Degree Sequence Conditions for Maximally Edge-Connected and Super Edge-Connected Hypergraphs

Abstract

Let H be a connected hypergraph with minimum degree $\delta$ and edge-connectivity $\lambda$. The hypergraph H is maximally edge-connected if $\lambda = \delta$, and it is super edge-connected or super-$\lambda$, if every minimum edge-cut consists of edges incident with some vertex. There are several degree sequence conditions, for example, Goldsmith and White (Discrete Math 23: 31–36, 1978) and Bollobás (Discrete Math 28:321–323, 1979) etc. for maximally edge-connected graphs and super-$\lambda$ graphs. In this paper, we generalize these and some other degree sequence conditions to uniform hypergraphs.

Appendix

If $r = 2$, then $t - 1 = \delta$ and $(t - 1)\big [\left( {\begin{array}{c}n - t\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}t\\ r - 1\end{array}}\right) \big ] = \delta n > \delta n - 1 = (t - 1)\big [\left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 1\end{array}}\right) \big ] + r(\delta - 1) + 1$. It remains to show that the inequality holds for the case $r \ge 3$. Since $n \ge 2t$, we have

$$\begin{aligned} \begin{aligned}&(t - 1)\bigg [\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 2\end{array}}\right) \bigg ]\\&\quad \ge 2(t - 1)\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) \\&\quad = t\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) + \left( {\begin{array}{c}t\\ r - 1\end{array}}\right) + (t - 3)\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) - \left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) . \end{aligned} \end{aligned}$$

(A1)

We proceed to show

$$\begin{aligned} \begin{aligned} (t - 3)\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) - \left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) > -r. \end{aligned} \end{aligned}$$

(A2)

If $r = 3$, $(t - 3)(t - 1) - \frac{(t - 1)(t - 2)}{2} + 3 = \frac{1}{2}t^{2} - \frac{5}{2}t + 5 > 0$ holds. If $r \ge 4$, by $\left( {\begin{array}{c}t\\ r - 1\end{array}}\right) > \delta \ge 1$, we have $t > r - 1 \ge 3$ and $\left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) - r = (\frac{t}{r - 1} - 1)\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) - r < (t - 3)\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right)$. Thus, the inequality (A2) holds. Therefore, combining the inequalities (A1) and (A2), we have

$$\begin{aligned} \begin{aligned}&(t - 1)\bigg [\left( {\begin{array}{c}n - t\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}t\\ r - 1\end{array}}\right) \bigg ]\\&\quad =(t - 1)\bigg [\left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 2\end{array}}\right) \bigg ]\\&\quad \ge (t - 1)\bigg [\left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 1\end{array}}\right) \bigg ] + t\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) + \left( {\begin{array}{c}t\\ r - 1\end{array}}\right) \\&\qquad + (t - 3)\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) - \left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) \\&\quad > (t - 1)\bigg [\left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 1\end{array}}\right) \bigg ] + t\left( {\begin{array}{c}t - 1\\ r - 2\end{array}}\right) + \left( {\begin{array}{c}t\\ r - 1\end{array}}\right) - r\\&\quad = (t - 1)\bigg [\left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 1\end{array}}\right) \bigg ] + r\left( {\begin{array}{c}t\\ r - 1\end{array}}\right) - r\\&\quad \ge (t - 1)\bigg [\left( {\begin{array}{c}t - 1\\ r - 1\end{array}}\right) + \left( {\begin{array}{c}n - t - 1\\ r - 1\end{array}}\right) \bigg ] + r(\delta - 1) + 1. \end{aligned} \end{aligned}$$