize this graph to a graph J_n, where the graph J above is J₃, so that J_n has (n+1)² nodes. Find and justify a formula for the weight of an MST for J_n.

    Again, Kruskal takes the n(n+1) 1-edges, since these do not form a cycle. These edges form a graph with n+1 connected components, so exactly n 2-edges are needed to form a spanning tree and Kruskal will find such a set. The total weight will be n(n+1) + 2n = n² + 3n.

• Question 7 (15): Consider a weighted bipartite (undirected) graph where the nodes are divided into two sets A and B, each of size n. We asserted but did not prove in lecture that there is an algorithm with time polynomial in n, that inputs such a graph a returns a matching of minimum possible weight. (A matching is a subset S of the edges such that each node is an endpoint of exactly one edge in S.)

  Suppose we instead wanted a matching with the maximum possible total weight. Indicate how we can use the given algorithm for minimum-weight matching to solve this problem in polynomial time.

  Let G be the weighted graph for which we want the maximum matching, and M be the largest weight of any of the edges of G. Define G' to be a weighted graph with the same edges as G, and edges given by taking each edge of weight w in G and replacing it with an edge of weight 2M - w.

  Use the given algorithm to find a minimum-weight matching in G'. Let the edges of this matching be e₁,..., e_n with weights w₁,..., w_n. The weight of this matching in G is 2Mm - (sum of w_i's). This is at least as great as the weight of any other matching in G, because such a matching of total weight t would have weight 2Mn - t in G', and this is greater than the sum of the w_i's because of the minimality of the matching chosen by the algorithm.

  There were a number of interesting wrong answers to this one:

  • Replacing weight w with weight -w rather than 2M - w, which doesn't work because the algorithm expects positive weights. (I didn't mind using weight M - w which gives 0 weight at least once.)

  • Making assumptions about how the minimum-weight algorithm works. You are only told to assume that it exists, which means you have no idea how it treats weights of individual edges.

  • Using a non-linear mapping of weights in G to weights in G', which does not guarantee that the maximum matching in G will be the minimum one in G'.

  • Many people gave the same algorithm of finding the minimum matching in G, deleting its edges, finding the minimum matching in the remaining edges, deleting those edges, and so on until (allegedly) only the edges of the maximum matching are left. This can't work in general, because there is no reason that a single edge can't occur in both the minimum and the maximum matching.

• Question 8 (15): Let G be a network with positive edge capacities, a source a, and a sink z. Assume that G has no edge from a to z. For any positive number c, let G_c be the network obtained from G by adding a new edge from a to z with capacity c.
  • (a, 10) State and prove a result relating the maximum possible flow in G to that in G_c.

    The result is that the maximum flow in G_c is exactly c greater than the maximum flow in G. There are two directions in the proof of this claim. First, given a maximum flow of value f in G, we can add a flow of c on the new edge and get a flow of f + c in G_c. Secondly, the maximum flow in G_c must have the new edge saturated, because otherwise it could be increased by increasing the flow on that edge. So it cannot be more than f + c, because removing the flow of c on the new edge gives us a valid flow in G and f is the maximum flow there.

    An equally good argument uses the min-cut max-flow theorem. Changing G to G_c adds exactly c to every cut in G, so it adds exactly c to the minimum cut, and thus adds exactly c to the maximum flow.

  • (b, 5) Explain why, given the result of part (a), we may assume in our analysis of maximum flows that there is no edge from a to z.

    Given any network with an edge of capacity c from a to z, we can delete the edge, analyze the resulting network, and increase the maximum flow we find by c to get the flow in the original network. So it suffices to be able to analyze networks with no edge from a to z.

Last modified 22 October 2016