# Equivalence classes of Niho bent functions

Equivalence classes of Niho bent functions are described for all known types of hyperovals.

## Authors

• 1 publication
07/17/2019

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## 1 Introduction

Bent functions were introduced by Rothaus [47] and then they were studied by Dillon [27]. A bent function is a Boolean function with an even number of variables which achieves the maximum possible distance from affine functions [14]. They are Hadamard difference sets in elementary abelian 2-groups. Bent functions have relations to coding theory, cryptography, sequences, combinatorics and design theory [1, 2, 14, 17, 28].

Dillon [27] introduced bent functions related to partial spreads of , . He constructed bent functions that are constant on the elements of a spread. Dillon also studied a class of bent functions that are linear on the elements of a Desarguesian spread. These functions were thoroughly studied in [11, 15, 29, 32, 36] as Niho bent functions. In [3, 4, 5, 18, 37] these investigations were extended to other types of spreads, and bent functions which are affine on the elements of spreads, were studied.

Carlet and Mesnager showed [15] that any bent function which is linear on the elements of a Desarguesian spread (they are equivalent to Niho bent functions in a bivariate form) determines an o-polynomial (oval polynomial). Every o-polynomial defines a hyperoval, therefore Carlet and Mesnager revealed a general connection between Niho bent functions and hyperovals in Desarguesian planes of even order.

However there are several equivalence classes of Niho bent functions associated with a fixed hyperoval. In [12, 13] there were constructed several new Niho bent functions associated with some o-polynomials and hyperovals. In this paper we address the question of finding all equivalence classes of Niho bent functions corresponding to a hyperoval. This question was mentioned as Open Problem 6 in [16]

. We describe equivalence classes of Niho bent functions for all known types of hyperovals. We show that the hyperconics, irregular translation hyperovals, Segre and Glynn hyperovals have respectively 2, 3, 4 and 4 equivalence classes of Niho bent functions (excluding some exceptional cases in small dimensions) and describe these classes. The Lunelli-Sce hyperoval has one class and the O’Keefe-Penttila hyperoval has 12 classes. For the Payne, Cherowitzo, Subiaco and Adelaide hyperovals the number of equivalence classes of associated Niho bent functions increases exponentially as the dimension of the underlying vector space grows. Note that hyperovals are not classified yet, and the list of known hyperovals can be found in

[19, 20, 21].

In [3, 6]

it was shown that bent functions linear on the elements of a Desarguesian spread are in one-to-one correspondence with line ovals in an affine plane. Points of the line oval completely define the dual bent function. More precisely, the zeros of the dual function of a Niho bent function are exactly the points of the line oval (in other words, the dual function of a Niho bent function is obtained from the characteristic function of the set of points of the line oval by adding all-one constant function). Hence this gives a general formula

[3, 6] for the dual function for any Niho bent function (this question was mentioned as Open Problem 10 in [16]). Furthermore, Niho bent functions are in one-to-one correspondence with ovals with nucleus at a designated point. Therefore, we have geometric characterization of Niho bent functions and of their duals.

In [3, 6] special -functions were introduced as a new analog of o-polynomials in case of univariate presentation of functions and hyperovals. Using -functions allow us to implement methods which do not employ the “unusual magic action” defined by O’Keefe and Penttila [42]. A criteria for existence of -functions is presented in [6].

We note that there are no analogs of such bent functions in case of odd characteristic: Çeşmelioğlu, Meidl and Pott

[18] showed that bent functions which are affine on elements of Desarguesian spreads over a field of odd characteristic will be constant on the elements of the spread.

The paper is organized as follows. We recall first in Section 2 definitions and notation concerning bent functions, ovals and line ovals. In Section 3 we study connections between -functions and o-polynomials, and establish general formulas for all -functions and Niho bent functions corresponding to a fixed hyperoval. Theorem 3.1 gives a formula for calculation of the -function corresponding to a given o-polynomial. In Section 4 we study equivalence classes of Niho bent functions corresponding to all known hyperovals. Finally, in dimensions up to the equivalence classes were considered in detail.

## 2 Preliminary considerations and notation

In this section we recall some definitions and notation.

### 2.1 Bent functions

Let and be finite fields of orders and respectively. Let be an -vector space of dimension . We shall endow with the structure of the field . A Boolean function on is a mapping from to the prime field .

If is a Boolean function defined on , then the Walsh transform of is defined as follows:

 Wf(b)=∑x∈F2n(−1)f(x)+b⋅x, (1)

where is a scalar product from to . A Boolean function on is said to be bent if its Walsh transform satisfies for all . Then is an even integer.

Given a bent function over , one can define its dual function, denoted by , by considering the signs of the values of the Walsh transform of . More precisely, is defined by the equation:

 (−1)~f(x)2n/2=Wf(x).

The dual of a bent function is bent again, and .

### 2.2 Polar representations

Let be a finite field of order . Consider as a subfield of , where , so is a two dimensional vector space over . Let be a prime field.

As usually, is the trace function with respect to a finite field extension . For particular field extensions we denote the corresponding trace functions by

 Tr(x)=TrK/F0(x),T(x)=TrK/F(x),tr(x)=TrF/F0(x).

The conjugate of over is

 ¯x=xq.

Then the trace and the norm maps from to are

 T(x)=TrK/F=x+¯x=x+xq,
 N(x)=NK/F(x)=x¯x=x1+q.

The unit circle of is the set of elements of norm :

 S={u∈K:u¯u=1}.

Therefore, is the multiplicative group of st roots of unity in . Since , each non-zero element of has a unique polar coordinate representation

 x=λu

with and . For any we have

 λ=√x¯x,
 u=√x/¯x.

One can define nondegenerate bilinear form by

 ⟨x,y⟩=T(x¯y)=x¯y+¯xy.

Then the form is alternating and symmetric:

 ⟨a,a⟩=0,
 ⟨a,b⟩=⟨b,a⟩.

### 2.3 Affine and projective planes

Consider points of a projective plane in homogeneous coordinates as triples , where , , and we identify with , . Then points of are

 {(x:y:1)∣x∈F, y∈F}∪{(x:1:0)∣x∈F}∪{(1:0:0)},

For , , the line in is defined as

 [a:b:c]={(x:y:z)∈PG(2,q)∣ax+by+cz=0}.

Triples and with define same lines. Then any line in is one of the following lines:

 [a:b:1] = {(x:y:1)∣ax+by+1=0}∪{(b:a:0)}, (a,b)≠(0,0), [0:0:1] = {(x:1:0)∣x∈F}∪{(1:0:0)}, [a:1:0] = {(x:ax:1)∣x∈F}∪{(1:a:0)}, a∈F, [1:0:0] = {(0:y:1)∣y∈F}∪{(0:1:0)}.

The point is incident with the line if and only if . The map defines a duality [33] for .

We shall call points of the form the points at infinity. Then indicates the line at infinity (it consists of all points at infinity). We define an affine plane , so points of this affine plane are

 {(x:y:1)∣x,y∈F}.

Associating with we can identify points of the affine plane with elements of the vector space

 V(2,q)={(x,y)∣x,y∈F},

and we will write .

Lines in are and , . These lines can be described by equations and .

We introduce now other representation of using the field . Consider pairs , where , , or , and we identify with , . Then points of are

 {(x:1)∣ x∈K}∪{(u:0)∣u∈S}.

For and we define lines in as

 [α:β]={(x:z)∈PG(2,q)∣⟨α,x⟩+βz=0}.

Pairs and with define same lines. Then any line in is one of the following lines:

 [α:1] = {(x:1)∣⟨α,x⟩+1=0}∪{(α:0)}, α≠0, [0:1] = {(u:0)∣u∈S}, [u:0] = {(λu:1)∣λ∈F}∪{(u:0)}, u∈S.

The point is incident with the line if and only if . The map defines a duality [33] for .

Element , , will be referred to as the point at infinity in the direction of . So indicates the line at infinity.

We define an affine plane , so points of this affine plane are

 {(x:1)∣x∈K}.

Associating with we can identify points of the affine plane with elements of the field , and we will write .

Lines of can be considered as the zeroes of an equation . Normalizing to , we see that lines of can be considered as the zeroes of an equation :

 L(u,μ)={x∈K:⟨u,x⟩+μ=0},

where and (see for details [7, subsection 2.1]). can be considered as a line in the direction of . Note that there are such lines. Lines and are parallel, and . Lines and are not parallel for distinct , since is a nondegenerate alternating bilinear form on , considered as a two dimensional vector space over .

Throughout the paper, we will consider these two representations of projective planes , and for each of such projective planes we consider a fixed affine plane , described above. They will be written as and .

### 2.4 Ovals and line ovals

Let be a finite projective plane of order . An oval is a set of points, no three of which are collinear. Dually, a line oval is a set of lines no three of which are concurrent. Any line of the plane meets the oval at either 0, 1 or 2 points and is called exterior, tangent or secant, respectively. All the tangent lines to the oval concur [33] at the same point , called the nucleus of . The set becomes a hyperoval, that is a set of points, no three of which are collinear. Conversely, by removing any point from hyperoval one gets an oval. If is a line oval, then there is exactly one line such that on each of its points there is only one line of . This line is called the (dual) nucleus of . The -set is a line hyperoval or dual hyperoval.

By a line oval in an affine plane we assume a set of nonparallel lines in , such that these lines, extended by the corresponding points at infinity, determine a line oval in whose nucleus is the line at infinity.

For any oval in there are secants and exterior lines. Let be a line oval in the affine plane and the set of points which are on the lines of the line oval :

 E(O)={(x,y)∈V(2,q)∣(x,y) is on a line of O}.

Then each point of belongs to two lines of ,

 |E(O)|=q(q+1)/2

and there are points in that do not belong to (see, for example, [25, 33]). Note that Kantor [34, Theorem 7] showed that is a difference set.

Any line oval in can be represented [6] as for some function .

### 2.5 Niho bent functions

Desarguesian spread on is a collection of one-dimensional subspaces , , and . So every nonzero element of lies in a unique subspace [25].

The field can be considered as a two dimensional vector space over . Then the set

 {uF:u∈S}

is a spread on this vector space. We consider Boolean functions , which are -linear on each element of the spread . Then for any , where , , the function can be defined by

 f(x)=tr(λg(u)) (2)

for some function . If such a function is bent then it is called Niho bent function. These functions can be written with the help of Niho power functions, so is the name.

Let Boolean function be linear on elements of the spread . Now we consider bent functions, and we use the scalar product in the formula (1) for the Walsh transform (as it was introduced in [6]). The following result was proved in [6]:

###### Theorem 2.1.

Let function be defined by , for some function . Then the following statements are equivalent:

1. The function is bent;

2. The set is a line oval in ;

3. The set is an oval in whose nucleus is the origin.

Here we assume that if then is an element at infinity in the direction .

Therefore, Niho bent functions are in one-to-one correspondence with the ovals in with nucleus at the origin. A function is said to be a -function if it satisfies conditions of Theorem 2.1.

Boolean functions are extended-affine equivalent if there exist an affine permutation of and an affine function such that . In our paper we will call such functions just equivalent. If Boolean functions and are equivalent and is bent then is bent too.

We recall that the automorphism (collineation) group of is and the automorphism group of is , where is the Galois group of , and is the affine group. Hyperovals (and ovals) are called (projectively) equivalent if they are equivalent under the action of the group .

Niho bent functions are equivalent if and only if the corresponding ovals (with nucleus at the origin) are projectively equivalent [3, 46]. Note also that functions and , where , lead to equivalent ovals [6].

###### Remark 2.2.

In thesis [26], following ideas from [30], the -polynomials were introduced. We note that the -polynomials and our -functions are connected in the following way: .

## 3 O-polynomials, g-functions and Niho bent functions

Following [30], consider an element with property . Then and is a root of a quadratic equation

 z2+z+δ=0,

where . Any element can be represented as , where . For we have and .

Note that if is odd then one can choose , . So if is a generator of then we can take .

Every hyperoval is equivalent to one, given by

 D(h)={(t:h(t):1)∣t∈F}∪{(1:0:0)}∪{(0:1:0)},

where is an o-polynomial. We can describe ovals in a similar way, but ensuring that the nucleus of the oval is the point :

 E(h)={(t:h(t):1)∣t∈F}∪{(0:1:0)}.

Now we would like to find an equivalent representation of this oval in using the field , namely, we would like to write the oval in the form

 {ug(u)∣u∈S}

for some function .

###### Theorem 3.1.

Let be an o-polynomial. Define an oval with nucleus by

 E(h)={(t:h(t):1)∣t∈F}∪{(0:1:0)}.

Then corresponding function can be determined by

 g(u)=h−1(⟨i,u⟩⟨1,u⟩)⟨1,u⟩+⟨i,u⟩,g(1)=1.
###### Proof.

We apply a collineation to the hyperoval in order to get a hyperoval without points on infinity, such that the point is mapped to and the point is mapped to . Define

 α((a:b:c))=(b:c:(a+b+cd))

for some . Then , and . We want to have for all . It happens if we choose in such a way that the line does not intersect the hyperoval .

Now we associate the point with the point . Then corresponds to and corresponds to . Assume that the oval corresponds to the oval . If then and . Considering points

 (tt+h−1(t)+d:1t+h−1(t)+d:1)

from the oval we see that

 tt+h−1(t)+d=1g(u)⟨i,u⟩,
 1t+h−1(t)+d=1g(u)⟨1,u⟩.

Therefore,

 t=⟨i,u⟩⟨1,u⟩,
 g(u)=(t+h−1(t)+d)⋅⟨1,u⟩=h−1(⟨i,u⟩⟨1,u⟩)⟨1,u⟩+⟨i,u⟩+⟨d,u⟩.

One has since and the equality implies , . Finally, we can subtract from the linear part , since adding linear function to will produce an equivalent function. ∎

###### Remark 3.2.

In place of the function one can consider , they produce equivalent ovals and equivalent Niho bent functions, but in this case one has .

###### Corollary 3.3.

Let be an o-polynomial. Then its corresponding -function and Niho bent function can be written respectively as

 g(u)=⟨i,u⟩s−1⋅⟨1,u⟩q−s−1,
 f(x)=tr(⟨i,x⟩s−1⋅⟨1,x⟩q−s−1)=tr((¯ix+i¯x)s−1⋅(x+¯x)q−s−1),

where is the inverse of modulo .

###### Proof.

We have

 g(u)=h−1(⟨i,u⟩⟨1,u⟩)⟨1,u⟩=(⟨i,u⟩⟨1,u⟩)s−1⟨1,u⟩=⟨i,u⟩s−1⋅⟨1,u⟩q−s−1.

Further, since , , , we have

 f(x) = tr(λg(u)) = tr(√x¯x⋅⟨i,√x/¯x⟩s−1⋅⟨1,√x/¯x⟩q−s−1) = tr(√x¯x⋅(¯i√x/¯x+i√¯x/x)s−1⋅(√x/¯x+√¯x/x)q−s−1) = tr⎛⎜⎝√x¯x⋅(¯ix+i¯x√¯xx)s−1⋅(¯ix+i¯x√¯xx)q−s−1⎞⎟⎠ = tr((¯ix+i¯x)s−1⋅(x+¯x)q−s−1),

which completes the proof. ∎

###### Remark 3.4.

In place of the oval one can consider the oval with nucleus . Then corresponding -function can be written as , . However this approach is less convenient for our reasonings in Section 4.

Next we derive formulas for Niho bent functions and -functions obtained from ovals in .

###### Theorem 3.5.

Let be an oval in whose nucleus is the origin. Then the associated Niho bent function is

 f(x)=m−1∑j=0q∑i=0(∑v∈O1vi(q−1)+2j)xi(q−1)+2j.
###### Proof.

By [3, Theorem 3.8] we have

It is clear that . Let . Then

 f(x) = ∑v∈O[(1xq−1+1vq−1)q2−1+1]m−1∑j=0(x/v)2j = ∑v∈O[(xq−1+vq−1xq−1vq−1)q2−1+1]m−1∑j=0(x/v)2j = ∑v∈O[(xq−1+vq−1)q2−1+1]m−1∑j=0(x/v)2j = ∑v∈O[q2−1∑i=0((xq−1)i(vq−1)q2−1−i+1]m−1∑j=0(x/v)2j.

We note that some terms of the second sum are the same:

 (xq−1)k(q+1)+i=(xq−1)i,(vq−1)k(q+1)+i=(vq−1)i.

Therefore,

 f(x) = ∑v∈O[q2−1∑i=0((xq−1)i(vq−1)q2−1−i+1]m−1∑j=0(x/v)2j = ∑v∈O[q∑i=0((xq−1)i(vq−1)q+1−i]m−1∑j=0(x/v)2j = ∑v∈Oq∑i=0(((x/v)q−1)im−1∑j=0(x/v)2j = m−1∑j=0q∑i=0(∑v∈O1vi(q−1)+2j)xi(q−1)+2j,

which establishes the formula. ∎

###### Theorem 3.6.

Let be an oval in whose nucleus is the origin. Then

 g(u)=q∑i=0∑v∈Ov(q−1)i/2−1ui+1.
###### Proof.

If then for some . Then we have

 v(q−1)/2=(z/g(z))(q−1)/2=z(q−1)/2=¯z.

Now fix . If , and then

 q∑i=0(uv(q−1)/2))i uv=q∑i=0(u¯z)i uv=uv=g(u).

If , and then

 q∑i=0(uv(q−1)/2))i uv=q∑i=0(u¯z)i uv=0.

Therefore,

 g(u) = ∑v∈Oq∑i=0(uv(q−1)/2))i uv = q∑i=0∑v∈Ov(q−1)i/2−1ui+1,

which proves the theorem. ∎

Given a hyperoval in a projective plane , an oval can be obtained by deleting one of the points of the hyperoval. This deleted point is the nucleus of the resulting oval. There are ovals which can be obtained in this way, but some of them will be equivalent under the action of the automorphism group of . If and are points of a hyperoval , then ovals and are equivalent if and only if and lie in the same orbit of the stabilizer of on . (The stabilizer of in is also called automorphism group of .) Therefore, the number of projectively inequivalent ovals obtained from hyperoval is equal to the number of orbits of under the action of the stabilizer of .

Assume that for all . Function determines an hyperoval

 H={ug(u)∣u∈S}∪{0}

in . Consider an oval , , whose nucleus is , and shift it by the element in order to get an oval , whose nucleus is the origin:

Any Niho bent function associated with the hyperoval is equivalent to a Niho bent function obtained from one of the ovals , and two such bent functions and are equivalent [3, 46] if and only if the points and are in the same orbit under action of automorphism group of the hyperoval .

###### Theorem 3.7.

Let be a hyperoval in defined by a function . Then the function associated with the oval , whose nucleus is the origin, is equal to

 gs(u)=q∑i=0aiui+1,
###### Proof.

From Theorem 3.6 we have

 ai = ∑z∈Osz(q−1)i/2−1 = (sg(s))(q−1)i/2−1+∑v∈S,v≠s(vg(v)+sg(s))(q−1)i/2−1 = (sg(s))(q−1)i/2−1+∑v∈S,v≠s(g(s)v+sg(v)g(s)g(v))(q−1)i/2−1 = g(s)s(q−1)i/2−1+g(s)∑v∈S,v≠sg(v)(g(s)v+sg(v))(q−1)i/2−1,

which is our claim. ∎

###### Theorem 3.8.

Let be a hyperoval in defined by a function . Then the Niho bent function associated with the oval , whose nucleus is the origin, is equal to

 fs(x)=m−1∑j=0q∑i=0⎛⎝g(s)2jsi(q−1)+2j+∑v∈S,v≠sg(s)2jg(v)2j(g(s)v+sg(v))i(q−1)+2j⎞⎠xi(q−1)+2j.
###### Proof.

From Theorem 3.5 we have

 fs(x) = m−1∑j=0q∑i=0(∑z∈O1zi(q−1)+2j