Excess Obstructions and Star-Isolated Certificates for the Hypergraph Nash–Williams–Tutte Conjecture


Abstract

Guo, Li, Shangguan, Tamo, and Wootters formulated in SIAM Journal on Computing a hypergraph Nash–Williams–Tutte conjecture: every \(k\)-weakly-partition-connected hypergraph on \(t\) vertices should admit a \(k\)-distinguishable tree assignment. We show that the conjecture, in its literal published form, is false for a sharp numerical reason. A tree assignment replaces every hyperedge \(e\) by a tree with \(|e|-1\) labelled edges, so its edge number is the excess \(\rho(H)=\sum_e(|e|-1)\). A \(k\)-tree decomposition, however, has exactly \(k(t-1)\) edges. Thus \(\rho(H)=k(t-1)\) is a necessary condition, whereas weak partition connectivity only implies \(\rho(H)\ge k(t-1)\). Consequently, for every \(t\ge2\), \(k\ge1\), and \(q\ge1\), the hypergraph consisting of \(k+q\) copies of the full hyperedge \(V\) is \(k\)-weakly-partition-connected but has no \(k\)-distinguishable tree assignment. We then isolate the critical corrected form, prove that its equality is exactly the equality required for using the full tree-assignment row set in the intersection-matrix route without pruning, and give an explicit infinite non-graphic class of critical positive instances. The positive construction uses layer-contained star realizations and extremal signature weights, producing weak partition connectivity by a quotient-rank argument and unique signatures under one-vertex sums and explicit two-sided star blocks.

Figure 1: image.

1 Problem, notation, and main results↩︎

The starting point of this note is the following conjecture from the Reed–Solomon list-decoding work of Guo, Li, Shangguan, Tamo, and Wootters [1]. It was introduced as a hypergraph version of the Nash–Williams–Tutte tree-packing theorem [2], [3] and is naturally adjacent to partition-connected hypergraph decomposition [4]; in the Reed–Solomon application it is a possible route to the nonsingularity of intersection matrices. We use the exact terminology needed in Section 6 of [1], but we restate the definitions in a form that keeps the edge-count parameter visible.

Let \(H=(V,E)\) be a finite labelled multihypergraph. Hyperedges are occurrences, not merely subsets; thus two equal subsets of \(V\) may carry two different labels. Throughout the paper every hyperedge occurrence has cardinality at least two. This is the convention under which tree assignments are defined; the support hypergraphs in Section 3 likewise discard supports of size less than two. We write \(t=|V|\) and assume throughout that \(t\ge2\) unless explicitly stated. For a partition \(\mathcal{P}\) of \(V\) and a hyperedge \(e\in E\), put \[\label{eq:pP} p_{\mathcal{P}}(e) \mathrel{:=}\left|\{P\in\mathcal{P}: P\cap e\ne\varnothing\}\right|.\tag{1}\] The weak partition excess of \(H\) across \(\mathcal{P}\) is \[\label{eq:weakcross} w_H(\mathcal{P}) \mathrel{:=}\sum_{e\in E}\bigl(p_{\mathcal{P}}(e)-1\bigr).\tag{2}\] The hypergraph \(H\) is \(k\)-weakly-partition-connected if \[\label{eq:kwpc} w_H(\mathcal{P}) \ge k\bigl(|\mathcal{P}|-1\bigr) \qquad\text{for every partition }\mathcal{P}\text{ of }V.\tag{3}\] For the discrete partition \(\mathcal{P}_0=\{\{v\}:v\in V\}\), the left hand side is the total excess \[\label{eq:rho-def} \rho(H) \mathrel{:=}w_H(\mathcal{P}_0) =\sum_{e\in E}\bigl(|e|-1\bigr).\tag{4}\] Thus every \(k\)-weakly-partition-connected hypergraph satisfies \[\label{eq:wpc-implies-excess} \rho(H)=w_H(\mathcal{P}_0)\ge k(t-1).\tag{5}\] The inequality 5 is the source of the obstruction below: it is a lower bound, while a \(k\)-tree decomposition demands exact equality.

A tree assignment of \(H\) chooses, for each hyperedge occurrence \(e\), a tree \(F_e\) on vertex set \(e\), and labels every edge of \(F_e\) by \(e\). The resulting labelled graph on the ambient vertex set \(V\) is \[\label{eq:tree-assignment} G=\biguplus_{e\in E}F_e, \qquad |E(G)|=\sum_{e\in E}|E(F_e)|=\rho(H).\tag{6}\] A \(k\)-tree decomposition of a labelled graph \(G\) on \(V\) is an ordered partition \[\label{eq:ktree-decomp} E(G)=E(T_0)\,\dot{\cup}\,E(T_1)\,\dot{\cup}\cdots\dot{\cup}\,E(T_{k-1}),\tag{7}\] where each \(T_i\) is a spanning tree of \(V\). Necessarily \[\label{eq:kdecomp-count} |E(G)|=\sum_{i=0}^{k-1}|E(T_i)|=k(t-1).\tag{8}\] If \(G\) is a tree assignment of \(H\) and \(\mathcal{T}=(T_0,\ldots,T_{k-1})\) is a \(k\)-tree decomposition of \(G\), its signature is the vector \[\label{eq:signature} \operatorname{sig}_{\mathcal{T}}(e) \mathrel{:=}\sum_{i=0}^{k-1} i\,\left|E(T_i)\cap E(F_e)\right| \qquad(e\in E).\tag{9}\] The graph \(G\) is \(k\)-distinguishable if it admits a \(k\)-tree decomposition \(\mathcal{T}\) such that every \(k\)-tree decomposition \(\mathcal{T}'\) with \(\operatorname{sig}_{\mathcal{T}'}=\operatorname{sig}_{\mathcal{T}}\) satisfies \(\mathcal{T}'=\mathcal{T}\) as an ordered tuple.

For positive integers \(t\) and \(k\), every \(k\)-weakly-partition-connected hypergraph \(H\) on \(t\) vertices has a \(k\)-distinguishable tree assignment.

The conjecture is a precise mathematical proposition, not an existentially vague question. The main point of this paper is that Conjecture [conj:literal] is incompatible with the equality forced by 8 . This gives a decisive negative resolution of the literal statement.

Theorem 1 (Sharp excess obstruction). Let \(H\) be a finite labelled multihypergraph on \(t\ge2\) vertices. If \(H\) has a \(k\)-distinguishable tree assignment, then \[\label{eq:critical-necessary} \rho(H)=k(t-1).\qquad{(1)}\] Consequently, every \(k\)-weakly-partition-connected hypergraph with \[\label{eq:overfull} \rho(H)>k(t-1)\qquad{(2)}\] is a counterexample to Conjecture [conj:literal].

The obstruction is sharp: weak partition connectivity always gives \(\rho(H)\ge k(t-1)\) by 5 , so 1 says that the only possible failure of the edge-count test inside the weakly connected class is overfullness.

Theorem 2 (Infinite overfull counterexample family). Fix integers \(t\ge2\), \(k\ge1\), and \(q\ge1\). Let \(H_{t,k,q}\) be the labelled multihypergraph on vertex set \(V\) with \[\label{eq:Htkq} E(H_{t,k,q})=\{e_1,\ldots,e_{k+q}\}, \qquad e_j=V\quad(1\le j\le k+q).\qquad{(3)}\] Then \(H_{t,k,q}\) is \(k\)-weakly-partition-connected and has no \(k\)-distinguishable tree assignment. The smallest member is the two-vertex hypergraph with two parallel \(2\)-edges at \(k=1\): \[\label{eq:smallest} V=\{a,b\},\qquad E=\bigl\{\{a,b\}_1,\{a,b\}_2\bigr\}.\qquad{(4)}\]

The natural repair is to add the missing critical-excess equality.

Let \(H\) be a \(k\)-weakly-partition-connected hypergraph on \(t\) vertices with \[\label{eq:critical-conj} \sum_{e\in E(H)}(|e|-1)=k(t-1).\tag{10}\] Then \(H\) has a \(k\)-distinguishable tree assignment.

The equality 10 is not an artificial addition. It is the equality appearing in the intersection-matrix conjecture of Shangguan–Tamo [5], and it is exactly what is needed when a full tree assignment is used as the labelled-edge row set of size \(k(t-1)\). Section 3 records this in a form that separates the partition-connectivity calculation from the square-determinant requirement.

We shall call a hypergraph \(H\) on vertex set \(V\) \(k\)-critical if \[\label{eq:k-critical-def} \rho(H)=k(|V|-1).\tag{11}\] This is a purely numerical convention; weak partition connectivity is always stated separately.

Our second contribution is a positive theorem for an explicit infinite non-graphic part of Conjecture [conj:critical]. The theorem is not a proof of the full critical conjecture; rather, it gives an explicitly checkable class where weak partition connectivity follows from a layer-wise quotient-rank argument and uniqueness is forced by an extremal signature mechanism. The class includes full hyperedges of arbitrary rank and is closed under one-vertex sums.

Let \(S_e(c)\) denote the star on a hyperedge \(e\) centered at \(c\in e\): \[\label{eq:star} E(S_e(c))=\bigl\{\{c,u\}:u\in e\setminus\{c\}\bigr\}.\tag{12}\] A tree assignment is star-minimal if every hyperedge of size at least \(3\) is assigned such a star and every hyperedge of size \(2\) is assigned its unique edge.

Definition 1 (Two-sided star certificate). A \(k\)-critical hypergraph \(H\) has a two-sided star certificate if there are a star-minimal tree assignment \(G=\biguplus_eF_e\), an ordered decomposition \(\mathcal{T}=(T_0,\ldots,T_{k-1})\) of \(G\), and a layer map \[\label{eq:layer-map} \ell:E(H)\longrightarrow \{0,1,\ldots,k-1\}\qquad{(5)}\] with the following properties: \[\begin{align} &E(F_e)\subseteq E(T_{\ell(e)}) &&(e\in E(H)),\label{eq:layer-containment}\\ &|e|=2 &&\bigl(e\in E(H)\text{ and }1\le \ell(e)\le k-2\bigr).\label{eq:interior-rank-two} \end{align}\] {#eq: sublabel=eq:eq:layer-containment,eq:eq:interior-rank-two} Equivalently, labels assigned to an interior layer must have rank two, while high-rank labels may occur only in the two extremal layers \(0\) and \(k-1\). When \(k\le2\) there are no interior layers, so the interior set is empty rather than unconstrained. For \(k\ge2\) one may write \[\label{eq:edge-partition-layer-map} E^-\mathrel{:=}\ell^{-1}(0),\qquad E^0\mathrel{:=}\ell^{-1}\bigl(\{1,\ldots,k-2\}\bigr),\qquad E^+\mathrel{:=}\ell^{-1}(k-1),\qquad{(6)}\] with \(E^0=\varnothing\) when \(k=2\). For \(k=1\) there is only the single extremal layer \(0\).

Theorem 3 (Star-isolated positive instances). If \(H\) has a two-sided star certificate, then \[\label{eq:star-positive-conclusion} H\text{ is }k\text{-weakly-partition-connected and }H\text{ has a }k\text{-distinguishable tree assignment}.\qquad{(7)}\] More precisely, the certified decomposition \(\mathcal{T}=(T_0,\ldots,T_{k-1})\) is the unique ordered \(k\)-tree decomposition with signature \(\operatorname{sig}_\mathcal{T}\).

The next theorem shows that the construction is stable under the main gluing operation for tree packings.

Theorem 4 (One-vertex sum closure). Let \(H^{(1)}\) and \(H^{(2)}\) be \(k\)-critical hypergraphs with two-sided star certificates, and suppose that \[\label{eq:one-sum-vertex} V(H^{(1)})\cap V(H^{(2)})=\{r\}, \qquad E(H^{(1)})\cap E(H^{(2)})=\varnothing.\qquad{(8)}\] Then their one-vertex sum \[\label{eq:one-sum-H} H=H^{(1)}\vee_r H^{(2)}\qquad{(9)}\] is \(k\)-critical and has a two-sided star certificate. Hence \(H\) is a positive instance of Conjecture [conj:critical].

The obstruction, the critical repair, and the positive mechanism fit into the following diagram.

Figure 2: Logical structure. Weak partition connectivity gives only \rho(H)\ge k(t-1); the equality \rho(H)=k(t-1) is mandatory before signature uniqueness can even be formulated for a full tree assignment.

2 The edge-excess obstruction↩︎

This section proves [thm:excess-obstruction,thm:full-edge-counterexamples]. The argument is short, but spelling it out is useful because it identifies the exact point where the literal conjecture and the determinant argument diverge.

Lemma 1 (Discrete partition lower bound). If \(H\) is \(k\)-weakly-partition-connected on \(t\) vertices, then \[\label{eq:discrete-bound} \rho(H)=\sum_{e\in E(H)}(|e|-1)\ge k(t-1).\qquad{(10)}\]

Proof. Apply 3 to the discrete partition \(\mathcal{P}_0=\{\{v\}:v\in V\}\). For this partition, \[\label{eq:discrete-p} p_{\mathcal{P}_0}(e)=|e|, \qquad w_H(\mathcal{P}_0)=\sum_{e\in E}(|e|-1)=\rho(H),\tag{13}\] while \(|\mathcal{P}_0|-1=t-1\). ◻

Lemma 2 (Tree-assignment edge count). Every tree assignment \(G=\biguplus_eF_e\) of \(H\) has exactly \(\rho(H)\) graph edges.

Proof. By definition, \(F_e\) is a tree on \(|e|\) vertices, so \(|E(F_e)|=|e|-1\). Therefore \[\label{eq:assignment-count-proof} |E(G)|=\left|\biguplus_{e\in E(H)}E(F_e)\right| =\sum_{e\in E(H)}|E(F_e)| =\sum_{e\in E(H)}(|e|-1)=\rho(H).\tag{14}\] The disjoint union is an edge-disjoint union of labelled copies even when two hyperedges have the same vertex set. ◻

Lemma 3 (Spanning-tree-packing edge count). If a labelled graph \(G\) on \(t\) vertices has a \(k\)-tree decomposition, then \[\label{eq:packing-count-proof} |E(G)|=k(t-1).\qquad{(11)}\]

Proof. If \(E(G)=E(T_0)\dot{\cup}\cdots\dot{\cup} E(T_{k-1})\) and each \(T_i\) is a spanning tree of the same \(t\)-vertex set, then \(|E(T_i)|=t-1\) for all \(i\). Summing gives ?? . ◻

Proof of 1. If \(H\) has a \(k\)-distinguishable tree assignment \(G\), then \(G\) has a \(k\)-tree decomposition by the definition of distinguishability. Hence, by Lemmas 2 and 3, \[\label{eq:excess-proof-chain} \rho(H)=|E(G)|=k(t-1).\tag{15}\] If \(H\) is \(k\)-weakly-partition-connected and \(\rho(H)>k(t-1)\), the equality 15 is impossible for every tree assignment, so \(H\) has no \(k\)-distinguishable tree assignment. ◻

Proof of 2. Let \(\mathcal{P}\) be any partition of \(V\). Since every hyperedge \(e_j\) is equal to \(V\), it meets every block of \(\mathcal{P}\), and therefore \[\label{eq:full-edge-wpc} p_{\mathcal{P}}(e_j)=|\mathcal{P}|, \qquad p_{\mathcal{P}}(e_j)-1=|\mathcal{P}|-1.\tag{16}\] Thus \[\label{eq:Htkq-wpc} w_{H_{t,k,q}}(\mathcal{P}) =\sum_{j=1}^{k+q}\bigl(p_{\mathcal{P}}(e_j)-1\bigr) =(k+q)(|\mathcal{P}|-1) \ge k(|\mathcal{P}|-1).\tag{17}\] This proves \(k\)-weak partition connectivity. On the other hand, \[\label{eq:Htkq-excess} \rho(H_{t,k,q}) =\sum_{j=1}^{k+q}(|V|-1) =(k+q)(t-1)>k(t-1),\tag{18}\] so 1 rules out a \(k\)-distinguishable tree assignment. The case \(t=2\), \(k=1\), \(q=1\) is exactly ?? . ◻

The relaxation proved from Cheriyan–Salavatipour and Calinescu–Chekuri–Vondrak is stated with a \(k\)-distinguishable subgraph, not necessarily with the full tree assignment. The edge-count obstruction does not apply to a subgraph: from an overfull tree assignment one may discard labelled edges. It applies exactly to Conjecture [conj:literal], whose phrase “tree assignment of \(H\)” replaces every hyperedge of \(H\) and whose definition of \(k\)-distinguishability requires a decomposition of the entire resulting graph.

For a weakly connected hypergraph, the inequality \(\rho(H)<k(t-1)\) is impossible by Lemma 1. Hence 1 is not merely a sufficient obstruction: it is the only numerical obstruction visible from the discrete partition. The unresolved content of the critical form is therefore purely structural: \[\label{eq:critical-structural} \left. \begin{array}{c} w_H(\mathcal{P})\ge k(|\mathcal{P}|-1)\quad(\forall\mathcal{P}),\\[1mm] \rho(H)=k(t-1) \end{array} \right\} \quad\Longrightarrow?\quad \exists\text{ signature-unique tree assignment.}\tag{19}\]

3 Critical normalization and the intersection-matrix route↩︎

This section records the exact relation between the missing equality and the Reed–Solomon application through intersection matrices. The equality is not needed merely to derive weak partition connectivity from the weight inequalities; it is needed to make the full tree-assignment row set square and compatible with a \(k\)-tree decomposition on exactly \(k(t-1)\) labelled edges.

Let \(I_1,\ldots,I_t\subseteq[n]\). For \(s\in[n]\) define the support hyperedge \[\label{eq:support-edge} e_s\mathrel{:=}\{j\in[t]:s\in I_j\}.\tag{20}\] The associated labelled multihypergraph has edge-occurrence set \[\label{eq:intersection-H} E_I\mathrel{:=}\{\varepsilon_s=(s,e_s):s\in[n],\;|e_s|\ge2\}, \qquad H(I_1,\ldots,I_t)=([t],E_I).\tag{21}\] We write the occurrence \(\varepsilon_s\) simply as \(e_s\) when no confusion can arise; the label \(s\) is part of the occurrence, so equal supports with different labels are distinct. The standard weight of a subfamily is \[\label{eq:wt-def} \operatorname{wt}(I_J) =\sum_{s=1}^n\max\bigl(0,|e_s\cap J|-1\bigr) =\sum_{j\in J}|I_j|-\left|\bigcup_{j\in J}I_j\right|.\tag{22}\] For \(J=[t]\), 22 gives \[\label{eq:wt-rho} \operatorname{wt}(I_{[t]}) =\sum_{s=1}^n\max(0,|e_s|-1) =\sum_{e\in E(H)}(|e|-1) =\rho(H).\tag{23}\] Thus the equality condition in the intersection-matrix conjecture, \[\label{eq:intersection-critical} \operatorname{wt}(I_{[t]})=k(t-1),\tag{24}\] is exactly the critical equality 10 .

Lemma 4 (Partition inequality from weights). Assume that \[\label{eq:weight-conditions} \operatorname{wt}(I_J)\le k(|J|-1)\quad(\varnothing\ne J\subsetneq[t]), \qquad \operatorname{wt}(I_{[t]})=k(t-1).\qquad{(12)}\] Then \(H(I_1,\ldots,I_t)\) is \(k\)-weakly-partition-connected and critical.

Proof. Criticality is 23 and 24 . Let \(\mathcal{P}=\{P_1,\ldots,P_s\}\) be a partition of \([t]\). For an edge occurrence \(e\in E(H)\), write again \(e\) for its support set. The identity \[\label{eq:edge-identity} |e|-1=(p_{\mathcal{P}}(e)-1)+\sum_{a=1}^s\max(0,|e\cap P_a|-1)\tag{25}\] holds by splitting the vertices of \(e\) among the blocks it meets. Summing 25 over \(e\in E(H)\) gives \[\begin{align} w_H(\mathcal{P}) &=\sum_{e\in E(H)}(p_{\mathcal{P}}(e)-1)\notag\\ &=\rho(H)-\sum_{a=1}^s\sum_{e\in E(H)}\max(0,|e\cap P_a|-1)\notag\\ &=k(t-1)-\sum_{a=1}^s\operatorname{wt}(I_{P_a}).\label{eq:weight-partition-exact} \end{align}\tag{26}\] If \(s=1\), this is zero and equals \(k(s-1)\). If \(s\ge2\), every \(P_a\) is a nonempty proper subset of \([t]\), so ?? implies \[\label{eq:weight-partition-proof} w_H(\mathcal{P}) \ge k(t-1)-\sum_{a=1}^s k(|P_a|-1) =k(s-1),\tag{27}\] which is 3 . ◻

Lemma 5 (GLSTW determinant expansion). Let \(G=\biguplus_{s}F_{e_s}\) be a tree assignment of \(H(I_1,\ldots,I_t)\) with exactly \(k(t-1)\) labelled graph edges. In the determinant polynomial used in [1], after restricting the row set to the labelled graph edges of \(G\), the Cauchy–Binet expansion has the grouping \[\label{eq:glstw-expansion} D_G(x)=\sum_{\mathcal{Q}=(Q_0,\ldots,Q_{k-1})} c(\mathcal{Q}) \prod_{s=1}^n x_s^{\sum_{i=0}^{k-1}i\,|Q_i\cap E(F_{e_s})|},\qquad{(13)}\] where \(\mathcal{Q}\) ranges over ordered partitions of \(E(G)\) into sets \(Q_i\) of size \(t-1\). Moreover, \(c(\mathcal{Q})=0\) unless each \(Q_i\) is a spanning tree of \([t]\), and \(c(\mathcal{Q})\ne0\) whenever \(\mathcal{Q}\) is an ordered \(k\)-tree decomposition of \(G\).

Proof. This is the determinant expansion from [1], specialized to the fixed row set \(E(G)\). The structural facts needed here are exactly these: choosing the columns belonging to the \(i\)th degree layer selects a set \(Q_i\) of \(t-1\) labelled graph edges and contributes the variable factor \[\label{eq:degree-layer-factor} \prod_{\{j,j'\}\in Q_i} x_{s(j,j')}^i,\tag{28}\] where \(s(j,j')\) is the label of the chosen graph edge; after these variable factors are removed, the remaining incidence determinant vanishes unless every \(Q_i\) is a spanning tree, and it is nonzero for every ordered spanning-tree decomposition. Grouping the resulting terms by the ordered partition \(\mathcal{Q}\) gives ?? . No further property of the GLSTW block is used below. ◻

Theorem 5 (Critical form is sufficient for the GLSTW implication). If Conjecture [conj:critical] holds for all \(k\) and \(t\), then the intersection-matrix nonsingularity conjecture used in [1] follows by the same signature-unique monomial argument as in Theorem 6.2 of that paper.

Proof. Let \(I_1,\ldots,I_t\) satisfy ?? . By Lemma 4, the support hypergraph \(H(I_1,\ldots,I_t)\) is critical and \(k\)-weakly-partition-connected. If Conjecture [conj:critical] holds, there is a tree assignment \(G\) and a decomposition \(\mathcal{T}=(T_0,\ldots,T_{k-1})\) with unique signature. Since \(\rho(H)=k(t-1)\), Lemma 2 gives \(|E(G)|=k(t-1)\), so the full labelled-edge set of the tree assignment is a square row set for the determinant block.

Apply Lemma 5. For an ordered \(k\)-tree decomposition \(\mathcal{Q}=(Q_0,\ldots,Q_{k-1})\), the exponent of \(x_s\) in the corresponding monomial is \[\label{eq:monomial-signature} \sum_{i=0}^{k-1} i\,\left|Q_i\cap E(F_{e_s})\right|.\tag{29}\] Thus the exponent vector is exactly the signature of \(\mathcal{Q}\). The decomposition \(\mathcal{T}\) contributes a nonzero coefficient to the monomial \(x^{\operatorname{sig}_\mathcal{T}}\). If any other nonzero grouped term contributed the same monomial, Lemma 5 would make it an ordered \(k\)-tree decomposition of \(G\) with the same signature as \(\mathcal{T}\), contradicting distinguishability. Hence the monomial \(x^{\operatorname{sig}_\mathcal{T}}\) has nonzero coefficient and cannot cancel. Therefore the determinant polynomial is nonzero. ◻

If 24 were replaced only by \(\operatorname{wt}(I_{[t]})\ge k(t-1)\), then the same calculation as in Lemma 4 would still give weak partition connectivity under the corresponding proper-subset inequalities. However, a full tree assignment would have \(\operatorname{wt}(I_{[t]})\) labelled graph edges rather than exactly \(k(t-1)\) labelled graph edges. The square determinant route can use a full assignment without row deletion only in the critical case; in the overfull case one must either prune labelled rows or exclude the case by equality. This is the same obstruction as 1.

4 Star-realized assignments and partition slack↩︎

We now prove the positive mechanism in 3. A star does not realize the hyperedge cut value for every partition; in general it has extra crossing edges. The point of a two-sided star certificate is different: every hyperedge label is contained in a single tree layer. This layer containment implies weak partition connectivity by a quotient-rank argument, while the extremal layers force signature uniqueness.

For a graph \(G\) on \(V\) and a partition \(\mathcal{P}\) of \(V\), let \[\label{eq:cross-def} \operatorname{cr}_G(\mathcal{P})=|\{xy\in E(G):x\text{ and }y\text{ lie in distinct blocks of }\mathcal{P}\}|.\tag{30}\] If \(F_e\) is a tree on \(e\), then the quotient of \(F_e\) by the blocks met by \(e\) is connected after loops are deleted, and hence \[\label{eq:tree-cross-lower} \operatorname{cr}_{F_e}(\mathcal{P})\ge p_\mathcal{P}(e)-1.\tag{31}\] For arbitrary trees this inequality can be strict. For stars the strictness is completely explicit.

Lemma 6 (Star crossing formula). Let \(e\subseteq V\), let \(c\in e\), and let \(S_e(c)\) be the star on \(e\) centered at \(c\). If \(P_c\) is the block of \(\mathcal{P}\) containing \(c\), then \[\begin{align} \operatorname{cr}_{S_e(c)}(\mathcal{P}) &=\sum_{P\in\mathcal{P}\setminus\{P_c\}} |P\cap e|\label{eq:star-cross-1}\\ &=p_\mathcal{P}(e)-1+\lambda_{e,c}(\mathcal{P}),\label{eq:star-cross-2} \end{align}\] {#eq: sublabel=eq:eq:star-cross-1,eq:eq:star-cross-2} where \[\label{eq:lambda-def} \lambda_{e,c}(\mathcal{P})= \sum_{P\in\mathcal{P}\setminus\{P_c\}}\max(0,|P\cap e|-1)\ge0 .\qquad{(14)}\] In particular, equality \(\operatorname{cr}_{S_e(c)}(\mathcal{P})=p_\mathcal{P}(e)-1\) holds whenever every noncenter block meets \(e\) in at most one vertex, and it always holds when \(|e|=2\).

Proof. Every edge of \(S_e(c)\) has the form \(cu\) with \(u\in e\setminus\{c\}\). Such an edge crosses \(\mathcal{P}\) exactly when \(u\notin P_c\). Therefore \[\label{eq:star-cross-proof-a} \operatorname{cr}_{S_e(c)}(\mathcal{P})=|e\setminus P_c| =\sum_{P\in\mathcal{P}\setminus\{P_c\}}|P\cap e| .\tag{32}\] The number of nonempty blocks of \(\mathcal{P}\) met by \(e\) outside \(P_c\) is \(p_\mathcal{P}(e)-1\), because \(c\in e\cap P_c\). Splitting each summand as \[\label{eq:star-cross-proof-b} |P\cap e|={\boldsymbol{1}}_{P\cap e\ne\varnothing}+\max(0,|P\cap e|-1)\tag{33}\] proves ?? . ◻

The formula shows that a general proof of the critical conjecture cannot proceed by assigning each hyperedge a fixed star and then replacing every hypergraph cut by the corresponding graph cut. The following definition records the stronger property that such an argument would need.

Definition 2 (Partition-tight realization). A tree assignment \(G=\biguplus_eF_e\) is partition-tight if \[\label{eq:partition-tight} \sum_{e\in E(H)}\operatorname{cr}_{F_e}(\mathcal{P})=\sum_{e\in E(H)}\bigl(p_{\mathcal{P}}(e)-1\bigr) \qquad\text{for every partition }\mathcal{P}\text{ of }V.\qquad{(15)}\] Equivalently, since each summand in 31 is nonnegative after subtracting \(p_\mathcal{P}(e)-1\), one has \(\operatorname{cr}_{F_e}(\mathcal{P})=p_\mathcal{P}(e)-1\) for every \(e\) and every \(\mathcal{P}\).

A single hyperedge of size at least \(3\) cannot be partition-tight for every partition: for any tree on at least three vertices, a bipartition obtained by separating a nonempty proper vertex set whose boundary in the tree has size at least two gives at least two crossing tree edges, while the corresponding hyperedge contribution is only one. The certificates below therefore do not rely on partition-tightness.

Lemma 7 (Layer quotient rank). Let \(T\) be a spanning tree of \(V\), let \(\mathcal{P}\) be a partition of \(V\) with \(s=|\mathcal{P}|\), and suppose that the edge set of \(T\) is partitioned into labelled classes \(A_e\) indexed by a set \(\mathcal{L}\) of hyperedges, where every edge in \(A_e\) has both endpoints in \(e\). Then \[\label{eq:layer-rank-bound} s-1\le \sum_{e\in\mathcal{L}}\bigl(p_\mathcal{P}(e)-1\bigr).\qquad{(16)}\]

Proof. Contract every block of \(\mathcal{P}\) in \(T\) and delete loops. The resulting multigraph \(Q\) is connected on the \(s\) blocks of \(\mathcal{P}\), hence its graphic rank is \(s-1\). Let \(Q_e\) be the set of nonloop edges of \(Q\) arising from \(A_e\). By subadditivity of graphic matroid rank, \[\label{eq:rank-subadditive} s-1=r_Q(E(Q))\le \sum_{e\in\mathcal{L}} r_Q(Q_e).\tag{34}\] The edges in \(Q_e\) are incident only with blocks that meet \(e\), so they live on at most \(p_\mathcal{P}(e)\) vertices of \(Q\). Therefore \(r_Q(Q_e)\le p_\mathcal{P}(e)-1\). Substituting this bound in 34 gives ?? . ◻

Proof of 3. Let \(G=\biguplus_eF_e\), \(\mathcal{T}=(T_0,\ldots,T_{k-1})\), and \(\ell:E(H)\to\{0,\ldots,k-1\}\) be the certified assignment, decomposition, and layer map. For each layer \(i\), put \[\label{eq:layer-label-set} E_i\mathrel{:=}\ell^{-1}(i).\tag{35}\] The sets \(E_0,\ldots,E_{k-1}\) partition \(E(H)\), and by layer containment the edge set of \(T_i\) is the disjoint union of the classes \(E(F_e)\) with \(e\in E_i\).

Fix a partition \(\mathcal{P}\) of \(V\), and put \(s=|\mathcal{P}|\). Applying Lemma 7 to each spanning tree \(T_i\) gives \[\label{eq:wpc-from-rank} s-1\le \sum_{e\in E_i}\bigl(p_\mathcal{P}(e)-1\bigr) \qquad(0\le i\le k-1).\tag{36}\] Summing over \(i\) and using that the sets \(E_i\) partition \(E(H)\), we obtain \[\label{eq:wpc-rank-sum} k(s-1) \le \sum_{i=0}^{k-1}\sum_{e\in E_i}\bigl(p_\mathcal{P}(e)-1\bigr) =w_H(\mathcal{P}).\tag{37}\] Thus \(H\) is \(k\)-weakly-partition-connected.

It remains to prove signature uniqueness. Let \(\mathcal{T}'=(T_0',\ldots,T_{k-1}')\) be a \(k\)-tree decomposition of \(G\) with \(\operatorname{sig}_{\mathcal{T}'}=\operatorname{sig}_{\mathcal{T}}\). For a label \(e\), write \(n_e=|E(F_e)|=|e|-1\).

If \(\ell(e)=0\), then all \(n_e\) labelled edges of \(F_e\) lie in \(T_0\) under \(\mathcal{T}\), so \[\label{eq:minus-sig-zero} \operatorname{sig}_\mathcal{T}(e)=0.\tag{38}\] Since every contribution to \(\operatorname{sig}_{\mathcal{T}'}(e)\) is nonnegative, equality of signatures forces \[\label{eq:minus-forced} E(F_e)\subseteq E(T_0').\tag{39}\] If \(\ell(e)=k-1\) and \(\ell(e)\ne0\), then all \(n_e\) labelled edges lie in \(T_{k-1}\) under \(\mathcal{T}\), so \[\label{eq:plus-sig-max} \operatorname{sig}_\mathcal{T}(e)=(k-1)n_e.\tag{40}\] Because each edge contributes at most \(k-1\), equality of signatures forces \[\label{eq:plus-forced} E(F_e)\subseteq E(T_{k-1}').\tag{41}\] Finally, if \(1\le\ell(e)\le k-2\), the certificate gives \(|e|=2\), so \(F_e\) consists of a single labelled graph edge. If this edge lies in \(T_{\ell(e)}\) under \(\mathcal{T}\), then \[\label{eq:middle-one-edge} \operatorname{sig}_\mathcal{T}(e)=\ell(e),\tag{42}\] and equality of signatures forces the same single edge to lie in \(T_{\ell(e)}'\).

Thus every labelled graph edge of \(G\) lies in the same tree of \(\mathcal{T}'\) as it does in \(\mathcal{T}\). Hence \(T_i'=T_i\) for all \(i\), and the signature is unique. ◻

The proof of 3 uses the star-minimal hypothesis only as a concrete way to exhibit layer-contained assignments in the explicit block constructions below. The quotient-rank argument and the signature-isolation argument remain valid for any tree assignment satisfying ?? and ?? . Thus the word “star” describes the explicit certificate class rather than an additional cut-counting mechanism.

The proof uses only the elementary inequalities \[\label{eq:signature-bounds} 0\le \sum_{i=0}^{k-1}i a_i\le(k-1)\sum_{i=0}^{k-1}a_i.\tag{43}\] Equality on the left forces \(a_i=0\) for all \(i>0\), and equality on the right forces \(a_i=0\) for all \(i<k-1\). For an interior index \(1\le c\le k-2\) and \(n_e\ge2\), the same arithmetic forcing is false: for example, \[\label{eq:interior-splitting} (a_{c-1},a_c,a_{c+1})=(1,n_e-2,1) \quad\Longrightarrow\quad \sum_i a_i=n_e,\qquad \sum_i i a_i=c n_e.\tag{44}\] This is why the certificate allows high-rank hyperedges only in the two extremal tree layers, while middle layers use rank-two labels.

The next definition gives a concrete supply of certified non-graphic blocks.

Definition 3 (Saturated two-sided star block). Fix \(t\ge2\) and \(k\ge2\), and let \(V=\{r,u_1,\ldots,u_{t-1}\}\). A saturated two-sided star block consists of two full hyperedges \(e^-,e^+\) equal to \(V\), assigned to stars \(S_V(c_-)\) and \(S_V(c_+)\), together with \((k-2)\) ordinary labelled spanning trees \(R_1,\ldots,R_{k-2}\) on \(V\), each represented by rank-two hyperedges. The certified layers are \[\label{eq:saturated-layers} T_0=S_V(c_-),\qquad T_i=R_i\;(1\le i\le k-2),\qquad T_{k-1}=S_V(c_+),\qquad{(17)}\] where the middle list is empty for \(k=2\).

Every saturated two-sided star block is \(k\)-critical, \(k\)-weakly-partition-connected, and has a \(k\)-distinguishable tree assignment.

Proof. Each layer in ?? is a spanning tree of \(V\). The excess is \[\label{eq:saturated-excess} \rho(H)=2(t-1)+(k-2)(t-1)=k(t-1),\tag{45}\] with the evident interpretation when \(k=2\). Define the layer map by \(\ell(e^-)=0\), \(\ell(e^+)=k-1\), and by assigning every rank-two label of \(R_i\) to layer \(i\) for \(1\le i\le k-2\). This is a two-sided star certificate. The conclusion follows from 3. ◻

5 One-vertex sums and explicit certified families↩︎

We prove the closure theorem. The operation is standard for tree packings and is particularly clean for ordered decompositions.

Let \(H^{(1)}\) and \(H^{(2)}\) have vertex sets \(V_1,V_2\) with \(V_1\cap V_2=\{r\}\). Their one-vertex sum has vertex set \(V=V_1\cup V_2\) and edge multiset the disjoint union of edge occurrences. If \(t_a=|V_a|\), then \[\label{eq:one-sum-size} |V|=t_1+t_2-1.\tag{46}\]

Lemma 8 (Critical excess under one-vertex sums). If \(H^{(a)}\) is \(k\)-critical on \(V_a\) for \(a=1,2\), then \(H^{(1)}\vee_rH^{(2)}\) is \(k\)-critical on \(V_1\cup V_2\).

Proof. Since the edge occurrences are disjoint, \[\begin{align} \rho(H^{(1)}\vee_rH^{(2)}) &=\rho(H^{(1)})+\rho(H^{(2)})\notag\\ &=k(t_1-1)+k(t_2-1) =k(t_1+t_2-2)\notag\\ &=k(|V_1\cup V_2|-1).\label{eq:critical-sum-proof} \end{align}\tag{47}\]  ◻

Lemma 9 (Tree decomposition under one-vertex sums). Let \(G^{(a)}=T^{(a)}_0\dot{\cup}\cdots\dot{\cup} T^{(a)}_{k-1}\) be ordered \(k\)-tree decompositions on \(V_a\), \(a=1,2\), with \(V_1\cap V_2=\{r\}\). Then \[\label{eq:T-sum} T_i\mathrel{:=}T^{(1)}_i\cup T^{(2)}_i\qquad(0\le i\le k-1)\qquad{(18)}\] are spanning trees of \(V_1\cup V_2\), and \[\label{eq:G-sum-decomp} G^{(1)}\cup G^{(2)}=T_0\dot{\cup}\cdots\dot{\cup} T_{k-1}.\qquad{(19)}\]

Proof. Each \(T_i\) is connected because \(T_i^{(1)}\) and \(T_i^{(2)}\) both contain \(r\). It is acyclic because a cycle in the union of two graphs meeting in the single vertex \(r\) would have to be contained entirely in one side. Equivalently, \[\label{eq:tree-edge-count-sum} |E(T_i)|=(t_1-1)+(t_2-1)=|V_1\cup V_2|-1,\tag{48}\] and \(T_i\) is connected, hence a tree. Edge-disjointness is inherited from the two decompositions. ◻

Lemma 10 (Two-sided star certificates glue). If \(H^{(1)}\) and \(H^{(2)}\) have two-sided star certificates, then \(H^{(1)}\vee_rH^{(2)}\) has a two-sided star certificate.

Proof. Take the disjoint union of the two tree assignments and use ?? for the target decomposition. If \(\ell_a:E(H^{(a)})\to\{0,\ldots,k-1\}\) is the layer map on side \(a\), define the layer map on the edge-disjoint union by \(\ell(e)=\ell_a(e)\) for \(e\in E(H^{(a)})\). Each hyperedge occurrence lies wholly on one side, so its assigned tree remains the same star or rank-two edge in the one-vertex sum. Layer containment and the rank-two condition for interior layers remain true in the corresponding glued tree layer. ◻

Proof of 4. Combine Lemmas 810 and then apply 3. ◻

The closure theorem gives infinite non-graphic examples because a single block may contain hyperedges of arbitrary size in the two extremal layers. Figure 3 illustrates one step of the construction.

Figure 3: One-vertex sum at the cut vertex r. Each certified tree layer is glued to the corresponding layer on the other side. The resulting layer is again a spanning tree, and label containment in layers is preserved.

6 Signature isolation as a labelled matroid statement↩︎

This section rewrites the uniqueness argument in the language of bases of the graphic matroid. The reformulation is useful because it separates two issues that are conflated in the literal conjecture: existence of a packing and uniqueness of the label signature.

Let \(M(G)\) be the graphic matroid of a graph \(G\) on \(V\). A spanning tree is a basis of \(M(G)\). A \(k\)-tree decomposition is an ordered basis partition \[\label{eq:matroid-basis-partition} E(G)=B_0\dot{\cup}\cdots\dot{\cup} B_{k-1}, \qquad B_i\in\mathcal{B}(M(G)).\tag{49}\] The signature map is the linear projection \[\label{eq:linear-sig} \Phi:\mathbb{Z}^{E(G)}\times\cdots\times\mathbb{Z}^{E(G)}\longrightarrow\mathbb{Z}^{E(H)}, \qquad \Phi(\mathbf{1}_{B_0},\ldots,\mathbf{1}_{B_{k-1}})_e =\sum_{i=0}^{k-1}i\,|B_i\cap E(F_e)|.\tag{50}\] Thus distinguishability asks for a fiber of \(\Phi\) over the set of ordered basis partitions to be a singleton.

For a fixed label \(e\), let \[\label{eq:label-multiplicity} n_e\mathrel{:=}|E(F_e)|=|e|-1, \qquad a_i(e)\mathrel{:=}|B_i\cap E(F_e)|.\tag{51}\] Then every decomposition satisfies \[\label{eq:a-constraints} a_i(e)\in\mathbb{Z}_{\ge0},\qquad \sum_{i=0}^{k-1}a_i(e)=n_e, \qquad \operatorname{sig}(e)=\sum_{i=0}^{k-1}i a_i(e).\tag{52}\] For the extremal layers \(0\) and \(k-1\), the signature determines the whole vector \((a_0(e),\ldots,a_{k-1}(e))\) when the target average is extremal: \[\begin{align} \operatorname{sig}(e)=0 &\Longleftrightarrow (a_0(e),a_1(e),\ldots,a_{k-1}(e))=(n_e,0,\ldots,0),\tag{53}\\ \operatorname{sig}(e)=(k-1)n_e &\Longleftrightarrow (a_0(e),a_1(e),\ldots,a_{k-1}(e))=(0,\ldots,0,n_e). \tag{54} \end{align}\] This is the algebraic core of 3. When \(k=1\) the two displayed extremal conditions coincide and are tautological.

For an interior layer \(c\), the real affine fiber is usually positive-dimensional, and even its integer points need not be unique. The affine fiber is \[\label{eq:interior-fiber} \mathcal{A}_{c,n} =\left\{a\in\mathbb{R}_{\ge0}^k: \sum_i a_i=n, \sum_i i a_i=cn\right\}.\tag{55}\] For \(1\le c\le k-2\) and \(n\ge2\), this polytope contains the nontrivial point \[\label{eq:interior-point} a_c=n-2, \qquad a_{c-1}=a_{c+1}=1, \qquad a_i=0\text{ otherwise},\tag{56}\] so the signature alone cannot force all \(e\)-labelled edges into \(B_c\). This explains why the conjecture is substantially harder than ordinary tree packing: uniqueness is not a matroid-union consequence but a labelled fiber-isolation problem.

Fix \(k\ge3\), \(1\le c\le k-2\), and \(n\ge2\). There are two distinct integer vectors \(a,b\in\mathbb{Z}_{\ge0}^k\) with \[\label{eq:same-average} \sum_i a_i=\sum_i b_i=n, \qquad \sum_i i a_i=\sum_i i b_i=cn,\tag{57}\] where \(b_c=n\) and \(a\) is not equal to \(b\). Hence no proof that uses only per-label signature arithmetic can force a high-rank label to remain in an interior layer.

Proof. Take \(b_c=n\) and use 56 for \(a\) when \(n\ge2\). Then \[\label{eq:average-check} \sum_i i a_i=(c-1)+c(n-2)+(c+1)=cn.\tag{58}\] Clearly \(a\ne b\). ◻

7 Critical examples and explicit calculations↩︎

We give three families to clarify the boundary between the negative theorem and the star-isolated positive theorem.

Example 1 (The minimal literal counterexample). Let \(V=\{a,b\}\), \(k=1\), and \(E=\{e_1,e_2\}\) with \(e_1=e_2=\{a,b\}\). For the only nontrivial partition \(\mathcal{P}=\{\{a\},\{b\}\}\), \[\label{eq:minimal-wpc} w_H(\mathcal{P})=(2-1)+(2-1)=2\ge1=k(|\mathcal{P}|-1).\qquad{(20)}\] Thus \(H\) is \(1\)-weakly-partition-connected. But every tree assignment is the graph with two parallel labelled edges from \(a\) to \(b\): \[\label{eq:minimal-G} |E(G)|=2, \qquad k(t-1)=1.\qquad{(21)}\] A \(1\)-tree decomposition would have one edge, so the full tree assignment cannot be decomposed. The conjecture fails before any signature issue arises.

Example 2 (A critical rank-two graph). If \(H\) is an ordinary multigraph with exactly \(k(t-1)\) edges and is \(k\)-partition-connected in the usual Nash–Williams–Tutte sense, then the identity tree assignment has a \(k\)-tree decomposition. Since each label occurs once, the signature records the tree index of every edge: \[\label{eq:rank-two-signature} \operatorname{sig}_\mathcal{T}(e)=i \quad\Longleftrightarrow\quad e\in T_i.\qquad{(22)}\] Thus every ordered decomposition is automatically signature-isolated. The hypergraph conjecture reduces to ordinary tree packing in rank two.

Example 3 (A non-graphic two-sided star certificate). Let \(k=2\) and let \(V=\{r,a,b\}\). Consider the critical hypergraph with one triple edge \(e^-=\{r,a,b\}\) and two rank-two edges \(f_a=\{r,a\}\), \(f_b=\{r,b\}\). Assign \[\label{eq:nongraphic-assignment} F_{e^-}=\{ra,rb\}, \qquad F_{f_a}=\{ra\}, \qquad F_{f_b}=\{rb\},\qquad{(23)}\] with the two edges of \(F_{e^-}\) placed in \(T_0\) and the rank-two edges placed in \(T_1\). Equivalently, the certificate has layer map \[\label{eq:nongraphic-layer-map} \ell(e^-)=0, \qquad \ell(f_a)=\ell(f_b)=1,\qquad{(24)}\] so there is no middle part when \(k=2\). The excess is \[\label{eq:nongraphic-excess} \rho(H)=(3-1)+(2-1)+(2-1)=4=2(|V|-1).\qquad{(25)}\] The weak partition inequalities are checked directly. For the discrete partition, \[\label{eq:nongraphic-discrete} w_H(\{r|a|b\})=(3-1)+(2-1)+(2-1)=4=2(3-1).\qquad{(26)}\] For a two-block partition, if the singleton is \(r\), then \[\label{eq:nongraphic-r} w_H(\{r|ab\})=1+1+1=3\ge2,\qquad{(27)}\] while if the singleton is \(a\) or \(b\), the rank-two edge contained in the opposite two-block contributes zero and the total is \(1+1+0=2\). Hence \(H\) is \(2\)-weakly-partition-connected. The signature of \(e^-\) is zero, so both triple-edge assigned edges are forced into \(T_0\); the rank-two labels have signature one, so their unique graph edges are forced into \(T_1\). This gives a non-graphic positive instance.

The last example is small but representative: high-rank labels can be made harmless when they sit in an extremal layer and the remaining cut inequalities are supplied by rank-two labels. One-vertex sums of this block yield arbitrarily large examples with high-rank labels.

8 Exact slack calculus and finite verification↩︎

Although the proof of 3 uses quotient rank rather than direct cut counting, it is useful to record the exact slack identity behind any fixed tree assignment and tree decomposition. This identity explains why stars may have extra graph-crossing edges without contradicting weak partition connectivity.

Let \(F=(F_e)_{e\in E(H)}\) be any tree assignment, and put \[\label{eq:global-lambda-def} \Lambda_F(\mathcal{P}) =\sum_{e\in E(H)}\left(\operatorname{cr}_{F_e}(\mathcal{P})-p_\mathcal{P}(e)+1\right)\ge0 .\tag{59}\] If \(G=\biguplus_eF_e\) is decomposed as \(G=T_0\dot{\cup}\cdots\dot{\cup} T_{k-1}\), define the graphical partition surplus \[\label{eq:graphical-surplus-def} B_{\mathcal{T}}(\mathcal{P}) =\sum_{i=0}^{k-1}\operatorname{cr}_{T_i}(\mathcal{P})-k(|\mathcal{P}|-1)\ge0 .\tag{60}\] The nonnegativity in 60 is the ordinary tree crossing inequality applied to each spanning tree \(T_i\).

For every partition \(\mathcal{P}\) of \(V\), \[\label{eq:slack-identity} w_H(\mathcal{P})-k(|\mathcal{P}|-1)=B_{\mathcal{T}}(\mathcal{P})-\Lambda_F(\mathcal{P}).\tag{61}\] Consequently, for a fixed tree assignment and a fixed \(k\)-tree decomposition of it, \[\label{eq:wpc-slack-equivalence} H\text{ is }k\text{-weakly-partition-connected} \quad\Longleftrightarrow\quad \Lambda_F(\mathcal{P})\le B_{\mathcal{T}}(\mathcal{P})\text{ for every }\mathcal{P}.\tag{62}\]

Proof. By the definition of \(\Lambda_F\), \[\begin{align} w_H(\mathcal{P}) &=\sum_{e\in E(H)}(p_\mathcal{P}(e)-1)\notag\\ &=\sum_{e\in E(H)}\operatorname{cr}_{F_e}(\mathcal{P})-\Lambda_F(\mathcal{P})\notag\\ &=\sum_{i=0}^{k-1}\operatorname{cr}_{T_i}(\mathcal{P})-\Lambda_F(\mathcal{P}),\label{eq:slack-identity-proof} \end{align}\tag{63}\] where the last equality uses the edge-disjoint union \(G=T_0\dot{\cup}\cdots\dot{\cup} T_{k-1}\). Subtracting \(k(|\mathcal{P}|-1)\) gives 61 . The equivalence 62 is exactly the definition of weak partition connectivity. ◻

If the assignment is star-realized with centers \(c(e)\in e\), Lemma 6 turns 59 into a closed expression: \[\label{eq:star-global-lambda} \Lambda_F(\mathcal{P})= \sum_{e\in E(H)}\lambda_{e,c(e)}(\mathcal{P}) =\sum_{e\in E(H)}\sum_{P\in\mathcal{P}\setminus\{P_{c(e)}\}} \max(0,|P\cap e|-1).\tag{64}\] Thus, for a fixed star assignment and fixed decomposition, weak partition connectivity is equivalent to the finite system \[\label{eq:finite-system} \sum_{e\in E(H)}\sum_{P\in\mathcal{P}\setminus\{P_{c(e)}\}} \max(0,|P\cap e|-1) \le \sum_{i=0}^{k-1}\operatorname{cr}_{T_i}(\mathcal{P})-k(|\mathcal{P}|-1) \qquad(\mathcal{P}\in\Pi(V)),\tag{65}\] where \(\Pi(V)\) denotes the set of all set partitions of \(V\). For a two-sided star certificate this system holds automatically by 3; the quotient-rank proof is a shorter certificate of the same inequalities.

The identity also isolates the role of the discrete partition. If \(\mathcal{P}_0=\{\{v\}:v\in V\}\), then every tree edge crosses \(\mathcal{P}_0\), and hence \[\label{eq:discrete-no-slack} \Lambda_F(\mathcal{P}_0)= \sum_e\left(|E(F_e)|-(|e|-1)\right)=0.\tag{66}\] Therefore the discrete partition cannot absorb any overfullness by changing the tree assignment. It only sees \[\label{eq:discrete-gap} w_H(\mathcal{P}_0)-k(|\mathcal{P}_0|-1)=\rho(H)-k(t-1).\tag{67}\] This is the numerical obstruction in 1 in its most rigid form.

Let \(H_{t,k,q}\) be the hypergraph on \(V\), \(|V|=t\), consisting of \(k+q\) parallel copies of the full hyperedge \(V\). If \(\mathcal{P}\) has \(s\) blocks, then \[\label{eq:bundle-surplus} w_{H_{t,k,q}}(\mathcal{P})-k(s-1)=q(s-1).\tag{68}\] In particular, the weak connectivity margin is positive on every nontrivial partition, while the full-assignment edge excess is \[\label{eq:bundle-excess-gap} \rho(H_{t,k,q})-k(t-1)=q(t-1).\tag{69}\]

Proof. Every copy of \(V\) meets all \(s\) blocks of \(\mathcal{P}\), so each contributes \(s-1\) to \(w_H(\mathcal{P})\). Hence \[\label{eq:bundle-surplus-proof} w_{H_{t,k,q}}(\mathcal{P})=(k+q)(s-1),\tag{70}\] which gives 68 . Taking \(\mathcal{P}=\mathcal{P}_0\) gives 69 . ◻

The smallest counterexamples are completely classified. This is useful because it shows that the obstruction persists even when no high-rank hyperedge is present.

Let \(L_m\) be the hypergraph on \(V=\{x,y\}\) consisting of \(m\) labelled copies of the rank-two hyperedge \(\{x,y\}\). For fixed \(k\ge1\), \[\begin{align} L_m\text{ is }k\text{-weakly-partition-connected} &\Longleftrightarrow m\ge k,\tag{71}\\ L_m\text{ has a }k\text{-distinguishable full tree assignment} &\Longleftrightarrow m=k.\tag{72} \end{align}\] Thus \(L_{k+q}\) with \(q\ge1\) is a counterexample to the literal conjecture, and \(L_{k+1}\) is deletion-minimal with respect to the edge-count obstruction.

Proof. There is only one nontrivial partition, namely \(\mathcal{P}=\{\{x\},\{y\}\}\). For this partition, \[\label{eq:line-wpc-proof} w_{L_m}(\mathcal{P})=m, \qquad k(|\mathcal{P}|-1)=k,\tag{73}\] which proves 71 . A tree assignment of \(L_m\) is the labelled multigraph with \(m\) parallel edges \(xy\). A spanning tree on two vertices has exactly one edge, so a \(k\)-tree decomposition of the full assignment exists only when \(m=k\). If \(m=k\), choose an ordering \(e_0,\ldots,e_{k-1}\) and put \(T_i=\{e_i\}\). The signature of \(e_i\) is \(i\), and since each label has only one graph edge, any decomposition with the same signature must place \(e_i\) in the same layer \(T_i\). Hence the signature is unique. ◻

For arbitrary fixed assignments, the slack identity separates two tasks: \[\label{eq:two-independent-tasks} \boxed{\text{signature isolation}} \qquad\text{and}\qquad \boxed{\Lambda_F(\mathcal{P})\le B_\mathcal{T}(\mathcal{P})\text{ for all }\mathcal{P}}.\tag{74}\] The two-sided certificate class handles them simultaneously: extremal signature weights isolate the labels, and the quotient-rank argument proves the partition inequalities. The literal conjecture fails before either task is relevant, because an overfull assignment has too many labelled graph edges to be a union of exactly \(k\) spanning trees.

9 Consequences for formulations of the conjecture↩︎

The negative theorem suggests three logically distinct statements.

  1. Literal full-assignment statement. \[\label{eq:form-A} \begin{align} &k\text{-WPC}(H)\\ &\qquad\Longrightarrow \exists\text{ a }k\text{-distinguishable full tree assignment of }H. \end{align}\tag{75}\] This is false by 2.

  2. Critical full-assignment statement. \[\label{eq:form-B} \begin{align} &k\text{-WPC}(H)\text{ and }\rho(H)=k(t-1)\\ &\qquad\Longrightarrow \exists\text{ a }k\text{-distinguishable full tree assignment of }H. \end{align}\tag{76}\] This is Conjecture [conj:critical]. It is not settled here in full generality, but it is the exact form compatible with the intersection-matrix application and with edge counting.

  3. Subassignment or row-pruning statement. \[\label{eq:form-C} \begin{align} k\text{-WPC}(H) \Longrightarrow{}& \exists\text{ a tree assignment }G\text{ of }H\\ &\text{and a labelled spanning subgraph }G'\subseteq G \end{align}\tag{77}\] with \(|E(G')|=k(t-1)\) such that \(G'\) has a signature-isolated \(k\)-tree decomposition, where the signature is computed from the label classes inherited by \(G'\). This avoids the overfull obstruction but adds a nontrivial pruning problem. The logarithmic relaxation in [6], [7] can be interpreted in this direction, but it does not imply the exact row-pruning statement.

The implications are \[\label{eq:form-implications} \text{(A)}\Longrightarrow\text{(B)}, \qquad \text{(C)}\Longrightarrow\text{a corrected overfull version},\tag{78}\] while 1 says that (A) is impossible. The formulation of (C) must allow pruning at the level of labelled graph rows or assigned edges, not only at the level of deleting whole hyperedge occurrences. Whole-hyperedge pruning has a separate arithmetic obstruction: for example, on \(V=\{1,2,3,4\}\) with \(k=1\), the two hyperedges \(\{1,2,3\}\) and \(\{1,2,4\}\) form a \(1\)-weakly-partition-connected hypergraph of excess \(4\), but no subhypergraph has excess \(3=k(|V|-1)\). Indeed, for any partition \(\mathcal{P}\) with \(s\) blocks, the two hyperedges cover \(V\) and intersect nontrivially, so their block-incidence counts satisfy \(p_\mathcal{P}(e_1)+p_\mathcal{P}(e_2)\ge s+1\), giving \(w_H(\mathcal{P})\ge s-1\).

From the viewpoint of Reed–Solomon codes, (B) is the appropriate full-assignment conjecture because the labelled-edge row set from a full tree assignment has exactly the required square size precisely when 24 holds.

For support hypergraphs coming from \(I_1,\ldots,I_t\subseteq[n]\), the following are equivalent: \[ &\rho(H(I_1,\ldots,I_t))=k(t-1),\label{eq:square-1}\\ &\operatorname{wt}(I_{[t]})=k(t-1),\label{eq:square-2}\\ &\begin{gather} \text{a full tree assignment supplies exactly }k(t-1)\text{ labelled-edge rows}\\ \text{for the tree-packing determinant block.} \end{gather} \label{eq:square-3}\] {#eq: sublabel=eq:eq:square-1,eq:eq:square-2,eq:eq:square-3}

Proof. The equivalence of ?? and ?? is 23 . Every full tree assignment has \(\rho(H(I_1,\ldots,I_t))\) labelled graph edges by Lemma 2. Thus the full labelled-edge row set has size \(k(t-1)\) exactly when ?? holds. If \(\rho(H)>k(t-1)\), one may still try to select a square submatrix by deleting rows, but that is a pruning operation rather than a full-assignment implication. ◻

10 Conclusion↩︎

The literal hypergraph Nash–Williams–Tutte conjecture has a decisive edge-count obstruction. Weak partition connectivity says \[\label{eq:conclusion1} \rho(H)\ge k(t-1),\tag{79}\] whereas a full \(k\)-distinguishable tree assignment requires \[\label{eq:conclusion2} \rho(H)=k(t-1).\tag{80}\] The family \(H_{t,k,q}\) with \(k+q\) copies of the full hyperedge gives counterexamples for every \(t\ge2\), \(k\ge1\), and \(q\ge1\). Thus the original statement must be replaced by either a critical equality hypothesis or by an explicit subassignment/row-pruning conclusion.

The critical form remains the mathematically meaningful version for the Reed–Solomon route through intersection matrices: the equality \(\operatorname{wt}(I_{[t]})=k(t-1)\) is exactly \(\rho(H)=k(t-1)\), and it is precisely the condition under which the full tree-assignment row set has the square size required by the determinant block without row pruning. Within the critical regime, the main structural problem is not merely the existence of spanning-tree packings but the isolation of a label signature. The two-sided star certificates in this paper provide an explicit infinite non-graphic family where weak partition connectivity follows from quotient rank, signature isolation is forced by extremal weights, and the construction remains stable under one-vertex sums.

The next natural target is an exact row-pruning problem: \[\label{eq:pruning-target} \begin{align} &k\text{-WPC}(H),\;\rho(H)>k(t-1)\\ &\qquad\Longrightarrow?\quad \exists\text{ a }k(t-1)\text{-edge labelled subgraph of a tree assignment}\\ &\qquad\text{with isolated signature.} \end{align}\tag{81}\] This pruning statement is independent of signature uniqueness for the critical full-assignment conjecture, and separating the two problems should make future work on the hypergraph Nash–Williams–Tutte program more precise.

Declaration of Generative AI and AI-Assisted Technologies in the Writing Process↩︎

During the preparation of this work, the authors used DeepSeek to build a specialized agent for solving mathematical problems, which was employed to generate an initial proof of the main theorem. After using this tool, the authors reviewed and edited the content as needed and take full responsibility for the content of the published article.

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  1. Corresponding author. School of Mathematics, Sichuan University, 24 First Loop Road South Section I, Chengdu, Sichuan 610064, China. Email: .↩︎

  2. School of Mathematics, Sichuan University, 24 First Loop Road South Section I, Chengdu, Sichuan 610064, China. Email: .↩︎