The problem
Draw four dots in a chain, with three lines joining consecutive dots. Delete one dot and its incident lines at a time. Deleting either end leaves a three-dot chain; deleting either middle dot leaves one line and an isolated dot. Keep all four little drawings, including repeats, but remove names. Would this collection of partial views always determine the original network up to renaming?[1]
Suppose every one-vertex deletion of a finite graph is available, but vertex names are absent from each resulting card. Is the original connection pattern uniquely determined? A deck is a multiset: a collection retaining the number of times each shape occurs. In a four-vertex path, two cards are three-vertex paths and two are one edge plus an isolated vertex.
The conjecture concerns simple undirected graphs, with no loops or repeated edges, on vertices. If , must there be a vertex renaming preserving every edge and nonedge between and ? Graphs related by such a renaming are called isomorphic. Labels used to describe an example are not extra information supplied with its cards.[1] [2]
Scope
Finite simple undirected graphs with at least three vertices; the multiset of all one-vertex-deleted subgraphs.
What remains open
The conjecture covers finite networks with at least three dots, undirected lines, no loops and no multiple lines between a pair. It asks for every such network. Working examples and recoverable statistics do not yet prove that two different networks can never share the same collection.[1]
Some information is already recoverable. Each original edge appears in exactly cards, since deleting either endpoint removes it. Hence
counting every repeated card. In the path example the card edge counts are , giving original edges. Subtracting each card's count from three recovers the deleted vertices' degrees: .
These totals do not specify every connection. For instance, a five-vertex path and a separate triangle plus an edge both have degrees , yet are different graphs. They do not have the same full deck. The general deck-uniqueness statement remains the gap.[1] [2]
Why it matters
Missing names make the information question sharp: there is no given way to match a dot in one view to a dot in another. The interest is whether their overlap nevertheless preserves every connection, rather than merely totals or summaries.[1]
A full answer would identify whether the cards lose any structural information once names are removed. With only two vertices, a joined pair and an unjoined pair both leave two single-dot cards; this explains the lower size restriction. The 2026 survey connects reconstruction to algebraic quantities unchanged by vertex renaming. Its proposed route seeks quantities that distinguish all original graphs from their decks, rather than only a few recoverable statistics.[1]
Definitions
3 terms
deck
The multiset of all one-vertex-deleted graphs, with vertex names forgotten. There is one card per deleted vertex; two identical shapes still count twice.[1] [2]
multiset
A collection that counts repetitions. The path example has four cards even though only two different shapes occur; discarding their frequencies would discard part of the supplied information.[1]
Sources
- Shuffling the Deck: Invariant Theory and the Graph Reconstruction Conjecture
Notes on Shuffling the Deck: Invariant Theory and the Graph Reconstruction Conjecture
Fresh April 17, 2026 version-1 expert survey read at deck definition, edge/degree recovery, special families and algebraic motivation. Toy deletion cards, edge counts and degree ambiguity independently checked; no complete reconstruction proof or survey of every subsequent claim is asserted.
- A congruence theorem for trees
Notes on A congruence theorem for trees
Fresh original journal PDF definition and Lemmas 1–2 describe vertex-deleted shapes with their frequencies, edge/degree recovery and vertex renaming. Its theorem is restricted to trees, not every graph. The cover gives January 1957, volume 7, number 1, pages 961–968; the received date in 1955 is not publication. No exact publication day or complete tree-proof audit is inferred.