The construction. A random (undirected) graph with n vertices is constructed in the following way: pairs of vertices are randomly selected one at a time in such a way that each pair has the same probability of being selected on any given occasion, and, each selection is made independently of previous selections. If the vertex pair is selected, then an edge is constructed which connects x and y.

Are multiple edges possible? Yes. For example, if the vertex pair were to be selected subsequently, there would now be more than one edge connecting x and y: a multiple edge or cycle of length 2.


Suppose that m edges have been selected. We shall be concerned with the behaviour of the graph in the limit as n and m become large, but in such a way that .

The problem. Our problem is to determine the limiting probability that the graph is acyclic.

Motivation. Havas and Majewskigif present an algorithm for minimal perfect hashing

(used for memory-efficient storage and fast retrieval of items from static sets) based on this random graph. Their algorithm is optimal when the graph is acyclic.


Consider a set W of m words (or keys). Every bijection , where , is called a minimal perfect hash function.

HM find hash functions of the form

map keys to integers (they identify the pair of vertices of the graph corresponding to the edge w) and g maps integers to I.

Given and , can g be chosen so that h is a bijection?

If the graph is acyclic then, yes, it is easy to construct g from h. Traverse the graph: if vertex w is reached from vertex u then set

where .


HM's algorithm generates and at random until an acyclic graph is found:

where and are tables of random integers and denotes the i-th character (an integer) of key i.

The efficiency of the algorithm is determined by the probability that the graph is acyclic: the expected number of iterations needed to find an acyclic graph will be (typically between 2 and 3).


Conjecture. If n and m tend to in such a way that n=cm, where c is a positive constant, the limiting probability, p, that the graph is acyclic is given by

``Proof''. Let be the number of cycles of length k and let . Following [HM] write

Now let , so that and

Erdös and Renyigif show that the distribution of is asymptotically Poisson: in particular,


It follows that

So, formally,

and hence


By Fatou's Lemma, we always have

from which it follows immediately that

this argument is valid even if the sum in (1) is divergent. Thus, we may deduce immediately that if , exists and equals 0.

The interesting case c>2 is also easily dealt with. When c>2, we have that and that

From Markov's inequality we have

and so

But, by Lemma 2 of [HM], we know have that, for each fixed , , as . In particular, for each fixed , the sequence is bounded above by . It follows that is bounded above by . Further, since , we have

Thus, by the Lebesgue Dominated Convergence Theorem, we have that

and, hence, that exists and is equal to .