MathLabs

Combinatorics and discrete mathematics

Ramsey theory

Shows that complete disorder is impossible: any large enough structure must contain an orderly pattern.

IntuitionThe party of six: strangers and mutual friends

Invite any 66 people to a party. Between each pair, either they have met before (draw a red edge) or they are strangers (draw a blue edge). No matter how the 1515 relationships are arranged, there will always be either 33 people who all know each other (a red triangle K3K_3) or 33 people who are all mutual strangers (a blue triangle K3K_3). With 55 people it is possible to avoid both, so R(3,3)=6R(3,3) = 6 is the exact threshold where order is forced out of chaos.

Complete graph network illustrating two-colored edges and monochromatic triangles in Ramsey theory.
In a complete graph K6K_6, any vertex vv has 55 incident edges in 22 colors; by the pigeonhole principle at least 33 share a color, forcing a monochromatic K3K_3.

UndergraduateRamsey numbers and the Erdős–Szekeres recursion

Definition: Two-color Ramsey number R(s, t)

For integers s,t≥2s, t \ge 2, the Ramsey number R(s,t)R(s, t) is the smallest integer NN such that every red–blue coloring of the edges of the complete graph KNK_N contains either a red complete subgraph KsK_s or a blue complete subgraph KtK_t.

R(s,t)≤R(s−1,t)+R(s,t−1)≤(s+t−2s−1)R(s, t) \le R(s - 1, t) + R(s, t - 1) \le \binom{s + t - 2}{s - 1}
2k/2<R(k,k)≤(2k−2k−1)<4k−12^{k/2} < R(k, k) \le \binom{2k - 2}{k - 1} < 4^{k - 1}
Exactly known small two-color Ramsey numbers R(s, t)
Pair (s,t)(s, t)Exact value or best known rangeLandmark
R(3,3)R(3, 3)66Putnam 1953
R(3,4)R(3, 4)99Greenwood–Gleason 1955
R(4,4)R(4, 4)1818Greenwood–Gleason 1955
R(5,5)R(5, 5)43≤R(5,5)≤4643 \le R(5, 5) \le 46Open (Angeltveit–McKay 2024)

UndergraduateCore theorems and proofs

The Ramsey number R(3,3)R(3, 3) equals 66: every 22-coloring of the edges of K6K_6 contains a monochromatic K3K_3, while K5K_5 admits a coloring with no monochromatic K3K_3.

Why is it true?

A single vertex in K6K_6 has 55 neighbors; splitting them into 22 colors puts at least 33 in the same class, and whether those 33 have an edge of that color among themselves or not, a monochromatic triangle appears.

Proof

**Step 1 (upper bound R(3,3)≤6R(3,3) \le 6).** Pick any vertex vv of K6K_6. Its 55 incident edges are colored red or blue. By the pigeonhole principle (⌈5/2⌉=3\lceil 5/2 \rceil = 3), at least 33 of these edges have the same color — say edges vu1,vu2,vu3vu_1, vu_2, vu_3 are all red.

**Step 2 (case split on {u1,u2,u3}\{u_1, u_2, u_3\}).** Look at the 33 edges inside the triangle {u1,u2,u3}\{u_1, u_2, u_3\}. If any one of them — say u1u2u_1u_2 — is red, then {v,u1,u2}\{v, u_1, u_2\} is a red K3K_3. Otherwise all 33 edges u1u2,u2u3,u3u1u_1u_2, u_2u_3, u_3u_1 are blue, so {u1,u2,u3}\{u_1, u_2, u_3\} itself is a blue K3K_3.

**Step 3 (lower bound R(3,3)>5R(3,3) > 5).** Label the vertices of K5K_5 by Z/5Z\mathbb{Z}/5\mathbb{Z}. Color edge ijij red if i−j≡±1(mod5)i - j \equiv \pm 1 \pmod 5 and blue if i−j≡±2(mod5)i - j \equiv \pm 2 \pmod 5. Both color classes form a 55-cycle, triangle-free, proving R(3,3)>5R(3,3) > 5 and hence R(3,3)=6R(3,3) = 6.

For all integers s,t≥2s, t \ge 2, the Ramsey number R(s,t)R(s, t) is finite and satisfies R(s,t)≤R(s−1,t)+R(s,t−1)R(s, t) \le R(s - 1, t) + R(s, t - 1). Consequently, R(s,t)≤(s+t−2s−1)R(s, t) \le \binom{s + t - 2}{s - 1}.

Why is it true?

Fix a vertex vv: either its red neighborhood is large enough to force a red Ks−1K_{s-1} or a blue KtK_t, or else its blue neighborhood is large enough to force a red KsK_s or a blue Kt−1K_{t-1}.

Proof

**Step 1 (pigeonhole split of the neighborhood of vv).** Let N=R(s−1,t)+R(s,t−1)N = R(s - 1, t) + R(s, t - 1) and fix a vertex v∈V(KN)v \in V(K_N). Partition the remaining N−1N - 1 vertices into VRV_R (red edge to vv) and VBV_B (blue edge to vv). Since ∣VR∣+∣VB∣=R(s−1,t)+R(s,t−1)−1|V_R| + |V_B| = R(s - 1, t) + R(s, t - 1) - 1, either ∣VR∣≥R(s−1,t)|V_R| \ge R(s - 1, t) or ∣VB∣≥R(s,t−1)|V_B| \ge R(s, t - 1).

Step 2 (inductive conclusion). If ∣VR∣≥R(s−1,t)|V_R| \ge R(s - 1, t), the subgraph on VRV_R contains a blue KtK_t or a red Ks−1K_{s-1} that joins vv to form a red KsK_s. Symmetrically for VBV_B.

Step 3 (binomial bound by induction). Base cases R(2,t)=tR(2, t) = t and R(s,2)=sR(s, 2) = s match (t1)\binom{t}{1} and (ss−1)\binom{s}{s-1}. For s,t≥3s, t \ge 3, Pascal's identity gives R(s,t)≤(s+t−3s−2)+(s+t−3s−1)=(s+t−2s−1)R(s, t) \le \binom{s + t - 3}{s - 2} + \binom{s + t - 3}{s - 1} = \binom{s + t - 2}{s - 1}.

AdvancedApplications and Worked Examples

Ramsey theory underpins communications network design, theoretical computer science (lower bounds via Dickson's lemma and Kruskal's tree theorem), and additive combinatorics (Schur's theorem and van der Waerden's theorem on monochromatic arithmetic progressions).

Example: Proving R(3, 4) = 9 using degree parity

The Erdős–Szekeres recursion gives R(3,4)≤R(2,4)+R(3,3)=4+6=10R(3, 4) \le R(2, 4) + R(3, 3) = 4 + 6 = 10. Improve this to R(3,4)≤9R(3, 4) \le 9 using the parity of red degrees in K9K_9.

Solution

In any coloring of K9K_9, if some vertex vv has red degree dR(v)≥4d_R(v) \ge 4, its 44 red neighbors either contain a red edge (forming a red K3K_3 with vv) or all blue edges between them (forming a blue K4K_4). If some vertex has dR(v)≤2d_R(v) \le 2, its blue degree dB(v)=8−dR(v)≥6=R(3,3)d_B(v) = 8 - d_R(v) \ge 6 = R(3, 3), so its blue neighborhood contains a red or blue K3K_3.

The only remaining case is dR(v)=3d_R(v) = 3 for every vertex. But the degree sum ∑vdR(v)\sum_v d_R(v) must be even, whereas 9×3=279 \times 3 = 27 is odd — impossible! Thus R(3,4)≤9R(3, 4) \le 9, and an explicit construction shows R(3,4)=9R(3, 4) = 9.

Example: Three-color Ramsey number R(3, 3, 3) = 17

Show that if the edges of K17K_{17} are colored with 33 colors, there must be a monochromatic triangle K3K_3.

Solution

Pick a vertex vv. Its 1616 incident edges are partitioned into 33 classes, so at least one has ≥⌈16/3⌉=6\ge \lceil 16/3 \rceil = 6 edges. Suppose vv has 66 green edges to a set UU.

If any edge inside UU is green, it joins vv to form a green K3K_3. Otherwise, edges inside UU use only the remaining colors on ∣U∣=6=R(3,3)|U| = 6 = R(3, 3) vertices, forcing a monochromatic K3K_3.

What is the exact value of the Ramsey number R(3,3)R(3, 3)?

What upper bound does R(s,t)≤R(s−1,t)+R(s,t−1)R(s, t) \le R(s-1,t) + R(s,t-1) give for R(4,4)R(4, 4), using R(3,4)=9R(3,4) = 9?

For any integer k≥2k \ge 2, what is the exact value of R(2,k)R(2, k)?

What did the 2023 breakthrough of Campos, Griffiths, Morris, and Sahasrabudhe establish for R(k,k)R(k, k)?

References

  1. Marcelo Campos, Simon Griffiths, Robert Morris, Julian Sahasrabudhe (2023). An exponential improvement for diagonal Ramsey · arXiv:2303.09521
  2. Sam Mattheus, Jacques Verstraëte (2024). The asymptotics of r(4,t) · DOI:10.4007/annals.2024.199.2.8
  3. Ronald L. Graham, Bruce L. Rothschild, Joel H. Spencer (1990). Ramsey Theory (2nd ed.)