1.8 Ramsey Theory
1.8.2 Exact Ramsey Numbers and Bounds
Take me to your leader.
— proverbial alien How many people are required at a gathering so that there must exist either three mutual acquaintancesorthree mutual strangers? We can rephrase this question as a problem in Ramsey theory: How many vertices do you need in an (edge) 2- colored complete graph for it to be necessary that there be either a redK3(people who know each other) or a blueK3(people who do not know each other)? As the next theorem states, the answer is 6.
Theorem 1.61. R(3,3) = 6.
Proof. We begin the proof by exhibiting (in Figure 1.122) a 2-coloring of the edges ofK5that produces neither a red (thick) K3 nor a blue (thin)K3. This
FIGURE 1.122. A 2-coloring of the edges ofK5.
2-coloring ofK5demonstrates thatR(3,3)>5. Now considerK6, and suppose that each of its edges has been colored red or blue. Letvbe one of the vertices of K6. There are five edges incident withv, and they are each colored red or blue, so it must be thatvis either incident with at least three red edges or at least three blue edges (think about this; it is called the Pigeonhole Principle—more on this in later chapters). Without loss of generality, let us assume thatvis incident with at least three red edges, and let us call themvx,vy, andvz(see Figure 1.123).
FIGURE 1.123.
FIGURE 1.124.
Now, if none of the edgesxy,xz,yzis colored red, then we have a blueK3
(Figure 1.124).
On the other hand, if at least one ofxy,xz,yzis colored red, we have a redK3
(Figure 1.125).
FIGURE 1.125.
Therefore, any 2-coloring of the edges ofK6produces either a redK3or a blue K3.
Let us determine another Ramsey number.
Theorem 1.62. R(3,4) = 9.
Proof. Consider the 2-coloring of the edges ofK8given in Figure 1.126.
A bit of examination reveals that this coloring produces no red (thick)K3and no blue (thin)K4. Thus,R(3,4) ≥ 9. We now want to prove thatR(3,4) ≤ 9, and we will use the facts thatR(2,4) = 4andR(3,3) = 6.
LetGbe any complete graph of order at least 9, and suppose that the edges of Ghave been 2-colored arbitrarily. Letvbe some vertex ofG.
Case 1. Suppose thatv is incident with at least four red edges. Call the end vertices of these edges “red neighbors” ofv, and letSbe the set of red neighbors ofv(see Figure 1.127).
FIGURE 1.126. A2-coloring of the edges inK8.
v
S FIGURE 1.127.
SinceS contains at least four vertices, and sinceR(2,4) = 4, the 2-coloring of the edges that are withinSmust produce either a redK2or a blueK4within S itself. If the former is the case, then we are guaranteed a red K3 inG (see Figure 1.128). If the latter is the case, then we are clearly guaranteed a blueK4
inG.
v
S FIGURE 1.128.
Case 2. Suppose thatv is incident with at least six blue edges. Call the other end vertices of these edges “blue neighbors” ofv, and letT be the set of blue neighbors ofv(see Figure 1.129).
SinceT contains at least six vertices, and sinceR(3,3) = 6, the 2-coloring of the edges that are withinT must produce either a redK3or a blueK3within T itself. If the former is the case, then we are obviously guaranteed a redK3inG.
T v
FIGURE 1.129.
If the latter is the case, then we are guaranteed a blueK4inG(see Figure 1.130).
T v
FIGURE 1.130.
Case 3. Suppose thatvis incident with fewer than four red edges and fewer than six blue edges. In this case there must be at most nine vertices inGaltogether, and since we assumed at the beginning that the order ofGis at least 9, we can say thatGhas order exactly 9. Further, we can say thatvis incident with exactly three red edges and exactly five blue edges. And since the vertexvwas chosen arbitrarily, we can assume that this holds true for every vertex ofG.
Now if we consider the underlying “red” subgraph ofG, we have a graph with nine vertices, each of which has degree 3. But this cannot be, since the number of vertices inGwith odd degree is even (the First Theorem of Graph Theory).
Therefore, this case cannot occur.
We have therefore proved that any 2-coloring of the edges of a complete graph on 9 vertices (or more) produces either a redK3 or a blue K4. Hence, R(3,4) = 9.
Some known Ramsey numbers are listed below.
R(1, k) = 1, R(2, k) =k,
R(3,3) = 6, R(3,4) = 9, R(3,5) = 14, R(3,6) = 18, R(3,7) = 23, R(3,8) = 28, R(3,9) = 36,
R(4,4) = 18, R(4,5) = 25.
Bounds on Ramsey Numbers
Determining exact values of Ramsey numbers is extremely difficult in general. In fact, the list given above is not only a list ofsomeknown Ramsey numbers, it is a list ofall known Ramsey numbers. Many people have attempted to determine other values, but to this day no other numbers are known.
However, there has been progress in finding bounds, and we state some im- portant ones here. The proofs of the first two theorems will be discussed in Sec- tion 2.10.2 (see Theorem 2.28 and Corollary 2.29). The first bound is due to Erd˝os and Szekeres [94], two major players in the development of Ramsey theory. Their result involves a quotient of factorials: Here,n!denotes the product1·2· · ·n.
Theorem 1.63. For positive integerspandq, R(p, q)≤ (p+q−2)!
(p−1)!(q−1)!.
The next theorem gives a bound onR(p, q)based on “previous” Ramsey num- bers.
Theorem 1.64. Ifp≥2andq≥2, then
R(p, q)≤R(p−1, q) +R(p, q−1).
Furthermore, if both terms on the right of this inequality are even, then the in- equality is strict.
The following bound is for the special casep= 3.
Theorem 1.65. For every integerq≥3, R(3, q)≤ q2+ 3
2 .
The final bound that we present is due to Erd˝os [90]. It applies to the special casep=q. In the theorem,xdenotes the greatest integer less than or equal to x.
Theorem 1.66. Ifp≥3, then
R(p, p)>2n/2.
A number of other specific bounds are known:
35≤R(4,6)≤41, 43≤R(5,5)≤49, 58≤R(5,6)≤87, 102≤R(6,6)≤165.
Even with the sophisticated computing power that is available to us today, we are not able to compute values for more than a handful of Ramsey numbers. Paul Erd˝os once made the following comment regarding the difficulty in finding exact values of Ramsey numbers [63]:
Suppose an evil alien would tell mankind “Either you tell me [the value ofR(5,5)] or I will exterminate the human race.” . . . It would be best in this case to try to compute it, both by mathematics and with a computer.
If he would ask [for the value ofR(6,6)], the best thing would be to destroy him before he destroys us, because we couldn’t [determine R(6,6)].
Exercises
1. Prove thatR(3,5)≥14. The graph in Figure 1.131 will be very helpful.
FIGURE 1.131. A 2-coloring ofK13.
2. Use Theorem 1.64 and the previous exercise to prove thatR(3,5) = 14.
3. Construct a graph and a 2-coloring that provesR(4,4)≥18.
4. Use Theorem 1.64 and the previous exercise to prove thatR(4,4) = 18.
5. Use Theorem 1.64 to prove Theorem 1.65.