Math 575 Problem Set 12

Transcription

1 Math 575 Problem Set Can you place a pencil on some point of the figure below and trace over all the lines without any repetition of a line and get back to where you started? Solution. No, this is not a Eulerian graph since it has vertices of degree Determine the closure of the graph G below. Solution. The red edges below show the edges that must be included in the closure of G. 3. (a). Show that the graph is not Hamiltonian. Hint: Can you locate some edges that must belong to every Hamiltonian cycle? Solution. All three edges of the vertex in the center would have to be on any Hamiltonian cycle since each of its neighbors is of degree 2.

2 (b). Show that the graph below is not Hamiltonian. Solution.The edges marked in green below must belong to any Hamiltonian cycle, and that forces the x ed out edge not to be any such cycle. But that in turn forces the two side edges to be in the cycle and so we get a 6-cycle which can not be part of a Hamiltonian cycle. 4. Is the Petersen graph Hamiltonian? Is the Petersen graph traceable? Answer: No it is not Hamiltonian can you prove it? This takes some work. The Petersen graph is traceable and spanning paths are abundant. 5. A graph G is Hamiltonian-connected if every two distinct vertices are joined by a Hamiltonian path. Prove: Let G be a graph on n vertices and suppose that for every two non-adjacent vertices v and u, deg(v) + deg(u) n + 1. Then G is Hamiltonian-connected. Hint: Let v and u be any two vertices in G. Form a new graph G * by adding a new vertex w to G with w adjacent to precisely v and u. Now, what is the closure of G *? Solution. Let v and u be any two vertices in G. We will be finished if we can show that there is a Hamiltonian path from v to u. Form a new graph G * by adding a new vertex w to G with w adjacent to precisely v and u. Then in G *, every two non-adjacent vertices x, y in G satisfy, deg G * x + deg G * y n + 1 = V(G * ), and so all the edges between vertices of G belong to the closure of G *. But after these edges are added to the closure of G *, we see that for any vertex x in G, deg G * x + deg G * w = (n 1) + 2 = n + 1, and so all edges from w to the vertices of G are also included in the closure of G *. Thus the closure of G * is complete and it follows that G * must have a Hamiltonian cycle, call it C. But since w is of degree 2, both of the edges xv and xu are in C. So it follows that C - w is a Hamiltonian path from v to u and the result follows.

3 6. (a). Is this graph Hamiltonian? Solution. No it is bipartite on an odd number of vertices. (b). Consider the checkerboard array below (with two missing squares) to be a graph where the squares are the vertices and two squares are adjacent if they share a common edge. Is this graph Hamiltonian? Solution. No. While this is bipartite on an even number of vertices, the number of vertices in the sets making up the bipartition are different; i.e., color the squares using two colors and you will get 30 of one color and 32 of the other color. 7. Find an example of a graph that is Eulerian but not Hamiltonian, and also find an example of a graph that is Hamiltonian but not Eulerian. Hint: There are lots and lots of examples of each. Solution. The graph on the left below is Eulerian but not Hamiltonian and the graph on the right is Hamiltonian but not Eulerian.

4 8. Prove: If G is a graph on n vertices in which every pair of non-adjacent vertices v and u satisfy, deg(v) + deg(u) n 1, then G contains a Hamiltonian Path (i.e., G is traceable). Hint: Form a new graph H by adding a new vertex to G that is adjacent to every vertex of G. Now apply a theorem from class to H. Solution. [This is much like the solution to problem 5.] Form a new graph H by adding a new vertex to G that is adjacent to every vertex of G. Then in H, every two non-adjacent vertices x, y in G satisfy, deg H x + deg H y = deg G x + 1 ( ) + ( deg G y + 1) n 1+ 2 = n + 1 = V(H ), and so all the edges between vertices of G belong to the closure of H. Thus, the closure of H is complete and it follows that H must have a Hamiltonian cycle, call it C. So it follows that C - w is a Hamiltonian path in G and so G is traceable. 9. Show that the graph below is not Hamiltonian by using Grinberg s Theorem. [Each region is labeled with the number of edges in its boundary.] Solution. Suppose that this plane graph is Hamiltonian. Then Grinberg s equation would apply to any Hamiltonian cycle, and for this graph the equation is simply, Δf 4 = 0. But for that to happen there must be an even number of 4-regions. However, this graph has nine 4-regions and thus it cannot be Hamiltonian.

5 10.Show that the graph below is not Hamiltonian by applying Grinberg s Theorem. [Each region is labeled with the number of edges in its boundary.] Solution. Suppose that this graph has a Hamiltonian cycle. Then for that cycle, Grinberg s equation 3Δf 5 + 6Δf 8 + 7Δf 9 = 0 would apply. However, the only 9-region is the exterior region and so, Δf 7 = 1. Thus the equation reduces to 3Δf 5 + 6Δf 8 = 7, which is impossible since the left hand side is a multiple of 3 and the right hand side is not a multiple of 3.