Perfect matchings in random uniform hypergraphs

Perfect Matchings in Random UniformHypergraphs

Jeong Han KimMicrosoft Research, One Microsoft Way, Redmond, Washington 98052; e-mail:

[email protected]

Received 23 April 2002; accepted 23 January 2003

DOI 10.1002/rsa.10093

ABSTRACT: Let �k(n, p) be the random k-uniform hypergraph on V � [n] with edge probabilityp. Motivated by a theorem of Erdos and Renyi [7] regarding when a random graph G(n, p) � �2(n,p) has a perfect matching, the following conjecture may be raised. (See J. Schmidt and E. Shamir[16] for a weaker version.) Conjecture. Let k�n for fixed k � 3, and the expected degree d(n, p) �p(k�1

n�1). Then

Pr��k�n, p� has a perfect matching� 3 �0 if d�n, p� � ln n 3 ��,e�e�c if d�n, p� � ln n 3 c,1 if d�n, p� � ln n 3 �.

(Erdos and Renyi [7] proved this for G(n, p).) Assuming d(n, p)/n1/ 2 3 �, Schmidt and Shamir[16] were able to prove that �k(n, p) contains a perfect matching with probability 1 � o(1). Friezeand Janson [8] showed that a weaker condition d(n, p)/n1/3 3 � was enough. In this paper, wefurther weaken the condition to

d�n, p�

n1/�5�2/�k�1�� 3 �.

A condition for a similar problem about a perfect triangle packing of G(n, p) is also obtained. Aperfect triangle packing of a graph is a collection of vertex disjoint triangles whose union is the

© 2003 Wiley Periodicals, Inc.

111

entire vertex set. Improving a condition p � cn�2/3�1/15 of Krivelevich [12], it is shown that if 3�nand p � n�2/3�1/18, then

Pr�G�n, p� has a perfect triangle packing� � 1 � o�1�.

© 2003 Wiley Periodicals, Inc. Random Struct. Alg., 23: 111–132, 2003

1. INTRODUCTION

A hypergraph is a pair � � (V, H) of a set V of vertices and a collection H of subsetsof V, which are called (hyper)edges. It is k-uniform if every edge is of size k. For example,the collection of all subsets of size k of V is a k-uniform hypergraph. This hypergraph iscalled the complete k-uniform hypergraph (on V). For V � [n], a random hypergraph�k(n, p) is a subhypergraph of the complete k-uniform hypergraph so that each edgeindependently belongs to the subhypergraph with probability p. A perfect matching of ahypergraph is a collection of pairwise disjoint edges whose union is the entire vertex set.

In 1966, Erdos and Renyi [7] found when a random graph1 G(n, p) � �2(n, p)contains a perfect matching (PM):

Theorem 1.1. If n is even, then

Pr�G�n, p� has a PM� 3 �0 if np � ln n 3 ��,e�e�c if np � ln n 3 c,1 if np � ln n 3 �.

Notice that np (n � 1) p is essentially the expected average degree. Let d(n, p) be theexpected average degree of the random hypergraph �(n, p), i.e.,

d�n, p� � p�n � 1k � 1� .

Motivated by the theorem, one may conjecture that the same result should be true for arandom hypergraph (see [16] for a weaker version):

Conjecture 1.2. If k�n for fixed k � 3,

Pr��k�n, p� has a PM� 3 �0 if d�n, p� � ln n 3 ��,e�e�c if d�n, p� � ln n 3 c,1 if d�n, p� � ln n 3 �.

Using the second moment method in a clever setting, Schmidt and Shamir [16] were ableto prove that

1Erdos and Renyi actually proved this for the G(n, m) model.

112 KIM

Theorem 1.3. If k�n for fixed k � 3 and d(n, p)/n1/2 3 �, then

Pr��k�n, p� has a PM� 3 1.

To improve this, Frieze and Janson [8] used the second moment method together with aconditioning on degree sequences:

Theorem 1.4. If k�n for fixed k � 3 and d(n, p)/n1/3 3 �, then


In this paper, we further weaken the condition. Let �k(k)(n, p) be a random k-partite

k-uniform hypergraph on kn vertices with edge probability p. That is, the vertex set of�k

(k)(n, p) is the union of pairwise disjoint sets X1, . . . , Xk of size n, and a k-tupleconsisting of one element from each Xi is an edge of �k

(k)(n, p) with probability p,independently of all other k-tuple.

Theorem 1.5. Let k�n for fixed k � 3. If

pnk�1/n1/�5�2/�k�1�� 3 �,

then

Pr��k�k��n, p� has a PM� 3 1.

Since �k(k)(n/k, p) � �k(n, p) for k�n, we have the same result for �(n, p).

Corollary 1.6. If k�n for fixed k � 3 and

d�n, p�/n1/�5�2/�k�1�� 3 �,

then


A condition for a similar problem regarding a perfect triangle packing of G(n, p) canalso be obtained. A perfect triangle packing of a graph is a collection of vertex disjointtriangles whose union is the entire vertex set. We use a coupling argument for this task.Let G3(n, p) be a random 3-partite graph on 3n vertices with edge probability p, i.e., thevertex set of G3(n, p) is the union of pairwise disjoint sets X1, X2, X3 of size n, and apair of vertices consisting of at most one vertex from each Xi is an edge of G3(n, p) withprobability p, independently of all other pairs. A coupling, or joint distribution, on (G3(n,p), �3

(3)(n, q)) is a probability space on the set of all pairs of 3-partite graphs and 3-partite3-uniform hypergraphs on 3n vertices. The marginal distributions of G3(n, p) and �3

(3)(n,q) are the same as those of the random 3-partite graph with edge probability p and therandom 3-partite 3-uniform hypergraph with edge probability q, respectively.

PERFECT MATCHINGS IN RANDOM UNIFORM HYPERGRAPHS 113

Theorem 1.7. If n�2/3log1/3n � p � n�3/5, then there is a coupling on (G3(n, 17p),�3

(3)(n, p3)) such that

Pr�all edges in �3�3��n, p3� are �vertex sets of� triangles in G3�n, 17p�� 1 � o�1�.

This theorem together with Theorem 1.5 yields

Pr�G3�n, p� has a perfect triangle packing� � 1 � o�1�,

provided p � n�2/3�1/18. Since G3(n/3, p) � G(n, p) for 3�n, we have:

Corollary 1.8. If 3�n and p � n�2/3�1/18, then

Pr�G�n, p� has a perfect triangle packing� � 1 � o�1�.

This improves the previous condition of Krivelevich [12] of p � cn�2/3�1/15. An earliercondition was p � cn�2/3�1/6(ln n)1/ 2 due to Rucinski [15], and Alon and Yuster [2].

The obvious conjecture is still open.

Conjecture 1.9. Let 3�n and the expected triangle degree �p � p3( 2n�1). Then

Pr�G�n, p� has a perfect triangle packing� 3 �0 if �p � ln n 3 ��,e�e�c if �p � ln n 3 c,1 if �p � ln n 3 �.

If �p � ln n 3 ��, it is easy to show that with probability 1 � o(1), or with highprobability, there is a vertex contained in no triangle. Thus, the first part is not actually apart of the conjecture.

The fractional version of Conjecture 1.2 has been considered too. Krivelevich [11]resolved the problem completely. Namely, considering a random hypergraph process, inwhich random edges are added one by one, he proved that, with high probability, a perfectfractional matching exists at the very moment when the last isolated vertex disappears.Similar results hold also for random k-partite k-uniform hypergraphs. In particular,�3

(3)(n, p3) contains a perfect fractional matching whenever p � n�2/3log1/3n. Thistogether with Theorem 1.7 yields the following corollary regarding the existence of aperfect fractional triangle packing.

Corollary 1.10. If p � n�2/3log1/3n, then G(n, p) contains a perfect fractional trianglepacking. That is, there are weights w(�) on triangles � of G(n, p) so that

��:v��

w�� 1 for all v � V and ��

w�� n/3.

The proof of Theorem 1.5 is partly inspired by the proof of a similar result for randomregular hypergraphs. Using ideas and techniques developed in the study of random regulargraphs (see, e.g., [3], [4], [5], and [17]) and small cycles conditioning method (see, e.g.,[13], [14], [9], and [10]), Cooper, Frieze, Molloy, and Reed [6] found the exact constant

114 KIM

ck (for a fixed k) so that with high probability the random d-regular k-uniform hypergraphhas a perfect matching if d � ck; otherwise, no perfect matching exists. Particularly, if d3 �, then the random d-regular k-uniform hypergraph for fixed k has a perfect matchingwith high probability. Roughly speaking, if one succeeded in showing the existence of arandom d-regular subhypergraph of �k(n, p) with d(n, p) slightly larger than d, say1.1d, then the result of Copper et al. [6] would imply that �k(n, p) had a perfect matchingwith high probability. Though it does not seem to be any easier to find such a regularsubhypergraph than to find a perfect matching (1-regular hypergraph), an almost regularsubhypergraph of �k(n, p) turns out to be enough to obtain an improved condition. In thenext section, we will precisely describe this in the form of two lemmas which will easilyyield Theorem 1.5.

In Sections 3 and 4, we prove the two lemmas. The last section is for the proof ofTheorem 1.7.

2. TWO LEMMAS

In this section, we introduce two lemmas which immediately yield Theorem 1.5.Let V be the underlying vertex set and {X1, X2, . . . , Xk} be a partition of V into k

parts of size n. An edge is, of course, a k-tuple consisting of one vertex from each part.Let e1, . . . , et, . . . be i.i.d. random edges each of which is chosen uniformly at randomfrom all possible nk edges. The random hypergraph �(t) has the vertex set V and the edgeset H(t) � {e1, . . . , et}. It is routine to check that �(t) is contained in �k

(k)(n, p) withhigh probability (in a suitable coupling) as long as pnk � t � t1/ 2. This is intuitivelyobvious since �k

(k)(n, p) would have more random edges than �(t) with high probability,and may be shown using a large deviation, or a Chernoff type, bound (see, e.g., [1], p. 268,Theorem A.1.13).

As usual in the study of random graphs with a given degree sequence, we consider aconfiguration (or pairing) model developed in [5]. For each vertex v � V and a positiveinteger d0, let (v, 1), . . . , (v, d0) be d0 copies of v. The set of those vertices is denotedby V0. Clearly, V0 has a k-partition X1 � [d0], X2 � [d0], . . . , Xk � [d0], where [d0]:� {1, . . . , d0}. An edge in V0 is a k-tuple consisting of one vertex from each part, anda matching of V0 is a collection of pairwise disjoint edges. A matching covers a vertex ifan edge in it contains the vertex. The projection of an edge {(v1, i1), . . . , (vk, ik)} in V0

is {v1, . . . , vk}. In this case, the edge {(v1, i1), . . . , (vk, ik)} is called a copy of theedge {v1, . . . , vk}. The projection M of a matching M in V0 is the hypergraph on Vconsisting of the projections of all edges in M.

We will construct a sort of a random almost perfect matching M(t) on V0 (meaning,without any mathematical sense, that M(t) leaves only few vertices uncovered), orequivalently, almost d0-regular hypergraph M(t) on V, such that M(t) � H(t) with highprobability. Therefore, if M(t) contains a matching which covers exactly one copy of eachvertex in V, called a V-PM, then �(t) contains a perfect matching with high probability.We use a coupling argument for this.

A sequence {(�(t), M(t))}t�1, where �(t) is a hypergraph of size t on V and M(t)is a matching of size at most t on V0, is nested if �(t) � �(t � 1) and M(t) � M(t �1) with �M(t � 1)� � �M(t)� � 1. A stopping time Ti of the sequence is the minimumt with M(t) � i. It is infinity if such t does not exist. If the sequence is nested, Ti is clearlyat least as large as i. We are looking for a coupling, or a joint distribution, or simply a


probability space, on the set of nested sequences {(�(t), M(t))}t�1 such that themarginal distribution of {�(t)} is the same as described above and the marginaldistribution of M(Ti) is uniform over all matchings of size i on V0, and importantly

M�t� � ��t� � t � 1, . . . .

A coupling satisfying these conditions is called good.

Lemma 2.1. Suppose d0 � n�, where � � �(n) � 1/5 is uniformly bounded below from0. Then there are a good coupling {(�(t), M(t))}t�1 and a constant c 0 such that forT � Td0n�d0

�1/�k�1�n

Pr�T � cd0n� � 1 � o�1�. (2.1)

Notice that if T � �, M(T) is the uniform random matching on V0 of size d0n �d0

�1/(k�1)n.

Lemma 2.2. Let M be the uniform random matching on V0 of size d0n � d0�1/(k�1)n.

If d0 � o(n) and

d0

n1/�5�2/�k�1�� 3 �, (2.2)

then

Pr�M contains a V-PM� � 1 � o�1�. (2.3)

Clearly, Lemmas 2.1 and 2.2 yield

Pr��cd0n� has no perfect matching� � Pr�T cd0n� Pr�M�T� has no V-PM� � o�1�,

provided (2.2). The constant c does not play any role here since cd0 satisfies (2.2) if andonly if d0 satisfies (2.2). It is also routine to check that the random k-partite hypergraph�k

(k)(n, p) with pnk � 1.1d0n contains �(d0n) with high probability, since �k(k)(n, p)

would have more random edges than �(d0n) with high probability. Hence Theorem 1.5follows.

3. COUPLING ON (�(t), M(t)): PROOF OF LEMMA 2.1

As defined before, let e1, . . . , et, . . . be i.i.d. random edges chosen uniformly at randomfrom all possible nk edges. The random hypergraph �(t) has a vertex set V and an edgeset H(t) � {e1, . . . , et}.

To construct M(t), let M(0) � A and L0(v) � [d0] for all v � V (L for list). For t �1, M(1) consists of a uniform random copy of e1 � { x1, x2, . . . , xk}. (It is crucial tonotice that the edge e1, or generally et, is the same edge used to construct �(1), or �(t),respectively.) Precisely, take a k-tuple (i1, i2, . . . , ik) � �j�1

k L0( xj) uniformly at

116 KIM

random and let M(1) consist of the single edge {( x1, i1), ( x2, i2), . . . , ( xk, ik)}. Ingeneral, Lt(v) is basically the set of uncovered copies of v:

Lt�v� � �i � �d0� : �v, i� is not covered by M�t��.

For an edge e � {v1, . . . , vk} in V, available copies (or edges) of e with respect to(w.r.t.) M(t) are edges in V0 of the form {(v1, i1), . . . , (vk, ik)} with ij � Lt(vj) for allj � 1, . . . , k. When we call an edge in V0 available, it means that the edge is an availablecopy of an edge in V. An edge in V0 will be sometimes called a copy. We will add toM(t � 1) a random available copy of et w.r.t. Mt�1 with a certain probability so thatM(t)�M(t � 1) is either empty or consists of a single edge chosen uniformly at randomamong all available copies w.r.t. M(t � 1). If we were able to add (or not add) et sucha way, then, conditioned on its size, M(t) would be a uniform random matching of V0.(See Lemma 3.1 below.)

This may be achieved by rejecting et with a certain probability. Let the rejectionprobability

pt�e� :� 1 ��v�e �Lt�1�v��

�t�1, where �t�1 � max

f:edge in V�v�f

�Lt�1�v��.

Clearly,

0 � pt�e� � 1.

for all edges e in V. Note that for any available copy f of e w.r.t. a matching M(t � 1),

Pr�M�t� � M�t � 1� � �f� � M�t � 1�� Pr�et � e��1 � pt�e��v�e

�Lt�1�v��1

� n�k��t�1��1,

which means that the probability that M(t) � M(t � 1) � { f} is independent of f. Inother words, if an edge is to be added to M(t � 1), then all available copies are equallylikely to be added. Since there are (d0n � �M(t � 1)�)k available edges, we also have that

Pr��M�t��M�t � 1�� 1�M�t � 1�� d0 � �M�t � 1��/n�k

�t�1. (3.1)

For completeness, we rigorously prove that M(Ti) is uniform.

Lemma 3.1. The matching M(Ti) is the uniform random matching among all matchingsof size i.

Proof. For any matching M of size i,


Pr�M�Ti� � M� � �f�M

Pr�M�Ti�1� � M��f�� Pr�M�Ti� � M�M�Ti�1� � M��f��.

For M(Ti�1) � M�{ f} and corresponding �t�1 � �t�1(M\{ f}), the time Ti � Ti�1

required to add an edge to M(t � 1) has a geometric distribution with success probabilityq :� (d0 � (i � 1)/n)k/�t�1 (see (3.1)). That is,

Pr�Ti � Ti�1 � t� � q�1 � q�t�1.

Once an edge is to be added to M(t � 1), each available edge is equally likely to beadded. Thus, conditioned that an edge is added, the probability that the added edge is f is(d0n � (i � 1))�k. Therefore,

Pr�M�Ti� � M � M�Ti�1� � M��f�� d0n � �i1��k �

t�1

�

q�1�q�t�1��d0n��i�1��k

and hence

Pr�M�Ti� � M� � �d0n � �i � 1��k �f�M

Pr�M�Ti�1� � M��f��.

Using the same argument inductively, we prove that M(Ti) is uniform. ■

To generate the uniform random matching of size i, one may choose i verticesuniformly at random from each part and generate a perfect matching on the chosenvertices. This together with routine large deviation type estimates yields the followingequation and bounds for �LTi

(v)�:

Corollary 3.2. For all v � V,

Pr��LTi�v�� d0 � m� ��md0�� i�m

d0�n�1��

� id0n�

.

In particular, with probability 1 � O(n�10),

�LTi�v�� d0 � i/n��1 O��log n��1/2�� for i � d0n � n log2n, (3.2)

�LTi�v�� 2 log2n for d0n � n log2n � i � d0n � d0�1/2kn, (3.3)

and

�LTi�v�� 20k

�for i � d0n � d0

�1/2kn, (3.4)

118 KIM

for all v � V. (Recall that d0 � n�.) ■

If (3.2) occurs for all i � d0n � n log2n, then

�t � �1 o�1��d0 � i/n�k,

and (3.1) implies that, at each time, an edge is added with probability 1 � o(1). Hence,it follows from the Chernoff type bound mentioned above (e.g., Theorem A.1.13 of [1])that

Td0n�n log2n � �1 o�1��d0n

with probability 1 � O(n�10).For d0n � n log2n � i � d0n � d0

�1/ 2kn, (3.3) yields that with probability 1 �O(n�10)

�t � 2klog2kn,

and an edge is added by (3.1), at any time t between Ti � 1 and Ti�1, with probability

�d0 � i/n�k

2knklog2kn�

d0�1/2

2klog2kn.

To add n log2n � d01/ 2kn edges, the number of required steps is at most

O��n log2n�d01/2�2klog2kn�� o�d0n�,

say with probability 1 � O(n�10) by the Chernoff type bound. For d0n � d0�1/ 2kn � i �

d0n � d0�1/(k�1)n/ 2, (3.4) yields

�t � �20k

� � k

.

If d0n � a � i � d0n � a/ 2, then an edge is added at each time with probability at least

� �

20k�k� a

2n�k

by (3.1). To add a/ 2 edges, O�((n/a)ka) steps are enough, say with probability 1 �O(n�11), again by the Chernoff type bound. Applying this to a � 2�jd0

�1/ 2kn, j � 0,1, . . . , until a � d0

�1/(k�1)n and adding O�((2jd01/ 2k)k�1n) over 1 � j � min{ j :

2�jd0�1/ 2k � d0

�1/(k�1)}, we deduce that the number of required steps is at most O(d0n)with probability 1 � O(n�10). Thus

Pr�T � cd0n� � 1 � O�n�10�. ■


4. PERFECT MATCHINGS IN M: PROOF OF LEMMA 2.2

We will use the second moment method to prove that M has a V-PM. For each v � V,the degree d(v) of v is the number copies of v covered by M. The average degree dsatisfies

d � d0 � d0�1/�k�1� or dn � d0n � d0

�1/�k�1�n �: l. (4.1)

It is also clear that d(v) � d0 � 1 for at most d0�1/(k�1)n vertices in each of Xj. Notice

that the random matching M of size l may be generated using a random labeling: Givenl unlabeled pairwise nonintersecting edges in V0, we label each vertex covered by M usingat most d0 copies of each vertex in V. Here we assume that vertices of each edge of M areordered and the jth vertex is to be assigned a copy of v � Xj. If the degree sequence{d(v)} is given, exactly d(v) copies of v must be used. Since (3.4) (and d(v) � d0 ��Lt(v)�) implies that

d0 � 20k/� � d�v� � d0 (4.2)

with probability 1 � O(n�10), it is enough to consider degree sequences satisfying (4.2).The quantity ¥v�Xj

1/d(v) turns out to play an important role in our estimates. (SeeLemma 4.5 and (4.9).) Since the function f( x) � 1/x, x 0, is convex, the conditions(4.1) and (4.2) imply that the quantity may be maximized when there are �d0

�1/(k�1)n/(20k) vertices of degree d0 � 20k/�, and all other vertices are of degree d0. Therefore,

� �v�Xj

1

d�v��

n

d� �d0

�2/�k�1�n

d3 �d0

�1/�k�1�n

20k�410k2

�2d3 � O�d�3�1/�k�1�� o��n/d�1/2�,

(4.3)

where in the last equality the hypothesis of Theorem 1.5 is necessary. We also easily have

�v�Xj

1

d�v�2 � o� �v�Xj

1

d�v�� . (4.4)

These conditions are needed to apply a lemma developed by Frieze and Janson [8] (seeLemma 4.5 below).

For a set R of n (unlabeled) edges of M, let ZR denote the indicator function of theevent that R is labeled to be V-PM. That is,

ZR � �1 if R is a V-PM �w.r.t. random label�,0 otherwise.

The number of V-PM’s in M is

Z :� �R

ZR.

120 KIM

4.1. Expectation

Lemma 4.1. Conditioned on {d(v)} with average degree d,

E�Z � �d�v�� dnn ��k�1 �

v�V

d�v�.

Proof. For given {d(v)} with average degree d, the number of ways to properly labelR (so that ZR � 1) is

�j�1

k �n!� �v�Xj

d�v�� dn � n�!�out of ((dn)!)k ways. Thus

E�ZR� � �dnn ��k �

v�V

d�v�.

For there are ( ndn) such R, the proof is complete. ■

4.2. Second Moment

Lemma 4.2. Conditioned on {d(v)} with average degree d,

E�Z2��d�v�� 0,1/n,. . .,1

��dnn �� d � 1�n

�1 � �n�� n n��k�1� �

v�V

d�v�� 2

� �j� �

Aj�Xj�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj�1 �

1

d�v�� .

Proof. For R, S with �R � S� � n, we take the vertices which will be used to labelR � S and let the set of such vertices be A1, . . . , Ak with �Aj� � n and Aj � Xj. Thenthere are

�j�1

k � �Aj�Xj

�Aj�� n

� n�!� �v�Aj

d�v�� n � n�!�2� �v�Xj�Aj

d�v��d�v� � 1�� dn � 2n n�!�ways to properly label both of R and S. Thus


E�ZRZS� � ��dnn �� d � 1�n

�1 � �n�� n n��k �

j�1

k � �Aj�Xj

�Aj�� n� �

v�Aj

d�v��

v�Xj�Aj

d�v��d�v� � 1�� dnn �� 1 � �n

�d � 1�n�� n n��k

� � �v�V

d�v�� 2 �j� �

Aj�Xj�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj�1 �

1

d�v�� .

Since there are

�dnn �� d � 1�n

�1 � �n�� n n�

such pairs, the proof is complete. ■

4.3. The Ratio E[Z2]/E[Z]2

Clearly,

E�Z2��d�v��

E�Z��d�v��2 � � �0,1/n,. . .,1

� �ndn�

��1� �n�d�1�n�� n

n �� k�1 �j�1

k � �Aj�Xj

�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj�1 �

1

d�v�� .

Remark. Let Y1, . . . , Ydn be i.i.d. random variables with 1 � Pr[Y1 � 0] � Pr[Y1 �1] � 1/d. Set S � Y1 � . . . � Y�dn and T � Y1 � . . . � Yn. Then the first factorof the summand is exactly

Pr�T � n�S � n�1�k.

Similarly, the second factor is the product of

Pr� �v�Xj

Wv � n,

where Wv’s are independent random variables with

1 � Pr�Wv � 0� � Pr�Wv � 1� �1

d�v�.

Large deviation inequalities might be useful to bound the above ratio. We will, however,use Stirling formula and a lemma of Frieze and Janson [8].

Notice that Theorem 2.2 follows from

122 KIM

E�Z2��d�v��

E�Z��d�v��2 � 1 o�1�

for all degree sequences satisfying (4.2) by Chebyschev Inequality. We will first considercases with � � 1/d� � 1/d log d. A more careful estimation will be required for � �1/d� � 1/d log d. The parameter � :� � 1/d is convenient since the maximum of thesummand turns out to occur when is around 1/d. (See also the above remark.) Thoughtedious, bounds for the terms with � � 1/d� � 1/d log d may be obtained by routineestimations using Stirling Formula and Taylor Theorem. Readers who are familiar withthis kind of estimation may skip or skim Lemma 4.3 and Corollary 4.4.

Applying Stirling formula, we have

� N�N� � ��2 o�1��1 � ��N��1/2eh��N,

where h(� ) � �� log � � (1 � �)log(1 � �) and o(1) goes to 0 as �N and (1 � � ) Ngo to infinity, and by Taylor expansion for � � �(1 � 1/k)/d,

h�1/d �� h�1/d� � log�d � 1� ��2d2

2�d � 1� O��3d2�. (4.5)

Thus

�dnn � � ��2 o�1��n��1/2edh�1/d�n,

� �d � 1�n

�1 � �n� � ��2 o�1��1 � �n��1/2e�d�1�h��1� �/�d�1��n,

and

� n n� � ��2 o�1�� 1 � �n��1/2eh� �n.

Using � � � 1/d and

h�1 �

d � 1� � h�1

d�

�

d � 1� � h�1

d� �� log�d � 1�

d � 1�

�2d2

2�d � 1�3 O��3

d �,

we obtain for � 1 � 1/ 2k

1

nlog� � n

dn�

��1� �n�d�1�n ��

log�1 � � o�1��

2n h�1

d� � log�d � 1� �2d2

2�d � 1�2 O��3�,

(4.6)


and for �� 1/d log d

1

nlog� � n

dn�

��1� �n�d�1�n �� n

n �� 1

2nlog�2�n o�n�

d � �d o�d��2

2. (4.7)

While (4.6) is more convenient when �� 1/d log d, (4.7) is needed in the case that �� 1/d log d.

For the second factor, an easy upper bound may be obtained. This bound will beenough if �� 1/d log d. Since there are at most �n :� d0

�1/(k�1)n vertices with d(v) �d, (4.2) yields 1/d(v) � 1/d except for at most �n vertices v for which

1

d�v��

1 21k/��d�

d�

1 O�1/d�

d,

and for all v � V

1 �1

d�v�� 1 �

1

d0� �1 �

1

d��1

2d0�1/�k�1�

d� � �1 �

1

d��1

2�

d� .

Thus

�v�Aj

1

d�v��

v�Xj�Aj

�1 �1

d�v�� d��Aj��1 �1

d�n��Aj�

eO��n/d�.

Moreover,

1

nlog�d� n�1 �

1

d��1� �n� � �� 1/d�log d �1 � 1/d � ��log�1 � 1/d�

� �� log�d � 1� � h�1/d�

implies that

�Aj�Xj

�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj

�1 �1

d�v�� n n�exp��n log�d � 1� � nh�1/d� O��n/d��.

(4.8)

Let

S� � �1

nlog � n

dn�

��1� �n�d�1�n �� n

n ��k�1

�j�1

k �Aj�Xj

�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj1 �

1

d�v��,

124 KIM

and T(�) � S(� � 1/d). We first show that T(�) is small enough for �� 1/d log dso that enT(�) is exponentially (in n/ 2d(log d)2) small.

Lemma 4.3. If �� 1/d log d, then

T�� 1 o�1�

2d�log d�2 .

Proof. If � � 1 � 1/ 2k � 1/d (or � 1 � 1/ 2k), then since (4.8) gives

� � ndn�

��1� �n�d�1�n�� n

n �� k�1

�j�1

k � �Aj�Xj

�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj

�1 �1

d�v��

� �dnn � k�1� n

�1/d ��n�exp��kn log�d � 1� � knh�1/d� O��n/d��

� �dnn �k�1� n

�1/d ��n�exp��kn log�d � 1� O��n/d��,

we use dh(1/d) � log d � O(1) and h( x) � 1 to have

T�� k log�d � 1� O��/d� 1

nlog��dn

n �k�1� n�1/d ��n��

� ��k log�d � 1� �k � 1�dh�1/d� h�1/d �� o�1� � ��1/3�log d.

Similarly, if � � 1 � 1/ 2k � 1/d and �� 1/d log d, then (4.6) and (4.8) yield

T�� k � 1�h�1

d� �k � 1�� log�d � 1� h�1

d ��

� k� log�d � 1� � kh�1

d� �k � 1��2d2

2�d � 1�2 O��3 �

d�� log�d � 1� h�1

d �� h�1

d� �k � 1��2d2

2�d � 1�2 O��3 �

d�.

Since all terms except the first term � log(d � 1) are bounded by a universal constant,if � � (log d)�1/ 2, then

T�� log d�1/2 O�1�.

Suppose � � (log d)�1/ 2. Take a constant 0 so that

T�� log�d � 1� h�1/d �� h�1/d� a�2/2 O��/d�


and denote

f�� log�d � 1� h�1/d �� h�1/d� a�2/2.

Then since

f�� log�d � 1� log1 � 1/d � �

1/d � a�

and

f�� 1

�1/d ��1 � 1/d � �� a � 0

for � � (log d)�1/ 2, f(�) is concave and has the maximum 0 at � � 0. If �� 1/d logd, then f has the maximum at � � 1/d log d or �1/d log d. Using (4.5), we know

h�1

d�

1

d log d� � h�1

d� � � log�d � 1� �1 o�1�

2�d � 1��log d�2

and obtain

f�� f��1

d log d� � �1 o�1�

2�d � 1��log d�2 .

Since � �� 1/(log n)2, the result follows. ■

Corollary 4.4.

� :

� �1/d� �1/d log d�� n

dn�

��1� �n�d�1�n �� n

n ��k�1 �j�1

k � �Aj�Xj

�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj�1 �

1

d�v�� o�1�.

It is now enough to show that

� :

� �1/d��1/d log d�� n

dn�

��1� �n�d�1�n �� n

n ��k�1 �j�1

k � �Aj�Xj

�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj�1 �

1

d�v�� 1 o�1�.

To do that, we introduce a lemma of Frieze and Janson [8, Lemma 3]. The proof uses agenerating function and a contour integral. For our purpose, the following weaker versionis enough.

Lemma 4.5 [8]. Suppose the (degree) sequence {d(v)}v�X with �X� � n and average(degree) d satisfies

126 KIM

d�v� 0, d � o�n�, � �v�X

1

d�v��

n

d� � o��n

d� 1/2� ,

and �v

1

d2�v�� o� �

v�X

1

d�v�� .

Then for � � 1/d� � 1/d log d and r � (1/n) ¥v�V 1/d(v), we have

�A�X

�A�� n

�v�A

1

d�v��

v�X�A

�1 �1

d�v�� 2 o�1��n

d ��1/2

exp� n � rn � n log�

r��.■

The conditions of the lemma are already verified in (4.2), (4.3), and (4.4). Let

g� � :� � r � log� /r�.

Then

g�r� � 0, g� � � �log� /r�, g�� 1/ , g�� 1/ 2

imply that for � � r� � o(r)

g� � ��1 o�1�� r�2

2r.

Using

r �1

d o�� 1

nd�1/2� , (4.9)

we have

�A�X

�A�� n

�v�A

1

d�v��

v�X�A

�1 �1

d�v��

� � �2 o�1��n

d ��1/2

exp�

nd�1 o�1�� 1

d� o� 1

�nd�1/2��2

2�.

Thus it is not hard to check that


� :

� �1/d��1/d log d�� n

dn�

��1� �n�d�1�n �� n

n ��k�1 �j�1

k � �Aj�Xj

�Aj�� n

�v�Aj

1

d�v��

v�Xj�Aj�1 �

1

d�v�� 1 o�1��

:� �1/d��1/d log d�

�2�n

d ��1/2

exp��nd� � 1/d�2

2 �� 1 o�1��

i:�i�n/d��n/d log d�

�2�n

d ��1/2

exp��d�i � n/d�2

2n �.

The last term except 1 � o(1) factor is an approximation of the Riemann integration 2

�2��1/2

�0� e�x2/ 2 � 1. For the sake of completeness, we rigorously show an upper bound, the

desired direction.

First, notice that we may replace n/d by n/d; for i with �i � n/d� � ndlog d

,

d�i � n/d�2

2n�

d�i � n/d�2

2n O�d�i � n/d�

n � �d�i � n/d�2

2n O� 1

log d�,

which yields


2n � � �1 o�1��exp��d�i � n/d�2

2n �.

We observe that

�i:

�i�n/d��n/d log d�

�2�n

d ��1/2


2n � � �i��

� �2�n

d ��1/2


2n ��

2

�2��1/2 �i�n/d

� �d

nexp��

d�i � n/d�2

2n ��

2

�2��1/2 �i�0

� �d

ne��i�n/d�2/2� 3

2

�2��1/2 0

�

e�x2/2 dx � 1

as a Reimann integral with mesh �dn 3 0. ■

5. PERFECT TRIANGLE PACKING: PROOF OF THEOREM 1.7

We first generate a sort of random partially d-regular 3-partite 3-uniform hypergraph�*3(n, d) on 3n vertices. The vertex set V of �*3(n, d) is the union of pairwise disjointsets X, Y, Z of size n, and d � 1.1p3n2 with p satisfying the condition in Theorem 1.7.It will be easy to check that it contains the random 3-partite 3-uniform hypergraph �3

(3)(n,

128 KIM

p3) with high probability: For each x � X, take d (uniform) random triples { x, y, z} �{ x} � Y � Z containing x with repetition. Let H( x) be the set of all such triples and�*3(n, d) � �x�X H( x). Then the degree of each vertex in X is exactly d. Each H( x)may be interpreted as a bipartite graph on Y � Z indexed by x in which { y, z} is an edgeif and only if { x, y, z} � H( x). We call H( x) a matching if the corresponding bipartitegraph is a matching and a near matching if all degrees are 1 except exactly one vertex ofdegree 2.

Lemma 5.1. We have

Pr�H�x� is a near matching� � O�d2/n�

and

Pr�H�x� is neither a matching nor a near matching� � ��d4/n2�.

Proof. Clearly,

Pr� � v � Y � Z, dH�x��v� � 2� � 2n�d2��1

n�2

�d2

n,

and

Pr� � v � Y � Z, dH�x��v� � 3� � 2n�d3��1

n�3

�d3

n2 .

For the probability that there are two distinct vertices v, w of degree 2, we consider twocases v H( x) w and v �H( x) w:

Pr� � v � w � Y � Z, dH�x��v� � dH�x��v� � 2, v H�x� w�

� n2�n � 1�2�d3�� 1

n2�3

� O�d3

n2�,

Pr� � v � w � Y � Z, dH�x��v� � dH�x��v� � 2, v �H�x� w�

� �2n2 ��n

2�2�d4�� 1

n2�4

� O�d4

n2�.

It is routine to check that this bound is tight, and the desired lower bound may be obtained.■

We now generate a partially 2d*-regular random 3-partite graph G3(n, d*), where d*� 1.05pn. Let X � { x1, . . . , xn}. For each i, choose uniform random subsets Yi andZi of size d* of Y and Z, independently of other i’s. The random 3-partite graph G( xi) on{ xi} � Yi � Zi has all edges containing xi, and each edge of the complete bipartite graphon Yi � Zi independently belongs to it with probability 1.1p. Let G* � �x�X G( x).


Again, we sometimes regard G( x) as a bipartite graph on Y � Z, and say it contains amatching or a near matching of a certain size if so does the corresponding bipartite graph.

Clearly, G( x)’s defined this way are mutually independent. Moreover, if each pair ( y,z) � Y � Z is in Yi � Zi for at most finite number of i’s, say 15, then G* may be regardedas a subgraph of the union of 15 i.i.d. random 3-partite graphs G3(n, 1.1p), which, in turn,may be regarded as a subgraph of the random 3-partite graph G3(n, 17p). This is possiblesince all edges containing a vertex in X may be regarded edges in one G3(n, 1.1p) andeach edge in Y � Z belongs to G* depending upon at most 15 independent experimentswith success probability 1.1p. We call a pair ( y, z) bad if it is in Yi � Zi for more than15 i’s. Then since Yi � Zi’s are mutually independent,

Pr� � a bad pair y, z� � n2 ( n

16)�(d*�1n�1 )

(d*n ) �32

�o�1

n�, (5.1)

(using p � n�3/5).Notice that the expected number of edges in G( x) is (d*)2(1.1p) � (1.05)21.1p3n2 �

(1.1)2p3n2 � 1.1d and the expected number of edges incident to another edge in G( x)is 1.1d � 2 � 1.1pd* � o(1). Thus it is routine by Chernoff type bounds (see, e.g.,Theorem A.1.11 and A.1.13 of [1]) to check that

P0 :� Pr�G�x� contains no matching of size d� � O�n�3�,

and a similar argument yields

cp5n3 � P1 :� Pr�G�x� contains a matching and a near matching of size both d� � 1,

for some constant c 0. It is worth pointing out that P1 is much larger than theprobability of H( x)’s being a near matching, and P0 is much less than the probability ofH( x)’s being neither a matching nor a near matching (see Lemma 5.1). This enables usto define the desired coupling.

Lemma 5.2. There is a coupling on (G(x), H(x)) with

Pr�H�x� � G�x�� 1 � O�d4/n2�.

Theorem 1.7 follows from the lemma since n � d4/n2 � o(1), and there is a probabilityspace in which

�3�3��n, p3� � �*3�n, d� � �

x�XH�x� � �

x�XG�x� � G* � G3�n, 17p�,

all with high probability and the second inclusion is in the sense that all edges in �*3(n,d) are triangles in G*.

Proof of Lemma 5.2. First, we generate G( x) without labeling vertices in Y and Z. If itcontains a matching and a near matching of size both d, take each of them and define

130 KIM

H0( x) without label to be the near matching with probability a and the matching withprobability 1 � a, where aP1 is exactly the probability of H( x)’s being a near matching.This is possible since P1 is larger than the probability of H( x)’s being a near matching.Then we use the same (random) labeling for G( x) and H0( x). If G( x) contains onlymatchings of size d, take one and define H0( x) to be the matching with a certainprobability, say 1 � b, so that the probability of H0( x)’s being a matching is exactly thesame as that of H( x)’s. (See below.) That is, take b satisfying

P1�1 � a� �1 � P1 � P0��1 � b� � Pr�H�x� is a matching�

or, equivalently,

1 � aP1 � P0 � b�1 � P1 � P0� � Pr�H�x� is a matching�.

Then we again use the same labeling. With probability b if G( x) has a matching of sized and with probability 1 if G( x) has no matching of size d, take independent H0( x) whichhas the same distribution as H( x) conditioned that H( x) is neither a matching nor a nearmatching. Notice that we took b so that

Pr�H0�x� is a matching� � P1�1 � a� �1 � P1 � P0��1 � b� � Pr�H�x� is a matching�.

Clearly, 0 � b � 1 since P0 � O(n�3), P1 � o(1), and Pr[H( x) is a matching] �1 � aP1 � �(d4/n2) by Lemma 5.1 and the definition of a. By the construction, H0( x)’sare i.i.d. with the same distribution as H( x), and H0( x) � G( x) as long as H0( x) is amatching or a near matching. That is,

Pr�H0�x� � G�x�� Pr�H0�x� is neither a matching nor a near matching� � O�d4/n2�. ■

REFERENCES

[1] N. Alon and J. Spencer, The probabilistic method. Second edition, Wiley-Interscience Seriesin Discrete Mathematics and Optimization, Wiley-Interscience, New York, 2000.

[2] N. Alon and R. Yuster, Threshold functions for H-factors, Combin Probab Comput 2 (1993),137–144.

[3] A. Bekessy, P. Bekessy, and J. Komlos, Asymptotic enumeration of regular matrices, StudiaSci Math Hungar 7 (1972), 343–353.

[4] E. Bender and R. Canfield, The asymptotic number of labeled graphs with given degreesequences, J Combin Theory Ser A 24 (1978), 296–307.

[5] B. Bollobas, A probabilistic proof of an asymptotic formula for the number of labelled regulargraphs, European J Combin 1 (1980), 311–316.

[6] C. Cooper, A. Frieze, M. Molloy, and B. Reed, Perfect matchings in random r-regular,s-uniform hypergraphs, Combin Probab Comput 5 (1996), 1–14.

[7] P. Erdos and A. Renyi, On the existence of a factor of degree one of a connected random graph,Acta Math Acad Sci Hungar 17 (1966), 359–368.

[8] A. Frieze and S. Janson, Perfect matchings in random s-uniform hypergraphs, RandomStructures Algorithms 7 (1995), 41–57.


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

[10] J. H. Kim and N. Wormald, Random matchings which induce Hamilton cycles and Hamilto-nian decompositions of random regular graphs, J Combin Theory Ser B 81 (2001), 20–44.

[11] M. Krivelevich, Perfect fractional matchings in random uniform hypergraphs, Random Struc-tures Algorithms 9 (1996), 317–334.

[12] M. Krivelevich, Triangle factors in random graphs, Combin Probab Comput 6 (1997),337–347.

[13] R. Robinson and N. Wormald, Almost all cubic graphs are Hamiltonian, Random StructuresAlgorithms 3 (1992), 117–125.

[14] R. Robinson and N. Wormald, Almost all regular graphs are Hamiltonian, Random StructuresAlgorithms 5 (1994), 363–374.

[15] A. Rucinski, Matching and covering the vertices of a random graph by copies of a given graph,Discrete Math 105 (1992), 185–197.

[16] J. Schmidt and E. Shamir, A threshold for perfect matchings in random d-pure hypergraphs,Discrete Math 45 (1983), 287–295.

[17] N. Wormald, Some problems in the enumeration of labelled graphs, Ph.D. thesis, Universityof Newcastle, 1978.

132 KIM

Documents

Perfect matchings in random uniform hypergraphs