Maximally distance-unbalanced trees

Abstract

For a graph G, and two distinct vertices u and v of G, let $n_{{G(u,v)}}$ be the number of vertices of G that are closer in G to u than to v. Miklavič and Šparl (arXiv:2011.01635v1) define the distance-unbalancedness ${{\mathrm{uB}}}(G)$ of G as the sum of $|n_G(u,v)-n_G(v,u)|$ over all unordered pairs of distinct vertices u and v of G. For positive integers n up to 15, they determine the trees T of fixed order n with the smallest and the largest values of ${\mathrm{uB}}(T)$, respectively. While the smallest value is achieved by the star $K_{1,n-1}$ for these n, which we then proved for general n (Minimum distance-unbalancedness of trees, J Math Chem, https://doi.org/10.1007/s10910-021-01228-4), the structure of the trees maximizing the distance-unbalancedness remained unclear. For n up to 15 at least, all these trees were subdivided stars. Contributing to problems posed by Miklavič and Šparl, we show

$$\begin{aligned} \max \Big \{{\mathrm{uB}}(T):T \text{ is } \text{ a } \text{ tree } \text{ of } \text{ order } n\Big \} =\frac{n^3}{2}+o(n^3) \end{aligned}$$

and

$$\begin{aligned} \max \Big \{{\mathrm{uB}}(S(n_1,\ldots ,n_k)):1+n_1+\cdots +n_k=n\Big \} =\left( \frac{1}{2}-\frac{5}{6k}+\frac{1}{3k^2}\right) n^3+O(kn^2), \end{aligned}$$

where $S(n_1,\ldots ,n_k)$ is the subdivided star such that removing its center vertex leaves paths of orders $n_1,\ldots ,n_k$.