Small maximal matchings of random cubic graphs

Small Maximal Matchingsof Random Cubic Graphs

H. Assiyatun1 and W. Duckworth2

1DEPARTMENT OF MATHEMATICSINSTITUT TEKNOLOGI BANDUNG

BANDUNG 40132, INDONESIAE-mail: [email protected]

2MATHEMATICAL SCIENCES INSTITUTEAUSTRALIAN NATIONAL UNIVERSITY, CANBERRA

ACT 2000, AUSTRALIAE-mail: [email protected]

Received August 13, 2002; Revised March 30, 2008

Published online 14 October 2009 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/jgt.20434

Abstract: We consider the expected size of a smallest maximal matchingof cubic graphs. Firstly, we present a randomized greedy algorithm forfinding a small maximal matching of cubic graphs. We analyze the average-case performance of this heuristic on random n-vertex cubic graphs usingdifferential equations. In this way, we prove that the expected size of themaximalmatching returned by the algorithm is asymptotically almost surely(a.a.s.) less than 0.34623n. We also give an existence proof which showsthat the size of a smallest maximal matching of a random n-vertex cubicgraph is a.a.s. less than 0.3214n. It is known that the size of a smallestmaximal matching of a random n-vertex cubic graph is a.a.s. larger than0.3158n. � 2009 Wiley Periodicals, Inc. J Graph Theory 62: 293–323, 2009

Keywords: matchings; random; cubic; graphs

Journal of Graph Theory� 2009 Wiley Periodicals, Inc.

293

294 JOURNAL OF GRAPH THEORY

1. INTRODUCTION

A maximal matching of a graph G is a set of edgesM⊆E(G) such that no two edges ofM share a common end-point and every edge in E(G)\M shares at least one end-pointwith an edge of M. The problem of finding a minimum maximal matching of a graphis encompassed by the minimum maximal flow problem which plays an important rolein the investigation of how inefficiently a network may be organized [14].

Yannakakis and Gavril [21] showed that the problem of finding a minimum maximalmatching of a graph is NP-hard, even when restricted to planar or bipartite graphs ofmaximum degree 3. In the same article they also gave a polynomial time algorithmthat finds a minimum maximal matching of trees. Horton and Kilakos [10] showedthat the problem of finding a minimum maximal matching of a graph remains NP-hard for planar bipartite graphs. They also gave a polynomial time algorithm thatfinds a minimum maximal matching for various classes of chordal graphs. Zito [22]extended these NP-hardness results to include (ks,3s)-graphs for every integer s>0 andfor k∈{1,2}. (A (�,�)-graph has minimum degree � and maximum degree �.)

A graph G is said to be d-regular if every vertex of G has degree d. We considersimple, connected, cubic (i.e. 3-regular) graphs on n vertices (nmust be even). For otherbasic graph theory definitions, not defined here, we refer the reader (as an example) toDiestel [5].

Duckworth [7] showed that the size of a minimum maximal matching of ann-vertex cubic graph is at most 9n/ 20+O(1). Their proof uses a linear programmingtechnique to analyze the worst-case performance of a simple greedy algorithm. Theuse of this technique also illustrates that there exist infinite families of cubic graphsfor which their algorithm only achieves this bound. In the same article they alsoshowed that there exist infinite families of n-vertex cubic graphs for which the size ofa minimum maximal matching is at least 3n/ 8. Note that it is simple to verify that thesize of a minimum maximal matching of an n-vertex cubic graph is at least 3n/ 10 andat most n/ 2.

As we consider random cubic graphs that are generated uniformly at random (u.a.r.),we need some notation. Throughout this article, we use the notation P (probability), E(expectation) and say that a property, B=Bn, of a random graph on n vertices holdsasymptotically almost surely (a.a.s.) if limn→∞P(Bn)=1. For other basic random graphtheory definitions not defined here, we refer the reader (as an example) to Jansonet al. [13].

Zito [23] showed that the size of a smallest maximal matching, M, of a randomn-vertex cubic graph, a.a.s., satisfies 0.3158n<|M|<0.47653n. The upper bound wasachieved by analyzing the performance of simple heuristic that is based, at each step, onrepeatedly choosing vertices of current maximum degree, choosing an edge to be partof the matching and deleting edges. The lower bound was achieved using a standarddirect expectation argument.

In the following section we present a simple, yet efficient, randomized greedy algo-rithm for finding a small maximal matching of cubic graphs. We introduce the modelwe use for generating cubic graphs u.a.r. in Section 3. In the same section, we also

Journal of Graph Theory DOI 10.1002/jgt

SMALL MAXIMAL MATCHINGS OF RANDOM CUBIC GRAPHS 295

describe a known technique, the differential equation method, for analyzing the perfor-mance of algorithms on random regular graphs. We analyze our algorithm on randomcubic graphs as detailed in Section 4. We prove that the expected size of the maximalmatching returned by the algorithm is a.a.s. less than 0.34623n in Section 5. We useanother known technique, the small subgraph conditioning method, to give an existenceproof that shows that the size of a minimum maximal matching of a random n-vertexcubic graph is a.a.s. less than 0.3214n. The details of this are given in Section 6.Although this upper bound is lower, the use of this method does not indicate an actualalgorithm that is expected to achieve this result. Both results are therefore of interest.

2. GREEDY ALGORITHMS

There are many ways in which one may design a greedy algorithm for finding a smallmaximal matching of a cubic graph. These differ in the way each subsequent edge isselected to be part of the matching. In all cases, once an edge has been selected to bepart of the matching, the graph may be updated by removing the selected edge andall edges incident with its end-points. The algorithm we introduce for finding a smallmaximal matching of cubic graphs is a greedy algorithm that is based, at each step,on selecting vertices of minimum degree u.a.r. from an ever-shrinking subgraph of theinput graph.

Each step of the algorithm involves the process of selecting an edge to add to a set,M, and the deletion of the edges incident with the end-points of the selected edge. Thedeletion of these edges ensures that the matching edges remain vertex disjoint which,in turn, ensures that once no edges remain, the set M is a maximal matching. Therationale behind selecting vertices of current minimum degree is that we do not wantto be in a situation where there are many vertices of small current degree. By selectingsuch a vertex at the start of each step we avoid this build-up of vertices of small degree.If we allowed this to happen, many vertices of small degree would eventually be chosenas an end-point of a matching edge which, in turn, removes a small number of edgesand thus causing the number of steps (and the size of the matching) to increase.

The algorithm, which we refer to as MMM Greedy, is given in Figure 1. Thealgorithm takes an n-vertex cubic graph G as input and returns a maximal matching,M,of G. We assume the input graph to be connected; otherwise, we apply our algorithmto each connected component. For a vertex v, deg(v) denotes the current degree of vin G and N(v) denotes the current set of neighbors of v. Also, Vi denotes the set ofvertices of current degree i in G and Yi denotes |Vi|.

The first step of the algorithm involves the process of selecting the first edge of thematching u.a.r. from all the edges of the input graph and deleting the edges incidentwith the end-points of this selected edge. As the input graph is assumed to be connected,we note that after the first step and before the completion of the algorithm, there mustalways exist a vertex of degree 1 or 2. For each subsequent step, if Y1 is non-zero, weselect a vertex, w, u.a.r. from V1. Otherwise, w is selected u.a.r. from V2. A vertex, u,is then selected u.a.r. from the neighbors of w. If u has degree 1, then so does w which



FIGURE 1. Algorithm MMM Greedy.

means that this edge is isolated in G so we simply add the edge uw to M. Otherwise, adifferent edge incident with u is selected to become part of M. We select a neighbor, v,of u u.a.r. from those vertices that have current maximum degree in the set {N(u)\w}.We then add the edge uv to M and delete all edges incident with u and v.

3. RANDOM GRAPHS AND DIFFERENTIAL EQUATIONS

In this section we introduce the model we use to generate a cubic graph u.a.r. and givean overview of an established method of analyzing the performance of randomizedalgorithms on regular graphs.

The standard model for random d-regular graphs is as follows. Take a set of dnpoints in n buckets labeled 1,2, . . . ,n, with d points in each bucket, and choose u.a.r. apairing P=p1, . . . ,pdn/2 of the points such that each pi is an unordered pair of pointsand each point is in precisely one pair pi. The resulting probability space of pairings isdenoted by Pn,d. Form a d-regular pseudograph (i.e. a non-simple graph in which loopsand multiple edges are allowed) on n vertices by placing an edge between vertices iand j for each pair in P having one point in bucket i and one in bucket j. In order toprove that a property is a.a.s. true of a uniformly distributed random d-regular (simple)graph, it is enough to prove that it is a.a.s. true of the pseudograph corresponding to arandom pairing (see, for example, Bollobas [4] and Wormald [19, Section 2]).

One method of analyzing the performance of a randomized algorithm is to use asystem of differential equations to express the expected changes in variables describingthe state of the algorithm during its execution. Wormald [20] gives an exposition of thismethod and this method has been applied to several other graph-theoretic optimizationproblems including (amongst others) independent dominating set [8], independent set[20] and 2-independent set [6].

In order to analyze the algorithm given in Figure 1, we incorporate the algorithmas part of a pairing process that generates a random regular graph. In this way, wegenerate the random graph in the order that the edges are examined by the algorithm.



We may consider the generation process as follows. Initially, all vertices have degreezero. Throughout the execution of the generation process vertices will increase in degreeuntil all vertices have degree 3. Once the degree of a vertex reaches 3, the vertex is saidto be saturated and the generation is complete when all vertices are saturated. Duringthe generation process, we refer to the graph being generated as the evolving graph asthe graph evolves from all vertices having degree zero at the start of the process to allvertices having degree 3 at the end of the process.

During the generation of a random cubic graph we choose the pairs sequentially.The first point, pi, of a pair may be selected by any rule but in order to ensure thatthe cubic graph is generated u.a.r., the second point, pj, of that pair must be selectedu.a.r. from all the remaining free (i.e. unpaired) points. This preserves the uniformdistribution of the final pairing. The freedom of choice of pi enables us to select it u.a.r.from the vertices of a particular degree in the evolving graph and we say that selectingpj u.a.r. from all the remaining free points denotes selecting a mate for pi. Using B(pk)to denote the bucket that the point pk belongs to, we say that the edge from B(pi) toB(pj) is exposed and B(pj) is hit by this edge. We may then determine the degree ofthe vertex represented by the bucket B(pj) without exposing further edges.

In what follows, we denote the set of vertices of current degree i of the evolvinggraph, at time t, by Vi=Vi(t) and let Yi=Yi(t) denote |Vi|. We may express the state ofthe evolving graph at any point during the execution of the algorithm by considering Y0,Y1 and Y2. In order to analyze our randomized algorithm for finding a small maximalmatching, M, of cubic graphs, we calculate the expected change in this state over apredefined unit of time in relation to the expected change in the size of M.

Let M=M(t) denote |M| at any stage of the algorithm (time t) and let E�X denotethe expected change in a random variable X conditional upon the history of the process.Equations representing E�Yi and E�M are then used to derive a system of differentialequations. The solutions to the differential equations describe functions which representthe behavior of the variables Yi. Wormald [20, Theorem 6.1] describes a generalresult which guarantees that the solutions of the differential equations almost surelyapproximate the variables Yi and M with error o(n). The expected size of M may bededuced from these results.

4. INCORPORATING THE PAIRING PROCESS

The incorporated algorithm and pairing process are given in Figure 2. In the algorithm,the function isolate(u,v) exposes all remaining edges incident with the vertices repre-sented by the buckets u and v. This ensures that the set of edges returned is vertexdisjoint.

The algorithm terminates when there are no vertices of degree 1 or 2 remainingwhich means that a connected component has been completely generated and a maximalmatching has been found in that component. It was shown independently by Bollobas [3]and Wormald [17] that for fixed d≥3, random d-regular graphs are a.a.s. d-connected,so the result is a.a.s. a maximal matching in the whole graph.



FIGURE 2. MMM Greedy incorporated with a pairing process.

We say that the combined process proceeds in operations where each operationdenotes the selection of an edge to add to the matching along with exposing all edgesincident with its end-points. The first operation of the algorithm involves randomlyselecting the first edge of M and exposing the appropriate edges. After this firstoperation we note that, before the end of the process, there always exists a vertex ofdegree 1 or 2. This follows as random regular graphs are a.a.s. connected.

There are two basic types of operation performed by the algorithm which we referto as Type 1 and Type 2 operations. A Type 1 operation refers to an operation whereY2=0 and a vertex is selected u.a.r. from V1. Similarly, a Type 2 operation refers toan operation where Y2>0 and a vertex is selected u.a.r. from V2.

Once w has been selected, we expose an edge incident with w and let u denote thevertex hit by this exposed edge. If both u and w are saturated after this edge is exposed,we add the edge uw to M. Otherwise, we expose all remaining edges incident with u.Using S to denote the set of vertices hit by these exposed edges, we select a vertex vu.a.r. from the vertices of minimum degree in S. We then add the edge uv to M andexpose all remaining edges incident with the vertices u and v.

5. ALGORITHM ANALYSIS

We analyze the average-case performance of MMM Greedy using the differential equa-tion method and in this way prove the following theorem.

Theorem 1. For a random cubic graph on n vertices, the size of a minimum maximalmatching is a.a.s. less than 0.34623n.

Proof. After the first operation, we split the remainder of the analysis into twodistinct phases. We informally define Phase 1 as the period of time where any verticesin V2 that are created are used up almost immediately and Y2 remains small. Once therate of generating vertices in V2 becomes larger than the rate that they are used up,



Phase 2 commences. The transition point between phases is not obvious but arises inour analysis. Define a clutch to be a series of operations in Phase i from an operationof Type i up to, but not including, the next operation of Type i. Increment time by oneunit for each clutch and at each time interval update all the variables. We calculateE(�Yi) and E(�M) for a clutch in each phase.

A. Preliminary Equations

The initial operation of Phase 1 is of Type 1 (at least a.a.s.) and a clutch of operationsin Phase 1 consists of one Type 1 operation and possibly some Type 2 operations.For Phase 2, a clutch consists of just one Type 2 operation. We develop equations torepresent the expected changes in the variables Yi and M for each type of operationand use these equations to define the expected changes in the variables for a clutchof operations in each phase. Due to their similarity, we consider both Type 1 and 2operations together and form a general equation for an operation of Type r∈{1,2}.

Let s denote the number of free points available at time t. Note that s=s(t)=3Y0+2Y2+Y1. For our analysis it is convenient to assume that s>�n for some arbitrarilysmall but fixed �>0. Operations where s≤�n will be discussed later.

The probability that a free point selected u.a.r. from all free points in the evolvinggraph (at time t) is selected from a vertex, v, of degree i is Pi where

Pi=Pi(t)= (3− i)Yis

, 0≤ i≤2. (1)

For any statement Q, let �Q=1 if Q evaluates to true and let �Q=0 otherwise. Then,the expected change in Yi due to increasing the degree of a vertex v from i to i+1 orfrom i−1 to i by exposing an edge to it (at time t) is �i where

�i=�i(t)=−Pi+Pi−1�i>0+o(1), 0≤ i≤2. (2)

To justify this, note that when we expose an edge to a vertex, v, we have selected thefirst point in a pair and we select its mate u.a.r. from all the remaining free points in theevolving graph. When the point in the bucket representing v was selected, the numberof points in the buckets corresponding to vertices currently of degree i is (3− i)Yi,and s is the total number of free points. In this case Yi decreases; it increases if thesecond point in the pair is selected from a bucket representing a vertex of degree i−1.These two quantities are added because expectation is additive. The term o(1) arisesas the values of all these variables may change by a constant during the course ofthe operation being examined (as only a constant number of edges are exposed in anyone operation and s is assumed to be larger than �n). Since s>�n the error is in factO(1 / n).

The probability that, after exposing m edges incident with a vertex, the minimumdegree of the vertices hit was j (before they had edges exposed to them) is �mj where

�mj =�mj (t)=(

2∑k=j

Pk

)2−m

−(

2∑k=j+1

Pk

)2−m

. (3)



The above definitions are sufficient to formulate an equation to represent the expectedchange in Yi (at time t) when performing an operation of Type r∈{1,2}. This Fr

i =Fri (t)

which is given by

−�i=r+�i=r+1−Pi+1∑

m=0Pm(2−m)�i+

1∑m=0

1∑j=0

�mj [−�i=j+1+(2− j)�i]+o(1) (4)

where 0≤ i≤2 and 1≤r≤2. The first term represents the removal of w from Vi

(if r= i) and the second term represents the addition of w to Vi (if r= i−1). The thirdterm denotes the removal of u from Vi (if u has degree i), and as all edges incidentwith u are exposed, u enters V3 (which we do not consider).

The remaining terms are sums over m where m represents the degree of u. We onlyconsider the cases when u has degree 0 or 1 as, clearly u was of degree less than 3 at thestart of the operation and if u had degree 2 at the start of the operation, exposing an edgefrom w to u would make u saturated and the operation would then be complete. Withprobability Pm we expose the remaining 2−m edges incident with u. The maximumdegree of the vertices hit by these edges is denoted by j. Summing over all relevantvalues for j (again, the only interesting values for j are 0 and 1) we remove one suchvertex and expose its remaining incident edges. Note that the term Pm(2−m)�i changesthe degree of v (the second end-point of the matching edge chosen) from j to j+1 andso −�i=j+1 is the correct term for the removal of v of (previous) degree j.

This completes the expected changes in the variables Yi for a single operation ina phase. In order to formulate the necessary equations for a clutch of operations, weneed a few more definitions.

We define a birth to be the generation of a vertex in V2 by performing an operationin Phase 1. The expected number of births from performing an operation of Type 2(at time t) is �2=�2(t) where

�2=P11∑

m=0

[Pm(2−m)+

1∑j=0

�mj (d− j−1)

]+o(1). (5)

Here, we consider the probability that vertices of degree 1 (in the evolving graph)become vertices of degree 2 by exposing an edge incident with the vertex. This isessentially F2

2 without considering vertices in V2 that enter V3. The number of births isgiven by the expected change in Y2 without taking into account the number of bucketsof degree 2 that becomes saturated, hence the similarity between Equations (4) and (5).

Similarly, the expected number of births from performing an operation of Type 1 (attime t) is �1=�1(t) where �1=�2+1+o(1) as the expected number of births is a.a.s.the same apart from the fact that w also attains degree 2 (in the evolving graph).

Consider the Type 1 operation at the start of the clutch to be the first generationof a birth–death process in which the individuals are the vertices in V2, each givingbirth to a number of children (essentially independent of the others since s>�n) withexpected number �2. Then, the expected number in the jth generation is �1�2j−1 andthe expected total number of births in the clutch is given by the limit of the sum of allthese generations which is �1 / (1−�2)+o(1).



For Phase 1, the equation giving the expected change in Yi for a clutch is thereforegiven by

E(�Yi)=F1i + �1

1−�2F2i , 0≤ i≤2. (6)

The first term represents the expected changes due to the Type 1 operation at the startof the clutch. For each birth generated by this operation, we have a Type 2 operation(at least a.a.s.) and the second term represents these expected changes.

As the expected increase in the size of the maximal matching is 1 for any operation,the equation giving the expected increase in M for a clutch in Phase 1 is given by

E(�M)=1+ �11−�2

(7)

representing one for the Type 1 operation and (a.a.s.) one for each birth.In Phase 2, all operations are considered to be of Type 2 and therefore a clutch

consists of just one operation. The expected change in the variables Yi is given by

E(�Yi)=F2i . (8)

As the expected increase in the size of the matching for any operation is 1 and inPhase 2, a clutch consists of a single Type 2 operation, the expected increase in thesize of the matching for a clutch in Phase 2 is just 1.

B. Differential Equations

The basic idea of the differential equation method is quite simple. Compute the expectedchanges in random variables of the process per unit time t (which we have done), andregarding the variables as continuous, write down the differential equations suggestedby the expected changes. Then, use large deviation theorems to show that, with highprobability, the solution of the differential equations is close to the values of thevariables. This approach is standard in the study of continuous time processes, suchas in the book by Kurtz [11]. Conversion to discrete can be done by simple relationsbetween the discrete time and continuous time processes (see [11, Proposition 4.5] and[20, Section 1.1]).

For Phase 1 (using Equation (6)) we define functions zi(x), 0≤ i≤2, by the systemof differential equations

dzidx

=�1i + �1

1−�2�2i , 0≤ i≤2, (9)

where differentiation is with respect to x (xn represents the number, t, of clutches) andwhere for 0≤ i≤2

=2∑

q=0(3−q)zq,

i =(3− i)zi

,



�i = −i+i−1�i>0,

�mj =(

2∑k=j

k

)2−m

−(

2∑k=j+1

k

)2−m

, 0≤ j≤1, m∈{0,1},

�ri =1∑

m=0

[m(2−m)�i+

1∑j=0

�mj [−�i=j+1+(d− j−1)�i]

]−�i=r+�i=r+1�i≤2−i, r∈{1,2},

�2 = 1

1∑m=0

[m(2−m)+

1∑j=0

�mj (d− j−1)

]and �1=1+�2.

Equation (7) suggests the differential equation for z as

dz

dx=1+ �1

1−�2(10)

where, again, differentiation is with respect to x.For Phase 2, Equation (8) suggests the differential equation

dzidx

=�2i , 0≤ i≤2 (11)

where xn represents the number t of clutches in Phase 2.The solution to these systems of differential equations represents the cardinalities of

the sets Vi and M (scaled by 1/n) for given t. For Phase 1, the equations are (9) and(10) with initial conditions z0(0)=1, z1(0)=0, z2(0)=0 and z(0)=0. For Phase 2, theinitial conditions are the final conditions of Phase 1 and the equation is (11).

We use a result from [20] to show that during each phase, the functions representingthe solutions to the differential equations almost surely approximate the variables Yi / nandM/ nwith error o(1). For this we need some more definitions. Consider a probabilityspace, S, whose elements are sequences (q0,q1, . . .) where each qt∈S. We use ht todenote (q0,q1, . . . ,qt), the history of the process up to time t, and Ht for its randomcounterpart. S(n)+ denotes the set of all ht= (q0, . . . ,qt) where each qi∈S, t=0,1, . . ..All these things are indexed by n and we will consider asymptotics as n→∞.

We say that a function f (u1, . . . ,uj) satisfies a Lipschitz condition on W⊆Rj if aconstant L>0 exists with the property that

|f (u1, . . . ,uj)− f (v1, . . . , vj)|≤Lmax1≤i≤j

|ui−vi|

for all (u1, . . . ,uj) and (v1, . . . , vj) in W. Note that

max1≤i≤j

|ui−vi|

is the distance between (u1, . . . ,uj) and (v1, . . . , vj) in the �∞ metric.



For variables Y1, . . . ,Ya defined on the components of the process, and W⊆Ra+1,define the stopping time TW =TW (Y1, . . . ,Ya) to be the minimum t such that

(t / n,Y1(t) / n, . . . ,Ya(t) / n) /∈W.

The following is a restatement of [20, Theorem 6.1] (which also appeared in [8]).We refer the reader to [20] for full details and explanations and to [18] for a similarresult with virtually the same proof.

Theorem 2 (Wormald [21]). Let W=W(n)⊆Ra+1. For 1≤ l≤a, where a is fixed,let yl :S(n)+ →R and fl :Ra+1→R, such that for some constant C0 and all l, |yl(ht)|<C0n for all ht∈S(n)+ for all n. Let Yl(t) denote the random counterpart of yl(ht).Assume the following three conditions hold, where in (ii) and (iii) W is some boundedconnected open set containing the closure of {(0, z1, . . . , za) :P(Yl(0)=zln,1≤ l≤a) �=0for some n}.

(i) For some functions = (n)≥1 and �=�(n), the probability that

max1≤l≤a

|Yl(t+1)−Yl(t)|≤ ,

conditional upon Ht, is at least 1−� for t<min{TW,TW}.(ii) For some function �1=�1(n)=o(1), for all l≤a

|E(Yl(t+1)−Yl(t) |Ht)− fl(t / n,Y1(t) / n, . . . ,Ya(t) / n) |≤�1

for t<min{TW,TW}.(iii) Each function fl is continuous, and satisfies a Lipschitz condition, on

W∩{(t, z1, . . . , za) : t≥0},with the same Lipschitz constant for each l.

The following are true:

(a) For (0, z1, . . . , za)∈W the system of differential equations

dzldx

= fl(x, z1, . . . , za), l=1, . . . ,a

has a unique solution in W for zl :R→R passing through zl(0)= zl, 1≤ l≤a,and which extends to points arbitrarily close to the boundary of W;

(b) Let �>�1+C0n� with �=o(1). For a sufficiently large constant C, with proba-bility

1−O

(n�+

�exp

(−n�3

3

)),

Yl(t)=nzl(t / n)+O(�n) uniformly for 0≤ t≤min{�n,TW} and for each l, wherezl(x) is the solution in (a) with zl= (1 / n)Yl(0), and �=�(n) is the supremum ofthose x to which the solution may be extended before reaching within �∞-distanceC� of the boundary of W.



First, we apply Theorem 2 to the process within Phase 1. Define W to be the vectorsfor which z1≥0, z2≥0 and z1+z2>0. Also, for arbitrary small �, define W to be theset of all (t, z0, z1, z2, z) for which

�2<1−�, zi<1+�, t>−�, z>−� and >�, 0≤ i≤2.

For part (i) of Theorem 2 we must ensure that Yi(t) does not change too quickly. Aslong as the expected number of births in a clutch is bounded above, the probability ofgetting say n� births is O(n−K) for any fixed K. This comes from a standard argumentas in [20, page 141]. So part (i) of Theorem 2 holds with =n� and �=n−K.

Near the start of the process, operations may be of Type 1 or Type 2. Equations (6)and (7) verify part (ii) for a function �1 which goes to zero sufficiently slowly. (Notein particular that since >� inside W, the assumption that s>�n used in deriving theseequations is justified. Also, since t<TW , it follows that Y1+Y2>0, so that the nextoperation is of Type 1 or Type 2.)

By the definition of the phase and the domainW, it may be verified that the functionsderived from Equations (6) and (7) are continuous on W and its boundary. This impliesthat the functions are uniformly continuous. From this, the Lipschitz property of thefunctions required by Theorem 2 part (iii) may be deduced.

The conclusion of Theorem 2 therefore holds for the process in Phase 1. This impliesthat with probability

1−O(n1−K+n��−1 exp(−n1−3��3)),

the random variables Yi and M a.a.s. remain within O(�n) of the corresponding deter-ministic solutions to the differential equations (9) and (10) until a point arbitrarilyclose to where it leaves the set W, or until t=TW if that occurs earlier. Note that thelatter may only occur when the algorithm has completely processed a component ofthe graph and a random cubic graph is a.a.s. connected. Choosing K=2 and �=n�−1/4,say, leads to a success probability of 1−o(1).

We compute the ratio dzi / dz and we have

dzidz

=�1i + �1

1−�2�2i

1+ �1

1−�2

, 0≤ i≤2 (12)

where differentiation is with respect to z and all functions may be taken as functionsof z.

By solving (numerically) this system of differential equations, we find that thesolution hits a boundary of the domain at �2=1−� (for �=0 this would be atz≥0.1805). At this point, we may formally define Phase 1 as the period of time fromtime t=0 to the time t0 such that z= t0 / n is the solution of �2=1.

Our next aim is to show that by the time �′n operations after the start of Phase 2(for some �′ >0), the variable Y2 is a.a.s. at least some constant times n. For this, themain requirement is that the variable �2 increases significantly above 1, since �2−1 isthe expected increase in Y2 when performing an operation of Type 2.



Unfortunately, the expected increase in �2 due to performing an operation of Type1 right near the end of Phase 1 is negative. So instead we consider the variable �2defined by setting Y2=0 in the definitions of all variables; that is, �2= �2(t) where

�2 = P11∑

m=0

[Pm(2−m)+

1∑j=0

�mj (d− j−1)

]

P1 = P1(t)= 2Y1s

,

� mj = � m

j (t)=(

2∑k=j

Pk

)2−m

−(

2∑k=j+1

Pk

)2−m

and

s= 3Y0+2Y1.

Regarding �2 as a function of Y0 and Y1 only, we may compute the expected increasein �2 due to an operation of Type 1 as

� �2�Y0

E0+ � �2�Y1

E1 (13)

where Ei is the expected increase in Yi in such an operation computed from (6).Plugging in the values of Y0 and Y1 at the end of Phase 1 gives a positive quantity,

approximately 2.16. For a Type 2 operation, the same calculation is used, but the valuesof E0 and E1 come from (8). The result is 3.92.

Since the formula given by (13) is Lipschitz, it must remain positive for at least �1noperations after reaching time t0−�n, for �1 sufficiently small. Subject to the choiceof �1, we may take � arbitrarily small. It now follows by the usual large deviationargument that the increase in �2 between time t0−�n and a time t1 when �1n operationshave occurred in Phase 2 is a.a.s. at least c for some positive constant c. By choosing� sufficiently small, �2 is a.a.s. arbitrarily close to 1 at time t0−�n, and so the samegoes for �2 since Y2 is a.a.s. very small in Phase 1. Thus �2>1+c1 a.a.s. at time t1 forsome c1>0.

Once this value of �2 is attained, since �2=�2 when Y2=0 we may choose a c>0such that either Y2>cn or �2>1+c. In the former case we are well into Phase 2 in theinformal sense. In the latter case, due to the Lipschitz property of �2, for the next �2noperations, performing a Type 2 operation produces an expected 1+c / 2 new verticesof V2. Again, using the usual large deviation argument, this ensures that with highprobability the process moves in the next �2n operations into a state where V2>c2n,and is thus, again, firmly entrenched in Phase 2 in the informal sense. Thus, in eithercase, there will be some time t2 which is followed by c2n consecutive operations ofType 2, which means that the equations for Phase 2 are valid.

For Phase 2 and for arbitrary small �, define W ′ to be the set of all (t, z0, z1, z2, z)for which t>t2−�, >�, z>−� and zi<1+� where 0≤ i≤2. Theorem 2 applies as inPhase 1 (with time shifted by subtracting t2) except that here, a clutch consists of just



one operation of Type 2. Note also that the starting point of the process is randomized,which is permitted in Theorem 2.

This implies that with probability 1−O(n1−K+n��−1 exp(−n1−3��3)), the randomvariables Yi and M a.a.s. remain within O(�n) of the corresponding deterministicsolutions to the differential equation (11) until a point arbitrarily close to where itleaves the set W ′. Choosing K=2 and �=n�−1/4, say, leads to a success probability of1−o(1). We have

dzidz

=�i,2, 0≤ i≤2

and by solving this, we see that the solution hits a boundary of W ′ at =�.From the point in Phase 2 after which Theorem 2 does not apply until the completion

of the algorithm, the change in each variable per step is bounded by a constant. Hence,letting � tend to 0 sufficiently slowly, in o(n) steps, the change in the random variablesYi and M is o(n). The differential equations were solved using a Runge–Kutta methodgiving =� at z<0.3462. This corresponds to the size of the maximal matching (scaledby 1 / n) when all vertices are used up, thus proving the theorem. �

6. SMALL SUBGRAPH CONDITIONING

The previous sections considered an algorithmic proof for an upper bound on the sizeof a minimum maximal matching of a random cubic graph. In this section we givean existence proof for the same bound using the small subgraph conditioning method.This method, which is a technique of analyzing variance, was introduced by Robinsonand Wormald [16] to prove the a.a. sure Hamiltonicity of random d-regular graphs. Themethod also provides a means of proving the existence, and the asymptotic distribution,of properties of random regular graphs.

Janson [12] and Robalewska [15] successfully used this method to determine the a.a.sure existence, and the asymptotic distribution, of the number of 1-regular and 2-regularspanning subgraphs in random regular graphs. Garmo [9] applied the method to thenumber of long cycles in random d-regular graphs. In more recent work, Assiyatun [1]have used the method to study independent sets and 2-independent sets.

Throughout this section, we use Gn,3 to denote the set of random cubic graphs onn vertices. We use Pn,3 to denote the set of pairings that generates a random n-vertexcubic graph. A pairing P∈Pn,3 corresponds to a cubic pseudograph G(P). The mainresults obtained of this section are encompassed by the following theorem and corollary.

Theorem 3. Let 310<�< 1

2 be fixed. Define Y ′� =Y ′

�(n) to be the number of maximalmatchings of size p=p(n)=��n� in G∈Gn,3 and let �=p/ n. If �=0.3214 then G∈Gn,3a.a.s. has a maximal matching of size �n. Moreover,

Y ′�

EY ′�

d→W=∞∏k≥3

(1+�k)Zk e−�k�k as n→∞,



where Zk are independent Poisson variables with EZk=�k for k≥3, with �k=2k−1 / kand �k=2Re(�k) where �= (2�−3+

√196�2−108�+9) / 16� and Re(x) denotes the

real part of x.

Note that �→� as n goes to infinity. As a simple implication of Theorem 3 weobtain the following corollary.

Corollary 1. For G∈Gn,3, the size of a minimum maximal matching is a.a.s. lessthan 0.3214n.

Let 0.3<�<0.5 be fixed. In what follows, we define Y� =Y�(n) to be the numberof maximal matchings of size ��n�=�n in P∈Pn,3. The proof of Theorem 3 usesthe small subgraph conditioning method which may be extracted in one theorem (seeJanson [12, Theorem 10] or Wormald [19, Theorem 4.1]) which we restate later inthe article. The bulk of the work in utilizing this method is the computation of theexpectation and variance of Y�, and the computation of the conditional expectation ofY� based on the distribution of short cycles.

Throughout this section we use the definitions N (2m)= (2m)! / (m!2m), which is thenumber of perfect matchings in the complete graph K2m and [n]m=n! / (n−m)! .

A. Expectation and Variance

The expectation part of the following theorem was first proven by Zito [23].

Theorem 4.

EY� ∼√

2

�n(1−2�)(10�−3)

(�3�32�−3/222�+3/2

(1−2�)1−2�(5�−3 / 2)5�−3/2

)n

.

Moreover, if p=�0.3214n� then

VarY� ∼(

16�2

10�−3

√2

4�2−12�+9−1

)(EY�)

2.

Proof. We consider the number of ways to select a given pairing that is associatedwith a minimum maximal matching of size p=�n. The number of ways to choosep independent edges, the buckets representing their end-points and the points used isgiven by (

n

2p

)N (2p)32p. (14)

From the definition of a maximal matching, we note that all edges other than thosein the matching must share at least one end-point with a matching edge. The n−2pvertices that are not end-points of matching edges are therefore independent. Thesevertices have 3(n−2p) incident edges and the number of ways to assign these edges is[4p]3(n−2p). As the remaining number of unused points in the pairing is now 10p−3n,



the number of ways to complete the pairing is N (10p−3n). Multiplying these togetherwe obtain the number of pairings containing a maximal matching of size p,

n! (4p)!32p

2pp! (n−2p)! (10p−3n)!N (10p−3n). (15)

Dividing this by N (3n) (the total number of pairings) we obtain

EY� = n! (4p)! (3n/ 2)!32p

(n−2p)!p! (5p−3n/ 2)! (3n)!26p−3n.

Approximating this using Stirling’s formula, n!∼√2�n(n/ e)n, we have

EY� ∼√

2

�n(1−2�)(10�−3)

(�3�32�−3/222�+3/2

(1−2�)1−2�(5�−3 / 2)5�−3/2

)n

.

This completes the proof of the first part of the theorem.

To prove the last part of the theorem we will first determine EY�(Y�−1). Fromthe assumption EY� →∞ as n→∞ we then obtain EY2

� ∼EY�(Y�−1). To computeEY�(Y�−1) we need to count the number of ways to lay down two copies of maximalmatchings of size p=�n. This number depends on the way in which the matchingsintersect each other, which is represented by a number of variables.

EY�(Y�−1) is expressed as the sum of a certain function, F, over all the variables.The most difficult part in the computation of this factorial moment is to prove that themain contribution to the sum comes from the maximum of F, which is denoted byxmax. The value of F at xmax is close to (EY�)2, as required by the theorem.

Let M1 and M2 denote two maximal matchings of size p. Denote the set of verticesof Mi by V(Mi). Given M1, let M={e∈E(G)|e∈M1∩M2} with |M|=x′. Recall thatthe other edges of M2 share at least one endpoint with edges in M1\M. Suppose thereare y′ edges in M2\M sharing two endpoints with edges in M1\M. Then in each Mi, theremaining edges (there are p−x′−y′ such edges) must share one endpoint (see Fig. 3).

The number of ways these two matchings may possibly intersect is given by(p

x′

)(p−x′

y′

)2p−x′−y′ . (16)

In M2 we choose 2y′ vertices from 2y′+p−x′−y′ =p−x′+y′ vertices availablein M1, while the completion of the p−x′−y′ edges requires p−x′−y′ extra vertices

FIGURE 3. Two intersecting maximal matchings. [Color figure can be viewed in theonline issue, which is available at www.interscience.wiley.com.]



selected from the remaining n−2p vertices. Thus, the number of ways to completeM2 is

(p−x′+y′)!2y′y′! (p−x′−y′)!

2p−x′+y′[n−2p]p−x′−y′3p−x′−y′ . (17)

Suppose there are z′ edges lying between V(M1)\V(M2) and V(M2)\V(M1). Thenumber of ways that the pairs of points corresponding to these edges may be selected is

(2(p−x′−y′)

z′

)2

z′! . (18)

At this stage we observe that there are 4(p−x′−y′)−2z′ points left in V(M1)\V(M2)∪V(M2)\V(M1) (with 2(p−x′−y′)−z′ points in each) and there are 3(n−3p+x′+y′)points left in V\(V(M1)∪V(M2)). All these points (3n−5p−x′−y′−2z′ in total) mustbe connected to the leftover points in V(M1)∩V(M2) (there are p+3x′+y′ such points).Therefore, the number of ways to assign these points is

[p+3x′+y′]3n−5p−x′−y′−2z′ . (19)

From Equation (19) we obtain that the number of unused points in the pairing is6p+4x′+2y′+2z′−3n. Thus, the number of ways to complete the pairing is

N (6p+4x′+2y′+2z′−3n). (20)

Hence, multiplying (14)–(20) and then dividing by N (3n), EY�(Y�−1) is given by

n! (3n/ 2)!33p23n−2p

(3n)!

∑R′

[(p−x′+y′)!

(p−x′−y′)!2 (n−3p+x′+y′)! (2(p−x′−y′)−z′)!2

× (2(p−x′−y′))!2 (p+3x′+y′)!(3p+2x′+y′+z′−3n/ 2)!x′!y′!2 z′!3x′+y′24x′+2y′+z′

],

where

R′ ={ (x′, y′, z′) :x′, y′, z′ ≥ 0,

n−3p+x′+y′ ≥ 0,

2(p−x′−y′)−z′ ≥ 0,

3p−3n/ 2+2x′+y′+z′ ≥ 0}.Set x=x′ / n,y=y′ / n, and z=z′ / n.We assume that all factorials above go to infinity

with n. This assumption is most easily justified after some intervening calculations.Hence by Stirling’s formula we have

EY�(Y�−1)∼ 1

8(�n)5/2(33�−3/223/2−2�)n

∑R

�(x,y, z)F(x,y, z)n (21)



where

R={ (x,y, z) :x,y, z ≥ 0,

1−3�+x+y ≥ 0,

2(�−x−y)−z ≥ 0,

3�−3 / 2+2x+y+z ≥ 0},

(22)

F(x,y, z)= f (�−x+y)

f (�−x−y)2f (1−3�+x+y)f (2(�−x−y)−z)2

× f (2(�−x−y))2f (�+3x+y)

f (3�+2x+y+z−3 / 2)f (x)f (y)2f (z)3x+y24x+2y+z(23)

with f (x)=xx and

�(x,y, z)=(

4(�−x+y)(�+3x+y)

(1−3�+x+y)(2(�−x−y)−z)2(3�+2x+y+z−3 / 2)xy2z

)1/2

.

We now determine the main contribution to the sum which comes from the maximumof F. We will show that for a specific value of �,

xmax=(2�2

3,10�2−3�

3,3(1−2�)2

)(24)

with

F(xmax)= �6�3�−3/226�+3/2

(1−2�)(2−4�)(5�−3 / 2)10�−3

being the local maximum point of F of greatest value in the interior of R.We look for all critical points of F in the interior of R. We set the partial derivatives

of logF, with respect to x,y and z, equal to 0, which results in the following threeequations:

(�+3x+y)3(2(�−x−y)−z)4

−283(�−x+y)(�−x−y)2(1−3�+x+y)(3�+2x+y+z−3 / 2)2x= 0,

(�−x+y)(�+3x+y)(2(�−x−y)−z)4

−263(�−x−y)2(1−3�+x+y)(3�+2x+y+z−3 / 2)y2 = 0,

(2(�−x−y)−z)2−2(3�+2x+y+z−3 / 2)z= 0.

We may simplify the above system by substituting the second equation into the firstand the third equation into the second consecutively.



The system now becomes

(�+3x+y)2y2−4(�−x+y)2(3�+2x+y+z−3 / 2)x= 0 (25)

(�−x+y)(�+3x+y)(3�+2x+y+z−3 / 2)z2

−48(�−x−y)2(1−3�+x+y)y2 = 0 (26)

(2(�−x−y)−z)2−2(3�+2x+y+z−3 / 2)z= 0. (27)

It is simple to verify that xmax satisfies the above system. Due to the complexity ofthe functions involved, finding all interior critical points of F is restricted to a particularvalue of �. The procedure to find the interior critical points of F may be describedas follows. Let fi, i∈{1,2,3}, be the functions on the left hand-side of Equations (25),(26) and (27), respectively. Solve f1 for z, as z appears as a linear term in f1. Denotethe result by fz. Substitute z= fz into f2 and f3. The resulting equations are

yfi,zx(�−x+y)

=0,

for i∈{2,3}, where fi,z are polynomials in x,y and � and are given in Appendix B.Since we are concerned only with the interior critical points, we will look for the

common roots of f2,z and f3,z. To do so we take the resultant of f2,z and f3,z with respectto x. This yields

Cy30(y+�)32(3y+3�−10�2)(2y+3−10�)2P2Q,

where C=39614081257132168796771975168, P=3y2+(6−20�)y+16�2−6� and Qis a large polynomial in y and � and is given in Appendix A. (The order of Q in y andk are both 18.) Therefore, the possible solutions come from zeros of 3y+3�−10�2,or 2y+3−10�, or P, or Q. We ignore the roots from factors y and y+� since they arenot in the interior of R.

Now fix �= =0.3214. Substituting back y= (10 2−3 ) / 3 into f2,z and f3,z yieldsx= (2 2) / 3 only. From fz, these values of y and x uniquely result in xmax as in Equation(24). On the other hand, we find that y1=5 −3 / 2 does not raise a feasible root sincef2,z(y1, ) and f3,z(y1, ) do not have any common root. From P we obtain two rootsy2<0 and y3> .

The remainder of the work requires finding the roots arising from Q. Letting int(R)be the interior of R, we look for YQ={yQ|yQ∈ int(R),Q( , yQ)=0}. Using back substi-tution, we find, if any, the common root, xQ, of f2,z( , yQ) and f3,z( , yQ) for yQ∈YQ.Then we obtain, if any, a unique zQ∈ int(R) by substituting {�= , x=xQ,y=yQ} into fz.From this procedure we obtain two other critical points in the interior of R, apart fromxmax, namely x1 and x2. These three critical points along with their nature (determinedfrom the Hessian of logF at the corresponding point for �= ) and values of F are



TABLE I. The stationary points of F.

x F (x) Hessian log(F (x))

xmax ≈ (0.068865,0.022927,0.382776) 1.112283 Negative definitex1 ≈ (0.156967,0.020342,0.214271) 1.108733 Indefinitex2 ≈ (0.239704,0.015200,0.069797) 1.112000 Negative definite

given in Table I. The given is the smallest � (to 4 decimal places) such that F attainsits maximum on xmax. It should be noted that the Hessian of logF at xmax is negativedefinite for �∈ ( 3

10 ,12 ). Thus xmax is a local maximum for � in the open interval. The

assertion follows.To show that F does not have any local maximum on the boundary of R (�R), we

use [1, Lemma 3.3] which is a generalization of the approach used by Garmo [9]. Themain idea is to show that for every point x0∈�R there exists a unit vector v0 suchthat the directional derivative of F at x0 in the direction of v0 is positive infinite. Thisimplies that F cannot reach any maximum on �R. The complete proof of this is givenin Lemma 2. For the sake of completeness we restate Lemma 3.3 in [1].

Lemma 1. Let R be a closed set in Rr and let �R be the boundary of R. Assumethat every point in �R is the endpoint of an interval in R\�R. Let fi(x)=bi+uixT fori=1, . . . ,m, where bi and ui are constant, such that fi(x)>0 for all i and all x∈R\�R.Define F to be a function on R such that

F(x)=g0(x)+m∑i=1

aigi(x)=g0(x)+m∑i=1

aifi(x) log fi(x)

with ai<0 for i≤m0≤m. Suppose that for every x∈R the directional derivativeof g0 at x in any direction is bounded. Let x0∈�R such that fi(x0)=0 for atleast one i≤m0 and fi(x0)>0 for all m0< i≤m. Then x0 is not a local maximumof F on R.

Lemma 2. Let 310<�< 1

2 be fixed. Let F and R be as in (22), and �R be the boundaryof R. Then F does not have any local maximum on �R.

Proof. Following the notation in Lemma 1, we have

logF(x)=g0(x)+9∑

i=1aifi(x) log fi(x)

where g0(x) is given by

(�+3x+y) log(�+3x+y)−(x−y) log2−(x+y) log3−(3x+3y+z) log2,



a1=a3=a4=a6=−1, a2=a5=a7=−2, a8=2, a9=1,

f1(x) = x,

f2(x) = y,

f3(x) = z,

f4(x) = 1−3�+x+y,

f5(x) = 2(�−x−y)−z,

f6(x) = 3�−3 / 2+2x+y+z,

f7(x) = �−x−y,

f8(x) = 2(�−x−y),

f9(x) = �−x+y,

u1 = (1, 0, 0),

u2 = (0, 1, 0),

u3 = (0, 0, 1),

u4 = (1, 1, 0)

u5 = (−2, −2, −1),

u6 = (2, 1, 1),

u7 = (−1, −1, 0),

u8 = (−2, −2, 0),

u9 = (−1, 1, 0).

For g0 we have

��tg0(x+ tv)

∣∣∣∣t=0

= (3v1+v2)[log(�+3x+y)+1]

−(v1−v2) log2−(v1+v2) log3−(3v1+3v2+v3) log2

which is bounded for all x∈R as the arguments of the log function are bounded awayfrom zero.

In view of Lemma 1, we need only to consider the part of �R that corresponds toai>0. In this case this occurs when i=8 or 9. Therefore to determine this part we mustobtain the solution to the following sets of equations:

fi(x) ≥ 0, i∈ [9]\{8}f8(x) = 0,

(28)

fi(x) ≥ 0, i∈ [8]

f9(x) = 0,(29)

fi(x) ≥ 0, i∈ [7]

f8(x) = 0

f9(x) = 0.

(30)

First we consider the system (28). Equation f8(x)=0 implies that fj(x)=0 simulta-neously for j=3,5 and 7. Thus for these conditions we obtain the solution set e1={�−x−y=0, z=0}. Next, we will check e1 for the conditions fj(x)≥0 for j=1,2,4,6,and 9. Given e1 it is easy to verify that f4(x)=1−2�>0. On the other hand, f1(x)=0and e1 yields c1= (0,�,0), whilst f2(x)=0 and e1 implies f9(x)=0, which in turn resultsin c2= (�,0,0). Furthermore, f6(x)=0 and e1 gives c3= ( 32 −4�,5�− 3

2 ,0). The lastcase of f9(x)=0 and e1 yields c2. Thus, the final solution to (28) is R0={c1,c2,c3,e1}.Arguing as above we also find that the solution to (29) and to (30) is c2.



Since for x0∈�R\R0, logF and x0 satisfy the hypotheses of Lemma 1, F does nothave any local maximum on �R\R0. The remaining work is to study the behavior of Fon R0. We observe that for v= (1,−2,1),∑

i=3,5,7,8ai

��tgi(x+ tv)=− log t−1+4 log2,

for x∈e1, ∑i=1,3,5,7,8

ai��tgi(c1+ tv)= −2 log t−2+2(d−1) log(d−1),

∑i=3,5,6,7,8

ai��tgi(c3+ tv)= −2 log t−2+4 log2,

whilst for v= (−3,1,1),∑i=2,3,5,7,8,9

ai��tfi(c2+ tv)=− log t−7 log3+8 log4+4 log2.

It is clear that each directional derivative above is positive infinite as t→0. Thus, Fdoes not have any local maximum on R0, and the assertion follows. �

At this stage, we have shown that the maximum of F is attained at xmax. In the nextlemma it will be proven that the sum in Equation (21) may be approximated within asmall region around xmax, B=B(xmax,n−2/5). By this we verify the assumption neededfor the use of Stirling’s formula.

Lemma 3. Let B=B(xmax,n−2/5) be a closed ball centered at xmax with diametern−2/5, with F and R be as in (22),∑

R�(x)F(x)n∼∑

B�(x)F(x)n,

with

∑B

�(x)F(x)n= 162�2(�n)3/2

(1−2�)(10�−3)2

√2

4�2−12�+9

(�6�3�−3/226�+3/2

(1−2�)(2−4�)(5�−3 / 2)10�−3

)n

.

Proof. Write ∑R

�(x)F(x)n=∑B

�(x)F(x)n+∑R\B

�(x)F(x)n.

It will be shown that ∑R\B

�(x)F(x)n=o(�(xmax)F(xmax)n),

whilst ∑B

�(x)F(x)n=O(n3/2�(xmax)F(xmax)n).



The Taylor expansion of F at xmax is

F(x)n=F(xmax)n×exp(−n(as21+bs22+cs23+es1s2+ fs1s3+gs2s3)+O(n−1/5))

where s1=x−2�2 / 3, s2=y−(10�2 / 3−�), s3=z−3(1−2�)2 and a=−h11 / 2, b=−h22 / 2, c=−h33 / 2, e=−h12, f =−h13 and g=−h23, where the hij (the (i, j)-elementof the Hessian of logF at xmax) are

a= −176�4+1792�3−1992�2+720�−81

16(10�−3)2(1−2�)2�2,

b= −2864�4−2048�3−168�2+432�−81

16(10�−3)2(1−2�)2�2,

c= − 52�2−36�+9

3(10�−3)2(1−2�)2,

e= −1808�4−1280�3+72�2+144�−27

16(10�−3)2(−1+2�)2�2,

f = − 16�

(10�−3)2(1−2�)and g=− 2(14�−3)

(10�−3)2(1−2�).

For x∗ ∈�B, where �B is the boundary of B, we note that the exponential factor is

O(e−n1/5 )=o(n−3).

Therefore

�(x∗)F(x∗)n∼�(xmax)F(xmax)no(n−3)=o(n−3�(xmax)F(xmax)

n)

for x∗ ∈�B.Since F attains its maximum uniquely at xmax then for x∈R\B

�(x)F(x)n=O(maxx∗∈�B�(x∗)F(x∗)n)=o(n−3�(xmax)F(xmax)n).

Thus, ∑R\B

�(x)F(x)n=O(n3) ·o(n−3�(xmax)F(xmax)n)=o(�(xmax)F(xmax)

n),

since the number of terms in this sum is O(n3).The remaining work is to determine

∑B �(x)F(x)n. Since the summation concentrates

near the maximum, each term �(x) can be taken as �(xmax) with

�(xmax)=16√6 / ((1−2�)3(10�−3)3).

Referring to the Taylor expansion of F above, we have∑

R �(x)F(x)n given by

�(xmax)F(xmax)n∑

Bexp(−n(as21+bs22+cs23+es1s2+ fs1s3+gs2s3)).



The summation is a Riemann sum for the triple integral

n3/2∫ n1/10

−n1/10

∫ n1/10

−n1/10

∫ n1/10

−n1/10exp(−(at21+bt22+ct23+et1t2+ ft1t3+gt2t3))dt1 dt2 dt3

where

t1 = (x−2�2 / 3)n√n

,

t2 = (y−(10�2 / 3−�))n√n

and

t3 = (z−3(1−2�)2)n√n

.

As n→∞, the range of integration may be extended to ±∞. Thus it is asymptoticto

n3/2∫ ∞

−∞

∫ ∞

−∞

∫ ∞

−∞exp(−(at21+bt22+ct23+et1t2+ ft1t3+gt2t3))dt1 dt2 dt3.

The evaluation of the triple integral gives

(2�)3/24�2(1−2�)2(10�−3)

√2

3(4�2−12�+9).

Recalling that F(xmax)= (�6�3�−3/226�+3/2) / ((1−2�)(2−4�)(5�−3 / 2)10�−3) we have

∑B

�(x)F(x)n ∼ 16√6 / ((1−2�)3(10�−3)3)

(�6�3�−3/226�+3/2

(1−2�)(2−4�)(5�−3 / 2)10�−3

)n

×(2�)3/24�2(1−2�)2(10�−3)

√2

3(4�2−12�+9)

∼ 162�2(�n)3/2

(1−2�)(10�−3)2

√2

4�2−12�+9

(�6�3�−3/226�+3/2

(1−2�)(2−4�)(5�−3 / 2)10�−3

)n

,

which is O(n3/2�(xmax)F(xmax)n). �

The result in the previous lemma transforms (23) so thatEY�(Y�−1) is approximately

32�2

�n(1−2�)(10�−3)2

√2

4�2−12�+9

(�6�34�−324�+3

(1−2�)(2−4�)(5�−3 / 2)10�−3

)n

. (31)

Noting that EY� →∞ implies EY�(Y�−1)∼EY2� and that VarY� =EY2

� −(EY�)2,Equation (31) gives the desired result and completes the proof of the theorem. �



B. Expectation Conditioned on Short Cycle Counts

For the proof of Theorem 3 we require the following lemma.

Lemma 4. Let 310<�< 1

2 be fixed and let Xk=Xk(n) denote the number of cycles oflength k in P∈Pn,3. Then for any finite sequence j1, . . . , jl of non-negative integers,

E(Y�[X1]j1 . . . [Xl]jl)

EY�→

l∏k=1

(�k(1+�k))jk as n→∞;

with �k=2k−1 / k and �k=2Re(�k) where �= (2�−3+√196�2−108�+9) / 16�.

Proof. We will show

E(Y�Xk)

EY�∼�k(1+�k). (32)

The number of ways to select an oriented cycle of length k in the pairing, with adistinguished point in a pair, is

n! (6)k

(n−k)!. (33)

This induces an orientation on the cycle and a distinguished vertex referred to as theroot vertex.

Let C denote the set of pairs corresponding to an oriented and rooted k-cycle and,similarly, define M to be the set of pairs corresponding to a maximal matching ofsize p=�n. Given a fixed C, suppose s0 vertices of C lie on I, the independent set ofvertices not in M, and suppose C∩M consists of s1 independent edges and k−s0−2s1vertices (these vertices are the endpoints of edges in M).

The vertices of C may be classified into 3 types. The first are the s0 vertices in I.We denote this type of vertices by 0. The remaining vertices in C are the endpointsof edges in M. They are either preceded by an edge in M or preceded by an edge notin M. We denote these by 1 and 2, respectively. If we walk along C from the rootvertex then we obtain a sequence S0∈{0,1,2}k.

For a fixed C and S0, the number of ways to choose the remaining p−s1 independentedges, together with the points used, is(

n−k

2p−k+s0

)(2p−2s1)! 32p−k+s0

2p−s1 (p−s1)!= (n−k)! (2p−2s1)!32p−k+s0

2p−s1 (p−s1)! (n−2p−s0)! (2p−k+s0)!, (34)

The number of ways to choose the points corresponding to the edges from M to I is

[4p−2k+2s0+2s1]3(n−2p)−2s0 = (4p−2k+2s0+2s1)!

(10p−3n−2k+4s0+2s1)!. (35)

The number of ways to complete the pairing is

N (10p−3n−2k+4s0+2s1). (36)



Multiply Equations (34)–(36), sum over all possible S0 and then multiply byEquation (33). This results in the number of pairings containing a maximal matchingof size p and an oriented and rooted cycle of length k,

∑S0

[n! (2p−2s1)! (4p−2k+2s0+2s1)!32p+s02kN (10p−3n−2k+4s0+2s1)

2p−s1 (p−s1)! (n−2p−s0)! (2p−k+s0)! (10p−3n−2k+4s0+2s1)!

]. (37)

Dividing this by the number of pairings containing a maximal matching of size p asin (15) and then evaluating this asymptotically we obtain(

10�−3

4�

)k

×∑S0

(24�(1−2�)

(10�−3)2

)s0 ( 8�

10�−3

)s1. (38)

To determine the sum, we follow an approach used in [12]. We may view 0, 1 and 2as three states in a process that is analogous to a Markov Chain where the final stateis equal to the initial state. We observe here that, in S0, each 0 must be followed by 2and contributes a factor of

24�(1−2�)

(10�−3)2.

Moreover, 2 followed by 1 indicates that, in the walk, we pass an independent edgein M and this contributes a factor of 8� / (10�−3). Hence the ‘transition’ matrix isgiven by

A=

⎛⎜⎜⎜⎜⎜⎝0 0

24�(1−2�)

(10�−3)2

1 0 1

18�

10�−31

⎞⎟⎟⎟⎟⎟⎠ .

Thus, we may rewrite

∑S0

(24�(1−2�)

(10�−3)2

)s0 ( 8�

10�−3

)s1=Tr

(Ak).

Note that the eigenvalues of A are �1=8� / (10�−3), �2=�1� and �3=�1�, where �is the conjugate of �. Hence

Tr(Ak)= (�1)k+(�2)

k+(�3)k.

Note that �k=�k+ �k. Equation (38) now becomes

2k(1+�k).

Finally, dividing by 2k to remove the orientation and rooting of the cycle we obtainEquation (32). The argument also works for higher moments. This completes the proofof the Lemma. �



C. Proof of Theorem 3

We first restate [19, Theorem 4.1] (for the convenience of the reader) and we will thenshow the corresponding result in Pn,3.

Theorem 5 (Wormald [20]). Let �i>0 and �i≥−1, i=1,2, . . . , be real numbers andsuppose that for each n there are random variables Xi=Xi(n), i=1,2, . . . , and Y=Y(n)defined on the same probability space G=G(n) such that Xi is non-negative integervalued, Y is non-negative and EY>0 ( for n sufficiently large). Suppose furthermore that

(a) For each k≥1 Xi, i=1,2, . . . , k, are asymptotically independent Poisson randomvariables with EXi→�i;

(b)E(Y[X1]j1 . . . [Xk]jk )

EY→

k∏i=1

(�i(1+�i))ji

for every finite sequence j1, . . . , jk of non-negative integers;

(c)∑

i �i�2i <∞;

(d) EY2n / (EYn)

2≤exp(∑

i �i�2i )+o(1) as n→∞.

Then

P(Yn>0)=exp

(− ∑

�i=−1�i

)+o(1),

and, provided∑

�i=−1<∞, G(Y)≈ G where G is the probability space obtained fromG due to conditioning on the event

∧�i=−1(Xi=0).

Theorem 6. If �=0.3214 then for P∈Pn,3, G(P) a.a.s. has a maximal matching ofsize �n. Moreover,

Y�

EY�

d→W=∞∏k=1

(1+�k)Zk e−�k�k as n→∞,

where Zk are independent Poisson variables with EZk=�k for k≥1, and where �k and�k are as in Lemma 4.

Proof. We only need to show that the random variable Y� satisfies the conditions(a) to (d) in Theorem 5. Since Xk is the number of short cycles of length k in apseudograph coming from Pn,3 then (a) is fulfilled by a result of Bollobas [2] with�k=2k−1 / k. The other parts are fulfilled by Lemma 4 and Theorem 4. The proof iscomplete. �

Since the assertion in Theorem 6 is a.a.s. true conditioned on no loops or multipleedges (see Bollobas [3]) Theorem 3 follows immediately. From the argument in



[12, page 375] we also obtain

EY ′�

EY�∼ exp(−�1�1−�2�2)=exp

(3−2�

8�− 100�2−60�+9

64�2

)and

EY ′�2

(EY ′�)2

∼ exp(−�1�21−�2�

22)=exp

(− (3−2�)2

64�2− (100�2−60�+9)2

4096�4

) EY2�

(EY�)2.

7. REMARKS

The extension of the algorithmic results in this article to d-regular graphs, where d>3,is not intuitively obvious. There are several directions that one may take. Moreover, theamount of work required to check that the differential equations behave in a mannerone would expect becomes much greater.

REFERENCES

[1] H. Assiyatun, Large subgraphs of regular graphs, Doctoral thesis, Departmentof Mathematics and Statistics, The University of Melbourne, Australia, 2001.

[2] B. Bollobas, A probabilistic proof of an asymptotic formula for the numberof labelled regular graphs, European J Combin 1 (1980), 311–316.

[3] B. Bollobas, Random graphs, In: Combinatorics, London MathematicalSociety Lecture Note Series 52 (H. N. V. Temperley, Ed.), CambridgeUniversity Press, Cambridge, 1981, 80–102.

[4] B. Bollobas, Random Graphs, Academic Press, New York, 1985.[5] R. Diestel, Graph Theory, Springer, Berlin, 1997.[6] W. Duckworth, Maximum 2-independent sets of random cubic graphs,

Australas J Combin 27 (2003), 63–80.[7] W. Duckworth, Greedy algorithms and cubic graphs, Doctoral thesis,

Department of Mathematics and Statistics, The University of Melbourne,Australia, 2001.

[8] W. Duckworth and N. C. Wormald, Minimum independent dominating setsof random cubic graphs, Random Struct Algorithms 21(2) (2002), 147–161.

[9] H. Garmo, Random railways and cycles in random regular graphs, Doctoralthesis, Uppsala University, Sweden, 1998.

[10] J. D. Horton and K. Kilakos, Minimum edge dominating sets, SIAM JDiscrete Math 6(3) (1993), 375–387.

[11] T. G. Kurtz, Approximation of Population Processes, SIAM, Philadelphia,1981.

[12] S. Janson, Random regular graphs: Asymptotic distributions and contiguity,Combin Probab Comput 4(4) (1995), 369–405.

[13] S. Janson, T. Łuczak and A. Rucinski, Random Graphs, Wiley, New York,2000.



[14] G. Jun-Ya, V. T. Nguyen and Y. Yoshitsugu, Global Optimization Method forsolving the minimum maximal flow problem, Optim Methods Softw 18(4)(2003), 395–415.

[15] H. D. Robalewska, 2-factors in random regular graphs, J Graph Theory 23(3)(1996), 215–224.

[16] R. W. Robinson and N. C. Wormald, Almost all regular graphs areHamiltonian, Random Struct Algorithms 5(2) (1994), 363–374.

[17] N. C. Wormald, The asymptotic connectivity of labelled regular graphs, JCombin Theory Ser B 31 (1981), 156–167.

[18] N. C. Wormald, Differential equations for random processes and randomgraphs, Ann Appl Probab 5 (1995), 1217–1235.

[19] N. C.Wormald, Models of random regular graphs, Surveys in Combinatorics,(Canterbury 1999), Cambridge University Press, Cambridge, 1999, 239–298.

[20] N. C.Wormald, The differential equation method for random graph processesand greedy algorithms, In: Lectures on Approximation and RandomizedAlgorithms (M. Karonski and H. J. Promel, Eds.), PWN, Warsaw, 1999,73–155.

[21] M. Yannakakis and F. Gavril, Edge dominating sets in graphs, SIAM J ApplMath 38(3) (1980), 364–372.

[22] M. Zito, Randomised techniques in combinatorial algorithmics, Doctoralthesis, Department of Computer Science, University of Warwick, UK, 1999.

[23] M. Zito, Small maximal matchings in random graphs, In: TheoreticalInformatics: 4th Latin American Symposium(LATIN 2000), Lecture Notesin Computer Science 1776 (G. H. Gonnet, D. Panario and A. Viola, Eds.),Springer, Berlin, 2000, 18–27.

APPENDIX A

The polynomial Q is given by16369457602560y3�14−12663624089600y3�15−68143048078080y10�8−5966329920y16�2

−50116540345280y12�6+16289088221952�10y2+357128352y6�−78118186163840y9�9

+174436056320y15�3+57600000000�18−67361583790080y4�14+266429088y7

+38213678448y4�4+1481421312y5�3−13591662624y6�2+245890608297984�13y4

−1640145024y7�+1980849180960�3y8−149717541024�7y4+9769600188000�6y5

−274382139744�5y6−3784207967928�4y7+554397326640�2y9−612935618400�y10

−1179090432y5�2−651900960y2�5−198060021982944�6y11−2397861792y3�4

+68024448�3y3+102036672�2y4−2107306040320y14�4−3015750528y4�3

+17006112y6+17006112�4y2+68024448�y5+140132882304�3y6−9825920y17�+37182054072320y5�13+10865645856y9+40128598152y10+665379458496�3y7

−28973376�9−48990115691520y7�11+905659287336�6y4+318160114396848�8y8

−19076501160�2y8−1600440161472�5y5−579774942216�4y6+147030449631040y8�10

+398488896�10+98398935529600y11�7−9676664780800y6�12−25730877312�y8

+130486007808�4y5+38301543360�5y3−260573316768�5y4−43556431968�2y7



−107075520�8y−158354616384�y9+869807201472�8y2+82317286679040�10y5

+2309716804032�7y3−121556539296�7y2+11443223808�6y2−365444970624�6y3

+1347891840�9y−3236060160�11−4414687297920�9y2+13736391413904�6y6

−10123072785792�8y3+31233479233536�9y3−14282427938688�8y4−9896297472�10y−36443372421888�7y5+177656003109y12+17178739200�12−66435941813472�7y6

−14104365713451�4y8+8619913265328�5y7+3112949169630y10�2−1517988987180y11�+3240684892260�3y9+46721691648�11y+43821586236204�5y8−1701792589698�3y10

−34911910385244�4y9+86217557473152�8y5+9349056215424�6y7+202857363738y13

+74342208592512�9y4+8126940646170�2y11−2363856241266y12�+149095583172y14

−44004845707776�11y2+297792000000�16−62270208000�13−67166431842816�10y3

−2266319087352y13�−147216476160�12y−15168335902200�3y11+160716989862144�7y9

−43139199686256�4y10+128016796137624�5y9−47547036800244�6y8

−108236408832576�7y7+165907899022464�8y6−123154043120640�9y5

−207394996624128�10y4+67301301384y15+96435649142784�11y3+86346240537600�12y2

+156038400000�14+11439591773832�2y12+310169088000�13y−20826386083440�4y11

−21351985294752�3y12−226358597240928�6y9−83650684021344�7y8

+182059162781328�5y10−236277311378688�9y6+3779136�7y+285652329003648�8y7

+361260657732096�11y4+8765333164512�2y13−1285518583800y14�−119963475339264�13y2−82233860179968�12y3−266803200000�15+16852748400y16

+23861712752640�11y5+177028580738304�10y6−423290880000�14y−11750651735040�3y13−323252694892512�6y10−2927936494368�4y12

+132640962478848�5y11+944784�8−369326021868288�9y7−393425392785408�12y4

+3397315286448�2y14+377920y18−392558595840y15�+111968331632640�14y2

+25548363104256�13y3+1777984128y17+57076376684640�5y12−363500408151360�9y8

+35915057356608�8y9+102757629672y11−1935122732352�3y14+259172297396448�7y10

−3417245193312�4y13+13640014206720y13�5−77137853847552�12y5

−45504833005056�11y6+229735182056448�10y7+2080414368y8+505081267008�2y15

−49334851968y16�−195840000000�17+16160530432000y2�16−135321600000y�16

+344217600000y�15+10752000000y�17−63009301708800y2�15.

APPENDIX B

The function f2,z is given by35�3y8−299408�3x8−118416�3x7+684�3x6+�7y4+7�6y5+21�5y6+35�4y7+144�9x2

+10128�8x3−144�8x2−66992�7x4−3888�7x3+210320�6x5+21456�6x4−390480�5x6

+36�7x2+180�6x3−108�5x4+440816�4x7+107760�4x6−1116�4x5+121392�2x9

+70128�2x8+2268�2x7−59760�5x5+21�2y9−30720�x10−17616�x9−2916�x8+7�y10

−2679x3y8+7827x7y4−12105x6y5+9559x5y6+21736x8y3+3828x8y2+9540x7y3

−16860x6y4+7499x4y7+10140x4y6+84275�3y4x4+24068�3y5x3−1846�3y6x2

−410736�3y3x5+4356�3y2x5−60240�3y3x4+3720�3y4x3−364�3y7x+420�3y6x+519936�3x7y+187104�3x6y−4464�3x5y−1080�3x4y2+3600�3x3y3−539�6y4x+3285�5y4x2−21�2xy8−882�5xy5−805�4xy6−112161�2x5y4−74407�2x4y5+22414�2x3y6

+297480�2x6y3+34068�2x6y2+120204�2x5y3−42120�2x4y4−726�2x2y7+324�2x2y6

−227184�2x8y−164880�2x7y+2052�2x6y−6696�2x5y2−1080�2x4y3−19415�4x3y4

−5688�2y5x3+252�2y7x+1104�8yx2+30784�7yx3−1056�7yx2−149520�6yx4

−14448�6yx3+383520�5yx5+54720�5yx4−600400�4yx6+252�6yx2+1080�5yx3

−540�4yx4−29880�2y2x7+2700�2y4x3−471�4y5x2+105�x2y8+37751�x6y4

+62510�x5y5−30175�x4y6−103360�x7y3−36924�x7y2−62136�x6y3+56316�x5y4

+228�x3y7−8388�x3y6+51360�x9y+58752�x8y+4536�x7y+2052�x6y2−4464�x5y3



+46�xy9+9456�x4y5+4800x11+480x10+972x9+y11+756�2y5x2+456�x2y7

−117456�4x5y+40920�x8y2−540�x4y4−24�8y2x+3384�7y2x2−176�7y3x+12�7y2x+16568�6y2x3+5080�6y3x2−3204�6y2x2+84�6y3x+14504�5y2x4−28064�5y3x3

−16596�5y2x3−5112�5y3x2+252�5y4x−72424�4y2x5+218040�4y3x4+14460�4y2x4

−2340�4y3x3−4380�4y4x2+420�4y5x+756�5x2y2+2700�4x3y2+1260�4x2y3

+1260�3x2y4+37704�3x6y2−1584�3x2y5−9600x10y−2736x9y−2916x8y+2268x7y2

+684x6y3+85x2y9−1836x5y5−2676x3y7−10752x9y2−1116x5y4+13xy10+108x2y8

−108x4y5+1080�x3y5+84�y8x+252�y6x2+12y9x+180y6x3+36y7x2.

The function f3,z is given by

−y8+36x6−y4�4−4y5�3−6y6�2−4y7�−209y4x4−140y5x3+42y6x2+160y2x6−112y3x5

−192y2x5+288y3x4−192y4x3+4y7x+48y5x2+400x2�6−1600x3�5−240x2�5+2400x4�4

+960x3�4−1600x5�3−1440x4�3+400�2x6+960�2x5+36�4x2−144�3x3+216�2x4−240�x6

−144�x5+48x6y−144x5y+216x4y2−144x3y3−44y4�3x−102y4�2x2−4y5�2x−16y2�5x+1728y2�4x2−48y3�4x−2848y2�3x3+640y3�3x2−1248y2�3x2+736y2�2x4+352y3�2x3

+2304y2�2x3−672y3�2x2+500y4�x3−12y5�x2+12y6�x+240y2�x5−832y3�x4−864y2�x4

+384y3�x3−48y4�x2+1440x2�5y−4160x3�4y−912x2�4y+3840x4�3y+2688x3�3y−960�2x5y−2592�2x4y+144�3x2y−432�2x3y+216�2x2y2−160�x6y+768�x5y+432�x4y−432�x3y2+144�x2y3+36x2y4.


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Small maximal matchings of random cubic graphs