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A hypergraph consists of a vertex set $V$ and a family $E$ of subsets of $V$, called hyperedges. Unlike a graph, a hyperedge may contain more than two vertices. A uniform hypergraph restricts all hyperedges to one fixed size.

For a uniform hypergraph with positive average hypergraph vertex degree $d$, the degree measure is $\mu (S)=(nd{)}^{-1}{\sum }_{u\in S}{d}_H(u)$. It is a probability measure and equals $|S|/n$ in a regular hypergraph. Every hypergraph independent set has degree measure at most $1-1/r$, even when its fraction of all vertices is close to one.

The degree of a vertex $u$ in a hypergraph is the number of hyperedges containing $u$. For a uniform hypergraph, summing degrees counts each hyperedge once for each of its vertices, so ${\sum }_u{d}_H(u)=re(H)$. In a multihypergraph, incidences are counted with multiplicity.

Assume the uniform hypergraph has positive average hypergraph vertex degree $d$, so $nd=re(G)$. Its hypergraph degree measure isIt is a probability measure on the vertex set; vertices of degree zero have measure zero. If the hypergraph is regular, it equals $|S|/n$. An edgeless hypergraph has no normalized degree measure of this form and must be treated separately.

Every edge entirely in $S$ contributes $r$ to the degree sum over $S$, giving the upper bound. For the lower bound, an edge entirely in $S$ contributes $r$, and any other edge contributes at most $r-1$. ThereforeTogether these give both degree-measure inequalities:In particular every hypergraph independent set has measure at most $1-1/r$, which explains why degree measure can control containers even in very nonregular hypergraphs.

A sufficient container theorem. For every fixed $r\ge 2$, there are positive constants ${\delta }_r,{\tau }_r,c_r,K_r$ such that if $0<\tau \le {\tau }_r$ andthere is a family of hypergraph containers covering all hypergraph independent sets, each with $\mu (C)\le 1-c_r$. Each container is determined by a container fingerprint $T$ with $|T|\le K_r\tau n$ and $\mu (T)\le K_r\tau$. In particular, for sufficiently small $\tau$, the number of containers is at most $\exp (O_r(\tau n\log (1/\tau )))$. We will give an algorithm and constants establishing this version.

The golden rule for container algorithms is that reconstruction may depend on the small container fingerprint, but never on unrevealed information about the independent set. When a deterministic test calls for inspecting a vertex, a positive membership answer is recorded in the container fingerprint, while a negative answer permits its exclusion from the container. All tests and all state updates must be reproducible using that container fingerprint. Choosing tests that permit only a few positive answers then yields few containers.

Here is an explicit implementation with controlled links. Use the fixed order $1,\dots ,n$, parameters $\tau ,\delta ,\zeta >0$, and a bound $d_G(\sigma )\le \delta d{\tau }^{|\sigma |-1}$ on nonsingleton codegrees. Let $d_s(\sigma )$ count incidences with multiplicity in the $s$-uniform multihypergraph $P_s$. Define thresholdsA saturated subset is one whose current degree has reached its threshold.

Every generated edge of $P_s$ extends to an original edge by adding $r-s$ earlier vertices from $T$. Thus, for a hypergraph independent set $I$, no singleton generated in $P_1$ can be a member of $I$. In particular an inspected vertex in ${\Gamma }_1$ cannot give a positive membership answer. All other positive answers are recorded. The golden rule for container algorithms now proves that the reconstructed container contains $I$ and that its state depends only on $T$.

Control the possible overshoot at saturation. The multisets are added in batches; merely claiming $d_s(\sigma )\le {\theta }_s(\sigma )$ would be wrong. In the last batch that increases $d_s(\sigma )$, the set $\sigma$ was not yet saturated, and the batch comes from some vertex $v\notin \sigma$. Its contribution is at most $d_{s+1}(\sigma \cup \{v\})$. HenceInduction downwards from $P_r$ yields, for $|\sigma |\ge 2$,Indeed the two terms in the last-batch bound have coefficients $1$ and $r-s$ with the same power of $\tau$. Applying this nonsingleton bound to $\{u,v\}$ gives the singleton estimateSumming over $U$, using $|U|\le n$, proves the requested estimate at every stage of the algorithm:It applies with multiplicities; replacing the multisets by sets in the accounting would invalidate the argument.

Check that the container fingerprint is small and the container loses positive measure. Assign each positive inspected vertex to one index $s$ witnessing its test. Its added link batch has size at least $\zeta {\tau }^{r-s-1}{d}_G(v)$. Distinct batches count separately in the multiset. Since $|P_s|=s^{-1}{\sum }_u{d}_s(u)$, the preceding bound givesThe second inequality uses $d_G(v)\ge \zeta d$ for every $v\in T$.

For completeness, the loss of measure can be verified by a short incidence count. Put $D=([n]\setminus C)\cup T\cup B$ and $a_s=|P_s|/({\tau }^{r-s}nd)$. Every saturated singleton belongs to $D$. Partition $P_{s+1}$ by its earliest vertex. An earliest vertex in $D$ accounts for at most ${\tau }^{r-s-1}nd(\mu (D)+r\delta )$ incidences. For any other earliest vertex, the unsaturated link has size less than $\zeta {\tau }^{r-s-1}{d}_G(v)$; the remaining edges meet some saturated subset of ${\Gamma }_s$.

For a saturated nonsingleton $\sigma$, the degree bounds imply $d_{s+1}(\sigma )\le (r/\tau )d_s(\sigma )$. For a saturated singleton the corresponding bound is $d_{s+1}(u)\le {\tau }^{-1}{d}_s(u)+r\delta d{\tau }^{r-s-1}$. Every multiedge of $P_s$ has fewer than $2^s$ nonempty subsets. These observations giveFor $s=1$, all saturated subsets are singletons in $D$, so directly summing their $P_2$ degrees givesSince $a_r=1/r$, these inequalities prevent $D$ from having arbitrarily small measure. To make the constants explicit, letIterating the displayed inequalities gives $a_r\le L(\mu (D)+\zeta +3r\delta )$, hence $\mu (D)\ge 3b/4$ whenever $\delta \le {\delta }_r$. Also $r\delta \le 1$, so $\tau \le {\tau }_r$ gives $\mu (T)\le 2r\tau /\zeta \le b/8$, while $\mu (B)\le \zeta =b/8$. ThereforeOne can take $K_r=2r/{\zeta }^2$ in the stated container fingerprint bounds. Counting subsets of size at most $K_r\tau n$ gives the asserted bound on the number of containers. Thus the explicit algorithm produces the family promised by the sufficient hypergraph container theorem, as well as the required intermediate degree estimate.

For the standard weak-threshold formulation, see Online containers for hypergraphs. The constants above come from the explicit incidence estimates given here.

For $r=1$, no multilevel construction is needed: take the sole container to be the vertices that are not singleton edges. It contains every hypergraph independent set and has degree measure zero when $d>0$; its fingerprint is empty. The requested degree-sum inequality follows directly from the definition of $\mu$.