Covering and tiling hypergraphs with tight cycles

0.67 MB·

by  Jie Han

COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES
JIE HAN, ALLAN LO, AND NICOLÁS SANHUEZA-MATAMALA
Abstract. Given 3 ≤ k ≤ s, we say that a k-uniform hypergraph Ck is a tight cycle on s
s
verticesifthereisacyclicorderingoftheverticesofCk suchthateveryk consecutivevertices
s
under this ordering form an edge. We prove that if k ≥3 and s≥2k2, then every k-uniform
hypergraphonnverticeswithminimumcodegreeatleast(1/2+o(1))nhasthepropertythat
everyvertexiscoveredbyacopyofCk. Ourresultisasymptoticallybestpossibleforinfinitely
s
many pairs of s and k, e.g. when s and k are coprime.
7
A perfect Ck-tiling is a spanning collection of vertex-disjoint copies of Ck. When s is
1 s s
0 divisible by k, the problem of determining the minimum codegree that guarantees a perfect
2 Csk-tilingwassolvedbyaresultofMycroft. Weprovethatifk≥3ands≥5k2 isnotdivisible
by k and s divides n, then every k-uniform hypergraph on n vertices with minimum codegree
n
atleast(1/2+1/(2s)+o(1))nhasaperfectCk-tiling. Againourresultisasymptoticallybest
a s
J possible for infinitely many pairs of s and k, e.g. when s and k are coprime with k even.

7
2 1. Introduction

] A hypergraph H = (V(H),E(H)) consists of a vertex set V(H) and an edge set E(H),
O where each edge e ∈ E(H) is a subset of V(H). Given a set V and a positive integer k, (cid:0)V(cid:1)
k
C
denotes the set of subsets of V with size exactly k. We say that a hypergraph H is k-uniform
h. if E(H) ⊆ (cid:0)V(H)(cid:1), and we abbreviate ‘k-uniform hypergraphs’ to k-graphs. Note that 2-graphs
k
t are usually known simply as graphs.
a
m Given a hypergraph H = (V(H),E(H)) and a set S ⊆ V(H), let the neighbourhood N (S)
H
[ of S be {T ⊆ V(H)\S : T ∪S ∈ E(H)} and let degH(S) = |NH(S)| denote the number of

edgesofH containingthesetS. Ifw ∈ V(H), thenwealsowriteN (w)forN ({w}). Wewill
1 H H
v omit the subscript if H is clear from the context. We denote by δs(H) the minimum s-degree
5 of H, that is, the minimum of deg (S) over all s-element sets S ∈ (cid:0)V(H)(cid:1). Note that δ (H) is
1 H s 0
equal to the number of edges of H. Given a k-graph H, δ (H) and δ (H) are referred to as
1 k−1 1
8 the minimum codegree and the minimum vertex degree of H, respectively.
0 A typical question in extremal graph theory is: what is the smallest δ such that every k-
.
1 graph H on n vertices with δs(H) ≥ δ has property P? In this paper we study three different
0 problems: the existence (or Turán) problem, the covering problem and the tiling problem.
7
1 1.1. Turán thresholds. Let F be a k-graph. We say a k-graph H is F-free if H does not
v: contain a copy of F. For all i,n ∈ Z with 0 ≤ i < k, define the ith-degree Turán number
i exk(n,F) of F to be the maximum of δ (H) over all F-free k-graphs H on n vertices. We also
X i i
write ex (n,F) if k is clear from the context. The quantity exk(n,F) is the maximum possible
r i 0
a number of edges in an F-free k-graph, and is usually known as the Turán number of F.
When F is a graph, there is a long and rich story of studying ex (n,F). The first result in
0
determining Turán numbers in graphs was due to Mantel [24], who proved that ex (n,K ) =
0 3
bn2/4c. For general complete graphs K with t ≥ 3, Turán [27] determined ex (n,K ). If F
t 0 t
is a graph with at least one edge, it was proven by Erdős, Stone and Simonovits [8,9] that
Instituto de Matemática e Estatística, Universidade de São Paulo, Rua do Matão 1010, 05598-
090, São Paulo, Brazil
School of Mathematics, University of Birmingham, Edgbaston, Birmingham, B15 2TT, UK
E-mail addresses: [email protected], [email protected], [email protected].
Date: January 30, 2017.
TheresearchleadingtotheseresultswaspartiallysupportedbyFAPESP(Proc. 2013/03447-6,2014/18641-
5, 2015/07869-8) (J. Han) EPSRC, grant no. EP/P002420/1 (A. Lo) and the Becas Chile scholarship scheme
from CONICYT (N. Sanhueza-Matamala).
1
2 COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES
lim ex (n,F)/(cid:0)n(cid:1) = (χ(F)−2)/(χ(F)−1), where χ(F) is the chromatic number of F. For
n→∞ 0 2
k-graphs with k ≥ 3, much less is known. The case ex (n,K3) is a long-standing problem of
0 4
Turán [28], where Kk denotes the complete k-graph on t vertices. For a summary of the known
t
results of Turán problems in hypergraphs, see [16].
1.2. Tiling thresholds. LetH andF bek-graphs. AnF-tiling inH isasetofvertex-disjoint
copies of F. An F-tiling is perfect if it spans the vertex set of H. Note that a perfect F-tiling
is also known as F-factor and perfect F-matching. For all i,n ∈ Z with 0 ≤ i < k, define
the ith-degree tiling threshold tk(n,F) to be the maximum of δ (H) over k-graphs H on n
i i
vertices without a perfect F-tiling. We write t (n,F) if k is clear from the context. Note that
i
if n 6≡ 0 mod |V(F)| then a perfect F-tiling cannot exist and so, tk(n,F) = (cid:0)n−i(cid:1). Hence we
i k−i
will always assume that n ≡ 0 mod |V(F)| whenever we discuss tk(n,F).
i
Regarding tiling thresholds, again there are more results in the graph case. A first result in
this sense comes from the celebrated theorem of Dirac [6] on Hamiltonian cycles, which easily
shows that t (n,K ) = n/2 − 1. Corrádi and Hajnal [4] proved that t (n,K ) = 2n/3 − 1,
1 2 1 3
and Hajnal and Szemerédi [12] generalized this result for complete graphs of any size, showing
that t (n,K ) = (1 − 1/t)n − 1. For a general graph F, Kühn and Osthus [22] determined
1 t
t (n,F) up to an additive constant depending only on F. This improved previous results due
1
to Alon and Yuster [2], Komlós, Sarközy and Szemerédi [20] and Komlós [19].
For k ≥ 3, Kühn and Osthus [22] determined t (n,Kk) asymptotically and Rödl, Ruciński
k−1 k
and Szemerédi [26] determined the exact value for sufficiently large n. Lo and Markström [23]
determined t (n,K3) asymptotically, and independently, Keevash and Mycroft [17] determined
2 4
t (n,K3) exactly for sufficiently large n.
2 4
We say that a k-graph H is s-partite (or that H is a (k,s)-graph, for short) if H has a vertex
partition V ,...,V such that |e∩V | ≤ 1 for all edges e ∈ E(H) and all i such that 1 ≤ i ≤ s.
1 s i
We say that a (k,s)-graph is complete if it has a vertex partition V ,...,V and E(H) consists
1 s
exactly of all e ∈ (cid:0)V(H)(cid:1) such that |e∩V | ≤ 1 for all 1 ≤ i ≤ s. We say that a (k,s)-graph
k i
is balanced if all sets in the vertex partition have the same size. Given integers a ,...,a ≥ 1,
1 s
let Kk(a ,...,a ) denote the complete (k,s)-graph with vertex partition V ,...,V such that
1 s 1 s
|V | = a for all 1 ≤ i ≤ s. For all k ≥ 2 and t ≥ 1, Kk(t) denotes the balanced complete
i i
(k,k)-graph whose vertex partition is made up of sets of size t.
Recently, Mycroft [25] determined the asymptotic value of t (n,K) for all complete (k,k)-
k−1
graphs K. For more results on tiling thresholds for hypergraphs, see the survey of Zhao [29].
1.3. Covering thresholds. Given a k-graph F, a k-graph H has an F-covering if for all
vertices v ∈ V(H), H contains a copy of F containing v. Similarly, for all i,n ∈ Z with
0 ≤ i < k, define the covering ith-degree threshold ck(n,F) of F to be the maximum of δ (H)
i i
over all k-graphs H on n vertices not containing an F-covering. We write c (n,F) if k is clear
i
from the context.
Trivially, every k-graph with a perfect F-tiling has an F-covering, and every graph with an
F-covering has a copy of F. Thus, for all k ≥ 2 and 0 ≤ i < k we have
ex (n,F) ≤ c (n,F) ≤ t (n,F).
i i i
In this sense, the covering problem is an intermediate problem between the Turán and the
tiling problems.
As for results on covering thresholds, for any non-empty graph F, we have c (n,F) =
1
(cid:16) (cid:17)
χ(F)−2 +o(1) n, see [13]. In the 3-graph and minimum vertex degree case, Han, Zang and
χ(F)−1
Zhao[13]obtainedasymptoticresultsforc (n,K)forcomplete(3,3)-graphsK. Inthe3-graph
1
and minimum codegree case, Falgas-Ravry and Zhao [10] studied c (n,F) when F is K3, K3
2 4 4
with one edge removed, K3 with one edge removed and other 3-graphs.
5
1.4. Cycles in hypergraphs. Given 1 ≤ ‘ < k, we say that a k-graph is an ‘-cycle if there
is a cyclic ordering of its vertices such that every edge consists of k consecutive vertices under
this order, and two consecutive edges (under the ordering of the vertices) intersect in exactly ‘
COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES 3
vertices. Note that an ‘-cycle on s vertices can exist only if k−‘ divides s. If ‘ = 1 we call the
cycle loose, if ‘ = k−1 we call the cycle tight. We write Ck for the tight cycle on s vertices,
s
or C if k is clear from the context.
s
Note that if 2‘ < k, i ≥ 2‘ and C is the ‘-cycle on s vertices, then c (n,C) ≤ s(cid:0)n−k+i(cid:1)+1 ≤
i k−i+1
snk−i−1 (by constructing C greedily).
Suppose that s ≡ 0 mod k. The tight cycle Ck is a spanning subgraph of Kk(s/k). Thus,
s
we obtain an upper bound for ex (n,Ck) from the following result by Erdős [7].
k−1 s
Theorem 1.1 (Erdős [7]). For all k ≥ 2 and s > 1, there exists n = n (k,s) such that
0 0
ex (n,Kk(s)) < nk−1/sk−1 for all n ≥ n .
0 0
We prove a sublinear upper bound for c (n,Ck) when s ≡ 0 mod k.
k−1 s
Proposition 1.2. For all 2 ≤ k ≤ s with s ≡ 0 mod k, there exists n (k,s) and c = c(k,s)
0
such that c (n,Ck) ≤ cn1−(k/s)k−1 for all n ≥ n .
k−1 s 0
We also determine bounds for c (n,Ck) and ex (n,Ck) whenever s 6≡ 0 mod k.
k−1 s k−1 s
Theorem 1.3. Let k,s ∈ N with k ≥ 2 and s ≥ 2k2. For all γ > 0, there exists n = n (k,s,γ)
0 0
such that for all n ≥ n , c (n,Ck) ≤ (1/2+γ)n.
0 k−1 s
Moreover, the bound above is asymptotically tight if k and s satisfy the following divisibility
conditions. Let 2 ≤ k < s and let d = gcd(k,s). We say that the pair (k,s) is admissible if
d = 1 or k/d is even.
Proposition 1.4. Let 3 ≤ k < s be such that (k,s) is an admissible pair. Then c (n,Ck) ≥
k−1 s
bn/2c−k+1. Moreover, if k is even, then ex (n,Ck) ≥ bn/2c−k+1.
k−1 s
There are some previously known results for tiling problems regarding ‘-cycles. Whenever C
is a 3-uniform loose cycle, t (n,C) was determined exactly by Czygrinow [5]. For general loose
2
cycles C in k-graphs, t (n,C) was determined asymptotically by Mycroft [25] and exactly
k−1
by Gao, Han and Zhao [11]. Note that C3 = K3, so t (n,C3) = (3/4+o(1))n [17,23]. For
4 4 2 4
tight cycles, whenever s ≡ 0 mod k, Mycroft [25] proved that t (n,Ck) ≤ (1/2+o(1))n.
k−1 s
We prove a lower bound on t (n,Ck) which shows that t (n,Ck) = (1/2+o(1))n if k
k−1 s k−1 s
dividessandsdividesn. Wealsoprovelowerboundsthatholdwhenever(k,s)isanadmissible
pair.
Proposition 1.5. Let 2 ≤ k < s ≤ n with n divisible by s. Then t (n,Ck) ≥ bn/2c−k.
k−1 s
Moreover, if (k,s) is an admissible pair, then
(cid:22)(cid:18)1 1 (cid:19) (cid:23)
 2 + 2s n −k if k is even,
t (n,Ck) ≥
k−1 s (cid:22)(cid:18)1 k (cid:19) (cid:23)
 2 + 4s(k−1)+2k n −k if k is odd.
The following upper bound on t (n,Ck) is our main technical result.
k−1 s
Theorem 1.6. Let 2 ≤ k < s be such that s ≥ 5k2 and s 6≡ 0 mod k. Then, for all γ > 0,
there exists n = n (k,s,γ) such that for all n ≥ n with n ≡ 0 mod s,
0 0 0
(cid:18)1 1 (cid:19)
t (n,Ck) ≤ + +γ n.
k−1 s 2 2s
Note that Theorem 1.6 is asymptotically sharp when s ≥ 5k2, k is even and (k,s) is an
admissible pair.
2. Notation and sketch of proofs
For a hypergraph H and S ⊆ V(H), we denote H[S] to be the subhypergraph of H induced
on S, that is, V(H[S]) = S and E(H[S]) = {e ∈ E(H) : e ⊆ S}. Let H \S = H[V(H)\S].
4 COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES
For hypergraphs H and G, let H −G be the subgraph of H obtained by removing all edges in
E(H)∩E(G).
Given a,b,c reals with c > 0, by a = b±c we mean that b−c ≤ a ≤ b+c. We write x (cid:28) y
to mean that for all y ∈ (0,1] there exists an x ∈ (0,1) such that for all x ≤ x the subsequent
0 0
statement holds. Hierarchies with more constants are defined in a similar way and are to be
read from the right to the left. We will always assume that the constants in our hierarchies are
reals in (0,1]. Moreover, if 1/x appears in a hierarchy, this implicitly means that x is a natural
number.
For all k-graphs H and all x ∈ V(H), define the link (k−1)-graph H(x) of x in H to be the
(k−1)-graph with V(H(x)) = V(H)\{x} and E(H(x)) = N (x).
H
For a family S of hypergraphs, an S-tiling is a set of vertex-disjoint copies of S ∈ S.
For distinct vertices v ,...,v , a k-graph P = v ···v is a tight path if all k consecutive
1 s 1 s
vertices form an edge. Throughout this paper, we assume that all tight paths come with their
ordering of vertices, and all paths are assumed to be tight. Hence, v ···v and v ···v are
1 s s 1
assumed to be different tight paths. For tight paths P , ..., P , we denote by P ···P , the
1 ‘ 1 ‘
concatenation of P ,...,P . For two paths P and P , we say that P extends P , if P = P P0
1 ‘ 1 2 2 1 2 1
for some tight path P0. Similarly, we might define a tight cycle C by writing C = v ...v ,
1 s
whenever v ...v v ...v is a tight path for all 1 ≤ i ≤ s.
i s 1 i−1
For all k ∈ N, let [k] = {1,...,k}. Let S be the symmetric group of all permutations of the
k
set [k], with the composition of functions as the group operation. Let id ∈ S be the identity
k
function that fixes all elements in [k]. Given distinct i ,...i ∈ [k], the cyclic permutation
1 r
(i i ···i ) ∈ S is the permutation that maps i to i for all 1 ≤ j < r and i to i , and fixes
1 2 r k j j+1 r 1
all the other elements; we say that such a cyclic permutation has length r. All permutations
σ ∈ S can be written as a composition of cyclic permutations σ ···σ such that these cyclic
k 1 t
permutations are disjoint, meaning that there are no common elements between all pairs of
these different cyclic permutations.
Let V ,...,V be disjoint vertex sets and let σ ∈ S . We say that a k-uniform tight path
1 k k
P = v ···v has end of type σ with respect to V ,...,V if for all 2 ≤ i ≤ k, v ∈ V .
1 ‘ 1 k ‘−k+i σ(i)
Similarly,wesayP hasstart of type σ with respect to V ,...,V ifv ∈ V forall1 ≤ i ≤ k−1.
1 k i σ(i)
If V ,...,V are clear from the context, we simply say that P has end of type σ and start of
1 k
type σ, respectively.
2.1. Sketch of proof of Theorem 1.3. The proof of Theorem 1.3 goes as follows. Let H
be a k-graph on n vertices with δ (H) ≥ (1/2+γ)n. Consider any vertex v ∈ V(H). We
k−1
first show that x is contained in a Kk(t) with classes V ,...,V , see Proposition 5.2. Thus x
k 1 k
is contained in a C with s ≡ 0 mod k. Hence, suppose that s ≡ r 6≡ 0 mod k with 1 ≤ r < k.
s
Let P be a tight path in Kk(t) containing x with start and end of type id. By using some
k
G-gadget (see Section 4), we extend P into a path P0 such that |P| ≡ |P0| mod k and P0 has
start and end of type id and (12···k)k−r, respectively. Then it is easy to extend P0 into a C
s
(by wrapping around V ,...,V ).
1 k
2.2. Sketch of proof of Theorem 1.6. TheproofofTheorem1.3usestheabsorbingmethod,
introduced by Rödl, Ruciński and Szemerédi [26]. We first find a small vertex set U ⊆ V(H)
such that H[U ∪ W] has a perfect C -tiling for all small sets W with |U| + |W| ≡ 0 mod s
s
(see Lemma 6.5). Thus the problem of finding a perfect C -tiling is reduced to finding a C -
s s
tiling covering almost all vertices. However, we do not find such C -tiling directly. Instead,
s
we show that there exists an {F ,E }-tiling T for some suitable k-graphs F and E , subject
s s s s
to the minimisation of some objective function φ(T). We do so by considering its fractional
relaxation, which we called a weighted fractional {F∗,K∗}-tiling (see Section 9.1), and further
s s
we use the hypergraph regularity lemma.
2.3. Organisation of the paper. WedescribetheextremalconstructionsinSection3, which
prove Propositions 1.4 and 1.5.
WeproveProposition1.2andTheorem1.3inSection5wherewedescribeafamilyofgadgets
that will be useful during the proofs in Section 4.
COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES 5
The rest of the paper is devoted to the proof of Theorem 1.6, bounding t (n,C ) from
k−1 s
above. In Section 6, we review the absorption technique for tilings, which we use in Section 7
to prove Theorem 1.6 under the assumption that we can find an almost perfect C -tiling
s
(Lemma 7.1). We prove Lemma 7.1 in the next two sections: in Section 8 we review tools of
hypergraph regularity and in Section 9 we introduce various auxiliary tilings that we use to
finish the proof.
We conclude with some remarks and open problems in Section 10.
3. Extremal graphs
In this section, we construct extremal graphs for the (k−1)-th degree Turán numbers and
covering and tiling thresholds for tight cycles, proving Proposition 1.4 and Proposition 1.5,
respectively.
Let A and B be disjoint vertex sets. Define Hk = Hk(A,B) to be the k-graph on A∪B
0 0
such that the edges of Hk are exactly the k-sets e of vertices that satisfy |e∩A| 6≡ k mod 2.
0
Note that δ (Hk(A,B)) ≥ min{|A|,|B|}−(k−1).
k−1 0
Proposition 3.1. Let 3 ≤ k ≤ s. Let A and B be disjoint vertex sets. Suppose that Hk(A,B)
0
contains a tight cycle C on s vertices with V(C )∩A 6= ∅. Let d = gcd(k,s). Then |V(C )∩
s s s
A| ≡ 0 mod s/d and (k,s) is not an admissible pair.
Proof. Let C = v ...v . For all 1 ≤ i ≤ s, let φ ∈ {A,B} such that v ∈ φ and let φ = φ .
1 s i i i s+i i
If two edges e and e0 in E(Hk(A,B)) satisfy |e ∩ e0| = k − 1, then |e ∩ A| = |e0 ∩ A| by
0
construction. Thus φ = φ for all 1 ≤ i ≤ s. Therefore, φ = φ for all 1 ≤ i ≤ s. Hence,
i+k i i+d i
|V(C)∩A| ≡ 0 mod s/d.
Let r = |{v ,...,v } ∩ A| = |{i : 1 ≤ i ≤ k,φ = A}|. Note that r > 0 and r ∈
1 k i
{k/d,2k/d,...,k}. Since {v ,...,v } ∈ Hk(A,B), it follows that r 6≡ k mod 2, implying
1 k 0
d ≥ 2 and k/d is odd. (cid:4)
Now we use Proposition 3.1 to prove Propositions 1.4 and 1.5.
Proof of Proposition 1.4. Let A and B be disjoint vertex sets of sizes |A| = dn/2e and |B| =
bn/2c. Consider the k-graph H = Hk(A,B). By Proposition 3.1, no vertex of A can be
0 0
covered with a copy of C . It follows that c (n,C ) ≥ δ (H ) ≥ bn/2c−k+1.
s k−1 s k−1 0
Moreover, if k is even, then Hk(A,B) = Hk(B,A). So no vertex of B can be covered by a
0 0
copy of C . Hence H is C -free. Therefore, ex (n,C ) ≥ bn/2c−k+1. (cid:4)
s 0 s k−1 s
Proof of Proposition 1.5. Let A and B be disjoint vertex sets whose size will be specified later
andconsiderthek-graphH = Hk(A,B). Lets0 = s/dandk0 = k/d. Notethatd ≤ k < s,thus
0 0
s0 > 1. Then,Proposition3.1impliesthatallcopiesC ofC inH satisfy|V(C)∩A| ≡ 0 mod s0.
s 0
Now choose A, B such that |A|+|B| = n, ||A|−|B|| ≤ 2 and |A| 6≡ 0 mod s0. It follows that
δ (H ) ≥ bn/2c−k. SinceallcopiesofC inH satisfy|V(C )∩A| ≡ 0 mod s0,itisimpossible
k−1 0 s 0 s
tocoverallverticesinAwithcopiesofC . Thisprovesthatt (n,C ) ≥ δ (H ) ≥ bn/2c−k,
s k−1 s k−1 0
as desired.
Now suppose that (k,s) is an admissible pair. Let H be the k-graph on n vertices with a
vertex partition {A,B,T} with |A| = d(n−|T|)/2e and |B| = b(n−|T|)/2c, where |T| will
be specified later. The edge set of H consists of all k-sets e such that |e∩A| 6≡ k mod 2 or
e∩T 6= ∅. Note that δ (H) ≥ min{|A|,|B|}+|T|−(k −1) ≥ b(n+|T|)/2c−k +1. We
k−1
separate the analysis in two cases depending on the parity of k.
Case 1: k even. Since H[A∪B] = Hk(A,B) = Hk(B,A), by Proposition 3.1, H[A∪B] is
0 0
C -free. Thus, allcopiesofC inH mustintersectT inatleastonevertex. Hence, allC -tilings
s s s
have at most |T| vertex-disjoint copies of C . Taking |T| = n/s−1 assures that H does not
s
contain a perfect C -tiling. This implies that t (n,C ) ≥ b(1/2+1/(2s))nc−k.
s k−1 s
6 COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES
Case 2: k odd. Since H[A ∪ B] = Hk(A,B), by Proposition 3.1 no vertex in A can be
0
covered by a copy of C . Hence, all copies of C in H with non-empty intersection with A
s s
must also have non-empty intersection with T. Moreover, all edges in H intersect A in at
most k − 1 vertices, so all copies of C contain at most bs(k − 1)/kc vertices in A. Thus
s
a perfect C -tiling would contain at least k|A|/(s(k − 1)) cycles intersecting A. Let |T| =
s
bnk/(2s(k−1)+k)c−1 < k|A|/(s(k−1)). Therefore H does not contain a perfect C -tiling.
s
j(cid:16) (cid:17) k
This gives t (n,C ) ≥ δ (H) ≥ 1 + k n −k. (cid:4)
k−1 s k−1 2 4s(k−1)+2k
4. G-gadgets
Throughout this section, we let τ = (123···k) ∈ S . Let H be a k-graph. Let V ,...,V be
k 1 k
a disjoint vertex partition of a complete (k,k)-graph K in H. Let P be a tight path in H with
end of type π ∈ S . For x ∈ V \V(P), Px is a tight path of H with end of type πτ. We call
k π(1)
such extension a simple extension of P. By repeatedly applying r simple extensions (which is
possible as long as there are available vertices), we may obtain an extension Px ···x of P
1 r
with end of type πτr, using r extra vertices in K.
Suppose K is a complete (k,k)-graph and P is a path with end of type π. We will find
a path P0 that extends P such that |V(P0)| ≡ |V(P)| mod k and P0 has end of type σ, for
arbitrary σ ∈ S , using G-gadgets defined as below.
k
Let G be a graph on [k]. Let H be a k-graph and let K be a complete (k,k)-graph in H
with vertex partition V ,...,V . For S ⊆ V(H), we say that K has a G-gadget W avoiding
1 k G
S if W = S W and for all ij ∈ E(G),
G ij∈E(G) ij
(W1) W ⊆ V(H) and |W | = 2k−1,
ij ij
(W2) |W \V(K)| = 1, W ∩S = ∅ and, for all 1 ≤ i0 ≤ k,
ij ij
(
1 if i0 ∈ {i,j},
|W ∩V | =
ij i0
2 otherwise,
(W3) for all σ ∈ S with σ(1) ∈ {i,j}, H[W ] contains a spanning tight path with start of
k ij
type στ and end of type (ij)σ, and
(W4) W are pairwise disjoint.
ij
For all edges ij ∈ E(G), we write w for the unique vertex in W \V(K).
ij ij
Suppose P is a tight path with end of type π and σ is a cyclic permutation. In the next
lemma, we show how to extend P into a tight path of type σπ using a G-gadget, where G is a
path.
Lemma 4.1. Let k ≥ 3 and r ≥ 2. Let σ = (i i ···i ) ∈ S be a cyclic permutation. Let G
1 2 r k
be a graph on [k] containing the path Q = i i ···i . Let H be a k-graph containing a complete
1 2 r
(k,k)-graph K with vertex partition V ,...,V . Suppose that P is a tight path in H with end
1 k
of type π ∈ S such that π(1) = i . Suppose that K has a G-gadget W avoiding V(P) and
k r G
|V \V(P)| ≥ 2|E(G)| for all 1 ≤ j ≤ r. Then there exists an extension P0 of P with end of
ij
type σπ such that
(i) |V(P0)| = |V(P)|+2k(r−1),
(ii) for all 1 ≤ i ≤ k,
(
2(r−1)−1 if i ∈ {i ,i ,··· ,i },
|V ∩(V(P0)\V(P))| = 1 2 r−1
i
2(r−1) otherwise,
(iii) K has a (G−Q)-gadget W avoiding V(P0) and
G−Q
(iv) V(P0)\V(P ∪K) = {w : 1 ≤ j < r}.
ijij+1
Proof. We proceed by induction on r. First suppose that r = 2 and so σ = (i i ). Consider a
1 2
G-gadgetW avoidingV(P). Asi i ∈ E(G),thereexistsasetW ⊆ W disjointfromV(P)
G 1 2 i1i2 G
suchthat|W | = 2k−1andH[W ]containsaspanningtightpathP00 withstartoftypeπτ
i1i2 i1i2
and end of type (i i )π = σπ. Note that |V ∩W | ≤ 2|E(G)|−1 as |V ∩W | = 1. Hence
1 2 i2 G i2 i1i2
V \(V(P)∪W ) 6= ∅. Takeanarbitraryvertexx ∈ V \(V(P)∪W )andsetP0 = Px P00.
i2 G i2 i2 G i2
COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES 7
As π(1) = i , it follows that P0 is a tight path with end of type σπ, and P0 satisfies properties
2
(i), (ii)and(iv). SetW = W \W . ThenW isa(G−i i )-gadgetforK avoiding
G−i1i2 G i1i2 G−i1i2 1 2
V(P0), so P0 satisfies property (iii), as desired.
Next, suppose r > 2. Define σ0 = (i i ···i ) and note that σ = (i i )σ0. Then σ0 is a cyclic
2 3 r 1 2
permutation of length r−1, with π(1) = i and the path Q0 = i ···i i is a subgraph of G.
r 2 r−1 r
By the induction hypothesis, there exists an extension P00 of P with end of type σ0π such that
|V(P00)| = |V(P)|+2k(r−2) and, for all 1 ≤ i ≤ k,
(
2(r−2)−1 if i ∈ {i ,i ,··· ,i },
|V ∩(V(P00)\V(P))| = 2 3 r−1
i
2(r−2) otherwise.
Moreover, K has a (G−Q0)-gadget W avoiding V(P00) and V(P00)\V(P ∪K) = {w :
G−Q0 ijij+1
2 ≤ j < r}.
Note that σ0π(1) = σ0(i ) = i and i i ∈ E(G − Q0). For all 1 ≤ i ≤ r, |V \ V(P0)| ≥
r 2 1 2 i
2|E(G−Q0)|. Again by the induction hypothesis, there exists an extension P0 of P00 with end
of type (i i )σ0π = σπ such that |V(P0)| = |V(P00)|+2k = |V(P)|+2k(r −1) and, for all
1 2
1 ≤ i ≤ k,
(
1 if i = i ,
|V ∩(V(P0)\V(P00))| = 1
i
2 otherwise.
and V(P0)\(V(P00 ∪K)) = {w }, so P0 satisfies properties (i), (ii) and (iv). Furthermore,
i1i2
set W = W −Sr−1W . Then W is a (G−Q)-gadget for K avoiding V(P0), so
G−Q G j=1 ijij+1 G−Q
P0 satisfies property (iii) as well. (cid:4)
In the next lemma, we show how to extend a path with end of type id to one with end of
arbitrary type. First we need the following definitions.
Consider an arbitrary σ ∈ S \{id}. Write σ in its cyclic decomposition
k
σ = (i i ···i )(i i ···i )···(i i ···i ),
1,1 1,2 1,r1 2,1 2,2 2,r2 t,1 t,2 t,rt
where σ is a product of t = t(σ) disjoint cyclic permutations of respective lengths r ,...,r so
1 t
that r ≥ 2 and i = min{i : 1 ≤ r0 ≤ r } for all 1 ≤ j ≤ t; and i < i < ··· < i .
j j,rj j,r0 j 1,r1 2,r2 t,rt
Define m(σ) = i . On the other hand, if σ = id, then define t(σ) = 0 and m(σ) = 1. Define
t,rt
G to be the graph on [k] consisting precisely of the (vertex-disjoint) paths Q = i i ···i
σ j j,1 j,2 j,rj
for all 1 ≤ j ≤ t(σ). So G is an empty graph. Note that for all σ,
id
t(σ)
X
2|E(G )|+t(σ) = 2 r −t(σ) ≤ 2k−1. (4.1)
σ j
j=1
For 1 ≤ i ≤ k and σ ∈ S \{id}, set X = 1 if i ∈ {i ,...,i } for some 1 ≤ t0 ≤ t, and
k i,σ t0,1 t0,rt0−1
X = 0 otherwise. Also, for 1 ≤ i ≤ k, set Y = 1 if i ∈ {σ(j) : 1 ≤ j < m(σ)} and Y = 0
i,σ i,σ i,σ
otherwise. If σ = id, then define X = Y = 0 for all 1 ≤ i ≤ k.
i,σ i,σ
Lemma 4.2. Let k ≥ 3. Let H be a k-graph containing a complete (k,k)-graph K with
vertex partition V ,...,V and a tight path P with end of type id. Let σ ∈ S and let G
1 k k
be a graph on [k] containing G . Suppose that K has a G-gadget W avoiding V(P), and
σ G
|V \V(P)| ≥ 2|E(G)|+2. Then there exists an extension P0 of P with end of type στm(σ)−1
i
such that
(i) |V(P0)| = |V(P)|+2k|E(G )|+m(σ)−1,
σ
(ii) for all 1 ≤ i ≤ k, |V ∩(V(P0)\V(P))| = 2|E(G )|−X +Y ,
i σ i,σ i,σ
(iii) K has a (G−G )-gadget avoiding V(P0) and
σ
(iv) V(P0)\V(P ∪K) = {w : ij ∈ E(G )}.
ij σ
Proof. Let
σ = (i i ···i )(i i ···i )···(i i ···i )
1,1 1,2 1,r1 2,1 2,2 2,r2 t,1 t,2 t,rt
8 COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES
as defined above. We proceed by induction on t = t(σ). If t = 0, then σ = id and m(σ) = 1,
so the lemma holds by setting P0 = P. Now suppose that t ≥ 1 and the lemma is true for all
σ0 ∈ S with t(σ0) < t. Let
k
σ = (i i ···i )(i i ···i )···(i i ···i )
1 1,1 1,2 1,r1 2,1 2,2 2,r2 t−1,1 t−1,2 t−1,rt−1
and σ = (i i ···i ), so σ σ = σ σ = σ. For 1 ≤ i ≤ 2, let G = G and m = m(σ ).
2 t,1 t,2 t,rt 1 2 2 1 i σi i i
Note that G = G ∪G . Let G0 = G−G . Since t(σ ) = t−1, by the induction hypothesis,
σ 1 2 1 1
there exists a path P1 that extends P with end of type σ1τm1−1 such that
(i0) |V(P )| = |V(P)|+2k|E(G )|+m −1,
1 1 1
(ii0) for all 1 ≤ i ≤ k, |V ∩(V(P )\V(P))| = 2|E(G )|−X +Y ,
i 1 1 i,σ1 i,σ1
(iii0) K has a G0-gadget W avoiding V(P ) and
G0 1
(iv0) V(P )\V(P ∪K) = {w : ij ∈ E(G )}.
1 ij 1
Note that for all 1 ≤ i ≤ k,
|V \(V(P )∪W )| ≥ 2|E(G)|+2−(2|E(G )|+1)−2|E(G0)| > 0.
i 1 G0 1
We extend P using m −m > 0 simple extensions, avoiding the set V(P )∪W in each step,
1 2 1 1 G0
to obtain an extension P2 of P1 with end of type σ1τm1−1τm2−m1 = σ1τm2−1 such that
|V(P )| = |V(P )|+m −m = |V(P)|+2k|E(G )|+m −1
2 1 2 1 1 2
and WG0 is a G0-gadget for K that avoids V(P2). As P1 has end of type σ1τm1−1, V(P2)\
V(P1) contains precisely one vertex in Vi for all i ∈ {σ1τm1−1(j) : 1 ≤ j ≤ m2 − m1} =
{σ (m ),...,σ (m −1)}. Since σ (i) = σ(i) for all m ≤ i < m and m = i , together with
1 1 1 2 1 1 2 2 t,rt
(ii0) we deduce that
|V ∩(V(P )\V(P))| = 2|E(G )|−X +Y . (4.2)
i 2 1 i,σ1 i,σ
Note that σ1τm2−1(1) = σ1(m2) = σ1(it,rt) = it,rt. Since G0 contains G2, by Lemma 4.1
there exists an extension P0 of P with |V(P0)| = |V(P )| + 2k|E(G )| and P0 has end of
2 2 2
type σ2σ1τm2−1 = στm(σ)−1, as m2 = m(σ). Moreover, as G0 − G2 = G − Gσ, K has a
(G−G )-gadget avoiding V(P0), implying (iii). Similarly, (iv) holds. Note that
σ
|V(P0)| = |V(P )|+2k|E(G )| = |V(P)|+2k|E(G )|+m(σ)−1,
2 2 σ
implying (i). Finally, for all 1 ≤ i ≤ k, we have
(
2|E(G )|−1 if i ∈ {i ,...,i },
|V ∩(V(P0)\V(P ))| = 2 t,1 t,rt−1
i 2
2|E(G )| otherwise.
2
So |V ∩(V(P0)\V(P ))| = 2|E(G )|−X . Note that X = X +X because σ and
i 2 2 i,σ2 i,σ i,σ1 i,σ2 1
σ are disjoint. Thus, together with (4.2), (ii) holds. (cid:4)
2
Let K be a complete (k,k)-graph with classes V ,...,V . Let P be a tight path with start
1 k
of type σ and end of type π. If π = σ, then there exists a tight cycle C containing P with
V(C) = V(P). Similarly if π = στ−r, then (by using r simple extensions) there exists a tight
cycle C on |V(P)|+r vertices containing P. In general, in order to extend P into a tight cycle
we use Lemma 4.2 to first extend P to a path P0 with end of type στ−r for some suitable r.
The next lemma formalizes the aforementioned construction of the tight cycle C containing P
and gives us precise bounds on the sizes of V ∩(V(C)\V(P)) in the case where σ = π, which
i
will be useful during Section 9.
Lemma 4.3. Let k ≥ 3. Let σ,π ∈ S and 0 ≤ r < k. Then there exists a graph G :=
k
G(σ,π,r) on [k] consisting of a vertex-disjoint union of paths such that the following holds for
all s ≥ k(2k−1) with s ≡ r mod k: let H be a k-graph containing a complete (k,k)-graph K
with vertex partition V ,...,V , and let P be a tight path with start of type σ and end of type π.
1 k
Suppose that K has a G-gadget W avoiding V(P) and |V \V(P)| ≥ bs/kc+1. Then there
G i
exists a tight cycle C on |V(P)|+s vertices containing P, such that
V(C)\(V(P ∪K)) = {w : ij ∈ E(G)}.
ij
COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES 9
Moreover, if σ = π, then for all 1 ≤ i,j ≤ k,
(cid:12) (cid:12)
(cid:12)|Vi∩(V(C)\V(P))|−|Vj ∩(V(C)\V(P))|(cid:12) ≤ 1.
Proof. Without loss of generality, we may assume that π = id. Define σ0 = στ−r ∈ S . Let
k
G = G . Note that |E(G)| ≤ k −1, t(σ0) ≤ k/2 and 2|E(G)|+t(σ0) ≤ 2k −1 by (4.1). Let
σ0
H,K,P be as defined in the lemma. By Lemma 4.2, there exists an extension P0 of P with
end of type σ0τm(σ0)−1 such that |V(P0)| = |V(P)|+2k|E(G)|+m(σ0)−1, for all 1 ≤ i ≤ k,
|V ∩(V(P0)\V(P))| = 2|E(G)|−X +Y .
i i,σ0 i,σ0
and V(P0)\(V(P ∪K)) = {w : ij ∈ E(G)}. We use k−m(σ0)+1 simple extensions to get
ij
an extension P00 of P0 of order
|V(P00)| = |V(P0)|+(k−m(σ0)+1) = |V(P)|+2k|E(G)|+k.
Note that V(P00) \ V(P0) uses precisely one vertex in each of the clusters V for all i ∈
i
{σ0τm(σ0)−1(j) : 1 ≤ j ≤ k − m(σ0) + 1} = {σ0(j) : m(σ0) ≤ j ≤ k} = {j : Y = 0}. It
j,σ0
follows that for all 1 ≤ i ≤ k,
|V ∩(V(P00)\V(P))| = 2|E(G)|+1−X .
i i,σ0
Note that P00 has end of type σ0τm(σ0)−1τk−m(σ0)+1 = σ0 = στ−r. For all 1 ≤ i ≤ k and
0 ≤ r < k, set Z = 1 if i ∈ {σ(j) : k−r+1 ≤ j ≤ k}, and set Z = 0 otherwise. We use
i,σ,r i,σ,r
r more simple extensions to get an extension P000 of P with end of type στ−rτr = σ of order
|V(P000)| = |V(P00)|+r = |V(P)|+2k|E(G)|+k+r,
such that for all 1 ≤ i ≤ k,
|V ∩(V(P000)\V(P))| = 2|E(G)|+1+Z −X .
i i,σ,r i,σ0
Since |E(G)| ≤ k − 1 and s ≡ r mod k, it follows that |V(P000)| ≤ |V(P)| + s. Also,
|V(P000)\V(P)| ≡ s mod k. For all 1 ≤ i ≤ k,
|V \V(P000)| ≥ |V \V(P)|−2|E(G)|−1+X −Z
i i i,σ0 i,σ,r
1
≥ bs/kc−2|E(G)|−1 = (kbs/kc−2k|E(G)|−k)
k
1 1
= (s−r−2k|E(G)|−k) = (cid:0)s−(|V(P000)|−|V(P)|)(cid:1).
k k
Since P000 has start of type σ and end of type σ, then we can easily extend P000 (using simple
extensions) into a tight cycle C on |V(P)|+s vertices. Note that V(C)\(V(P ∪K)) = {w :
ij
ij ∈ E(G)}, as desired.
Moreover, for all 1 ≤ i,j ≤ k,
(cid:12) (cid:12)
(cid:12)|Vi∩(V(C)\V(P))|−|Vj ∩(V(C)\V(P))|(cid:12)
= (cid:12)(cid:12)|Vi∩(V(P000)\V(P))|−|Vj ∩(V(P000)\V(P))|(cid:12)(cid:12)
= |(Z −X )−(Z −X )|.
i,σ,r i,σ0 j,σ,r j,σ0
Suppose now that σ = π = id. We will show that −1 ≤ Z −X ≤ 0 for all 1 ≤ i ≤ k,
i,σ,r i,σ0
(cid:12) (cid:12)
implying that for all 1 ≤ i,j ≤ k, (cid:12)|Vi∩(V(C)\V(P))|−|Vj ∩(V(C)\V(P))|(cid:12) ≤ 1. It suffices
to show that if Z = 1, then X = 1. If r = 0 then it is obvious, so suppose that 1 ≤ r < k.
i,σ,r i,σ0
Let 1 ≤ i ≤ k such that Z = 1. Since σ = π = id, then σ0 = τ−r. So if Z = 1, then
i,σ,r i,σ,r
k−r+1 ≤ i ≤ k. To show that X = 1, we need to show that i is not the minimal element
i,τ−r
in the cycle that it belongs in the cyclic decomposition of τ−r, that is, there exists m < i
such that i is in the orbit of m under τ−r. Let d = gcd(r,k). Choose 1 ≤ m ≤ d such that
m ≡ i mod d. The order of τ−r is exactly k/d and the orbit of m has exactly k/d elements.
There are exactly k/d elements i0 satisfying 1 ≤ i0 ≤ k and i0 ≡ m mod d, and all elements i0
in the orbit of m also satisfy i0 ≡ m mod d, so it follows that i is in the orbit of m under τ−r.
Finally, m ≤ d ≤ k−r < i. This proves that X = 1, as desired. (cid:4)
i,τ−r
10 COVERING AND TILING HYPERGRAPHS WITH TIGHT CYCLES
Let K be a (k,k)-graph with vertex partition V ,...,V . Consider a graph G on [k] with
1 k
E(G) = {a b : 1 ≤ i ≤ ‘}. Define F = F(K,G) to be the k-graph such that V(F) =
i i
V(K)∪{w ,...,w } and F = K ∪S (H(w ,a )∪H(w ,b )), where H(v,j) is a complete
1 ‘ 1≤i≤‘ i i i i
k-partite k-graph with partition {v},V ,V ,...,V ,V ,...,V . Note that if |V | ≥ 2‘ for
1 2 j−1 j+1 k i
all 1 ≤ i ≤ k, then K has a G-gadget in F. We write X for {w ,...,w }.
G 1 ‘
Consider s ≥ 2k2. Providing |V | is sufficiently large for all 1 ≤ i ≤ k, by Lemma 4.3 (taking
i
P to be any edge in K), there exists a graph G such that there exists a copy C of C such
s s
that C ⊆ F(K,G ). For 1 ≤ i ≤ k, let a = |V(C)∩V |. We get the following corollary from
s s,i i
Lemma 4.3.
Corollary 4.4. Let k ≥ 3, s ≥ 2k2 and s 6≡ 0 mod k. Then there exists a graph G on [k]
s
that is a disjoint union of paths, and a ,...,a ,‘ ∈ N such that |a − a | ≤ 1 for all
s,1 s,k s,i s,j
1 ≤ i,j ≤ k and ‘ = |E(G )| ≤ k − 1 and if K = Kk(a ,...,a ), then F = F(K,G )
s s,1 s,k s s
contains a spanning C and |V(F )\V(K)| = ‘.
s s
Note that the statement of Lemma 4.3 does not imply the corollary but its proof does.
The graph G , the integers a ,...,a ,‘ and the k-graph F will be useful in Section 9.
s s,1 s,k s
5. Covering thresholds for tight cycles
Inthissection,weprovetheupperboundsforthecoveringcodegreethresholdfortightcycles,
proving Proposition 1.2 and Theorem 1.3. We first prove Proposition 5.2, which immediately
implies Proposition 1.2 since Kk(s) contains a Ck-covering for all s0 ≡ 0 mod k with s0 ≤ sk.
k s0
We will use the following classic result of Kővári, Sós and Turán [18].
Theorem 5.1 (Kővári, Sós and Turán [18]). Let z(m,n;s,t) denote the maximum possible
number of edges in a bipartite graph G with parts U and V for which |U| = m and |V| = n,
which does not contain a K subgraph with s vertices in U and t vertices in V. Then
s,t
z(m,n;s,t) < (s−1)1/t(n−t+1)m1−1/t+(t−1)m.
Proposition 5.2. For all k ≥ 3 and s ≥ 1, let n ≥ 0 and c > 0 such that 1/n,1/c (cid:28) 1/k,1/s.
Then c (n,Kk(s)) ≤ cn1−1/sk−1.
k−1 k
Proof. Let H be a k-graph on n vertices with δ (H) ≥ cn1−1/sk−1. Fix a vertex x ∈ V(H)
k−1
and consider the link graph H(x) of x. Note that
(cid:0)n−1(cid:1)δ (H)
|E(H(x))| ≥ k−2 k−1 ≥ c1/2nk−1−1/sk−1. (5.1)
k−1
Let U = V(H)\{x}. Consider the bipartite graph B with parts U and E(H(x)), where
u ∈ U is joined to e ∈ E(H(x)) if and only if {u}∪e ∈ E(H). By the codegree condition of
H, all (k−1)-sets e ∈ E(H(x)) have degree at least δ (H)−1 in B. Hence
k−1
|E(B)| ≥ |E(H(x))|(δ (H)−1) ≥ |E(H(x))|(cn1−1/sk−1 −1). (5.2)
k−1
We claim there is a K as a subgraph in B, with s − 1 vertices in U and
s−1,nk−1−1/sk−2
nk−1−1/sk−2 vertices in H(x). Suppose not. Then, by Theorem 5.1,
|E(B)| ≤ z(|E(H(x))|,n−1,nk−1−1/sk−2,s−1) ≤ z(|E(H(x))|,n,nk−1−1/sk−2,s−1)
< (cid:16)nk−1−1/sk−2(cid:17)s−11 n|E(H(x))|1−s−11 +(s−1)|E(H(x))|
 nk−1−1/sk−2!s−11 
= |E(H(x))|n +s−1
|E(H(x))|
(5.1) (cid:18) − 1 1− 1 (cid:19) 1− 1
≤ |E(H(x))| c 2(s−1)n sk−1 +s−1 < |E(H(x))|n sk−1.
This contradicts (5.2).