Combinatorics and Graph Theory 2 Midterm, scoring guide.

Combinatorics and Graph Theory 2 Midterm, scoring guide.

1. There are 55 dierent discs on the plane. Any two of them are either disjoint or one of them contains the other one. We know that no point is contained in 7 discs. Prove that there are 10 pairwise disjoint discs.

Solution: We dene a poset (H,≤H), where H is the set of discs and A ≤H B if and only if eitherB contains A or A=B. (1 point)

≤_H is a partial orderring, because

ˆ reexive: A≤H B by def (1 point)

ˆ anstisymmertic: A ≤H B and B ≤H B can happen if and only A = B. It is not possible that A containsB and B contains A if they are dierent discs. (1 point)

ˆ transitive: A ≤_H B and B ≤_H C in the general case means that B contains A and C contains B, therefore C contains A so A ≤_h C. If either A =B or B = C, then we trivially getA≤_H C. (1 point)

No point is contained in 7 discs means that there is no chain of size 7. (1 point) So the maximum chain has at most 6 elements. (1 point)

By Dilworth's theorem H can be partitioned into 6 antichains. (2 points)

Therefore there is an antichain which contains at least d⁵⁵₆e= 10 elements. (1 points) The elements of this antichain are pairwise disjoint discs, so we are done. (1 point) 2. Gis a plane graph and it has 50 faces. The dual graphG^∗ is simple. Show thatGhas at

most 144 edges.

Solution

G^∗ is also a plane graph. (2 point)

Since G^∗ is planar and simple, e^∗ ≤3n^∗−6. (2 points)

Each face of G correspond to a vertex ofG^∗, therefore n^∗ =f = 50. (2 points) Soe^∗ ≤3∗50−6 = 144(1 point)

Each edge of Gcorrespond to an edge of G^∗, therefore e=e^∗. (2 points) Soe^∗ ≤144. (1 point)

3. Determine the list chromatic number (ch(G)) of the following graph:

2 1

4 5

6 3

Solution:

χ(G)≤ch(G)(1 point)

χ(G)≥3 because G contains a clique of size 3, for example: 2,3,4. (3 points) Thereforech(G)≥3. (1 point)

To prove that ch(G) ≤ 3 we show that G can be colored from any lists of length 3. (2 points)

If we move throug the vertices from 1 to 6 and pick a color from its list, then at each step, at most two neighbors have already received a color, so one is always available from the list. (2 points)

So we can color this graph from any list of length 3. (1 point) 3≤ch(G)≤3, therefore ch(G) = 3.

4. Show that the Ramsey number R(16,17) ≥241. Solution I.

R(16,17)≥R(16,16) (3 points)

Because if we have an R(16,17) vertex complete graph, then it does not matter how we color the edges by red end blue, there is always either a red clique of size 16, or a blue clique of size 17. In the later case, we also have a blue clique of size 16. (2 points) R(k, k)≥2^k/2 (3 points)

Therefore:

R(16,17)≥R(16,16)≥2⁸ = 256≥241 (2 points) Solution II.

To show that R(16,17) ≥ 241 it is enough to give a 240 vertex graph which neither contain a clique of size 16 nor an independent set of size 17. (Or we can talk about an edge coloring of K₂₄₀ by two colors.) (3 points)

Let graph G be 16 disjoint copies ofK₁₅. (3 points).

G does not contain a clique of size 16, because each connected component contains at most 15 vertices. (1 point)

Gdoes not contain an independent set of size 17, because an independent set can contain at most one vertex from a clique and the vertex set is covered by 16 cliques. (2 points).

G contains 16*15=240 vertices. (1 point) So this Gis a good example.

5. In a football legaue any two teams play with each other at most once. We know that there are 12 teams in the league and 29 matches have been played. Show that there is a group of three teams, such that they have not played with each other. (If A, B and C are these teams then the following matches have not been played: A-B, A-C, B-C.) Solution

LetG be a graph whose vertices are the teams, and two vertices are adjacent if and only if the two teams have played with each other. (1 point)

We need to show that there is an independent set of size 3 in G. (1 point)

An independent set of size 3 in G correspond to a clique of size 3 inG, the complement of G. (1 point)

G has 12 vertices. (1 point)

G has ^12∗11₂ −29 = 66−29 = 37edges. (2 points)

The 2-partite Turan graph T(12,2) has 6∗6/2 = 36 edges. (2 points)

So by Turán's theorem, any 12 vertex graph which has 37 edges contain a K₃. (1 point) ThereforeG contains a clique of size 3, soGhas an independent set of size 3, the corres- ponding three teams have not played with each other. (1 point)

6. There is a 100 element set. We choose some of its 40 and 70 element subsets in such a way that any two choosen subsets intersect each other. At most how many such subsets can be choosen?

Solution:

Any two 70 element subsets of a 100 element set intersect each other. (1 point) A 70 element subset intersect any 40 element subset. (1 point)

Therefore we can take all the 70 element subsets and an intersecting family of the 40 element subsets. (2 points)

40≤100/2, (1 point) Therefore the maximum size of an intersecting family containing 40 element subsets is ⁹⁹₃₉

by Erd®s-Ko-Rado Theorem. And this can be attained. (2 points) The number of 70 element subsets is ¹⁰⁰₇₀

. (1 point) So the nal answer is ¹⁰⁰₇₀

+ ⁹⁹₃₉

. (2 points)