The vertex Folkman number $F_v(3,3;5)$ equals~$8$

TL;DR

This paper proves that the vertex Folkman number $F_v(3,3;5)$ equals 8 by constructing a specific graph and verifying its properties through exhaustive enumeration and SAT checks.

Contribution

It establishes the exact value of $F_v(3,3;5)$ using a combination of graph construction, theoretical arguments, and computational verification.

Findings

The graph $K_1 \vee \overline{C_7}$ is $K_5$-free and satisfies the Folkman property.

No smaller $K_5$-free graph on 7 or fewer vertices has the property.

The proof combines theoretical reasoning with exhaustive computational methods.

Abstract

The vertex Folkman number $F_v(s,t;k)$ is the smallest $n$ for which there exists a $K_k$-free graph on $n$ vertices whose vertices cannot be $2$-colored without producing a monochromatic copy of $K_s$ or $K_t$. We show $F_v(3,3;5)=8$. The witness is the cone $K_1 \vee \overline{C_7}$, a single universal vertex joined to the complement of a $7$-cycle. That this graph is $K_5$-free and arrows $(3,3)^v$ follows from a short independence-number argument. The matching lower bound -- no $K_5$-free graph on $7$ or fewer vertices works -- comes from exhaustive enumeration via nauty and a SAT check using Glucose\,4. The appendix has a self-contained Python script for verification.