Reconstruction conjecture

A graph and the associated deck of single-vertex-deleted subgraphs. Note some of the cards show isomorphic graphs.

Given a graph ${\displaystyle G=(V,E)}$, a vertex-deleted subgraph of ${\displaystyle G}$ is a subgraph formed by deleting exactly one vertex from ${\displaystyle G}$. By definition, it is an induced subgraph of ${\displaystyle G}$.

For a graph ${\displaystyle G}$, the deck of G, denoted ${\displaystyle D(G)}$, is the multiset of isomorphism classes of all vertex-deleted subgraphs of ${\displaystyle G}$. Each graph in ${\displaystyle D(G)}$ is called a card. Two graphs that have the same deck are said to be hypomorphic.

With these definitions, the conjecture can be stated as:

• Reconstruction Conjecture: Any two hypomorphic graphs on at least three vertices are isomorphic.

(The requirement that the graphs have at least three vertices is necessary because both graphs on two vertices have the same decks.)

Harary[4] suggested a stronger version of the conjecture:

• Set Reconstruction Conjecture: Any two graphs on at least four vertices with the same sets of vertex-deleted subgraphs are isomorphic.

Given a graph ${\displaystyle G=(V,E)}$, an edge-deleted subgraph of ${\displaystyle G}$ is a subgraph formed by deleting exactly one edge from ${\displaystyle G}$.

For a graph ${\displaystyle G}$, the edge-deck of G, denoted ${\displaystyle ED(G)}$, is the multiset of all isomorphism classes of edge-deleted subgraphs of ${\displaystyle G}$. Each graph in ${\displaystyle ED(G)}$ is called an edge-card.

• Edge Reconstruction Conjecture: (Harary, 1964)[4] Any two graphs with at least four edges and having the same edge-decks are isomorphic.

In context of the reconstruction conjecture, a graph property is called recognizable if one can determine the property from the deck of a graph. The following properties of graphs are recognizable:

• Order of the graph – The order of a graph ${\displaystyle G}$, ${\displaystyle |V(G)|}$ is recognizable from ${\displaystyle D(G)}$ as the multiset ${\displaystyle D(G)}$ contains each subgraph of ${\displaystyle G}$ created by deleting one vertex of ${\displaystyle G}$. Hence ${\displaystyle |V(G)|=|D(G)|}$ [5]
• Number of edges of the graph – The number of edges in a graph ${\displaystyle G}$ with ${\displaystyle n}$ vertices, ${\displaystyle |E(G)|}$ is recognizable. First note that each edge of ${\displaystyle G}$ occurs in ${\displaystyle n-2}$ members of ${\displaystyle D(G)}$. This is true by the definition of ${\displaystyle D(G)}$ which ensures that each edge is included every time that each of the vertices it is incident with is included in a member of ${\displaystyle D(G)}$, so an edge will occur in every member of ${\displaystyle D(G)}$ except for the two in which its endpoints are deleted. Hence, ${\displaystyle |E(G)|=\sum {\frac {q_{i}}{n-2}}}$ where ${\displaystyle q_{i}}$ is the number of edges in the ith member of ${\displaystyle D(G)}$.[5]
• Degree sequence – The degree sequence of a graph ${\displaystyle G}$ is recognizable because the degree of every vertex is recognizable. To find the degree of a vertex ${\displaystyle v_{i}}$—the vertex absent from the ith member of ${\displaystyle D(G)}$—, we will examine the graph created by deleting it, ${\displaystyle G_{i}}$. This graph contains all of the edges not incident with ${\displaystyle v_{i}}$, so if ${\displaystyle q_{i}}$ is the number of edges in ${\displaystyle G_{i}}$, then ${\displaystyle |E(G)|-q_{i}=\deg(v_{i})}$. If we can tell the degree of every vertex in the graph, we can tell the degree sequence of the graph.[5]
• (Vertex-)Connectivity – By definition, a graph is ${\displaystyle n}$-vertex-connected when deleting any vertex creates a ${\displaystyle n-1}$-vertex-connected graph; thus, if every card is a ${\displaystyle n-1}$-vertex-connected graph, we know the original graph was ${\displaystyle n}$-vertex-connected. We can also determine if the original graph was connected, as this is equivalent to having any two of the ${\displaystyle G_{i}}$ being connected.[5]
• Tutte polynomial
• Characteristic polynomial
• Planarity
• The types of spanning trees in a graph
• Chromatic polynomial
• Being a perfect graph or an interval graph, or certain other subclasses of perfect graphs[6]

Both the reconstruction and set reconstruction conjectures have been verified for all graphs with at most 11 vertices by Brendan McKay.[7]

In a probabilistic sense, it has been shown by Béla Bollobás that almost all graphs are reconstructible.[8] This means that the probability that a randomly chosen graph on ${\displaystyle n}$ vertices is not reconstructible goes to 0 as ${\displaystyle n}$ goes to infinity. In fact, it was shown that not only are almost all graphs reconstructible, but in fact that the entire deck is not necessary to reconstruct them — almost all graphs have the property that there exist three cards in their deck that uniquely determine the graph.

The reconstruction conjecture is true if all 2-connected graphs are reconstructible. [11]

The vertex reconstruction conjecture obeys the duality that if ${\displaystyle G}$ can be reconstructed from its vertex deck ${\displaystyle D(G)}$, then its complement ${\displaystyle G'}$ can be reconstructed from ${\displaystyle D(G')}$ as follows: Start with ${\displaystyle D(G')}$, take the complement of every card in it to get ${\displaystyle D(G)}$, use this to reconstruct ${\displaystyle G}$, then take the complement again to get ${\displaystyle G'}$.

Edge reconstruction does not obey any such duality: Indeed, for some classes of edge-reconstructible graphs it is not known if their complements are edge reconstructible.

It has been shown that the following are not in general reconstructible:

• Digraphs: Infinite families of non-reconstructible digraphs are known, including tournaments (Stockmeyer[12]) and non-tournaments (Stockmeyer[13]). A tournament is reconstructible if it is not strongly connected.[14] A weaker version of the reconstruction conjecture has been conjectured for digraphs, see new digraph reconstruction conjecture.
• Hypergraphs (Kocay[15]).
• Infinite graphs. Let T be a tree on an infinite number of vertices such that every vertex has infinite degree, and let nT be the disjoint union of n copies of T. These graphs are hypomorphic, and thus not reconstructible. Every vertex-deleted subgraph of any of these graphs is isomorphic: they all are the disjoint union of an infinite number of copies of T.
• Locally finite graphs. The question of reconstructibility for locally finite infinite trees (the Harary-Schwenk-Scott conjecture from 1972) was a longstanding open problem until 2017, when a non-reconstructible tree of maximum degree 3 was found by Bowler et al.[16]

• New digraph reconstruction conjecture
• Partial symmetry

For further information on this topic, see the survey by Nash-Williams.[17]