The probabilistic lower bound argument for R(k,k) colors each edge of K_n red or blue at random and shows that the probability of any monochromatic k-clique existing is less than 1. What does this conclude?
AMost random colorings will avoid monochromatic k-cliques
BAt least one 2-coloring of K_n exists with no monochromatic k-clique, so R(k,k) > n
CThe expected number of monochromatic k-cliques is less than 1 for all n
DThe probability decreases exponentially, so R(k,k) grows faster than any polynomial
Probabilistic existence proofs work as follows: if the probability that a random object has a bad property is less than 1, then with positive probability it does not have the bad property — meaning at least one good object must exist. Showing P(monochromatic k-clique in K_n) < 1 proves a 'good' coloring of K_n exists, which means R(k,k) > n. The argument is purely non-constructive: it establishes existence without exhibiting the coloring. Option A is wrong — the argument says nothing about 'most' colorings, only that some good one exists.
Question 2 Multiple Choice
The recurrence R(s,t) ≤ R(s-1,t) + R(s,t-1) is derived by considering a fixed vertex v in K_n. What is the key pigeonhole argument?
AAny k-clique must include v, so we count cliques containing v and those avoiding v
BVertex v has n-1 edges; if n is large enough, v must have many red or many blue neighbors
Cv has at least R(s-1,t) red neighbors or at least R(s,t-1) blue neighbors, because otherwise n is smaller than required
DThe neighborhood of v must contain a clique by the inductive hypothesis on smaller graphs
Fix any vertex v. Suppose v has fewer than R(s-1,t) red neighbors AND fewer than R(s,t-1) blue neighbors. Then the total neighbors is at most R(s-1,t) + R(s,t-1) − 2, giving n − 1 < R(s-1,t) + R(s,t-1) − 2. But we assumed n = R(s-1,t) + R(s,t-1), a contradiction. So one condition must hold. If v has R(s-1,t) red neighbors, those neighbors either contain a red K_{s-1} (giving red K_s with v) or a blue K_t. Either way we get the desired monochromatic clique.
Question 3 True / False
R(5,5) has been determined exactly through exhaustive computer search of most edge colorings of graphs on 43 to 48 vertices.
TTrue
FFalse
Answer: False
R(5,5) is only known to lie between 43 and 48, despite decades of effort. Exhaustive search is infeasible: the number of 2-colorings of K_n is 2^(n(n-1)/2), which for n = 43 exceeds 10^270. Even with modern computing power, this search space cannot be explored. Erdős's famous quip about R(6,6) captured the computational intractability: improving Ramsey bounds requires genuinely new mathematical ideas, not just more compute.
Question 4 True / False
The probabilistic lower bound for R(k,k) is non-constructive: it proves a good 2-coloring exists without exhibiting one.
TTrue
FFalse
Answer: True
This is a defining feature of the probabilistic method as pioneered by Erdős. By showing that a randomly chosen coloring avoids monochromatic k-cliques with positive probability, we prove existence without construction. The argument gives no hint of what such a coloring looks like. Finding explicit constructions that match probabilistic lower bounds is a major open problem in combinatorics — for most values, explicit constructions fall far short of the probabilistic bound.
Question 5 Short Answer
Explain why computing R(6,6) exactly is considered computationally infeasible, even though R(3,3) = 6 and R(4,4) = 18 were determined exactly.
Think about your answer, then reveal below.
Model answer: The number of 2-colorings of K_n grows as 2^(n(n-1)/2), which increases explosively with n. R(3,3) = 6 is verified by small case analysis; R(4,4) = 18 required significant work but is manageable. R(5,5) is only known to lie between 43 and 48; R(6,6) between 102 and 165. The search space for K_102 through K_165 — needing to verify all 2-colorings — is astronomically large, far beyond any feasible computation. Progress requires new mathematical bounds, not brute force.
The key is the doubly exponential growth of the search space. Each additional vertex v adds n-1 new edges, multiplying the search space by 2^(n-1). Even checking 10^30 colorings per second, verifying R(6,6) would take far longer than the age of the universe. This is why Ramsey theory combines deep mathematics with fundamental computational limits — and why the probabilistic bounds, despite being non-constructive, represent a genuine mathematical achievement.