Stanley’s tree-isomorphism conjecture

For all finite trees T1T_1 and T2T_2, if their chromatic symmetric functions are equal, XT1=XT2X_{T_1}=X_{T_2}, then the trees are isomorphic, T1T2T_1\cong T_2. Here, for a finite graph GG, XG(x)=κvV(G)xκ(v)X_G(\mathbf{x})=\sum_{\kappa}\prod_{v\in V(G)}x_{\kappa(v)}, where the sum is over all proper colorings κ:V(G)N\kappa:V(G)\to\mathbb{N}; equivalently, the chromatic symmetric function distinguishes all nonisomorphic finite trees.

Progress summary

Partially solved

The conjecture remains open, but new work proves it for broader classes of trees and leaves only difficult cases unresolved.

Stanley asked whether two trees with the same chromatic symmetric function must be isomorphic. The unrestricted conjecture remains unresolved.

Known results

  • Caterpillars: established in 2018 via reconstruction from the UU-polynomial.
  • Trees of diameter at most 55: reconstructed from their chromatic symmetric functions by an explicit algorithm (2025).
  • Trees with exactly two vertices of degree at least 33: settled affirmatively (2023).
  • Further classes generalizing proper caterpillars: established in 2023.

August 2026 reconstruction criterion

Zeng, Zijian established a criterion that handles repeated leaf-component orders, extending the verified range beyond earlier distinct-order conditions. The result advances reconstruction but explicitly leaves arbitrary diameter-66 trees and the general conjecture open.

Current status (as of August 2026): The conjecture is proved for several substantial classes, including all trees of diameter at most 55, but arbitrary diameter-66 trees and the unrestricted case remain open.

Sources
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Primary source

arXiv

Additional references

Solutions 0

No solutions have been posted yet.