a) Draw a graph that illustrates which exams have students in common with other exams.
b) What is the minimum number of time slots that will be needed to schedule the six exams? Which exams should be given in each of these time slots?
Solution: 3 time slots. The minimum number of time slots corresponds to the chromatic number of the above graph. Since the graph contains a triangle (Math, Reading, Science is one), no less than three colors will do. The coloring given above shows that it can be done in three colors, so the chromatic number of the graph is 3. In the "black" time slot, Math and Art can be given, in the "red" time slot, Reading and History and in the "blue" time slot, French and Science.
2. Is the graph drawn below planar? If it is redraw it so that no edges cross at a non-vertex, if it isn't explain why not.
Solution: The graph is not planar. It contains a K3,3 as a subgraph (shown below) and by Kuratowski's theorem, it can not be planar.
3. There are 3 women and 5 men who will split up into two 4-person teams. How many ways are there to do this so that there is (at least) one woman on each team?
Solution: The total number of ways to split the 8 people into two 4-person teams is C(8,4)/2 = 35. We must divide by two here because as soon as one 4-person team is chosen, the second team is determined (the complement). C(8,4) counts both a set and its complement, since they both have size 4, but these two choices lead to the same split into two teams. If we count the splits that have no women on one team, we get 5 (all the women are on the other team - no choices, plus one man - 5 choices). Thus, the number of splits with at least one woman on each team is 35 - 5 = 30.
4. A graph is critical if the removal of any one of its vertices (and the edges adjacent to that vertex) results in a graph with a lower chromatic number.
a) Show that Kn is critical for all n > 1.
Solution: The chromatic number of Kn is n. Removing one vertex and all of its edges from a Kn, gives a Kn-1, with chromatic number n-1. Since it doesn't matter which vertex is removed, the chromatic number will always go down and so, Kn is critical.
b) Show that Cn (the circuit graph with n vertices) is critical if and only if n is odd.
Solution: The chromatic number of Cn for odd n is 3 (for even n the chromatic number is 2). Removing a vertex and its two edges in any Cn will give a chain. Since any chain has chromatic number 2, the chromatic number will go down if and only if we start with a Cn with odd n.
5. Show that if a graph is not connected then its complement is connected. The complement of a graph is a graph on the same set of vertices, but in which two vertices are adjacent if and only if they are not adjacent in the original graph.
Solution: Let G be the non-connected graph and CG be its complement. Consider two vertices x and y in CG. If x and y are not adjacent in G, then they are adjacent in CG, so there is a chain (of length 1) joining them in CG. If x and y are adjacent in G, then they must be in the same connected component of G. Since G is not connected, there must be a vertex, say z, which is not in the same component as x and y. Now, x and z are not adjacent in G, so there is an edge between them in CG. Also, y and z are not adjacent in G, so there is an edge between them in CG. But then, there is an edge from x to z and another from z to y in CG, so x and y are joined by a chain of length 2. Thus, any pair of vertices is joined by a chain in CG, so CG is connected.