(a) Prove that if G has an Eulerian circuit then L(G) has a hamiltonian circuit.
Consecutive edges of the eulerian circuit in G correspond to adjacent vertices in L(G). Since every edge of G appears once, and only once in the eulerian circuit, every vertex will appear once and only once in a a circuit in L(G), thus L(G) has a hamiltonian circuit.
(b) The complete bipartite graph Kn,m is not regular if n does not equal m, but prove that the line graph L( Kn,m ) is regular.
The degree of a vertex in L( Kn,m ) is the number of edges in Kn,m that meet the edge corresponding to this vertex. Every edge of Kn,m has one vertex of degree n and the other vertex of degree m, so the number of edges meeting a given edge is n - 1 + m - 1 = n + m - 2. Since this number does not depend upon the edge chosen, every vertex of L( Kn,m ) has the same degree, i.e., L( Kn,m ) is regular.
2. Consider a job which can be broken up into 15 tasks, some of which can be carried out
only after others have been finished, as indicated below:
|1 <-- 2, 7, 13||2 <-- 3, 8, 14||3 <-- 9, 15||4 <-- 3|
|5 <-- 4, 11||6 <-- 5, 12||7 <-- 6||8 <-- 7, 9, 14|
|9 <-- 15||10 <-- 4, 9||11 <-- 10||12 <-- 11|
|13 <-- 7, 12||14 <-- 13, 15|
To answer the question, construct a digraph with vertices representing the tasks and arcs (i, j) meaning that task i must be performed before task j. We wish to determine if a topological order exists for this digraph. If none exists it will not be possible to complete the job. If one does exist, then this gives the order of the tasks. For the digraph constructed for this problem, a topological order does exist, and the tasks should be performed in the following order: 1, 2, 8, 14, 13, 7, 6, 5, 12, 11, 10, 4, 3, 9, 15.
3. Prove that a graph having a unique spanning tree is a tree.
We will prove the contrapositive of this statement. Suppose that G is not a tree. Either G is disconnected or G is connected but contains circuits. If G is disconnected, then it has no spanning tree, hence it does not have a unique one. If G contains a circuit, then a spanning tree for G can not contain all the edges of this circuit. By putting a missing edge in and removing an edge used in the spanning tree that is in this circuit, we can form a second spanning tree for G. Thus, G would have more than one spanning tree. So, if G is not a tree then G does not have a unique spanning tree.
4. Petersen's graph (below) does not have a hamiltonian circuit. What is the minimum number of edges that you must add to it (and where are they to be added) to obtain a graph with a hamiltonian circuit?
One edge must be added, and it can be added between any two non-adjacent vertices. An example is given below:
5. The diagram below represents a floor plan with the doors between the rooms and the outside indicated. The real estate agent would like to be able to tour the house, starting and ending outside, by going through each door exactly once. What is the fewest number of doors that should be added, and where should they be placed in order to make this tour possible? Give reasons for your answer.
Letting the rooms be vertices (with the outside as a vertex also) and the doors be edges between these vertices, the tour corresponds to an eulerian circuit in this graph. This is possible if and only if every vertex has even degree. Since rooms B, C, D and the outside have odd degree, the tour is not possible. We would have to add at least two edges between these vertices (i.e., at least 2 new doors) to make it possible. Since we can not join C to D (there is no common wall to put a door in), we could join C to B and D to the outside. Thus, it is possible to take the tour with the addition of two doors, a second one between rooms C and B and a second door from D to the outside.
6. In the graph below, the marked edges form a matching. Show that this is a maximum matching, or find a maximum matching if it is not.
This is not a maximum matching since there exist augmenting chains. The matching obtained after switching the edges in the augmenting chain is a maximum matching since there is only one unsaturated vertex in the graph, so no further edges can be added to the matching.