Báo cáo toán học: "Holes in graphs" potx

In this paper we show that every graph with density d contains a large, relatively dense -regular pair.. The goal of this paper is to estimate the size of the largest such pair that can

Holes in graphs

Yuejian Peng

Department of Mathematics and Computer Science

Emory University, Atlanta, USA peng@mathcs.emory.edu

Vojtech R¨ odl∗

Department of Mathematics and Computer Science

Emory University, Atlanta, USA rodl@mathcs.emory.edu

Andrzej Ruci´ nski †

Department of Discrete Mathematics Adam Mickiewicz University, Pozna´n, Poland

rucinski@amu.edu.pl Submitted: November 7, 2000; Accepted: October 14, 2001

MR Subject Classifications: 05C35

Abstract

The celebrated Regularity Lemma of Szemer´edi asserts that every sufficiently

large graph G can be partitioned in such a way that most pairs of the partition sets span -regular subgraphs In applications, however, the graph G has to be dense and the partition sets are typically very small If only one -regular pair is needed,

a much bigger one can be found, even if the original graph is sparse In this paper

we show that every graph with density d contains a large, relatively dense -regular pair We mainly focus on a related concept of an (, σ)-dense pair, for which our

bound is, up to a constant, best possible

Szemer´edi’s Regularity Lemma is one of the most powerful tools in extremal graph theory

It guarantees an -regular partition of every graph G with n vertices, but the size of each

∗Research supported by NSF grant DMS 9704114.

†Research supported by KBN grant 2 P03A 032 16 Part of this research was done during the author’s

visit to Emory University.

-regular pair is at most n/T , where T is the tower of 2’s of height (1/)16 ([4]) However,

in some applications, only one pair is needed That was already observed and explored

by Koml´os (see [8]) and Haxell [6] The goal of this paper is to estimate the size of the largest such pair that can be found in any graph of given size and density The density

may decay to 0 with n → ∞.

The density of a bipartite graph G = (V1, V2, E) is defined as

|V1||V2| ,

and the density of a pair (U1, U2), where U1 ⊆ V1 and U2 ⊆ V2, is defined as

d(U1, U2) = e(U1, U2)

|U1||U2| ,

where e(U1, U2) is the number of edges of G with one endpoint in U1 and the other in U2

Definition 1.1 Let G = (V1, V2, E) be a bipartite graph and 0 <  < 1 A pair (U1, U2),

where U1 ⊆ V1 and U2 ⊆ V2, is called -regular if for every W1 ⊆ U1 and W2 ⊆ U2 with

|W1| ≥ |U1| and |W2| ≥ |U2|, we have

(1− )d(U1, U2)≤ d(W1, W2)≤ (1 + )d(U1, U2).

Our first result states that in every bipartite graph one can find a reasonably large

and relatively dense -regular pair.

Theorem 1.1 Let 0 < , d < 1 Then every bipartite graph G = (V1, V2, E) with |V1| =

|V2| = n and d(G) = d contains an -regular pair (U1, U2) with density not smaller than

(1− 

3)d and |U1| = |U2| ≥ n

2d c/

2

, where c is an absolute constant.

The constant c in Theorem 1.1 is determined by inequality (40) For instance, one can take c = 50.

In most applications the whole strength of -regular pairs is not used Instead, it is only required that d(W1, W2) is not much smaller than d(U1, U2) whenever W1 ⊆ U1 and

W2 ⊆ U2 are large enough This observation leads to the following definition.

Definition 1.2 Let G = (V1, V2, E) be a bipartite graph and 0 < , σ < 1 A pair

(U1, U2), where U1 ⊆ V1 and U2 ⊆ V2, is called (, σ)-dense if for every W1 ⊆ U1 and

W2 ⊆ U2 with |W1| ≥ |U1| and |W2| ≥ |U2|, we have e (W1, W2) ≥ σ|W1||W2| The graph G itself is called (, σ)-dense if (V1, V2) is an (, σ)-dense pair.

Now, let us consider the following problem For a bipartite graph G with n vertices in each color class and density d, we want to find an (, d/2)-dense pair as large as possible (The choice of σ = d/2 is not essential here.)

Definition 1.3 For any given 0 < , d < 1 and a positive integer n, f (, d, n) is the

largest integer f such that every bipartite graph G with n vertices in each color class and density at least d contains an (, d/2)-dense subgraph with f vertices in each color class.

As for  ≤ 2 − √ 2.5, every -regular pair with density at least (1 − /3)d is (,

d/2)-dense, Theorem 1.1 immediately implies that f (, d, n) ≥ n

2d c/

2

In 1991, Koml´os stated the following lower bound for f (, d, n).

Theorem 1.2 [8] For all 0 <  ≤ 0, 0 < d < 1 and for all integers n,

f (, d, n) ≥ nd (3/) ln(1/) .

In Section 2 of this paper we prove a different bound which is better for small values

of .

Theorem 1.3 For all 0 <  < 1, 0 < d < 1, and for all integers n,

f (, d, n) ≥ 1

2nd

12/ .

We also prove the following upper bound on f (, d, n), which shows that, up to a

constant, Theorem 1.3 is best possible

Theorem 1.4 For all 0 <  ≤ 0 and 0 < d ≤ d0, there exists n0 < (1/d) 1/(12) such that

for all n ≥ n0,

f (, d, n) < 4nd c/ , where c is an absolute constant.

In fact, we prove a stronger result than Theorem 1.4

Definition 1.4 Let G = (V1, V2, E) be a bipartite graph and 0 <  < 1 A pair (U1, U2),

where U1 ⊆ V1, U2 ⊆ V2, is said to contain an -hole if there exist W1 ⊆ U1 and W2 ⊆ U2 with |W1| ≥ |U1| and |W2| ≥ |U2| such that e(W1, W2) = 0.

By definition, if a pair contains an -hole, then it cannot be (, σ)-dense for any σ > 0.

Definition 1.5 For any given 0 < , d < 1 and a positive integer n, let h(, d, n) be the

largest integer h such that, every bipartite graph G with n vertices in each color class and density at least d contains a subgraph with h vertices in each color class and with no

-hole.

Clearly, f (, d, n) ≤ h(, d, n).

Theorem 1.5 For all 0 <  ≤ 0 and 0 < d ≤ d0 there exists n0 < (1/d) 1/(12) such that for all n ≥ n0,

h(, d, n) < 4nd c/ , where c is an absolute constant.

With no effort to optimize, it follows from the proofs of Theorems 1.4 and 1.5 that

the constant c appearing in them can be equal to 1/2000.

In this section we prove the lower bound given in Theorem 1.3 That is, we show that any

bipartite graph G = (V1, V2, E) with n vertices in each color class and density d contains

an (, d/2)-dense bipartite subgraph with at least 12nd c1/ vertices in each color class We then show that Theorem 1.1 the proof of which is a refinement of the proof of Theorem 1.3

Before giving the proof of Theorem 1.3, we prove the following claim which plays a crucial role

Claim 2.1 Every bipartite graph H = (V1H , V2H , E) with |V H

1 | = |V H

pair (U1, U2) satisfying one of the following conditions:

1 (U1, U2) is an (, d(H)/2)-dense pair and |U1| = |U2| ≥ m/2,

2 |U1| = |U2| ≥ m/4 and d(U1, U2)≥ (1 + /8)d(H).

that H contains a pair satisfying condition 2 For simplicity, we assume that 1/ is an

integer

Since, in particular, H itself is not (, d(H)/2)-dense, there exist A 01 ⊂ V H

1 , B10 ⊂ V H

2

with |A 0

1| = |B 0

1| ≥ m and e (A 0

1, B10 ) < d(H)2 |A 0

1||B 0

1| By an averaging argument, we can

take A1 ⊂ A 0

1, B1 ⊂ B 0

1 satisfying |A1| = |B1| = 

2m and e (A1, B1) < d(H)2 |A1||B1| (For

simplification, we assume that 2m is an integer Later we will make similar assumption

which are not essential but simplify our presentation.) Let F1 be the graph obtained by

removing A1 from V H

1 and B1 from V2H .

By the assumption, F1 is not an (, d(H)/2)-dense graph, and we apply the same argument as above to F1

In general, after l steps, l < 1/, we define l disjoint pairs (A1, B1) , · · · , (A l , B l) of size

|A i | = |B i | = 

2m for 1 ≤ i ≤ l Assume that F l is obtained by removingSl

j=1 A j from V H

1

and Sl

j=1 B j from V H

2 By assumption, F l is not (, d(H)/2)-dense, therefore there exists

A 0 l+1 ⊂ V H

1 \Sl

j=1 A j , B l+1 0 ⊂ V H

2 \Sl j=1 B j of size |A 0

l+1 | = |B 0

l+1 | ≥  (1 − l/2) m ≥ 

2m

and e A 0 l+1 , B l+1 0

< d(H)2 |A 0

l+1 ||B 0 l+1 | Take A l+1 ⊂ A 0

l+1 , B l+1 ⊂ B 0

l+1 , |A l+1 | = |B l+1 | =



2m and e (A l+1 , B l+1 ) < d(H)2 |A l+1 ||B l+1 |.

After 1/ steps the sets S1/

j=1 A j cover a half of V1H, and the sets S1/

j=1 B j cover

a half of V H

2 . Denote ¯V1 = S1/

j=1 A j and ¯V2 = S1/

j=1 B j Set e0 = e ¯ V1, ¯ V2

, e1 =

e ¯ V1, V2H \ ¯ V2

, e2 = e V1H \ ¯ V1, ¯ V2

, e3 = e V1H \ ¯ V1, V2H \ ¯ V2

.

Now we claim that there exists a pair satisfying condition 2 Indeed, if

e0 ≤ (1 − 3/8) d(H)m2/4,

then

e1+ e2+ e3 = d(H)m2− e0 ≥ 31 + 

8



2

4 .

Therefore, there exists i ∈ {1, 2, 3} satisfying

e i ≥1 + 

8



2

4

and we find a pair satisfying condition 2.

If e0 > (1 − 3/8) d(H)m2/4, we define e

ij = e (A i , B j ) Then

X

i

X

j6=i

e ij = e0−

1/

X

i=1

e (A i , B i ) >



1− 3

8



2

4 −1



d(H)

2

m 2

2

=



1− 7

8



2

4 .

For any I ⊂ {1, , 1/} of size |I| = 1/(2), we define

i∈I

X

j∈{1, ,1/}\I

e ij

Then P

I e (I) counts each e ij exactly 1/(2)−1 1/−2

times, where i 6= j Thus, there exists I0

such that

e(I0)≥

P

I e (I)

1/

1/(2)

=

1/−2 1/(2)−1

1/

1/(2)

X

i

X

j6=i

e ij > (1− 7/8) d(H)m2/4

4 (1− ) ≥



1 +  8



2

16,

and consequently the pair (S

i∈I0A i ,S

j∈{1, ,1/}\I0B j ) satisfies condition 2.

Proof of Theorem 1.3 Let G = (V1, V2, E) be any bipartite graph with n vertices in

each color class and density d If G contains a pair satisfying condition 1 in Claim 2.1, then we are done Otherwise, by Claim 2.1, there exists an induced subgraph G1 ⊂ G

with at least n/4 vertices in each color class and d(G1) ≥ (1 + /8)d Applying Claim

2.1 to G1, if G1 contains a pair satisfying condition 1 in Claim 2.1, then we have an

(, d(G1)/2)-dense pair, which is also an (, d/2)-dense pair, with at least n/8 vertices in each color class, and we are done again Otherwise we find an induced subgraph G2 ⊂ G1

with at least n/16 vertices in each color class and d(G2)≥ (1 + /8)2d.

Suppose we have iterated this process s times, obtaining a subgraph G s of G with at least n/4 s vertices in each color class and density at least (1 + /8) s d If the (s + 1)-th

iteration cannot be completed, it means that G s contains an (, d/2)-dense subgraph with

at least n/(2 · 4 s) vertices in each color class Because the density of any graph is not

larger than 1, we can only iterate this process at most t times, where t is the smallest

integer such that



1 +  8

t+1

d > 1.

Hence, at some point an (, d/2)-dense subgraph with at least n/(2 · 4 t) vertices in each

color class must be found It remains to estimate t from above By the choice of t, we have (1 + /8) t d ≤ 1, or, equivalently,

t ≤ log2(1/d)

log2(1 + /8) ,

and so

4t= 22t ≤ (1/d) log2(1+/8)2 .

Notice that log2(1 + /8) ≥ /6 for 0 <  < 1 Indeed, it follows from the facts that g(x) = log2(1 + /8) − /6 is concave in [0, 1], g(0) = 0 and g(1) > 0 Therefore

1 2

n

4t ≥ 1

2nd

12/ ,

and consequently we have proved the existence of an (, d/2)-dense subgraph of G with

at least 12nd 12/ vertices in each color class This completes the proof of Theorem 1.3

Proof of Theorem 1.1 (Sketch) The proof of Theorem 1.1 is similar to the proof of

Theorem 1.3; the only modification is to replace Claim 2.1 by Claim 2.3 below

The first alternative of Claim 2.3, rather than asking for a large -regular pair, demands

a stronger property which is however easier to analyze

Definition 2.1 Let G = (V1, V2, E) be a bipartite graph, 0 <  < 1 A pair (U1, U2),

where U1 ⊆ V1 and U2 ⊆ V2, is called (, G)-regular if for every W1 ⊆ U1 and W2 ⊆ U2 with |W1| ≥ |U1| and |W2| ≥ |U2|, we have

Fact 2.2 Every (, G)-regular pair (U1, U2) is -regular.

Claim 2.3 Every bipartite graph H = (V1H , V2H , E) with |V H

1 | = |V H

pair (U1, U2) satisfying one of the following conditions:

1 |U1|, |U2| ≥ m/2 and (U1, U2) is (, H)-regular,

2 |U1|, |U2| ≥ m/2 and d(U1, U2)≥ (1 + /3)d(H),

3 |U1|, |U2| ≥ m/4 and d(U1, U2)≥ (1 + 2/12)d(H).

Assuming that H contains no pair satisfying conditions 1 or 2, and using the same technique as in the proof of Claim 2.1, we can prove that H must contain a pair satisfying condition 3.

Applying Claim 2.3, one can prove Theorem 1.1 in the same way as we derived Theorem

1.3 from Claim 2.1 (see the Appendix for details) Note that the obtained -regular pair (U1, U2) has density at least (1− /3)d.

In this section we prove the upper bound for h(, d, n) given in Theorem 1.5 To prove that h(, d, n) < u, we need to find a bipartite graph G with n vertices in each color class and density at least d such that every subgraph of G with u vertices in each color class contains an -hole The following construction will be central for the proof.

Let k and t be positive integers, and [t] denote {1, 2, , t} Let G (k, t) = (V1, V2, E)

be the bipartite graph with

V1 ={x = (x1, x2, , x t) : 1≤ x s ≤ k, 1 ≤ s ≤ t.},

V2 ={y = (y1, y2, , y t) : 1≤ y s ≤ k, 1 ≤ s ≤ t.},

and xy ∈ E if and only if x s 6= y s for each s ∈ [t], where x = (x1, x2, , x t) ∈ V1 and

y = (y1, y2, , y t)∈ V2.

Observe that G(k, t) is a bipartite graph with k tvertices in each color class and density

k−1

k

t

For G (k, t) we prove the following property From now on we set n1 = k t

Lemma 3.1 Let k and t be positive integers and let 0 <  ≤ 1/4k For every U1 ⊆ V1,

U2 ⊆ V2 such that

min{|U1|, |U2|} ≥ n1

 e

2k

2kt1 + 4k

√

2

2t

, there exists an -hole in the subgraph of G (k, t) induced by the sets U1 and U2.

Proof: Suppose that there is no -hole in the subgraph of G(k, t) induced by the sets

U1, U2 We will estimate min {|U1|, |U2|} from above.

For each s = 1, 2, , t, the integer i ∈ [k] is called rare with respect to s in U1 if

|{x ∈ U1 : x s = i }| < |U1|.

Otherwise i is called frequent with respect to s Let R1s be the set of all rare values i ∈ [k]

with respect to s in U1 and F s1 be the set of all frequent values i ∈ [k] with respect to s

in U1 Similarly, let F s2 be the set of all frequent values i ∈ [k] with respect to s in U2

Note that F s1∩ F2

s = ∅ for each s ∈ [t], since otherwise the vertices x ∈ U1 and y ∈ U2

with x s = y s = i ∈ F1

s ∩ F2

s would form an -hole between U1 and U2

Next we are going to prove that more than half of the vertices in U1 have each less

than 2k rare coordinates At the same time we give an upper bound on the number of

such vertices which enables us to estimate |U1|.

For every x = (x1, , x s , , x t)∈ V1, define Sx ={s : x s ∈ R1

s } Let V 0

1 ={x ∈ V1 :

|Sx| < 2kt} and U 0

1 = U1∩ V 0

1.

Claim 3.2

|U 0

1| > 1

|U 0

1| ≤ |V 0

1| ≤ 2kt e

2k

2kt2k2t +Pt

s=1 |F1

s | t

t

Proof of Claim 3.2: To prove (2), we use a standard double counting argument

Con-sider an auxiliary bipartite graph M = (U1, [t], E(M )) in which a pair {x, s} ∈ E(M) if

and only if x s ∈ R1

s , where x = (x1, x2, , x t)∈ U1 and s ∈ [t] By the definition of R1

s,

it is easy to see that deg M (s) < k |U1| for any s ∈ [t] Therefore there are fewer than

1

2|U1| vertices x ∈ U1 which satisfy |Sx| = deg M(x) ≥ 2kt.

Now we prove (3) Let L ⊂ [t] with |L| < 2kt Then by the definition of Sx,

|{x ∈ V1 : Sx = L }| ≤Y

q∈L

|R1

s∈[t]\L

|F1

s |.

Hence

|V 0

1| ≤ X L⊂[t],|L|<2kt

s∈[t]\L

|F1

Since the geometric mean is not larger than the arithmetic mean, we obtain

|U 0

1| ≤ |V 0

1| ≤ X l<2kt



t l

 

kl +Pt

s=1 |F1

s | t

t

Since l < 2kt ≤ t/2, we have t

l

2kt

2k

2kt , and

|V 0

1| ≤ 2kt e

2k

2kt2k2t +Pt

s=1 |F1

s | t

t

which completes the proof of the claim.

Now we continue the proof of Lemma 3.1 By Claim 3.2

|U1| < 2|U 0

1| ≤ 2|V 0

1| ≤ 4kt e

2k

2kt2k2t +Pt

s=1 |F1

s | t

t

Similarly,

|U2| < 4kt e

2k

2kt2k2t +Pt

s=1 |F2

s | t

t

Since F s1∩ F2

s =∅ for each s ∈ [t], we have Pt

s=1 |F1

s | ≤ tk

2 or

Pt

s=1 |F2

s | ≤ tk

2 Therefore,

min{|U1|, |U2|} < 4kt e

2k

2kt 2k2t + kt

2

t

!t

(9)

= 4kt

 e

2k

2kt

(1 + 4k) t



k

2

t

Applying the inequality 4kt < (1 + 4k) t, we finally obtain that

min{|U1|, |U2|} <  e

2k

2kt

(1 + 4k) 2t



k

2

t

(11)

= n1

 e

2k

2kt1 + 4k

√

2

2t

which completes the proof

Now for any n ≥ n1, let r and q, where 0 ≤ q < n1, be the positive integers such that

n = rn1 + q We “blow up” the graph G(k, t) in the following sense: fix any q vertices

in each color class, and replace each of them by r + 1 new vertices At the same time replace every other vertex by r new vertices Finally, replace every edge of G(k, t) by the corresponding complete bipartite graph (K r,r , K r+1,r , or K r+1,r+1) Denote this new

graph by G n (k, t) = (V n

1 , V2n , E) It is easy to see that

r

r + 1



k

t

≤ d(G n (k, t)) ≤ r + 1

r



k

t

For this graph we now prove the following lemma which is very similar to Lemma 3.1

Recall that n1 = k t

Lemma 3.3 Let k and t be positive integers and let 0 <  ≤ 1/4k For every n ≥ n1, and for all U1 ⊆ V n

1 , U2 ⊆ V n

min{|U1|, |U2|} ≥ 2n e

2k

2kt1 + 4k

√

2

2t

, there exists an -hole in the subgraph of G n (k, t) induced by the sets U1 and U2.

Proof: Assume that there is no -hole in the subgraph of G n (k, t) induced by the sets

U1, U2 For each s ∈ [t] , define rare and frequent values i ∈ [k] with respect to s, for U1

and U2, in the same way as in the proof of Lemma 3.1 We follow the lines of the proof

of Lemma 3.1 The only novelty is to multiply the right hand side of equations (4) – (11)

by r + 1 Therefore, we have

min{|U1|, |U2|} < (r + 1)n1

 e

2k

2kt1 + 4k

√

2

2t

.

Since (r + 1)/r ≤ 2, and thus (r + 1)n1 ≤ 2rn1 ≤ 2(rn1+ q) = 2n, we obtain

min{|U1|, |U2|} < 2n e

2k

2kt1 + 4k

√

2

2t

The goal of blowing up G(k, t) was to obtain graphs with more vertices than n1 and

still having -holes in large subgraphs Next we consider a random “contraction” of G(k, t)

to obtain graphs with fewer than n1 vertices and with the same property

From now on, to make our description simpler, we set

2k −2 log k (1+4k), n0 = max{n 3α/21 , n 3δ/21 }.

Note that n0 ≤ n1 when k ≥ 3, and under this notation,

n −α1 = d(G(k, t)) =



k − 1 k

t

and

n −δ1 =

 e

2k

2kt1 + 4k

√

2

2t

.

Lemma 3.4 Let k ≥ 3 be a positive integer, 0 <  ≤ 1/4k, and t > t0 = t0(k, ) Then,

for every n0 ≤ n < n1, there exists a graph G n = (V n

1 , V2n , E n ) with n vertices in each

color class such that

k − 1

−α

and for all U1 ⊆ V n

1 , U2 ⊆ V n

2 with

min{|U1|, |U2|} ≥ 4n e

2k

2kt1 + 4k

√

2

2t

, there exists an -hole in the subgraph of G n induced by the sets U1 and U2 .