For every finite connected graph G, the stacking and clearing thresholds are determined by almost stacked configurations: operatornamestack(G)=mint∈N: every configuration C:V(G)→N with ∑_v∈ V(G)C(v)=t and |v∈ V(G):C(v)≥ 2|≤ 1 is stackable, and operatornameclear(G)=mint∈N: every configuration C:V(G)→N with ∑_v∈ V(G)C(v)=t and |v∈ V(G):C(v)≥ 2|≤ 1 is clearable.

Almost Stacked Hypothesis and the tree stacking-number conjecture

For every finite connected graph $G$ , the stacking and clearing thresholds are determined by almost stacked configurations: $\operatorname{stack}(G)=\min\{t\in\mathbb{N}:\text{ every configuration }C:V(G)\to\mathbb{N}\text{ with }\sum_{v\in V(G)}C(v)=t\text{ and }|\{v\in V(G):C(v)\ge 2\}|\le 1\text{ is stackable}\}$ , and $\operatorname{clear}(G)=\min\{t\in\mathbb{N}:\text{ every configuration }C:V(G)\to\mathbb{N}\text{ with }\sum_{v\in V(G)}C(v)=t\text{ and }|\{v\in V(G):C(v)\ge 2\}|\le 1\text{ is clearable}\}$ .

References

Primary source

arXiv

Additional references

The Almost Stacked Hypothesis: A Conjectural Analogue of Sjöstrand's Cover Pebbling Theorem — arXiv — Tamás Csernák, Lajos Soukup

graph-pebbling graph-theory

Posted 11 hours ago · edited 11 hours ago

Progress summary

Refreshed 11 hours ago

Claimed progress

A new paper claims to settle the tree case, but the broader hypothesis remains open and the claim has not been independently checked.

Csernák and Soukup’s work proposes the Almost Stacked Hypothesis and a tree conjecture asserting that the stacking number equals a rooted estimator.

Known results

For cycles, the paper proves lower bounds and obtains exact values only assuming ASH.
For finite trees, it proves an upper bound conditional on ASH and reports computational support through $7$ vertices.
The paper gives exact computed values $\operatorname{stack}(C_3),\operatorname{stack}(C_5),\operatorname{stack}(C_7),\operatorname{stack}(C_9),\operatorname{stack}(C_{11})=4,8,17,37,77$ .

October 2026 tree-case claim

A separate arXiv paper claims $\operatorname{stack}(T)=\operatorname{estim}(T)$ for every finite tree, using a recursive characterization, a zero-score obstruction, weighted cancellation, and Lean formalization. No independent mathematical assessment or verification was found; it does not prove ASH.

Current status (as of October 2026): ASH remains open; the tree equality has a claimed but unverified proof, while the cycle results and conditional implications are established.

Sources

No solutions have been posted yet.