Sidon sets

# On Sidon sets in a random set of vectors

Sang June Lee Department of Mathematics, Duksung Women’s University, South Korea
July 19, 2019, \currenttime
###### Abstract.

For positive integers and , let be the set of all vectors , where is an integer with . A subset of is called a Sidon set if all sums of two (not necessarily distinct) vectors in are distinct.

In this paper, we estimate two numbers related to the maximum size of Sidon sets in . First, let be the number of all Sidon sets in . We show that , where the constants of depend only on . Next, we estimate the maximum size of Sidon sets contained in a random set , where denotes a random set obtained from by choosing each element independently with probability .

E-mail: sanglee242@duksung.ac.kr
The author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (2013R1A1A1059913).
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## 1. Introduction

For positive integers and , let be the set of all vectors , where ’s are integers with . A subset of is called a Sidon set if all sums of two (not necessarily distinct) vectors in are distinct. A well-known problem on Sidon sets in is the determination of the maximum size of Sidon sets in . For , Erdős and Turán [4] showed in 1941 that . Then, Lindström [8], in 1969, improved the bound to . On the other hand, in 1944, Chowla [1] and Erdős [3] observed that a result of Singer [12] implies that . Consequently, we know . For a general , Lindström [9] showed in 1972 that . On the other hand, in 2010, Cilleruelo [2] proved that Therefore,

 F([n]d)=nd/2(1+o(1)). (1)

For more information, see the classical monograph of Halberstam and Roth [5] and a survey paper by O’Bryant [11].

In this paper we consider two numbers related to the number . The first one is the number of all Sidon sets contained in . The second one is the maximum size of Sidon sets contained in a random subset of instead of .

We first start with the problem of estimating . Recalling that , one can easily see that

 2F([n]d)≤Zn,d≤F([n]d)∑k=1(ndk)≤F([n]d)(ndF([n]d)).

This implies the following.

###### Fact 1.1.
 2nd/2(1+o(1))≤Zn,d≤n(d/2)nd/2(1+o(1)).

In this paper, we improve the above upper bound as follows.

###### Theorem 1.2.

For a positive integer , there exists a positive constant such that, for any sufficiently large ,

 Zn,d≤2cnd/2.

Note that this upper bound matches the lower bound in Fact 1.1 up to a multiplicative constant factor in the exponent. Our proof of Theorem 1.2 will be provided in Subsection 3.1. The case of Thereom 1.2 was also proved in [6].

Next, we deal with the maximum size of Sidon sets contained in a random subset of . Let be a random set obtained from by choosing each element independently with probability . Let be the maximum size of Sidon sets in a random set . Our result about is as follows.

###### Theorem 1.3.

For a positive integer , let be a constant with , and let . Then, there exists a constant such that, asymptotically almost surely (a.a.s.), that is, with probability tending to as ,

 F([n]dp)=nb+o(1). (2)

Moreover,

 b(a)=⎧⎪⎨⎪⎩a+dif −d

The graph of is given in Figure 1. A refined version of Theorem 1.3 is stated in Theorems 2.42.7 in Subsection 2.2. Theorems 2.42.7 will be proved in Sections 4 and 5. The case of Theorem 1.3 was also proved in [6].

### 1.1. Remark and Notation

From now on, let be a fixed positive integer. Constants in , and may depend on . We write if goes to as . We also write if .

## 2. Main Results

### 2.1. The number of Sidon sets of a given size

We will obtain an upper bound on the number of Sidon sets in of a given size. For a positive integer , let be the number of Sidon sets in of size . Observe that the following result applies when .

###### Lemma 2.1.

Let be a positive integer. For a sufficiently large integer , the following holds: If is a positive integer with , where , then

Our proof of Lemma 2.1 will be given in Subsection 3.2. Lemma 2.1 will be used in order to prove Theorem 1.2 (see Subsection 3.1 for its proof) and the upper bounds in Theorems 2.6 and 2.7 (see Section 4 for its proof).

The next lemma provides an upper bound on the number for . Observe that the range of here is a bit wider than the range of in Lemma 2.1.

###### Lemma 2.2.

Let and be real numbers and let , and be positive integers satisfying that

 0<γ
 ω≥4, and t=ωs∗. (5)

Then,

 Zn,d(t)≤(4endtγ1−2/ω)t.
###### Remark 2.3.

For , a version of Lemma 2.2 was given in Lemma 3.3 in [6], but we improve the previous one as follows:

1. We have a better upper bound on by removing the multiplicative factor in the base in Lemma 3.3 of [6].

2. We remove the variable used in Lemma 3.3 of [6].

Our proof of Lemma 2.2 will be given in Subsection 3.3. Lemma 2.2 will be applied to our proof of the upper bound in Theorem 2.5. (See Section 4 for the proof.)

### 2.2. The maximum size of Sidon sets in a random set [n]dp

Recall that is a random set obtained from by choosing each element independently with probability . Also, recall that denotes the maximum size of Sidon sets in a random set . We state our results on the upper and lower bounds of in Theorems 2.42.7 in full. Recall that if .

###### Theorem 2.4.

The following holds a.a.s.:
If   , then

 F([n]dp)=(1+o(1))ndp. (6)

If   , then

 (1/3+o(1))ndp≤F([n]dp)≤(1+o(1))ndp. (7)
###### Theorem 2.5.

Let . If   , then there exist a positive absolute constant and a positive constant such that a.a.s.

 c1nd/3(log(n2dp3))1/3≤F([n]dp)≤c2nd/3(log(n2dp3))1/3.
###### Theorem 2.6.

Let . If   , then there exist a positive absolute constant and a positive constant such that a.a.s.

 c3nd/3(logn)1/3≤F([n]dp)≤c4nd/3(logn)4/3.
###### Theorem 2.7.

If   , then there exist a positive absolute constant and a positive constant such that a.a.s.

 c5nd/2p1/2≤F([n]dp)≤c6nd/2p1/2.

### 2.3. Organization

In Section 3, we prove Theorem 1.2 and Lemmas 2.1 and 2.2. Our proof of the upper bounds in Theorems 2.42.7 will be provided in Section 4. In Section 5, we prove the lower bounds in Theorems 2.42.7.

## 3. The number of Sidon sets in [n]d of a given size

### 3.1. The number of Sidon sets

Now we show Theorem 1.2 by using Lemma 2.1.

###### Proof of Theorem 1.2.

We have that

 Zn,d = nd∑t=1Zn,d(t)=F([n]d)∑t=1Zn,d(t),

where the second equality holds since is the maximum size of Sidon sets in . Since , we have that

 Zn,d=nd/3logn∑t=1Zn,d(t)+F([n]d)∑t=nd/3logn+1Zn,d(t). (8)

The first sum of (8) is estimated by

 nd/3logn∑t=1Zn,d(t) ≤ nd/3logn∑t=1(ndt)≤nd/3logn⋅(endnd/3logn)nd/3logn ≤ n(2d/3)nd/3logn(1+o(1))≤2c1nd/3(logn)2,

where is a positive constant depending only on . Next, it follows from Lemma 2.1 that the second sum of (8) is estimated by

 F([n]d)∑t=nd/3logn+1Zn,d(t) ≤ F([n]d)⋅nc2nd/3(logn)1/3(c3nd(F([n]d))2)F([n]d) ≤ 2c4nd/2,

where , and are positive constants depending only on . Therefore, in view of identity (8), the above estimates of the first and second sums of (8) imply Theorem 1.2. ∎

### 3.2. The number of Sidon sets of a larger size

Recall that is the number of Sidon sets in of size . Now we show Lemma 2.1 which gives an upper bound on for . Our proof uses the following strategy from [6]. Let be an integer with , and let be a seed Sidon set in of size . For such a Sidon set , we estimate the number of extensions of to larger Sidon sets of size containing . Then, by summing over all Sidon sets of size , we will obtain an upper bound on . In order to bound the number of extensions, we define the following graph.

###### Definition 3.1.

For a Sidon set in , let be the graph on in which is an edge of if and only if there exist some such that .

Observe that if is a Sidon set in of size containing , then the set is an independent set in of size . Hence, the number of extensions of to larger Sidon sets of size is bounded above by the number of independent sets in of size .

In order to bound the number of independent sets in of a given size, we will use the following result from [6].

###### Lemma 3.2 (Lemma 3.1 of [6]).

For positive integers and and a positive real number , let be a graph on vertices such that for every vertex set with , the number of edges in the subgraph of induced on satisfies

 e(U)≥β(|U|2). (9)

If is a positive integer satisfying

 q≥β−1log(N/R), (10)

then, for all positive integers , the number of independent sets in of size is at most

 (Nq)(Rr). (11)

Next we show that the graph with a Sidon set satisfies condition (9) with suitable and .

###### Lemma 3.3.

For a Sidon set in of size , the graph on vertices satisfies the following: For every vertex set with

 |U|≥(2d+1/s)nd, (12)

the number of edges in the subgraph of induced on satisfies

 e(U)≥s22d+1nd(|U|2). (13)
###### Proof.

Let be an arbitrary vertex set of with . We define an auxiliary bipartite graph with disjoint vertex classes and in which a vertex is adjacent to a vertex if and only if there exists such that . Observe that distinct vertices and in have a common neighbor if and only if is an edge of the subgraph of induced on . Hence, we infer that , where denotes the degree of in .

Now we claim that

 e(U)=∑w∈[2n]d(dB(w)2). (14)

In order to prove (14), we need to show that contains no -cycle, i.e., that two distinct vertices in do not have two distinct common neighbors in . Towards contradiction, suppose that there is a -cycle in , that is, both and in are adjacent to both and in . From the definition of , there exist some such that , , and . Thus, and , and hence, we have that , that is, . Since is a Sidon set, we infer that . However, by the assumptions and , we have that and , which contradicts to . Therefore, there is no 4-cycle in , and hence, identity (14) holds.

It follows from (14) that

 e(U)=∑w∈[2n]d(dB(w)2)≥(2n)d(1(2n)d∑w∈[2n]ddB(w)2),

where the inequality follows from the convexity of . Since is a bipartite graph in which for all , we have that

 e(U) = (2n)d(1(2n)d∑u∈UdB(u)2)=(2n)d(s|U|/(2n)d2) = (2n)d⋅12s|U|(2n)d(s|U|(2n)d−1).

Under the assumption (12), that is, , we infer that

 e(U)≥s|U|2⋅12s|U|(2n)d=s22d+1nd|U|22≥s22d+1nd(|U|2).

This completes the proof of Lemma 3.3. ∎

Now we are ready to bound the number of Sidon sets of a larger size by applying Lemmas 3.2 and 3.3 as follows.

###### Lemma 3.4.

Let , , and be positive integers satisfying

 s2q≥d2d+1ndlogn. (15)

Then, for any integer , we have

 Zn,d(s+q+r)≤Zn,d(s)(ndq)(2d+1nd/sr). (16)
###### Proof.

Fix as an arbitrary Sidon set in of size . We first consider the number of Sidon sets of size containing . Recall that if is a Sidon set of size containing , then the set is an independent set in of size . Hence, in order to bound the number of Sidon sets of size containing , we are going to estimate the number of independent sets of size in . To this end, we will apply Lemma 3.2 to the graph .

We first check conditions (9) and (10) of Lemma 3.2. First, Lemma 3.3 implies that (9) holds with and . Next, condition (10) follows from ineqality (15) because

 q ≥ d2d+1ndlogns2=2d+1ndlog(nd)s2≥β−1log(NR).

Thus, Lemma 3.2 with , and gives that for any integer , the number of independent sets in of size is at most Consequently, the number of Sidon sets of size containing is at most By summing over all Sidon sets of size , we infer that the number of all Sidon sets of size is at most which completes the proof of Lemma 3.4. ∎

Now we show Lemma 2.1 by applying Lemma 3.4 iteratively.

###### Proof of Lemma 2.1.

Since , we infer that if for a sufficiently large , depending only on . Hence, let be an integer satisfying

 2s0≤t≤1.1nd/2, (17)

where . Let be the largest integer satisfying . We define three sequences , , and as follows: for ,

 (18)

Under the assumption and the definition of and , we have that, for ,

 s2kqk=s21q1≥s30=d2d+1ndlogn.

Equivalently, condition (15) holds with and . Thus, Lemma 3.4 with , , and gives that for ,

 Zn,d(sk+1)=Zn,d(sk+qk+rk)≤Zn,d(sk)(ndqk)(2d+1nd/skrk).

Consequently,

 Zn,d(t)=Zn,d(sK+1)≤(ns1)⋅K∏k=1(ndqk)⋅K∏k=1(2d+1nd/skrk). (19)

Now we estimate three parts of the right-hand side of (19) separately. The first part is estimated by

 (ns1)≤(n2s0)≤n2s0. (20)

Next, for the second part of (19), we have that

 K∏k=1(ndqk) ≤ K∏k=1(nd)qk=(nd)∑Kk=1qk(nd)s0∑Kk=14−k+1 (21) ≤ (nd)s0(4/3)≤n2ds0,

where the second inequality follows from For the last part of (19), we first have that

 K∏k=1(2d+1nd/skrk) ≤ K∏k=1(2d+1nd/skrk+qk)

since

 rk+qk2d+1nd/sk≤sk+1−sk2d+1nd/sk=sk2d+1nd/sk=s2k2d+1nd≤t22d+1nd12.

We further have that

 K∏k=1(2d+1nd/skrk) ≤ K∏k=1(2d+1nd/sksk+1−sk)=K∏k=1(2d+1nd/sksk) (22) ≤ K∏k=1(e2d+1nds2k)sk=K∏k=1(e2d+1nds2K−k+1)sK−k+1 K∏k=1(e2d+122kndt2)t2−k = (e2d+1ndt2)t∑Kk=12−k22t∑Kk=1k2−k ≤ (e2d+1ndt2)t24t=(e2d+5ndt2)t.

In view of (19), combining (20)–(22) yields that for , which completes our proof of Lemma 2.1. ∎

### 3.3. The number of Sidon sets of a smaller size

Now we show Lemma 2.2 which gives an upper bound on for . Our proof of Lemma 2.2 is similar to the proof of Lemma 3.4, and hence, we only give a sketch. By weakening condition (12) of Lemma 3.3 into , where , we clearly have the following corollary of Lemma 3.3. (We omit the proof.)

###### Corollary 3.5.

Let be an arbitrary real number with . For a Sidon set in of size , the graph on vertices satisfies the following: For every vertex set with the number of edges in the subgraph of induced on satisfies

 e(U)≥s22d+1nd(|U|2). (23)

Combining Lemma 3.2 and Corollary 3.5 implies Lemma 2.2 as follows.

###### Proof of Lemma 2.2.

Before applying Lemma 3.2 with , we first check conditions (9) and (10) in Lemma 3.2 with . First, by Corollary 3.5, the graph satisfies (9) with and . Next, condition (10) holds by setting where .

Now Lemma 3.2 with and implies that, for ,

 Zn,d(t) ≤ Zn,d(s∗)(nds∗)(nd/γt−2s∗)≤(nds∗)(nds∗)(nd/γt−2s∗) (24) ≤ (ends∗)2s∗(endγ(t−2s∗))t−2s∗=(ends∗)t(1γ(ω−2))t−2s∗ = (eωndt)t(1γ(ω−2))t(1−2/ω)=(eωndt[γ(ω−2)]1−2/ω)t = (Cωndtγ1−2/ω)t,

where . We also have that

 Cω=eω(ω−2)1−2/ω≤eω(ω/2)1−2/ω=e21−2/ωω2/ω≤2eω2/ω≤2e41/2=4e, (25)

where the first inequality follows from the assumption , and the last inequality follows from the fact that is a decreasing function for . Combining (24) and (25) completes our proof of Lemma 2.2. ∎

## 4. Upper bounds on F([n]p)

In this section we prove the upper bounds in Theorems 2.42.7. We first provide our proof of the upper bound in Theorem 2.4. In the proof, we will use the following version of Chernoff’s bound.

###### Lemma 4.1 (Chernoff’s bound, Corollary 4.6 in [10]).

Let be independent random variables such that and , and let . For ,

###### Proof of the upper bound in Theorem 2.4.

We clearly have that Hence, in order to show the upper bound in Theorem 2.4, it suffices to show that a.a.s.

 X:=∣∣[n]dp∣∣≤ndp(1+o(1)). (26)

By the definition of , we have that the expectation is . Then, Lemma 4.1 implies that a.a.s. provided that . It gives (26), and hence, it completes our proof of the upper bound in Theorem 2.4. ∎

Next, we prove the upper bound in Theorem 2.5 using Lemma 2.2 as follows.

###### Proof of the upper bound in Theorem 2.5.

For the upper bound, it suffices to show that there exists a positive constant such that

 Pr[[n]dp contains a Sidon set of size cnd/3(log(n2dp3))1/3]→0 as n→∞. (27)

The first moment method gives that the probability that contains a Sidon set of size is at most . We will use Lemma 2.2 in order to bound . Now we define suitable numbers , and satisfying both (4) and (5) in Lemma 2.2. For a positive constant , we consider two cases separately: the first case is when and the second case is for the remaining range of , that is, .

• Case 1: This case is when . Let . Under the assumption , we have that , and hence, the second inequality of (4) holds. Let , where is a sufficiently large positive constant depending only on . Then, the inequality of (5) holds.

With the choice of and , Lemma 2.2 implies that

 Pr[[n]dp contains a Sidon set of size t]≤ptZn,d(t)≤(4endptγ1−2/ω)t. (28)

The base in the right-hand side of (28) is

 4endptγ1−2/ω ≤ 4endpCnd/3(n2dp3)0.99≤(n2dp3)