    # Digital almost nets

Digital nets (in base 2) are the subsets of [0,1]^d that contain the expected number of points in every not-too-small dyadic box. We construct sets that contain almost the expected number of points in every such box, but which are exponentially smaller than the digital nets. We also establish a lower bound on the size of such almost nets.

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## 1 Introduction

We call a subinterval of basic (in base ) if it is of the form . A basic box is a product of basic intervals, i.e., a set of the form . If , a basic interval is called a dyadic interval, and a basic box is called a dyadic box.

We say that a set is a -almost net in base if it is of size for some natural number and

 (1−ε)m≤|β∩P|≤(1+ε)m

for every basic box of volume .

The case has been well studied. If , almost nets are known as -nets in the literature. They are used extensively in discrepancy theory and numerical integration algorithms, and are subject to numerous works, including a book devoted exclusively to them . It is known [9, Theorem 3] that, for each , there exist arbitrarily large -almost nets with . On the other hand, must grow exponentially with for large enough nets  (see also 

for asymptotic analysis of the bound in

).

In contrast to these results, for , we construct -almost nets with being only polynomial in .

###### Theorem 1.

For any prime , any , and any positive integers satisfying , there exists a set of size such that, for any basic box of volume ,

 ⎛⎝1−10√dlog(dq)m⎞⎠m≤|β∩P|≤⎛⎝1+10√dlog(dq)m⎞⎠m.

In particular, for every and every , there exist arbitrarily large -almost nets in base with .

Using this result, one can construct sets in without convex holes of size rather than that can be obtained from the conventional nets. See  for details.

The construction behind Theorem 1 is a minor modification on the construction in . Whereas the construction in  uses primes in , this construction uses irreducible polynomials in . The reason for this change is to make the denominators be powers of the same prime . Furthermore, because addition in satisfies ultrametric inequality, and because we do not need to worry about boxes that are not basic, several details in the new construction are simpler. As such, we do not make any claims about the novelty. Our purpose in writing the present note is to record the details of the construction for its application to convex holes. We also hope that almost nets will find applications in many other areas that currently use the conventional nets.

We do not know when the bound in Theorem 1 is sharp. The following is the best lower bound we were able to prove. Its dependence on is close to optimal, as long as is not too small, but the dependence on is poor. In the special case , we recover the lower bound in -nets via a proof different than those in [7, 8].

###### Theorem 2.

Assume that there exists an -net in base , then the following holds.

If , then

If , then

 m≥Ω(q−2k−2ε−2),

where .

In particular, if for some constant , then we have .

If for some constant such that , then we get an exponential lower bound , where .

### Open problem.

It would be interesting to prove a result similar to Theorem 1 which applies to all boxes, not only to basic boxes.

### Acknowledgment.

We are thankful to Ron Holzman for useful discussions.

## 2 Proof of Theorem 1

Let . Since the number of irreducible polynomials of degree in is

 1t(∑i|tμ(i)qt/i)≥1t(qt−qt/2+1)≥d,

we may pick distinct irreducible polynomials of degree in . We associate each of these polynomials to the respective coordinate direction. We will be interested in canonical boxes, which are the boxes of the form

 B=d∏i=1[aiqkit,ai+1qkit).

for some integers .

We say that a polynomial is a basic polynomial if and all of its coefficients are in .

For an irreducible polynomial of degree and a polynomial , we define the base- expansion of to be , where each is a basic polynomial. Put , where we view the basic polynomials as polynomial functions on . Note that . Define the function by .

###### Definition 3.

We say that a box is good if is a basic box of volume . Let be the smallest canonical box containing . We call a good pair.

Note that .

Suppose is a canonical box. Write it as , and consider . The set consists of the solutions to the system

 f ≡a′1(modpk11), f ≡a′2(modpk22), \makebox[7.499886pt]⋮ f ≡a′d(modpkdd),

where and are the unique basic polynomials satisfying .

By the Chinese Remainder theorem, the set is of the form where and is the unique element in of degree less than . Note that .

Given a good pair , define

 LB(β)\tiny def={g∈Fq[x]:r(A(B)+gD(B))∈β}.
###### Claim 1.

The set is of size at most .

###### Proof.

Let be a good pair. Write and in the form

 B=d∏i=1[aiqkit,ai+1qkit),β=d∏i=1[aiqkit+biq(ki+1)t,aiqkit+ciq(ki+1)t).

The condition is equivalent to

 A(B)+gD(B) ∈a′1+pk11J1(modpk1+11), A(B)+gD(B) ∈a′2+pk22J2(modpk2+12), \makebox[7.499886pt]⋮ A(B)+gD(B) ∈a′d+pkddJd(modpkd+1d),

where the sets consist of the basic polynomials such that .

On the other hand,

 A(B)+gD(B) ≡a′1+(α1+gδ1)pk11(modpk1+11), A(B)+gD(B) ≡a′2+(α2+gδ2)pk22(modpk2+12), \makebox[7.499886pt]⋮ A(B)+gD(B) ≡a′d+(αd+gδd)pkdd(modpkd+1d)

for some . There are at most different choices for . Also, there are at most different choices for satisfying . Since is determined by , the claim is true. ∎

To each canonical box of volume between and inclusive we assign a type, so that boxes of the same type behave similarly. Formally, let be the polynomial obtained from the polynomial by setting the coefficients of to zero. Similarly, let be the polynomial obtained from by setting the coefficients of to zero. The type of is then the pair .

Note that, from and it follows that

 n−dt+1≤degD(B)≤n. (1)
###### Claim 2.

The number of types is at most .

###### Proof.

Since , only the (resp. ) leading coefficients of (resp. ) may be non-zero. Hence, the number of types is at most . ∎

For a type , let . Note that if , then is an approximation to . That is to say, the respective elements of and of differ only in low-degree coefficients.

Let denote polynomials of degree less than in . Our construction will be a union of sets of the form where .

We first prove that there is no difference between the behaviors of and intersecting are the same.

###### Claim 3.

Suppose . Then for any polynomial and any polynomial , if and only if .

###### Proof.

If , then . Since and ,it follows that . From the definition of and , the coefficients of in are the same as the respective coefficients in . The opposite direction is similar. ∎

For a type and that satisfy and for some good pair , define

 YT(L)\tiny def={A+gD:g∈L}.

With this definition, is the approximation to induced by the approximation to .

###### Claim 4.

The set is of size exactly .

###### Proof.

Let be a good pair such that and . From the previous claim, we know that the size of is the same as the size of . By the Chinese remainder theorem, each of the canonical boxes of volume contains equally many points from . Since , the number of points in is equal to . ∎

###### Claim 5.

Let be chosen uniformly from . Then is

with probability

and is otherwise.

###### Proof.

Let be arbitrary. Clearly . The events of the form are mutually disjoint as ranges over . Indeed, suppose and are such that for some . We may write and . Then . Since , this implies that and hence .

In the combination with Claim 4, this implies that

Sample elements uniformly at random from , independently from one another. Let be the resulting multiset, and consider the multiset . For a type and that satisfy and for some good pair

, define the random variable

. This random variable is distributed according to the binomial distribution

.

Let . Note that , and in particular . Hence, . By the tail bounds for the binomial distribution [3, Theorems A.1.11 and A.1.13] we obtain

 Pr[NT,L−m>εm]
 Pr[NT,L−m<−εm]

From Claims 2 and 1 and the union bound it then follows that there exists a choice of such that is bounded between and whenever , and is a good pair. By Claim 3, this implies that the number of points in any good box of volume , the size is bounded between and .

Hence the multiset in is of size exactly and satisfies the conclusion of the theorem. To obtain a set satisfying the same conclusion, we may perturb the points of slightly to ensure distinctness.

## 3 Proof of Theorem 2

We shall derive Theorem 2 from the following lemma.

###### Lemma 4.

For any positive integers , prime , and positive real numbers with , and , if there exists an -almost net in base of size , then

 m≥Ω(log((dk))q2kε2log(1/ε)),

for any integer such that and

 2ε≥(dk)−1/2 (2)

holds.

###### Proof.

Let be the box . For any point , write its coordinates in base as . Noting that the first digits of are zero, we let be the first non-trivial digit of , i.e., . Similarly, let for .

The proof idea is to use almost independence of functions for a randomly chosen point of . However, we do not directly appeal to the known bound on the size of probability spaces supporting almost independent random variables (see e.g. [1, 2]) because those bounds are formulated for -valued random variables, whereas take distinct values.

Let , and . Since is an -almost net, it follows that is between and . Assume are all the points in .

For , let . Let be a -by- matrix, where the rows are indexed by and the columns are indexed by . The general entry of is

 Ui,J\tiny def=eq(∑j∈JXj(vi)).

Also, define .

###### Claim 6.

The diagonal terms in are all . The off-diagonal terms are, in absolute value, bounded above by .

###### Proof.

The general term of is given by

 AJ1,J2=1tt∑i=1eq(∑j∈J1Xj(vi)−∑j∈J2Xj(vi)).

If , this is clearly .

Suppose . Note that, for any choice of , the set

 {v∈B:Xj(v)=αj for j∈J1ΔJ2}

is a basic box of volume . Thus, for any , the region

 Bτ\tiny def={v∈B:∑j∈J1Xj(v)−∑j∈J2Xj(v)≡τ(modq)}

can be partitioned into many basic boxes of volume each. Since we have , it follows that the number of such that is bounded between and . Thus,

 |AJ1,J2|= 1t∣∣ ∣∣q∑τ=1|Bτ∩S|eq(τ)∣∣ ∣∣ ≤ = εq2kmt≤2ε.\qed

We apply [1, Theorem 2.1] to the matrix . We obtain that, if then . Therefore,

 m≥Ω(log((dk))q2kε2log(1/ε)).\qed

The right hand side of lemma 4 is a decreasing function of for . Therefore, we shall pick as small as possible. If , then we may set and get

If , then we may set . From the assumption on , we have . Therefore,

 (dk)≥ (d/k)k ≥ exp(2log(1/ε)logd−loglog(1/ε)(logd−logk)) ≥ exp(2log(1/ε)logd−loglog(1/ε)(logd−loglog(1/ε))) ≥ 1ε2,

and so (2) holds. Hence, we may apply Lemma 4 with in place of and obtain

 m≥Ω(q−2k−2ε−2).

In particular, if for some constant , then is also a constant, and so in this case.

If for some constant such that , then the -net is also an -net, when is large enough. We may apply the result above with in place of . In this case, the calculations above yield , and we get where .

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