COMPSCI 575/MATH 513: Combinatorics and Graph Theory

Solutions to Practice First Midterm Exam, Fall 2016

David Mix Barrington

6 October 2016

Directions:

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

Question text is in black, solutions in blue.

  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

For Questions 1-4, the graph H has node set {a, b, c, d, e} and nine edges, consisting of all possible edges except (c, e).

• Question 1 (10): Are all graphs with five nodes and nine edges isomorphic to H? Justify your answer.

  They are. Such a graph has all but one possible edge, and so has three vertices of degree 4 and two of degree 3. Any bijection of the nodes that preserves degree is an isomorphism, since each degree-4 node has an edge to every other node, and each degree-3 node has an edge to every other node except the other degree-3 node.

• 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?

  The graph H is planar. One possible embedding has nodes a, b, c, d, and e arranged in a circle in that order, with chords (a, c) and (a, d) inside the circle and chords (b, d) and (b, e) outside it. Since there are five nodes and nine edges, the number of regions is 9 - 5 + 2 = 6. The drawing described above has three regions inside the circle and three outside.

• 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 there are exactly two odd-degree vertices, H has an Euler path from c to e or vice versa, but no Euler circuit and no Euler path with any other endpoints. The cycle a-b-c-d-e-a is a Hamilton cycle, and the path a-b-c-d-e is a Hamilton path -- there are many others.

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

  H has a four-coloring by making a red, b blue, c and e yellow, and d green. It has no 3-coloring because the subgraph obtained by deleting node c (or by deleting node e) is a K₄ and requires four colors in itself. (That is, nodes a, b, d, and e must necessarily use four different colors.) Hence the chromatic number is 4.

• Question 5 (15): Prove that if G is any graph, either G is connected or G's complement G^C is connected. (G^C is a graph with the same vertices as G and exactly the edges that are not in G.) (Hint: Use induction on n, the number of vertices, and consider the possible cases.)

  Induction on n. Base case, n = 1, and both G and its complement are connected. Assume that all graphs of n nodes have the property that either they or their complements are connected. Let G be an arbitrary graph with n+1 nodes and let x be an arbitrary node in G. Let H be the graph obtained from G by deleting node x. By the IH, either H or its complement is connected.

  If H is connected, consider whether x has degree 0 in G. If it doesn't, G is connected because any node in H now also has a path to x. If it does, every node of H has an edge to x in G^C, so every node has a path in G^C to every other node going through x, and G^C is connected.

  If H^C is connected, consider whether x has degree n in G. If it does, G is connected because every node in G has a path to every other node through x. If it doesn't, then x has an edge into H in G^C, and thus x has a path to every other vertex in G^C and G^C is connected.

• Question 6 (15): Let K be a weighted (undirected) complete graph with n vertices, where every edge weight is either 1 or 2. Let K' be the ordinary graph with the same vertices as K and an edge wherever the corresponding edge of K has weight 1. Suppose that K' has a Hamilton circuit. What is the weight of a minimum spanning tree of K? Prove your answer (that is, prove both that it is not too high and not too low).

  If we delete any single edge from the Hamilton circuit of K', we get a path of weight-1 edges in K that includes all the nodes, and this path is a spanning tree for K of total weight n-1. Any spanning tree for K would have to include n-1 edges, and each edge has weight at least 1, so there can be no spanning tree of total weight less than 1 and the tree we have constructed is an MST.

• Question 7 (15): Suppose we had an approximation algorithm for TSP that, on graphs with n nodes, always found a Hamilton circuit whose weight was at most n times the weight of the optimal circuit. Indicate how we could use this algorithm to solve the NP-complete Hamilton circuit problem. (Hint: Your argument should involve weighted graphs that do not obey the triangle inequality.)

  Let G be an arbitrary (undirected) graph with N nodes. Let H be a weighted complete graph on the same n nodes where edges corresponding to edges of G get weight 1 and other edges get weight n³. (Note that this graph violates the triangle inequality if, say, (a, b) and (a, c) are edges of G but (b, c) is not.)

  If G has a Hamilton circuit, then H has a Hamiton circuit of total weight n, consisting of the corresponding edges. If G has no Hamilton circuit, then any Hamilton circuit in H contains at least one edge not in G, and its total weight must be at least n³.

  So to test whether G has a Hamilton circuit, we run the given approximation algorithm on H and see whether the result has total weight less than n². If it does, this can only come from a Hamilton circuit in G, and if it doesn't, no Hamilton circuit in G can exist.

• Question 8 (15): Suppose we have a network with one or more sources, one or more sinks, positive integer capacities on the directed edges, and intermediate nodes with zero net flow. The value of a flow in such a network is the sum of the flows out of the sources. Describe how we can use the results about ordinary networks (with one source and one sink) to find the maximum possible flow in this new network.

  Given such a network N, we make a network N' with the nodes and edges of N plus two more nodes, a source a and a sink z, and new edges of very high capacity from a to each source of N and from each sink of N to z. Now any flow in N' must correspond to a flow in N of the same value, because the flow out of a must correspond to flows out of the sources of N that add up to the value of the flow in N', and similarly for the sinks.

Last modified 6 October 2016