The Easiest Unsolved Problem in Graph Theory
By: Sergei Ivanov
Tags: Graph, Conjecture, Graph Theory, Mathematics, Math

Good Will Hunting character, a janitor and amateur mathematician draws all homomorphically irreducible trees in response to the call by the professor at MIT. © 1997 Miramax Pictures

This post is written together with Ekaterina Vorobyeva.

At ten I once said to my friend how great it would if our teacher presented to us unsolved math problems. It would be much more fun to approach these in class instead of the textbook exercises. It was the time when we had no easily available internet, and the information on unsolved problems was scarce. Today, 20 years after, when I google unsolved problems in mathematics I get a huge list of problems. But, sadly, most of these are beyond my comprehension. Still, there are unsolved problems that are both easy to understand and provide a chance to an enthusiast to solve them.

I have been practicing in this field for long enough. I have solved puzzles, published papers, and even got a Ph.D. And, I am quite aware that there are people outside the profession of mathematics, who are interested in solving yet-unsolved math problems. This post is about an interesting math problem that even a 10-year-old can grasp.

Amateur mathematician Anthony Hill presents the graph drawings with minimal crossing number.

The problem I will be talking about is the graph-theoretical problem. Graph theory has a long history of problems being solved by keen amateurs. In 1879, a lawyer Alfred Kempe attempted to solve the four-color theorem introducing the concepts that were used a hundred years later to finally prove the theorem by computers. In the 1950s, a painter Anthony Hill discovered the minimum crossing number for any drawing of complete graphs and conjectured general formula that still remains unproved. In 2018, a gerontologist Aubrey de Grey improved the lower bound for coloring arbitrary graphs — a 60-year open challenge in graph theory.

Graph theory has a wealth of open problems. The one I will describe here is a specific “easy” case of reconstruction conjecture (RC), also known as Kelly-Ulam conjecture. Despite many online “proofs”, this problem remains unsolved for about 80 years, presenting a grand challenge. So while it’s tempting to resolve this problem as a whole, it may be too complicated. I will explain it step by step, outlining the graphs for which the reconstruction conjecture has been proved already, and finish with a particular “easy” set of graphs for which RC is still open.

Оne can try to solve a specific case first, before approaching the conjecture in its general formulation.

The Holy Grail problem

A reconstruction conjecture is very easy to formulate: do different graphs have different collections of vertex-deleted subgraphs? Let’s elaborate on what that means.

In the figure below, we have a graph G. If we remove a red vertex, we will get some subgraph on the left. Or, we can remove the blue vertex instead of the red one, and get a different subgraph. We can do so for any node in the graph and get n different subgraphs, one for each deleted vertex. Each of the obtained subgraphs is called a card and their collection is called a deck.

An undirected graph G and its deck of cards. Each card corresponds to a vertex-deleted subgraph from the original graph.

The problem, formulated by Kelly and his supervisor Ulam in 1942 is what can be considered as Holy Grail problem in graph theory:

Problem 1 [Reconstruction Conjecture, RC]: Two undirected graphs G and H with at least three nodes are the same if and only if their decks are the same.

Note: in mathematical lingo, two graphs being the same are isomorphic, i.e. the underlying connectivity of nodes remains the same regardless of the ordering of the nodes.

The reason why RC is formulated for graphs with at least three nodes is that for graphs with two nodes RC does not hold. Figure 4 illustrates this.

The only known exception for the reconstruction conjecture. Different graphs G and H have the same decks.

How to approach the reconstruction conjecture? One way is to understand it is to consider simple families of graphs where RC holds. A simple example is regular graphs, i.e. graphs where each vertex has the same degree. Suppose we have a deck for some graph G. It is easy to show that it’s possible to obtain a degree sequence of a graph from its deck. Hence, we can determine if the graph is regular by its deck. If it’s regular, we can take any card from the deck and find nodes with a degree either of d or a degree of d-1. So, to reconstruct the original graph, we just need to add one more vertex and connect it with vertices that have degree d-1.

To reconstruct a regular graph, pick any card and add one vertex that connects to vertices with degree 1.

Other families of reconstructible graphs include, among others, trees and disconnected graphs, i.e. graphs with multiple connected components.

Reconstruction conjecture has been solved for trees, disconnected graphs, and more. Yet, many interesting cases still have not been tackled.

Falling down the rabbit’s hole

One step further would be to consider graphs with just two possible node degree values. Such graphs are called bidegreed. For example, a graph below has vertices with either 2 or 4 neighbors.

Example of bidegreed graph, where each vertex has a degree either 4 or 2.

The bidegreed graphs with degrees a and b have a certain feature. If b-a>1, they are reconstructable. In other words, if the graph has degrees 2 and 4, as in the example above, the reconstruction conjecture holds. This result follows the same argument as for regular graphs.

The bidegreed graphs with consecutive degrees a and a+1 are called bidegreed consecutive. Here, the previous argument does not hold but a few results are known. First, one can see that if a vertex in a bidegreed graph is connected to only vertices with a degree of a, then such graph is also reconstructible. Similarly, if there is only one vertex of degree a or only one vertex of degree a+1 the graph is also reconstructable. And finally, Kocay proved that if a=2, then a bidegreed graph with two vertices of degree a and the remaining n-2 of degree a+1 is reconstructible. No more general rule is known, yet. So the easiest version of RC would be to restrict ourselves for bidegreed consecutive graphs. After all, it seems they are very simple instances of graphs.

Problem 2 (an easier one) [Reconstruction conjecture for bidegreed graphs]: Two undirected bidegreed consecutive graphs, G and H, with degrees $a$ and $a+1$ are the same if and only if their decks are the same.

A needle in a haystack

The experts in reconstruction conjecture have split in their opinions. The majority consider RC to be correct and we just have not discovered additional tools and theories that are needed to prove that. Other experts believe that counterexamples to RC exist, but the field of search is so big that it’s almost impossible to find these. So let’s consider the reasons for RC to be true or not.

Reasons to accept RC

The strong argument to believe reconstruction conjecture is true is the Bollobás theorem. It states that almost all graphs are reconstructible using only three cards of the deck. In other words, as the number of vertices in a random graph grows, the probability tends to one, that knowing three cards are enough to reconstruct the graph.

Another good reason is that RC has been verified algorithmically for all graphs with up to n=13 vertices. If you think that these are very small graphs, consider the overall number of such graphs. For example, for n=13, there are approximately 50 trillion graphs and one may need to search across all possible pairs, which can grow proportional to the square of that number. Algorithmic results also confirmed the above-mentioned Bollobás theorem, showing that most of the reconstruction numbers are 3 for graphs with up to 11 vertices.

Finally, as shown in the figure before, there are many more families of graphs for which RC is known to be true. This includes trees, complete multipartite graphs, unit interval graphs, unicyclic, cactus graphs, and more. While this does not cover the space of all possible graphs, the growing number of such families is an encouraging sign that RC will be proved one day.

Reasons to reject RC

Despite the general belief that the reconstruction conjecture holds, there are valid reasons to believe counterexamples to RC exist. First, it has been famously shown by Stockmeyer that a version of RC for directed graphs does not hold. In particular, for any given number n it’s possible to construct a pair of non-isomorphic directed graphs, known as Stockmeyer tournaments, of order 2^n +1 such that their decks are equivalent. This result was later extended to non-tournaments, showing that in general, it’s possible to construct a directed graph with a sufficiently large order such that it is not uniquely determined by its deck.

Stockmeyer tournaments are the counterexamples to RC in the directed case. A subgraph with black edges is embedded into a bigger graph by adding one extra vertex in two different ways. We can verify that the decks of these graphs are the same, even though the graphs are not isomorphic.

Similarly, it has been shown that generalizations of reconstruction conjecture for 3-hypergraphs (where each edge connects 3 vertices) as well as for infinite graphs (including those with finite degrees) have counterexamples.

So to summarize here are different reasons to accept or reject the reconstruction conjecture:

Reasons to accept or reject reconstruction conjecture. While there are good reasons to believe RC holds, similar versions of RC for other types of graphs exhibit counterexamples.

Reducing the space of candidates

It is tempting to transfer known counterexamples to the domain of undirected graphs. After all, Stockmeyer’s tournaments and undirected graphs are closely related (the total number of the tournaments and undirected graphs is the same), yet no straightforward adaptations have been found so far. For example, Stockmeyer tournaments have a recursive property that allows the construction of a pair of similar graphs from a smaller base graph. Finding undirected graphs with similar properties presents a key challenge to disprove an RC, and much ink has been spilled on this problem.

One of the most successful attempts has been done by Kimble et al. who have shown how to construct undirected graphs, with all vertices grouped in pairs of pseudosimilar vertices, i.e. vertices that produce the same card after its removal. Similar property has a Stockmeyer counterexample. Allen Schwenk, a professor at Western Michigan University, describes that such a graph is the closest we got to disprove a conjecture for undirected graphs. However, after almost 40 years of extensive research, the counterexamples have not been found.

Interestingly, the graphs with pseudosimilar vertices constructed by Kimble et al. are also bidegreed consecutive, defined above in Problem 2, and the authors suggest that this is the right neighborhood for the search of a counterexample. Indeed, graphs with the nodes, having diverse degree values, are easily reconstructable since vertices are easily identifiable. At the other extreme, regular graphs, as we have seen, are also easily reconstructible. On the other hand, the bidegreed graphs present the most confusion. As the smallest graph that Kimble et al. discovered has 20 vertices, the easiest unsolved case of reconstruction conjecture is the following.

Problem 3 (finally, an easy one!) [Reconstruction conjecture for small bidegreed graphs]: Two undirected bidegreed consecutive graphs G and H with 20 vertices and degrees $a$ and $a+1$ are the same if and only if their decks are the same.

Because of the relatively small number of vertices, one can try to search through the space of the bidegreed graphs with 20 vertices and check if there are two non-isomorphic graphs with the same decks. How much time it could take? Well, Brendan McKay recently checked reconstruction conjecture for all graphs with 13 vertices and it took him 1.5 years. So maybe, at least for small degree a, one can check bidegreed graphs with 20 vertices too.

So who knows, maybe one day some genius will prove reconstruction conjecture entirely or find a counterexample to it, as it happened recently to another graph problem called Hedetniemi conjecture. But at least now we know the easiest unsolved case of the 80-year old grand challenge in mathematics and any math enthusiast can try to prove it.

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