Improved Universal Graphs for Trees
Julian Becker, Konstantinos Panagiotou, Matija Pasch
TL;DR
The paper resolves, up to an improved constant, the Chung–Graham question on the minimal number of edges in an $n$-vertex graph universal for all $n$-vertex trees by constructing explicit universal graphs based on ternary trees. The core idea is a divide-and-conquer embedding into a ternary-tree-derived host, facilitated by refined separator lemmas and an eating-order framework that preserves admissibility during recursive embeddings. The main result tightens the upper bound to $s(n)\le\frac{19}{6\ln 3}\,n\ln n+O(n)$ and extends the approach to graphs of treewidth $w$, obtaining $nw\ln(n/w)-O(nw)\le s_w(n)\le\frac{19}{6\ln3}(w+1)n\ln(n/w)+O(nw)$. The technique relies on hierarchical graph-buildings $U_{n,d}$, precise edge-counting, and a blow-up argument to handle width-$w$ graphs, offering a robust path toward asymptotics for the tree-universal problem and its near-tree generalizations.
Abstract
A graph $G$ is universal for a class of graphs $\mathcal{C}$, if, up to isomorphism, $G$ contains every graph in $\mathcal{C}$ as a subgraph. In 1978, Chung and Graham asked for the minimal number $s(n)$ of edges in a graph with $n$ vertices that is universal for all trees with $n$ vertices. The currently best bounds assert that $n\ln n-O(n)\le s(n) \le C n\ln n+O(n)$, where $C = \frac{14}{5\ln 2} \approx 4.04$. Here, we improve the upper bound to $c n\ln n + O(n)$, where $c = \frac{19}{6\ln 3} \approx 2.88$. We develop in the proof a strategy that, broadly speaking, is based on separating trees into three parts, thus enabling us to embed them in a structure that originates from ternary trees. Our method also applies to graphs that are close to being trees, measured by their treewidth. Let $s_w(n)$ be the minimum number of edges in a $n$-vertex graph that is universal for graphs with treewidth $w$. By performing a graph blow-up to our universal structure and counting necessary edges carefully, we establish that $nw \ln(n/w) -O(nw) \leq s_w(n) \leq \frac{19}{6\ln3} n (w+1) \ln(n/w) + O(nw)$.
