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Math 462 HW 7 Solutions
Problem 1: Let {an}∞ n=0 be the sequence such that when the complete graph Kan is edge colored with n + 2 colors there always exists a monochromatic triangle. Find a recurrence relation for an and solve the recurrence to get a closed form for an.
Proof: We first claim we have the recurrence relation
an = (n + 2)(an− 1 − 1) + 2
with base cases a 0 = 6, a 1 = 17. The base cases come from a problem on the last midterm. So let G be the complete graph Kan and color the edges with n + 2 colors. Then for any vertex v of G, the degree of v is an − 1, but by the Pigeonhole Principle there must be at least an− 1 edges starting at v colored the same color. Let’s assume this color is red. If any of these vertices are connected by a red edge, then we must have a red monochromatic triangle. If not, then the induced complete subgraph on these an− 1 vertices is colored with n + 1 colors, which by the definition of an− 1 must have a monochromatic triangle. Writing out the first few terms of this recurrence we might guess the closed form
f (n) = (n + 2)!
n∑+
m=
m!
Then f (0) = 6 and f (1) = 17 as desired. We then show that this closed form satisfies the same recurrence relation and hence an = f (n). We have
(n + 2)(f (n − 1) − 1) + 2 = (n + 2)
(n + 1)!
n∑+
m=
m!
= (n + 2)!
n∑+
m=
m!
(n + 2)! (n + 2)!
= (n + 2)!
n∑+
m=
m!
Note: For this problem it seemed a little bit easier to just guess the closed form for the recurrence relation and prove it satisfied the same recurrence and initial condi- tions than to solve the recurrence explicitly using exponential generating functions. Of course, both methods are equally valid though.
Problem 2: Prove that if we color the edges of K 6 with two colors, there will al- ways be at least two monochromatic colors.
Proof: We have already seen that K 6 must have at least one monochromatic tri- angle. Suppose this triangle is on the vertices a 1 , a 2 , a 3 and is colored red. The remaining three vertices b 1 , b 2 , b 3 must have a total of 9 edges connecting to the monochromatic triangle.
a 1 • • a 2
??? ??? ??
At most three of these edges can be red however, or else two red edges must connect to the same bi, resulting in a second red triangle. So at least 6 of these 9 edges must be colored blue. In particular, each bi must have at least 2 blue edges connecting it to the aj ’s. However, the only way to avoid a blue triangle amongst the bi is to color all the 3 edges connecting them to each other all red. Thus we have a second monochromatic triangle.
Problem 3: There are n guests at a house party. Assume the following hold.
- In any group of 3 people, there are two people who do not know each other.
- In any group of 7 people, there are two people who know each other.
At some point a person looks around and claims she knows 20 of the people in the room, and doesn’t know the other 9. Prove that she must be mistaken.
Proof: Among the 20 people that she does know, no two of them can know each other. This is because of the first condition, where we take the group of three to be the woman and the two others. But then if we take any 7 of the people that she does know, since none of them know each other, we have a contradiction to the second assumption.
