How to determine if a random graph with a fixed degree sequence has a giant component
Abstract
For a fixed degree sequence , let be a uniformly chosen (simple) graph on where the vertex has degree . In this paper we determine whether has a giant component with high probability, essentially imposing no conditions on . We simply insist that the sum of the degrees in which are not 2 is at least for some function going to infinity with . This is a relatively minor technical condition, and when does not satisfy it, both the probability that has a giant component and the probability that has no giant component are bounded away from .
1 Introduction
The traditional Erdős-Rényi model of a random network is of little use in modelling the type of complex networks which modern researchers study. Such a graph can be constructed by adding edges one by one such that in every step, every pair of non-adjacent vertices is equally likely to be connected by the new edge. However, 21st century networks are of diverse nature and usually exhibit inhomogeneity among their nodes and correlations among their edges. For example, we observe empirically in the web that certain authoritative pages will have many more links entering them than typical ones. This motivates the study, for a fixed degree sequence , of a uniformly chosen simple graph on where the vertex has degree . In this paper, we study the existence of a giant component in .
A heuristic argument suggests that a giant component will exist provided that the sum of the squares of the degrees is larger than twice the sum of the degrees. In 1995, Molloy and Reed essentially proved this to be the case provided that the degree sequence under consideration satisfied certain technical conditions [24]. This work has attracted considerable attention and has been applied to random models of a wide range of complex networks such as the World Wide Web or biological systems operating at a sub-molecular level [1, 2, 4, 27, 28]. Furthermore, many authors have obtained related results which formalize the Molloy-Reed heuristic argument under different sets of technical conditions [5, 16, 18, 21, 25].
Unfortunately, these technical conditions do not allow the application of such results to many degree sequences that describe real-world networks. While these conditions are of different nature, here we exemplify their limitations with a well-known example, scale-free networks. A network is scale-free if its degree distribution follows a power-law, governed by a specific exponent. It is well-known that many real-world networks are scale-free and one of the main research topic in this area is to determine the exponent of a particular network. It has been observed that many scale-free networks have a fat-tailed power-law degree distribution with exponent between and . This is the case of the World Wide Web, where the exponent is between and [9], or the Movie Actor network, with exponent [3]. In scale-free networks with exponents between and , the vertices of high degree (called hubs) have a crucial role in several of the network properties such as in the “small-world” phenomenon. However, one of the many technical conditions under the previous results on the existence of a giant component in hold, is that the vertices of high degree do not have a large impact on the structure of the graph. (In particular, it is required that there is no mass of edges in vertices of non-constant degree.) Hence, often these results cannot be directly applied to real-world networks where hubs are present and for each particular network ad-hoc approaches are needed (see e.g. the Aiello-Chung-Lu model for the case of scale-free networks [1]).
Another problem is that all the previous results apply to a sequence of degree sequences satisfying various technical conditions instead of a single degree sequence . Finally, all the previous results on the existence of a giant component in do not cover degree sequences where most of the vertices have degree .
In this paper we characterize when has a giant component for every feasible111A degree sequence is feasible if there is a graph with the given degree sequence. degree sequence of length . We only require that the sum of the degrees in the sequence which are not is at least for some arbitrary function going to infinity with . Besides the fact that it is a relatively minor technical condition, we also show that if it is not satisfied, both the probability that has a giant component and the probability that has no giant component are bounded away from .
It turns out that the heuristic argument which was used in [24] to describe the existence of a giant component in for degree sequences satisfying some technical conditions and that was generalized in the subsequent papers [5, 16, 18, 21, 25], actually suggests the wrong answer for general degree sequences. Precisely, if we let be a smallest set such that (i) no vertex outside of has degree bigger than a vertex in , and (ii) the sum of the squares of the degrees of the vertices outside of is at most twice the sum of their degrees, then whether or not a giant component exists depends on the sum of the degrees of the vertices in , not on the sum of the squares of the degrees of the vertices in as suggested by this heuristic argument
This new unified criterion on the existence of a giant component in is valid for every sequence and implies all the previous results on the topic both for arbitrary degree sequences [5, 18, 24] or for particular models [1].
1.1 The Molloy-Reed Approach
Let us first describe the result of Molloy and Reed [24]. For every , one can explore the component containing a specific initial vertex of a graph on via breadth-first search. Initially we have “open” edges out of . Upon exposing the other endpoint of such an open edge, it is no longer open, but we gain open edges out of . Thus, the number of open edges has increased by (note that this is negative if ).
One can generate the random graph for and carry out this exploration at the same time, by choosing each vertex as with the appropriate probability.
Intuitively speaking, the probability we pick a specific vertex as the other endpoint of the first exposed edge is proportional to its degree. So, the expected increase in the number of open edges in the first step is equal to . Thus, it is positive essentially if and only if the sum of the squares of the degrees exceeds twice the sum of the degrees.
Suppose that this expected increase remains the same until we have exposed a linear number of vertices. It seems intuitively clear that if the expected increase is less than , then the probability that initial vertex is in a linear order component is , and hence the probability that has no linear order component is . If for some positive constant , the expected increase is at least , then there is some such that the probability that is in a component with at least vertices exceeds .
In [24], Molloy and Reed proved, subject to certain technical conditions which required them to discuss sequences of degree sequences rather than one single degree sequence, that we essentially have that (i) if for some , then the probability that has a giant component is , and (ii) if for some , then the probability that has no giant component is . We present their precise result and some of its generalizations later in this introductory section.
1.2 Our Refinement
It turns out that, absent the imposed technical conditions, the expected increase may change drastically during the exploration process. Consider for example the situation in which for some large odd integer , and . Then , and so the Molloy-Reed approach would suggest that with probability there is a giant component. However, with probability , is the disjoint union of a star with leaves and components of order and hence it has no giant component. The problem is that as soon as we explore vertex , the expected increase drops from roughly to , so it does not stay positive throughout (the beginning of) the process.
Thus, we see that the Molloy-Reed criterion cannot be extended for general degree sequences. To find a variant which applies to arbitrary degree sequences, we need to characterize those for which the expected increase remains positive for a sufficiently long time.
Intuitively, since the probability that we explore a vertex is essentially proportional to its degree, in lower bounding the length of the period during which the expected increase remains positive, we could assume that the exploration process picks at each step a highest degree vertex that has not been explored yet. Moreover, note that vertices of degree have a neutral role in the exploration process as exposing such a vertex does not change the number of open edges, provided we assume that our component locally looks like a tree (which turns out to be a good approximation around the critical window). These observations suggest that we should focus on the following invariants of defined by considering a permutation of the vertices that satisfies :
-
-
,
-
-
, and
-
-
.
We emphasize that these invariants are determined by the multiset of the degrees given by and are independent from the choice of subject to .
Our intuition further suggests that in the exploration process, the expected increase in the number of open edges will be positive until we have explored edges and will then become negative. Thus, we might expect to explore a component with about edges, and indeed we can show this is the case.
This allows us to prove our main result which is that whether has a giant component essentially depends on whether is of the same order as or not. There is however a caveat, this is not true if essentially all vertices have degree .
For any function , we say a degree sequence is -well-behaved or simply well-behaved if is at least . Our main results hold for any function as .
Theorem 1.
For any function as , for every , if is a well-behaved feasible degree sequence with , then the probability that has a component of order at least is .
Theorem 2.
For any positive constant , there is a , such that if is a well-behaved feasible degree sequence with , then the probability that has a component of order at least is .
As we shall see momentarily, previous results in this field apply to sequences of degree sequences, and required that both the proportion of elements of a given degree and the average degree of an element of the degree sequences in the sequence approaches a limit in some smooth way. We can easily deduce results for every sequence of (feasible) degree sequences from the two theorems above, and from our results on degree sequences which are not well-behaved, presented in the next section.
We denote by a sequence of degree sequences where has length . We say that is well-behaved if for every , there is an such that for all , we have ; is lower bounded if for some , there is an such that for all , we have ; and is upper bounded if for every , there is an such that for all , we have .
The following is an immediate consequence of Theorem 1 and 2, and Theorem 6, which will be presented in the next section.
Theorem 3.
For any well-behaved lower bounded sequence of feasible degree sequences , there is a such that the probability that has a component of order at least is .
For any well-behaved upper bounded sequence of feasible degree sequences and every , the probability that has a component of order at least is .
If a sequence of feasible degree sequences is either not well behaved or neither upper bounded nor lower bounded, then for every sufficiently small positive , there is a such that there are both arbitrarily large for which the probability that has a component of order at least is at least , and arbitrarily large for which the probability that has a component of order at least is at most .
1.3 The Special Role of Vertices of Degree 2
At first glance, it may be surprising that the existence of a giant component depends on the ratio between and rather than the ratio between and . It may also be unclear why we have to treat differently degree sequences where the sum of the degrees which are not 2 is bounded.
To clarify why our results are stated as they are, we now highlight the special role of vertices of degree 2.
The non-cyclic components222A component is cyclic if it is a cycle and non-cyclic if it is not. of can be obtained by subdividing the edges of a multigraph none of whose vertices have degree 2 so that every loop is subdivided at least twice, and all but at most one edge of every set of parallel edges is subdivided at least once. Note that can be obtained from by deleting all cyclic components and suppressing all vertices of degree .333Here and throughout the paper, when we say we suppress a vertex of degree , this means we delete and we add an edge between its neighbours. Observe that this may create loops and multiple edges, so the resulting object might not be a simple graph. Clearly, is uniquely determined by . Moreover, the degree sequence of is precisely that of without the vertices of degree .
The number of vertices of a non-cyclic component of equals the sum of the number of vertices of the corresponding component of and the number of vertices of degree 2 used in subdividing its edges. Intuitively, the second term in this sum depends on the proportion of the edges in the corresponding component of . Subject to the caveat mentioned above and discussed below, if the number of vertices of degree 2 in is much larger than the size444As it is standard, we use order and size to denote the number of vertices and the number of edges of a graph, respectively. of , then the probability that has a giant component is essentially the same as the probability has a component containing a positive fraction of its edges. The same is true, although not as immediately obvious, even if the number of vertices of degree 2 is not this large.
Theorem 4.
For every , there exists a such that for every well-behaved feasible degree sequence , the probability that has a component of order at least and has no component of size at least is .
Theorem 5.
For every , there exists a such that for every well-behaved feasible degree sequence , the probability that has no component of order at least and has a component of size at least is .
As we mentioned above, degree sequences which are not well behaved behave differently from well-behaved degree sequences. For instance, suppose that . Then is empty and is a uniformly chosen disjoint union of cycles. In this case it is known that the probability of having a giant component is bounded away both from and . Indeed, the latter statement also holds whenever is at most for any constant .
Theorem 6.
For every and every , there exist a positive integer and a such that if and is a degree sequence with , then the probability that there is a component of order at least in lies between and .
1.4 Previous Results
The study of the existence of a giant component in random graphs with an arbitrary prescribed degree sequence555Random graphs with special degree sequences had been studied earlier (see, e.g. [22, 33])., started with the result of Molloy and Reed [24]. Although they define the concept of asymptotic degree sequences, we will state all the previous results in terms of sequences of degree sequences . Using a symmetry argument, one can easily translate results for sequences of degree sequences to asymptotic degree sequences, and vice versa. For every , we define . Recall that we only consider degree sequences such that .
Before stating their result, we need to introduce a number of properties of sequences of degree sequences. A sequence of degree sequences is
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feasible, if for every , there exists at least one simple graph on vertices with degree sequence .
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smooth, if for every nonnegative integer , there exists such that .
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sparse, if there exists such that .
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-bounded, for some function , if for every .
In particular, observe that random graphs arising from a sparse sequence of degree sequences have a linear number of edges, provided that is large enough.
Given that is smooth, we define the following parameter
Note that is very close to the notion of initial expected increase described in Section 1.1.
We say a sequence of degree sequences satisfies the -conditions if
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(a.1)
it is feasible, smooth and sparse,
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(a.2)
it is -bounded, for some ,
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(a.3)
for every , converges uniformly to , and
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(a.4)
exists and converges uniformly to .
For a precise statement of the uniform convergence on conditions (a.3)–(a.4), we refer the reader to [24]. Note that these conditions imply that .
Now we can precisely state the result of Molloy and Reed [24].
Theorem 7 (Molloy and Reed [24]).
Let be a sequence of degree sequences that satisfies the -conditions. Then,
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1.
if , then there exists a constant such that the probability that has a component of order at least is .
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2.
if and the sequence is -bounded for some , then for every constant , the probability that has no component of order at least is .
Note that the case is not considered in Theorem 7, since implies .
Theorem 7 has been generalized to other sequences of degree sequences, which in particular include the case . In Section 8, we show that Theorem 3 implies all the criteria for the existence of a giant component in introduced below.666Note that some of these results give a more precise description on the order of the largest component. Our results only deal with the existential question.
We say a sequence of degree sequences satisfies the -conditions if
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(b.1)
is feasible, smooth, and sparse,
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(b.2)
, and
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(b.3)
.
Observe that if satisfies the -conditions, then, by (b.2), it is also -bounded. Moreover, they also imply that . Janson and Luczak in [18] showed that one can prove a variant of Theorem 7 obtained by replacing the -conditions by the -conditions.777Their result gives convergence in probability of the proportion of vertices in the giant component and they also consider the case . They also note that if , then the criterion based on does not apply. Our results completely describe the case .
We say a sequence of degree sequence satisfies the -conditions if
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(c.1)
it is feasible, smooth and sparse,
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(c.2)
, and
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(c.3)
.
Bollobás and Riordan in [5] proved a version of Theorem 7 for sequences of degree sequences obtained by replacing the -conditions by the -conditions.888They also proved some results on the distribution of the order of the largest component and also consider the case .
Theorem 7 and its extensions provide easy-to-use criteria for the existence of a giant component and have been widely used by many researchers in the area of complex networks [2, 4, 27]. However, the technical conditions on to which they can be applied, restrict its applicability, seem to be artificial and are only required due to the nature of the proofs. It turns out to be the case that many real-world networks do not satisfy these conditions. For this reason, researchers have developed both ad-hoc approaches for proving results for specific types of degree sequences and variants of the Molloy-Reed result which require different sets of technical conditions to be satisfied.
An early example of an ad-hoc approach is the work of Aiello, Chung and Lu on Power-Law Random Graphs [1]. They introduce a model depending on two parameters that define a degree sequence satisfying . One should think about these parameters as follows: is typically large and determines the order of the graph (we always have ), and is a fixed constant that determines the power-decay of the degree distribution. Among other results, the authors prove that there exists , such that if the probability that there is a component of linear order is and if the probability there is a component of linear order is . Here, the previous conditions are only satisfied for certain values of and the authors need to do additional work to determine when a giant component exists for other values of . In Section 8 we will show how Theorem 3 also implies Aiello-Chung-Lu results on the existence of a giant component in the model of Power-Law Random Graphs.
1.5 Future Directions
Beginning with the early results of Molloy and Reed, the study of the giant component in random graphs with prescribed degree sequence has attracted a lot of attention. Directions of study include determining the asymptotic order of the largest component in the subcritical regime or estimating the order of the second largest component in both regimes [5, 16, 18, 20, 21, 25, 30]. It would be interesting to extend these known results to arbitrary degree sequences.
For example, Theorem 1 and 2 precisely describe the appearance of a giant component when the degree sequence is well-behaved. While bounds on the constant in terms of and respectively, may follow from their respective proofs, these bounds are probably not of the right order of magnitude. Molloy and Reed in [25], precisely determined this dependence for sequences of degree sequences that satisfy the -conditions. Precise constants are also given in [5, 14, 18]. We wonder whether it is possible to determine the precise dependence on the parameters for arbitrary degree sequences. It is likely that our methods can be used to find this dependence and to determine the order of the second largest component when a giant one exists.
Another direction is the study of site and bond percolation in for arbitrary degree sequences . This problem has been already approached for sequences of degree sequences that satisfy certain conditions similar to the ones presented in Section 1.4 [5, 13, 17, 30].
Motivated by some applications in peer-to-peer networks (see, e.g. [6]), one can study efficient sampling of the random graph . Cooper et al. [8] showed that the switching chain rapidly mixes for -regular graphs for every . Greenhill [15] recently extended this result to , but, due to some technical reasons, this result only holds if the maximum degree in is small enough.
2 A Proof Sketch
2.1 The Approach
The proofs of Theorem 4, 5 and 6 are simpler than the remaining proofs and we delay any discussion of these results to Section 6 and 7. By applying them, we see that in order to prove Theorem 1 and 2 it is enough to prove the following results:
Theorem 8.
For any function as and for every , if is a well-behaved degree sequence with , then the probability that has a component of size at least is .
Theorem 9.
For any positive constant , there is a such that if is a well-behaved degree sequence with , then the probability that has a component of size at least is .
The proofs of both theorems analyse an exploration process similar to the one discussed in Section 1.1 by combining probabilistic tools with a combinatorial switching argument. However, we will focus on the edges of rather than the ones of . Again, we will need to bound the expected increase of the number of open edges throughout the process and prove that the (random) increase is highly concentrated around its expected value. In order to do so, we will need to bound the probability that the next vertex of explored in the process, is a specific vertex . One of the key applications of our combinatorial switching technique will be to estimate this probability and show that it is approximately proportional to the degree of .
Crucial to this approach is that the degrees of the vertices explored throughout the process are not too high. Standard arguments for proving concentration of a random variable require that the change at each step is relatively small. This translates precisely to an upper bound on the maximum degree of the explored graph. Furthermore, without such a bound on the maximum degree, we cannot obtain good bounds on the probability that a certain vertex is the next vertex explored in the process. So, a second key ingredient in our proofs will be a preprocessing step which allows us to handle the vertices of high degree, ensuring that we will not encounter them in our exploration process.
2.2 The Exploration Process
We consider a variant of the exploration process where we start our exploration at a non-empty set of vertices of , rather than at just one vertex.
Thus, we see that the exploration takes steps and produces sets
where is either a neighbour of a vertex of or is a randomly chosen vertex in if there are no edges between and .
To specify our exploration process precisely, we need to describe how we choose and . To aid in this process, for each vertex we will choose a uniformly random permutation of its adjacency list in . For this purpose, an input of our exploration process consists of a graph equipped with an ordering of its adjacency lists for all vertices . Applying the method of deferred decisions (cf. Section 2.4 in [26]), we can generate these random linear orders as we go along with our process. We note that this yields, in a natural manner, an ordering of the non-loop edges of which have the vertex as an endpoint. If there are no edges between and , we choose each vertex of to be with probability proportional to its degree. Otherwise we choose the smallest vertex of (with respect to the natural order in ), which has a neighbour in . We expose the edge of from to which appears first in our random ordering and let be its other endpoint. Furthermore, we expose all the edges of from to as well as the loops incident to . Finally, we expose the paths of corresponding to the edges of which we have just exposed and the position in the random permutation of the adjacency list of in of the edges we have just exposed.
Thus, after iterations of our exploration process we have exposed
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•
the subgraph of induced by ,
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•
the paths of corresponding to the exposed edges of , and
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•
where each initial and final edge of such a path appears in the random permutation of the adjacency list of any of its endpoints which is also an endpoint of the path.
We refer to this set of information as the configuration at time . A configuration can also be understood as a set of inputs. During our analysis of the exploration process, we will consider all the probabilities of events conditional on the current configuration.
An important parameter for our exploration process is the number of edges of between and . We note that if , then is the union of some components of containing all of . We note that if , then every is a lower bound on the maximum size of a component of (not necessarily the one containing the vertex in ).
We prove Theorem 8 by showing that under its hypotheses for every vertex of , there is a set containing such that, given we start our exploration process with , the probability that there is a with for which the number of edges within is at most , is . Since has at most vertices, it follows that the probability that has a component of size at least is . The set is a set of highest degree vertices the sum of whose degrees exceeds . By the definition of and , this implies that, unless , the expectation of is negative. We show that, as the process continues, the expectation of becomes even smaller. We can prove that the actual change of is highly concentrated around its expectation and hence complete the proof, because contains all the high degree vertices and so in the analysis of our exploration process we only have to deal with low degree vertices.
We prove (a slight strengthening of) Theorem 9 for graphs without large degree vertices by showing that under its hypotheses and setting to be a random vertex chosen with probability proportional to its degree, with probability , there exists some such that (and hence there is a component of of size at least ). Key to doing so is that the expected increase of is a positive fraction of the increase in the sum of the degrees of the vertices in until this sum approaches . To handle the high degree vertices, we expose the edges whose endpoints are in components containing a high degree vertex. If this number of edges is at least a constant fraction of , then we can show that in fact all the high degree vertices lie in one component, which therefore contains a constant fraction of the edges of . Otherwise, we show that the conditions of Theorem 9 (slightly relaxed) hold in the remainder of the graph, which has no high degree vertices, so we can apply (a slight strengthening of) Theorem 9 to find the desired component of .
3 Switching
As mentioned above, the key to extending our branching analysis to arbitrary well-behaved degree sequences is a combinatorial switching argument. In this section, we describe the type of switchings we consider and demonstrate the power of the technique.
Let be a multigraph. We say a multigraph is obtained by switching from on a pair of orientated and distinct edges and if can be obtained from by deleting and as well as adding the edges and . Observe that switching and in yields . Observe further that if is simple and we want to ensure that is simple, then we must insist that , and, unless or , the edges and are not edges of .
Switching was introduced in the late 19th century by Petersen [29]. Much later, McKay [23] reintroduced the method to count graphs with prescribed degree sequences and, together with Wormald, used it in the study of random regular graphs. We refer the interested reader to the survey of Wormald on random regular graphs for a short introduction to the method [34].
In this paper we will consider standard switchings as well as a particular extension of them. This extension concerns pairs consisting of a simple graph and the multigraph obtained from by deleting its cyclic components and suppressing the vertices of degree 2 in the non-cyclic ones. For certain switchings of which yield , our extension constructs a simple graph from such that . We now describe for which switchings in we can obtain such an and how we do so.
Our extension considers directed walks (either a path or a cycle) of which correspond to (oriented) edges in , (note that an edge of corresponds to exactly two such directed walks, even if it is a loop and hence has only one orientation). We can switch on an ordered pair of such directed walks in , corresponding to an ordered pair of orientated distinct edges and of , such that none of the following hold:
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(i)
there is an edge of between and which forms neither nor , and the walk corresponding to has one edge,
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(ii)
there is an edge of between and which forms neither nor and the walk corresponding to has one edge,
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(iii)
and the directed walk corresponding to has at most two edges, or
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(iv)
and the directed walk corresponding to has at most two edges.
To do so, let be the directed walk corresponding to and let be the directed walk corresponding to . We delete the edges and and add the edges and .
We note that (i)-(iv) ensure that we obtain a simple graph . Furthermore, we have that is obtained from by switching on and . We remark further that if we reverse both the ordering of the edges and the orientation of both edges, we always obtain the same graph ; that is, it is equivalent to switch the ordered pair or the ordered pair . Therefore, given two walks between and and between and (either paths or cycles) of , we always consider the four following possible switches: , , and . We note that some of these choices might give rise to the same graph . However, we consider each of them as a valid switch since it will be simpler to count them considering these multiplicities.

Given any two disjoint sets of (multi)graphs and , we can build an auxiliary bipartite graph with vertex set where we add an edge between and for every (extended) switching that transforms into , or equivalently, into . We can also consider subgraphs of this auxiliary graph where we only add an edge if the switching satisfies some special property. Given a lower bound on the degrees in and an upper bound on the degrees in , we obtain immediately that . We frequently use this fact without explicitly referring to it.
To illustrate our method, we show here that if is large with respect to the number of vertices, then there exists a component containing most of the vertices.
Lemma 10.
If , then the probability that has a component of order is .
In proving the lemma, we will need the following straightforward result on -edge cuts of graphs. We defer its proof to the end of the section.
Lemma 11.
The number of pairs of orientations of edges in a graph of order such that by switching on and we obtain a graph with one more component than is at most .
Proof of Lemma 10.
We can assume is large enough to satisfy an inequality stated below since the lemma makes a statement about asymptotic behaviour. Let . For every integer , let be the event that has exactly components and let be the event that is in and that all components of have order at most . Denote by . Our goal is to show that . If so, with high probability has a component of order larger than . Observe that if one proves for some function with as that for every , then . We adopt this approach with .
Fix . Now suppose that there exist and such that for every in , there are at most switchings that transform into a graph in , and for every graph in , there are at least switchings that transform into a graph in . Then,
Let us now obtain some values for and . On the one hand, applying Lemma 11, we can choose
On the other hand, if is in , in order to merge two components it is enough to perform a switching between an oriented non-cut edge (at least choices) and any other oriented edge not in the same component as the first one (since has minimum degree at least and the largest component has order at most , there are at least choices). Since is large, we can choose
From the previous bounds, we obtain the desired result
∎
Proof of Lemma 11.
Any such pair of oriented edges must lie in the same component and swapping on them within the component must yield a disconnected graph. So, as the function is convex, we may assume that is connected.
First, suppose that at least one of the oriented edges, say , is an edge cut of the graph. Then, if is not an edge cut of , the switching does not disconnect the graph. Since there are at most cut-edges, there are at most switchings using at least one (and hence two) oriented cut-edges.
Otherwise is a proper 2-edge cut (that is, both and are connected). Consider an arbitrary spanning tree of . This tree contains at least one edge of every -edge cut of . Thus, select among these edges (exactly choices). Observe now that, in order to construct the proper 2-edge cut, we need to select as a cut-edge in (at most choices).
Therefore, in total there are at most switchings in which disconnect it. ∎
4 The Proof of Theorem 8
Theorem 8 follows immediately from the following result.
Lemma 12.
For every sufficiently small and every degree sequence such that and is sufficiently large in terms of , for every vertex of , the probability that lies in a component of of size larger than is less than .
In order to prove this lemma, we analyse our random exploration process on using a set of vertices including and show that the probability that there is a with and such that there are at most edges in the graph induced by in , is at least .
Since it is difficult to keep track of , we will instead focus on the random variable (defined below), which overestimates until . Clearly, is at most
Provided that , the edge is an edge of , and we can upper bound by
Observe that, provided that , the process coincides with if the explored components are trees and is a stable set. More importantly, we observe that if , then there is a for which . We note further that the number of edges in the graph induced by in is at most , so we only need to show that the probability that there is no for which is less than .
As suggested by our introductory discussion and proven below, the probability that is a specific vertex in is essentially proportional to the degree of . Therefore, the expected value of is with high probability close to
By putting the high degree vertices of into , we can ensure that the expectation of is negative. The fact that, if the process has not died out by time , then the sum of the degrees of the vertices we picked cannot be much less than , allows us to obtain a bound on the expectation of which decreases as increases. Having all the high degree vertices in facilitates our analysis of the exploration process, and allows us to show that the probability that drops to before the number of edges in the graph induced by in is at least , is more than . Forthwith the details.
We use for , for , for , for and for . We implicitly assume that is small enough and is large enough in terms of to ensure various inequalities scattered throughout the proof are satisfied.
Let be a smallest set of vertices of such that and there is no vertex outside of with degree bigger than a vertex in . Since the sum of the degrees of the vertices in is , such a set exists. Furthermore, since , the definition of implies that . It is straightforward to prove, as we do below, the following strengthening, which is important for our analysis.
Lemma 13.
We have
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(a)
, and
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(b)
there is a vertex of with degree at most .
The sum of the degrees in is at most the sum of and the minimum degree in . Let . Since every vertex not in has degree at most the minimum degree in and is the sum of the degrees of the vertices in , the following observation holds.
Observation 14.
We have
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(a)
for every , and
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(b)
.
As we carry out our process, we let , where the expectation is conditional on . By construction and by Lemma 13 (b), the absolute value of is bounded from above by . As we explain below, by applying Azuma’s Inequality we immediately obtain:
Lemma 15.
The probability that there is a such that is less than .
Thus, in order to bound from above, we need to bound from above for each . Letting be the sum of the degrees of the vertices of which are not in and using our swapping arguments, we can prove the following result, which will be useful to give a precise estimation for .
Lemma 16.
If and , and for all , then the following statements hold:
-
(a)
If and , then
-
(b)
If , then
Iteratively applying this result to bound , we obtain:
Lemma 17.
Letting be the minimum of and the first for which or , we have the following for all :
Lemma 18.
With probability greater than , there exists such that .
Proof.
4.1 The Details
We start with some simple observations.
Observation 19.
The maximum degree of is at most .
Proof.
By definition, , which implies . ∎
We let be the number of vertices of degree 1 in .
Observation 20.
We have .
Proof.
By the definition of , we obtain
Hence, , which implies . ∎
If there is a vertex in of degree , then every vertex outside of has degree . Thus every edge in the components of containing is incident to a vertex of , hence there are at most such edges and we are done. So every vertex in has degree at least .
Proof of Lemma 13.
Let be such that . Since , we have . By the definition of , we have . Now, since contains only vertices of degree at least , the definition of implies , and the first statement follows.
If every vertex in has degree at least then, since for sufficiently small , we have that . With the above observation, we have . Since there are at most values of for which , it follows that , which is a contradiction to the choice of , and the second statement follows. ∎
For the proof of Lemma 15 we recall a standard concentration inequality.
Lemma 21 (Azuma’s Inequality (see, e.g. [26])).
Let be a random variable determined by a sequence of random experiments such that for every and any possible sequences and :
then
Proof of Lemma 15.
Recall that , which implies . By Lemma 13, we have . Azuma’s inequality applied to with and gives us
where we used that . A union bound over all suffices to obtain the desired statement. ∎
Proof of Lemma 16.
We can assume that there is an edge of from to as otherwise the probability that is is exactly , by construction of the exploration process, and we are done.
It is enough to prove this result conditioned on the current configuration . In doing so, we partition the inputs within (a graph with an ordering of the adjacency list of each vertex) into different equivalence classes. All the inputs in the same equivalence class share the same underlying graph , a partial order of the adjacency list of each vertex in and a specification of the first edge from to in the ordering of the adjacency list for . Observe that each equivalence class corresponds to the same number of inputs; these arise from each other by suitably reordering some of the adjacency lists. More precisely, every equivalence class contains inputs, where if , and is the number of edges of from to if and is one less than this number if .
We generalise the definition of extended switching from graphs to equivalence classes of inputs, provided that the switching neither uses nor creates an edge of within . This ensures that after the switching, the set of edges of within (and the set of edges of corresponding to them) remains unchanged. Consider switching the pair of edges as explained in Section 3. Let be the directed walk in corresponding to and let be the directed walk in corresponding to . When switching, we delete the edges and and add the edges and . This naturally preserves the ordering of the adjacency lists of and . The orderings of the adjacency lists of and are obtained by preserving the position of the edges and in them, respectively. If , after the switching, the ordering of the adjacency list of is obtained by preserving the position of the edges incident to and different from . If , after the switching, the position of the edges and in the adjacency list of is swapped. This works analogously for . Note that this generalisation preserves the equivalence class of the input: if a switching is performed to an input, first, no edges within are modified (neither their position on the corresponding adjacency lists) and, second, the specification of the first edge from to in the adjacency list of is preserved.
For any vertex , we let be the set of equivalence classes for which is , and let be the set of equivalence classes for which is not . In order to bound the probability that is chosen to be the vertex that will be added to , we consider switchings between elements of and , which allow us to relate and .
Since , the set contains at least of the at least vertices of degree and so .
Proof of (a). A switching from to involves the oriented edge and one of the remaining oriented edges with first endpoint in or and one of the remaining oriented edges with last endpoint in . This implies that there are less than switchings from to .
Next, we prove a lower bound for the number of switchings from to . The choice of an equivalence class in fixes a choice of the edge . A vertex in is bad if it either has a neighbour in , has a common neighbour with or is adjacent to . If is not bad, then for the unique neighbour of there exists four switchings using and (or their reverses), from to . By the hypothesis of the lemma, at most vertices in have a neighbour in . By Observation 14, the degrees of the vertices in are at most , which implies that there are at most bad vertices. Since sampling an element of uniformly at random, and applying a random permutation to the at least vertices of degree in , yields an equivalence class of that is still sampled uniformly at random, the proportion of equivalence classes in in which a specific vertex of degree is not bad, is at least . This implies that there are at least switchings from to .
Double-counting these switchings yields . Since, , and is large in terms of , it follows that .
Proof of (b). Let . For a switching from to , we have to switch the ordered edge with one of the ordered edges or the ordered edge with one of the ordered edges . Therefore, the number of switchings from to is at most .
Next, we prove a lower bound for the number of switchings from to . We have such a switching involving the ordered edge , and some ordered edge provided is in , and and are not edges. We have such a switching involving the ordered edge , and some ordered edge provided is in , and and are not edges. Since has degree at most , and the maximum degree of a vertex in is , the number of choices for is at least .
Double-counting the number of switchings, we obtain . As before , and hence ∎
Proof of Lemma 17.
The proof is by induction on . Recall that there are between and vertices of degree in .
First, let . Observation 14 (b) gives us , hence Lemma 16 implies
Thus, by Lemma 13 and the fact that is at most (recall that there are no vertices of degree in ), we have .
Now, let . By induction, we obtain
Claim 1.
For any sequence of positive integers distinct from 2 and a nonnegative integer such that , we have .
Proof.
The proof is by induction on . The statement is trivially true if .
If all are at least , then . Hence, we may assume without loss of generality that . Since , if , we obtain by induction that .
So, we can assume that . If for all , the integer , then the claim also follows easily, as then . Since , the negative contribution of a is compensated by a .
Hence, without loss of generality, and . If then by decreasing by 1 and then applying induction we are done. So, we may assume , and thus there must be at least values of for which . The sum of over these elements and is . So, deleting these elements from the set and again applying induction, we are done. ∎
5 The Proof of Theorem 9
Again, we use for , for , for , for and for . Since is well-behaved, we can also assume that is sufficiently large to satisfy various inequalities scattered throughout the proof.
As we have already mentioned, in the preprocessing step we need to handle the vertices of large degree. We let be the set of vertices of degree exceeding . We will prove the following result using the combinatorial switching approach:
Lemma 22.
The probability that the number of edges in the components of intersecting is at least and there exists no component containing at least edges is .
We also need the following straightforward result:
Lemma 23.
Let be a set of vertices of that contains all vertices of degree exceeding and let be such that . Then
Proof.
The assumptions imply . Letting and , the definition of implies
∎
Starting with the set , we first explore all the components in that contain at least one vertex in . Let be the set of all vertices in such components. If , then by Lemma 22 the probability that there does not exist a component in with at least edges is . So, in what follows, we condition on .
We let , where is a random vertex selected with probability proportional to . Since there are no edges from to , we have . Note that this implies that, for every , the edges counted by belong to the same component (not necessarily the one of ). In addition, the maximum degree of a vertex in is at most and .
For each vertex , we let be the sum of the number of loops of at and the number of edges of between and . Observe that we can control the number of edges between and as follows
(1) |
The next lemma shows that the probability of selecting a vertex at time is essentially proportional to its degree.
Lemma 24.
Let be a fixed constant. If and , then for every ,
and,
We are now in a position to carry out our exploration process. Let and let . By our convention, both expectations are conditional on . We let be the set of those inputs such that for some either or .
Lemma 25.
.
Let , , and be the smallest for which either or .
Lemma 26.
For any , , , and .
Proof.
We have
By the definition of , the hypotheses of Lemma 24 are satisfied. Using Lemma 23 as well as Lemma 24 we conclude
since . This proves the first statement. Again, by Lemma 24, we obtain
where the last inequality follows from the first statement of this lemma.
Now, since the third statement follows directly from the first and second one. ∎
Since all the edges counted by are in the same component of , this next lemma proves Theorem 9.
Lemma 27.
With probability , we have .
Proof.
We show that if our configuration is not in then . By Lemma 25, the result follows.
Applying (1) recursively, we have
By adding , subtracting the expectation of the right hand side in the previous equation and since , we obtain that
If , then Lemma 26 implies , and we are done. Now, let . If , then, by the definition of , . Note that , because . Using Lemma 26 as before, we obtain
a contradiction. Thus, in both cases. ∎
5.1 The Details
We start this section with a result showing that if there are many vertices in , then they all lie in the same component of .
Lemma 28.
If , then the probability the vertices of lie in the same component of is .
Proof.
Let and be the vertices of degree at least and , respectively. We divide the proof into two cases depending on the size of .
Case 1: .
We begin with a claim which shows that every vertex in is adjacent in to a large number of vertices in .
Claim 2.
For every , the probability that is adjacent to at most vertices in is at most .
Proof.
Let . Assume for a contradiction that the claim fails for . For every , let be the event that is adjacent to exactly vertices in . By our assumption, there is some such that .
Suppose that is in . We consider switchings that lead to a multigraph in either or . We stress here that we will use a specially adjusted version of switchings. Consider edges such that or is not an edge in . We have at least choices for such an edge. Moreover, there are at least vertices . Now we discuss different switching situations depending on the structure of .
First, suppose that there are at least edges such that . Choose such an oriented edge . Then, for each , there are at least edges such that and either is not an edge in or . In both cases we get at least switchings which increase the degree of in .
Otherwise, there are at least edges that are not edges in . Choose such an edge . Next, suppose that there are at least vertices such that there are at least edges with . As before this give rise to at least switchings. (Observe that if , then the obtained graph is in . Observe also that we chose to switch so that the new edge between and corresponds to the edge between and and hence has an internal vertex). Otherwise, there are at least vertices such that there are at least edges with . Choose such an that it is not an edge in (all but at most one of them are not edges of ). If either or corresponds to a path of length at least in , then there exists at least one switching (the one that switches such an edge to a new loop in ) that transforms into a graph in . If both and correspond to paths of length , then we perform a special type of switching. Let and be the corresponding paths in . Then, we obtain the switched graph by deleting the edges , , and and by adding the edges , , and . This gives a graph in . In this case, there are also at least switchings.
Now, for any in either or , consider the switchings that transform it into a multigraph in . We must use an edge for which is not a parallel edge in . While there might be many edges between and , note that there are at most of this type. We can select in at most ways. Thus there are at most switchings leading to a multigraph in . The factor comes from the fact that we performed the special type of switching introduced above, that given two edges in a graph can give rise to at most graphs.
Hence , and in particular . Using that , we have , which is a contradiction. ∎
Now we use Claim 2 to show that any two vertices in whose degree is not extremely large, lie in the same component.
Claim 3.
For every each of degree at most , the probability that they are not in the same component is at most .
Proof.
Let . For every , we define the following events,
-
and have no common neighbour in .
-
there are edges between and in .
-
has an edge-cut of size at most separating and .
Let be the event . Observe that if is not satisfied, then there exists a path between and . Thus, it suffices to show .
Here, we will show that for every satisfying , we have
This implies that and proves the claim.
So, suppose . Let be the event that are in addition adjacent to at least vertices in . By Claim 2 and since , we obtain that .
We consider switchings from a graph in to or , which use no edge incident to or . We are going to switch using edges from two sets which we now define.
Fix an edge cut of size at most separating and and let be the set of edges between and . These two sets of edges exist by and , respectively.
Given that is in , there are at least vertices and for each such , there are at least edges . Since , essentially all such satisfy . Indeed, we can find a set of vertices such that is not an endpoint of and there are at least edges with . Let be the set of edges such that , and either is not an edge of or is not an endpoint of any edge of . Since , .
In the same vein, we can obtain a set of vertices such that for each , we have , is not an endpoint of and there are at least edges with . Moreover, we can also obtain a set of edges with such that for each , , and either is not an edge of or is not an endpoint of any edge of .
Observe that for any and any , we have . Otherwise, if , there exists a path non of whose edges are in the edge cut , getting a contradiction.
If is an edge of , then and both and are not edges of . Note that . Thus, we can always switch and to obtain a new graph which is or (it only belong to if and ). There are at least switchings.
Given a graph in , there are at most switchings which yield a graph in .
We conclude that , as desired. ∎
A union bound over all pairs together with Claim 3 suffices to show that in Case 1, with probability all the vertices in with degree at most lie in the same component of .
Now we consider a set consisting of all the vertices of of degree at least and one other vertex of , if there are any more. Therefore, .
For any , we let be the event that and are in the same component and be the event that they are in different components. We will use switchings involving and . On the one hand, for any graph in , there are switchings which yield a graph in in which is an edge of . On the other hand, for any graph in there are at most switchings using the edge and another edge, that transform the graph into a graph in . Therefore, . So, with probability all the vertices in lie in the same component. This implies that with probability , all the vertices in lie in the same component of and this completes the proof of the lemma for Case 1.
Case 2: .
Let be the size of . By the hypothesis of the lemma and of this case, we have
(2) |
The following claim shows that with high probability, the multigraph induced in by the vertices in , has large minimum degree.
Claim 4.
With probability at least , every vertex is incident to at least edges of which join it to other vertices of .
Proof.
Fix a vertex . For every , let be the event that is incident to exactly edges joining it to vertices of in . Using (2), we have that .
Let be a graph in . We will count how many (extended) switchings lead to a graph in . By the hypothesis of Case 2, there are at least edges such that either or and corresponds to a path of length at least in .
For any such edge , we can switch with any edge disjoint from such that and is not adjacent to unless one of the following situation happens:
-
(i)
and both correspond to edges of and there is an edge corresponding to an edge of between and , or
-
(ii)
.
There are at least choices for . Given the choice of and , since and if is an edge of then , there are at least choices for an edge that do not satisfy . Since , there are in total at most edges satisfying (ii) for .
Thus, there are at least choices for an edge that give a valid switching with . So, in total there are at least switchings.
If is in , then there are at most switchings that lead to a multigraph in .
Using (2), we conclude,
In particular, for every , we have
A union bound for all and all vertices now yields to the desired result. ∎
To complete Case , we will use the minimum degree within the vertices in to show that the multigraph induced by in , denoted by , is connected. For every , let be the event that has exactly components. We show that there is an which is such that for every that satisfies , we have . If so, , or in other words, with probability the multigraph is connected. This proof follows the same lines as the one in Lemma 10.
Fix such that . Suppose is in . Any (extended) switching from that leads to a graph in creates a new component and hence either uses two cut edges or uses a 2-edge cut which does not contain a cut edge. By Lemma 11, there are at most switchings leading to a multigraph in .
For every , let be the event with the additional restriction that has minimum degree at least . Since , by Claim 4, . Suppose now that is in . We will lower bound the number of (extended) switchings to graphs in . In order to merge two components it is enough to select non-cut edge and an edge in another component. By the definition of , there are at least choices for . Given the choice of , there is at least one vertex in another component, and hence, there are at least choices for . The total number of switchings is at least .
We proceed with the proof of Lemma 22.
Proof of Lemma 22.
If , we can use Lemma 28 to show that with probability , all the vertices in lie in the same connected component, and hence the statement of the lemma holds in this case.
Therefore, we can assume that . Since , this implies that if the union of the components intersecting have at least edges there is a component of size at least which contains a vertex of . So, it is enough to prove that for every pair , the probability that is in a component with at least edges not containing is .
Fix . Let be the event that the component of in has at least edges and are in different components. Let be the event that are in the same component of . We will show that .
Let be a graph in . Since , there are at least oriented edges (i.e. we count loops twice) and at least oriented edges in the same component as ordered in such a way that is at least as close to , as . Thus, the total number of switchings, using an edge ordered in such a way leading to a multigraph in with an edge is at least .
Consider in obtained by such a swap. If and , then there exists a unique edge in any shortest path from to . Otherwise we can find two edges incident to , such that every shortest path from to contains one of them. So, the total number of such switchings leading to is at most .
We conclude the desired result, . ∎
Proof of Lemma 24.
Recall that for every , we have . Moreover, by construction of the exploration process, the component we are exploring at time has no vertices in . Thus any vertex that belongs either to the current explored component or to , has degree at most . This implies that, for every vertex that plays a role in the exploration process, the number of edges incident to a neighbour of is at most . This property will be crucial in our analysis.
In this proof we consider inputs (graphs equipped with an order of each adjacency list) instead of graphs. As in the proof of Lemma 16, we will perform our switchings between equivalence classes of inputs.
We start by proving the second part of the lemma. We fix and condition on the configuration at time . For every , we let be the equivalent class of inputs such that and that the sum of the number of loops on and edges from to is . For any equivalence class in , there are at least switchings that lead to one in . For any equivalence class in , there are at most switchings that lead to . It follows that for , we have . The second statement of the lemma follows from applying the last inequality recursively.
In the same vein, one can obtain that conditional on , the probability that is incident to more than edges connecting to , is at most . We omit the details. This fact will be used at the end of the proof.
As before, we fix and condition on the configuration at time . We let be the union of the equivalence classes of inputs consistent with this configuration where and we let be those where .
Let be the elements of such that there are edges between and . For each equivalence class in , we can switch any edge with some other ordered edge with to get an element of unless is a neighbour of or is a neighbour of . Thus, given a graph in , there are at least switchings that lead to . On the other hand, given an element of , there are at most switchings that lead to . Since , we have that . This implies, . Thus,
(3) |
We let be the elements of such that there are edges between and the neighbours of (recall that ). Given an equivalence class in , there exist at least switchings that lead to and that do not use any edge incident to . On the other hand, given an equivalence class in , there are at most switchings that lead to .
Given that , we have that,
Using (3), it follows that conditional on our input being in , the probability that has no edge to and has at most edges incident to is at least .
Combining the statements above, we obtain that the proportion of elements of such that in the corresponding multigraph we have that has no edge to , is incident to at most edges that are also incident to and is incident to at most edges that are also incident to , is at least . Note that this implies that is not an edge of .
We consider switching using an ordered pair of oriented edges and of such that is not in . For inputs as in the last paragraph, there are at least choices for an oriented edge to switch with to construct an input in (we simply cannot choose x in or such that is an edge of and is an edge of ) . Clearly, for any input in , there are at most oriented edges to switch that lead to . For any equivalence class in , similarly as before, there are between and such switchings that lead to ( we pick an oriented edge of with both endpoints outside and say we produced and by swapping on and , this will work for certain provided is not incident to a neighbour of and is not incident to a neighbour of ).
Straightforward computations give,
∎
Proof of Lemma 25.
Recall that and that . Note that and since the maximum degree of the vertices in is at most , we have . We can apply Azuma’s Inequality (see Lemma 21) to with and , to obtain
since and is large enough. A union bound over all suffices to obtain that the probability there exists a such that is . The same argument can be used for . Thus, we obtain . ∎
6 Handling Vertices of Degree 2
6.1 Disjoint Unions of Cycles
The graph partitions into a set of cyclic components and a subdivision of . We consider first the structure of the graph formed by its cyclic components. We let be the set of vertices in these components and let be a union of cycles chosen uniformly at random among all -regular graphs with vertex set . We emphasize that is a simple graph so these cycles have length at least .
Fix some vertex . Let be the probability that is in a cycle of length in if . Let be the number of configurations of vertices into disjoint cycles of length at least . We will use the following result on the asymptotic enumeration of -regular graphs (see, e.g. Example VI in [12]).
Theorem 29 ([12]).
We have
Corollary 30.
For every integer and every , we have
Proof.
If belongs to a cycle of length , then there are ways to select the remaining vertices in its cycle. In addition, there are possible configurations for the cycle containing given we have selected the vertices in this cycle. Hence . The desired results follow from straightforward computations using the bounds from Theorem 29. ∎
Corollary 31.
For every and for every sufficiently large , the probability that lies on a cycle of of length at least but less than is at least .
Proof.
Corollary 32.
For every and for every sufficiently large , the probability that contains a cycle of length at least is at least .
Proof.
The probability there exists a cycle of length at least is at least the probability that is in a cycle of length . Using Corollary 31, we conclude that the probability that is contained in a cycle of length at least (and at most ) is at least . ∎
Corollary 33.
For every , there exists a such that for any sufficiently large , the probability that contains no cycle of length at least is at least .
Proof.
It suffices to prove the statement for .
We let be the event that the sum of the lengths of the cycles of which have length at least but less than exceeds . Clearly it is enough to prove a lower bound on .
For , we let be the event that contains cycles of length such that for each , if , then is disjoint from for all and the lowest indexed vertex in is in , and .
We set . Observe that , so it suffices to lower bound . Clearly, . For , we have if the number of vertices not in the union of the for is less than as we can simply set . Given that , this conditional probability is at least the probability the vertex with lowest index in a uniformly random disjoint union of cycles of total length is in a cycle of length with . By applying Corollary 31 with the parameters and , we have . Hence . ∎
6.2 The order of the components
In this section we study the number of vertices in the union of some components of conditioned on an (at least partial) choice of . As before, we write . Observe that the degree sequence fixes the number of edges of and also , the number of vertices of degree 2 in , while a choice of fixes the number of edges of in each of its components.
For a given choice of , we expose edges and vertices of degree in in two phases. In the first phase we do the following.
-
(1.1)
For each set of parallel edges between two vertices in , we expose at most one of them as an edge of . The parallel edges of that we exposed as edges of will be called fixed edges of .
-
(1.2)
For each non-fixed edge in with distinct endpoints which is parallel to some other edge (so it corresponds to a path in of length at least ), we expose the edge on the corresponding path of which is incident to its endpoint of lowest index; that is, we expose the other endpoint of such an edge.
-
(1.3)
For each loop of rooted at a vertex, we expose the two edges of incident to this vertex in the cycle of corresponding to the loop; that is, we expose the other endpoints of these two edges.
We note that if an edge of is neither a loop nor parallel to another edge, then we do not expose whether it corresponds to an edge or a non-trivial path of .
Let be the number of edges of that are non-fixed. We let be the number of vertices of degree which have been exposed in (1.2) or (1.3). Let be the vertices of degree that have not been exposed yet.
Observation 34.
We observe that
-
(a)
for every component of , there are at most fixed edges, and
-
(b)
for every component of , the number of vertices of degree exposed in (1.2) or (1.3) inside , is at most , and the sum of the number of such vertices and is at most .
Suppose that we also condition on the set of the vertices of degree that have not been exposed yet and which lie in the non-cyclic components of . Then we can specify the non-cyclic components of the graph in a second phase.
-
(2.1)
Fix an arbitrary ordering of the edges of that are non-fixed and a direction on each of these edges.
-
(2.2)
Choose a uniformly random permutation of length of and a set of indistinguishable delimiters. We let be the position of the -th delimiter in the permutation and add a delimiter at the start of the permutation and a delimiter at its end.
-
(2.3)
For every , let be the -th non-fixed edge of , with the corresponding direction. We let (resp. ) be the neighbour of (resp. ) on the path of corresponding to if we have exposed it, otherwise we set (resp. ). We expose the vertices with to construct a path in connecting and . We do this by starting at and by following the order induced by .
Now, conditional on the choice of of size , the number of choices for the non-cyclic components of is exactly
Recall that, if we only condition on the information exposed in (1.1)–(1.3), we have non-fixed edges in and vertices of degree which were exposed in non-cyclic components. Also recall, . For each with and every , we let be the number of graphs with vertices of degree in cyclic components and vertices of degree in non-cyclic components given our exposure of .
By the previous observations, , where has been defined in the previous section as the number of configurations of disjoint cycles using vertices. Now, . Theorem 29 allows us to estimate the ratio . We thus obtain that there exists some function such that and , and such that for every ,
(4) |
It is also not hard to see that for non-negative we have:
(5) |
We let . For every , let
That is, equals the probability that there are vertices in the cycle components given our choices of and the exploration explained above. Observe that if is a function such that for each and we have , then there are two positive constants such that for every integer , the product lies between them. Using (4) and letting , we have that for ,
(6) |
We now provide the proof of Theorem 6.
Proof of Theorem 6.
Define . Using (6) and the fact that , we can find positive constants and such that for every , we have
(7) |
where we use that is large in terms of and . From (5), we obtain that , provided that is large enough. Recall that . We also observe that for every by (4). Thus, for every . Observe that for all . Since is a probability distribution, it follows from (7) that
from where we obtain that , for some positive constant . Therefore, we conclude that the probability that is at least
Now we are able to conclude the proof. Recall that . Observe that , since and . It follows that with probability we have . If this is the case, there are at most vertices in non-cyclic components, and thus, there is no component of order at least in such components. First, we apply Corollary 32 with and , and obtain that the probability that there exists a cycle of length at least is at least . Second, we apply Corollary 33 with and , and obtain that the probability that there exists no cycle of length at least is at least . Let . Finally, since this holds for every conditioning on , and , it also holds for the unconditioned statement and we have proved the theorem for . ∎
Lemma 35.
For any positive constant , if , then the probability that there are more than vertices in cyclic components of is .
Proof.
Recall that by Observation 34 (a), we have . We split the proof into two cases.
First, suppose that . We use (6) to upper bound and obtain the desired probability. There exists some constant such that
since .
Lemma 36.
For every positive constant , if the following is satisfied. Fix a choice of and let be a union of some components of with . Let be the union of the corresponding components of . Then, with probability ,
Proof.
Observe that the choice of determines , the number of edges that have not been fixed in (1.1), and , the number of vertices of degree 2 which have been exposed in (1.2) or (1.3). We will also condition on , the number of vertices of degree that have been exposed in (2.3). Since , by Lemma 35, the probability that is . Hence, we may assume that .
We denote by the number of vertices of degree exposed in (1.2) or (1.3) in a component of . By Observation 34 (b), we have
We let be the number of non-fixed edges in . By Observation 34 (a), we have . Similarly, . Thus,
Let be the number of vertices of degree which have been exposed in (2.3) to the edges of . Since the ordering of the edges in (2.1) was arbitrary, symmetry amongst the non-fixed edges yields:
Since the minimum degree in is at least one, there are at most vertices in . So, the number of vertices in satisfies
Now, we will use our random permutation model to show that the random variable is concentrated around its expected value, which by the previous equation implies that also is. In (2.1) we insist on choosing an ordering of the non-fixed edges of in such a way that the first ones correspond to .
The probability that conditional on the value of , is the same as the probability that in choosing elements in , we choose less than of them that are smaller than . So, since , letting be the number of elements chosen from the ones which are smaller than , a standard concentration argument on for 999Observe that follows a hypergeometric distribution., shows that with probability we have . The same holds for the probability of the event , concluding the proof of the lemma. ∎
7 Relating the size of a component of to the order of a component of
As usual we use and throughout this section. We start with the proof of Theorem 4 and conclude the section with the proof of Theorem 5.
Proof of Theorem 4.
By Theorem 9 with , there is a such that if , then the probability that has no component of size is . We will choose to be the minimum of and . Hence we can assume that .
To complete the proof, we show that under this assumption, the probability has a component of order given has no component of size is . Under this hypothesis, for every component of , we have . Since each component is connected, this also implies that .
The following claim allow us to bound in terms of .
Claim 5.
If is the number of vertices of degree in , then
Proof.
Suppose that . By the definition of , we obtain . Since the function is convex,
which is a contradiction to the choice of . Thus . ∎
Since , Claim 5 implies .
Condition now on the choice of and on the set of fixed edges and let be the number of non-fixed ones. These choices determine .
Suppose that and recall that by Observation 34. Then, deterministically, each component of has at most vertices.
Otherwise, . Group the components of in sets such that . Clearly such a collection exists and . For every , we apply Lemma 36 with and . Since is well-behaved, the conclusion of Lemma 36 holds with probability . Using a union bound over all the sets , we obtain that with probability , for , we have
∎
Proof of Theorem 5.
It is enough to prove the theorem for sufficiently small positive , for which we will prove it with .
We can use Lemma 10 to show that if , then the probability has no component of order is .
Next, suppose that . We show that in this case, the probability that has no component of order given that contains a component of size is . We have that . Letting be the component of of size at least and be the component of corresponding to , and applying Lemma 36 with and , we conclude that with probability (since is well-behaved)
Thus, it remains to prove the theorem for
For any subgraph of , let be the excess of . We also define the near-excess of as , where is the set of vertices of degree at least in . Let be the set of vertices of that correspond to a component in with .
The following claim is crucial to finish the proof of the theorem.
Claim 6.
We have
We now conclude the proof of the theorem, assuming that the claim is true. Suppose contains a component that satisfies . If the corresponding component in satisfies , there is nothing to prove. So, suppose . Then
(8) |
and is non-empty.
Since the total excess of the graph is at most , the total near-excess is at most and there are at most components in . Hence, there exists a component in with at least vertices, which by Claim 6 implies that with probability , has a component of order at least .
So, it is indeed enough to prove Claim 6. To do so, we need the following:
Claim 7.
We have
Proof of Claim 7.
We let be the event that , and for any vertex , we let be the event that , but .
Say a graph is in . We only consider switchings between ordered pairs of oriented edges and in , where is an edge in a component of whose vertex set is in , which is not a cut-edge and we allow that or .
Since is in a component of and , there are at least choices for . Clearly, there are choices for the (directed) edge .
We show below that there are at most such switchings from any element of to . Thus
So, we have,
Since , Markov’s inequality implies that
and the claim follows.
It remains to prove the mentioned bound on the number of switchings between and . In doing so, we note that (i) if a connected subgraph contains a subgraph of near-excess , then also has near-excess at least , and (ii) the near-excess of the disjoint union of and is at least the sum of the near-excesses of and .
Consider a graph that is in . Let be the component of containing and let be the components of . For each neighbour of , let be the graph induced by and all components in in which has a neighbour.
In the following, we consider a triple of vertices such that switching leads to a graph in such that the edge is not a cut-edge and the component containing and has near-excess at least . We note that unless , this implies there is at most one edge between and . We note further that are distinct from but and may coincide as may and .
If , then and are both neighbours of in and there are exactly two edges of from to . Furthermore, there is a path of from to and the near-excess of is at least . So, we see that there are at most two choices for the pair and at most eight choices for switchings of this type.
If , then must be in and in there can be no path from to any of . Thus, there is a unique choice of for each such ordered and at most total choices for such switchings. ∎
Proof of Claim 6.
Let be the number of sets of vertices that satisfy
-
(1)
,
-
(2)
the sum of the degrees of the elements in is at least ,
-
(3)
,
-
(4)
there are no edges between and .
Observe that as defined satisfies (2), since it is non-empty and each component in has at least edges. Since is a union of components, it also satisfies (4).
We will show that . This directly implies that the probability satisfies (1) and (3) is at most , which combined with Claim 7 yields Claim 6.
Fix a set of vertices that satisfies (1), (2) and (3). For every , let be event that there are exactly edges that connect with .
There are at most switchings from a graph in which yields a graph in .
We now lower bound the number of switchings from a graph in , to a graph . To do so we consider pairs consisting of (i) an edge such that and is not a neighbour of a vertex in , and (ii) an edge with both endpoints in such that is not adjacent simultaneously to both and . For such a pair, we can always switch with at least one of or . There are at least choices for , since satisfies (3) and . Since , there are at least the same number of choices for the edge . Since , we have which implies that there are at most edges within that have both endpoint adjacent to . Using that satisfies (2) and that , we conclude that there are at least choices for the edge , given the choice of . The total number of switchings is at least .
So, using that and that
Therefore,
Since satisfies (1), there are at most choices for . Provided that is small enough, we conclude that . ∎
This completes the proof of Theorem 5. ∎
8 Applications of Theorem 3
We briefly show that Theorem 3 implies the results mentioned in Section 1.4. We consider a sequence of degree sequence such that
-
(d.1)
it is feasible, smooth, and sparse,
-
(d.2)
, and
-
(d.3)
.
Conditions (d.1)–(d.3) are essentially -conditions and they are weaker than -conditions and -conditions. An interesting consequence of them is the following: for every , there exist positive integers and such that if , then
Therefore, any sequence of degree sequences that satisfies (d.1)–(d.3) has almost all the edges incident to vertices of bounded degree.
The results of Molloy and Reed [24], Janson and Luczak [18], and Bollobás and Riordan [5] on the existence of a giant component, are of the following form: provided that satisfies certain conditions, if then has a linear order component with probability , and if , then has no linear order component with probability .101010In fact, the Molloy-Reed result does not discuss the case . We will show that if satisfies (d.1)–(d.3), then Theorem 3 implies the same statement.
Theorem 37.
Let be a sequence of degree sequences that satisfies conditions (d.1)–(d.3). Then,
-
1.
if , then there exists a constant such that the probability that has a component of order at least is .
-
2.
if , then for every constant , the probability that has no component of order at least is .
Proof.
First of all, observe that (d.2) implies that is well-behaved.
Fix small enough. Since , there is an integer such that . Also, since is smooth and provided that is large enough, we have
and hence, as is sparse,
(9) |
Recall . Assume first that . We will show that is lower bounded. Fix such that . Now note that there exists a positive integer such that . Since is smooth and provided that is large enough, we have that for every
Therefore,
Since every vertex of degree contributes in at most to the previous sum, this implies Using that is sparse and since is large enough, we have that and hence . Therefore,
and thus, the sequence is lower-bounded with . Since it is also well-behaved, Theorem 3 implies that there exists a constant such that the probability that has a component of order at least is .
Now suppose that . We aim to show that is upper bounded; that is, for every sufficiently small and large enough , we have . We fix an arbitrary and sufficiently small . Observe first that for any positive integer , as for every .
Furthermore, for sufficiently large , the number of vertices of degree different than is at least . Since the minimum degree is at least one, .
Using (9), we can choose large enough such that
As before, since is smooth, we can consider large enough such that for every , we have . Therefore,
Because for every , this implies
Note that the choice of was arbitrary and thus is upper-bounded. Since it is also well-behaved, Theorem 3 implies that for every constant , the probability that has no component of order at least is . ∎
Theorem 3 can be also applied to obtain results on the existence of a giant component in specific models of random graphs. Here, as an example, we will study the case of the Power-Law Random Graph introduced by Aiello, Chung and Lu [1]. Let us first recall its definition. Choose two parameters and consider the sequence of degree sequences where has vertices of degree , for every . We should think about these parameters as follows: is typically large and determines the order of the graph (we always have ), and is a fixed constant that determines the power-decay of the degree sequence. The total number of vertices, can be determined in terms of and ,
and similarly for the number of edges
where is the standard zeta function. Moreover, the maximum degree is .
The Power-Law Random Graph, denoted by , is a graph on vertices chosen uniformly at random among all such graphs with degree sequence . There are some values of (for instance ) for which does not satisfy the conditions (d.1)–(d.3). Thus, Theorem 37 cannot be used in general.
Nevertheless, observe that for every , we have , and thus, the asymptotic degree sequence is well-behaved. This in particular implies that .
Let be a solution to the equation . This is the important threshold for the appearance of the giant component since if and only if
(10) |
We refer the reader to the beginning of Section 3 in [1] for a formal proof of this fact. Thus for , the parameter is simply the maximum degree of , which is . Since , we have .
Theorem 38 (Aiello, Chung and Lu).
Let be a Power-Law random graph. Then,
-
1.
if , then there exists a constant such that the probability that has a component of order at least is .
-
2.
if , then for every constant , the probability that has no component of order at least is .
We want to emphasize that the structural description the authors give in their paper is more precise than the just mentioned result.
Proof.
We already addressed the case before the theorem. In such a case, the sequence is upper-bounded and it follows from Theorem 3 that for every constant , the probability that has no component of order at least is .
Now, we consider . We will show that in this case is lower-bounded. We split its proof into three cases,
Case : Suppose . The number of edges is of order and thus the average degree is of order , which tends to as . Thus, provided that is large enough, we have . The same argument applies for , as the average degree is of order .
Case : Let be the smallest integer such that . Thus
Therefore,
for some small constant . Here we used that .
Case : Suppose now that (the case is similar to what follows). Let and be the smallest integer such that . Using an integral approximation of the sum, we obtain . Thus
As in the previous case, we have for some constant .
Since is well-behaved and lower-bounded for , we can use Theorem 3 to conclude that there exists a constant such that the probability that has a component of order at least is . ∎
References
- [1] W. Aiello, F. Chung, and L. Lu, A random graph model for massive graphs, Proceedings of the thirty-second annual ACM Symposium on Theory of Computing (STOC), ACM, 2000, pp. 171–180.
- [2] R. Albert and A.-L. Barabási, Statistical mechanics of complex networks, Reviews of modern physics 74 (2002), 47.
- [3] A.-L. Barabási and R. Albert, Emergence of scaling in random networks, science 286 (1999), no. 5439, 509–512.
- [4] S. Boccaletti, V. Latora, Y. Moreno, M. Chavez, and D.-U. Hwang, Complex networks: Structure and dynamics, Physics reports 424 (2006), 175–308.
- [5] B. Bollobás and O. Riordan, An old approach to the giant component problem, J. Combin. Theory (Series B) 113 (2015), 236–260.
- [6] V. Bourassa and F. Holt, SWAN: Small-world wide area networks, Proceeding of International Conference on Advances in Infrastructures (SSGRR 2003w), L’Aquila, Italy, 2003.
- [7] C. Cooper, The cores of random hypergraphs with a given degree sequence, Random Structures Algorithms 25 (2004), 353–375.
- [8] C. Cooper, M. Dyer, and C. Greenhill, Sampling regular graphs and a peer-to-peer network, Comb., Probab. Comput. 16 (2007), 557–593.
- [9] M. Faloutsos, P. Faloutsos, and C. Faloutsos, On power-law relationships of the internet topology, ACM SIGCOMM computer communication review, vol. 29, ACM, 1999, pp. 251–262.
- [10] D. Fernholz and V. Ramachandran, Cores and connectivity in sparse random graphs, The University of Texas at Austin, Department of Computer Sciences, technical report TR-04-13 (2004).
- [11] , The diameter of sparse random graphs, Random Structures Algorithms 31 (2007), 482–516.
- [12] P. Flajolet and R. Sedgewick, Analytic combinatorics, Cambridge University Press, 2009.
- [13] N. Fountoulakis, Percolation on sparse random graphs with given degree sequence, Internet Mathematics 4 (2007), 329–356.
- [14] N. Fountoulakis and B. Reed, A general critical condition for the emergence of a giant component in random graphs with given degrees, Electronic Notes in Discrete Mathematics 34 (2009), 639–645.
- [15] C. Greenhill, The switch markov chain for sampling irregular graphs, Proceedings of the Twenty-Sixth Annual ACM-SIAM Symposium on Discrete Algorithms, SIAM, 2015, pp. 1564–1572.
- [16] H. Hatami and M. Molloy, The scaling window for a random graph with a given degree sequence, Random Structures Algorithms 41 (2012), 99–123.
- [17] S. Janson, On percolation in random graphs with given vertex degrees, Electron. J. Probab. 14 (2009), 87–118.
- [18] S. Janson and M. J. Luczak, A new approach to the giant component problem, Random Structures Algorithms 34 (2009), 197–216.
- [19] S. Janson and M.J. Luczak, A simple solution to the k-core problem, Random Structures Algorithms 30 (2007), 50–62.
- [20] A. Joseph, The component sizes of a critical random graph with pre-described degree sequence, The Annals of Applied Probability 24 (2014), 2560–2594.
- [21] M. Kang and T. G. Seierstad, The critical phase for random graphs with a given degree sequence, Comb., Probab. Comput. 17 (2008), 67–86.
- [22] T. Luczak, Sparse random graphs with a given degree sequence, Proceedings of the Symposium on Random Graphs, Poznan, 1989, pp. 165–182.
- [23] B. D. McKay, Subgraphs of random graphs with specified degrees, vol. 33, 1981, pp. 213–223.
- [24] M. Molloy and B. Reed, A critical point for random graphs with a given degree sequence, Random Structures Algorithms 6 (1995), 161–180.
- [25] , The size of the giant component of a random graph with a given degree sequence, Comb., Probab. Comput. 7 (1998), 295–305.
- [26] , Graph colouring and the probabilistic method, vol. 23, Springer Science & Business Media, 2013.
- [27] M. E. J. Newman, The structure and function of complex networks, SIAM review 45 (2003), 167–256.
- [28] M. E. J. Newman, D.J. Watts, and S.H. Strogatz, Random graph models of social networks, Proceedings of the National Academy of Sciences 99 (2002), no. suppl 1, 2566–2572.
- [29] J. Petersen, Die Theorie der regulären graphs, Acta Math. 15 (1891), 193–220.
- [30] O. Riordan, The phase transition in the configuration model, Comb., Probab. Comput. 21 (2012), 265–299.
- [31] H. van den Esker, R. van der Hofstad, G. Hooghiemstra, and D. Znamenski, Distances in random graphs with infinite mean degrees, Extremes 8 (2005), no. 3, 111–141.
- [32] R. van der Hofstad, G. Hooghiemstra, and P. van Mieghem, Distances in random graphs with finite variance degrees, Random Structures Algorithms 27 (2005), 76–123.
- [33] N. C. Wormald, Some problems in the enumeration of labelled graphs, Doctoral Thesis (1980).
- [34] , Models of random regular graphs, Surveys in combinatorics, 1999 (Canterbury), London Math. Soc. Lecture Note Ser., vol. 267, Cambridge Univ. Press, Cambridge, 1999, pp. 239–298.