The theory of three-way decision is built on the philosophy of thinking in threes. The essence of three-way decision is trisecting the whole and taking different strategies for different parts accordingly. The theory of three-way decision has been successfully implemented to diverse fields since it provides an elegant and efficient solution for solving complicated problems. In this paper, a useful representation for hesitant fuzzy sets is obtained by means of canonical soft sets. We also define unit interval parameterized soft sets and their derived hesitant fuzzy sets. Mutual representations and inner connections between hesitant fuzzy sets and soft sets are examined. With the help of soft rough sets, a generalized rough model based on hesitant fuzzy sets is established. A novel three-way decision method is presented for solving decision-making problems by means of hesitant fuzzy sets and canonical soft sets. Finally, a numerical example regarding peer review of research articles is given to illustrate the validity and efficacy of the proposed method.
Citation: Feng Feng, Zhe Wan, José Carlos R. Alcantud, Harish Garg. Three-way decision based on canonical soft sets of hesitant fuzzy sets[J]. AIMS Mathematics, 2022, 7(2): 2061-2083. doi: 10.3934/math.2022118
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The theory of three-way decision is built on the philosophy of thinking in threes. The essence of three-way decision is trisecting the whole and taking different strategies for different parts accordingly. The theory of three-way decision has been successfully implemented to diverse fields since it provides an elegant and efficient solution for solving complicated problems. In this paper, a useful representation for hesitant fuzzy sets is obtained by means of canonical soft sets. We also define unit interval parameterized soft sets and their derived hesitant fuzzy sets. Mutual representations and inner connections between hesitant fuzzy sets and soft sets are examined. With the help of soft rough sets, a generalized rough model based on hesitant fuzzy sets is established. A novel three-way decision method is presented for solving decision-making problems by means of hesitant fuzzy sets and canonical soft sets. Finally, a numerical example regarding peer review of research articles is given to illustrate the validity and efficacy of the proposed method.
The key to solving the general quadratic congruence equation is to solve the equation of the form x2≡amodp, where a and p are integers, p>0 and p is not divisible by a. For relatively large p, it is impractical to use the Euler criterion to distinguish whether the integer a with (a,p)=1 is quadratic residue of modulo p. In order to study this issue, Legendre has proposed a new tool-Legendre's symbol.
Let p be an odd prime, the quadratic character modulo p is called the Legendre's symbol, which is defined as follows:
(ap)={1, if a is a quadratic residue modulo p;−1, if a is a quadratic non-residue modulo p;0, if p∣a. |
The Legendre's symbol makes it easy for us to calculate the level of quadratic residues. The basic properties of Legendre's symbol can be found in any book on elementary number theory, such as [1,2,3].
The properties of Legendre's symbol and quadratic residues play an important role in number theory. Many scholars have studied them and achieved some important results. For examples, see the [4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21].
One of the most representative properties of the Legendre's symbol is the quadratic reciprocal law:
Let p and q be two distinct odd primes. Then, (see Theorem 9.8 in [1] or Theorems 4–6 in [3])
(pq)⋅(qp)=(−1)(p−1)(q−1)4. |
For any odd prime p with p≡1mod4 there exist two non-zero integers α(p) and β(p) such that
p=α2(p)+β2(p). | (1) |
In fact, the integers α(p) and β(p) in the (1) can be expressed in terms of Legendre's symbol modulo p (see Theorems 4–11 in [3])
α(p)=12p−1∑a=1(a3+ap)andβ(p)=12p−1∑a=1(a3+rap), |
where r is any integer, and (r,p)=1, (rp)=−1, (∗p)=χ2 denote the Legendre's symbol modulo p.
Noting that Legendre's symbol is a special kind of character. For research on character, Han [7] studied the sum of a special character χ(ma+ˉa), for any integer m with (m,p)=1, then
|p−1∑a=1χ(ma+ˉa)|2=2p+(mp)p−1∑a=1χ(a)p−1∑b=1(b(b−1)(a2b−1)p), |
which is a special case of a general polynomial character sums ∑N+Ma=N+1χ(f(a)), where M and N are any positive integers, and f(x) is a polynomial.
In [8], Du and Li introduced a special character sums C(χ,m,n,c;p) in the following form:
C(χ,m,n,c;p)=p−1∑a=0p−1∑b=0χ(a2+na−b2−nb+c)⋅e(mb2−ma2p), |
and studied the asymptotic properties of it. They obtained
p−1∑c=1|C(χ,m,n,c;p)|2k={p2k+1+k2−3k−22⋅p2k+O(p2k−1),ifχ is the Legendre symbol modulo p;p2k+1+k2−3k−22⋅p2k+O(p2k−1/2),ifχ is a complex character modulo p. |
Recently, Yuan and Zhang [12] researched the question about the estimation of the mean value of high-powers for a special character sum modulo a prime, let p be an odd prime with p≡1mod6, then for any integer k≥0, they have the identity
Sk(p)=13⋅[dk+(−d+9b2)k+(−d−9b2)k], |
where
Sk(p)=1p−1p−1∑r=1Ak(r), |
A(r)=1+p−1∑a=1(a2+rˉap), |
and for any integer r with (r,p)=1.
More relevant research on special character sums will not be repeated. Inspired by these papers, we have the question: If we replace the special character sums with Legendre's symbol, can we get good results on p≡1mod4?
We will convert β(p) to another form based on the properties of complete residues
β(p)=12p−1∑a=1(a+nˉap), |
where ˉa is the inverse of a modulo p. That is, ˉa satisfy the equation x⋅a≡1modp for any integer a with (a,p)=1.
For any integer k≥0, G(n) and Kk(p) are defined as follows:
G(n)=1+p−1∑a=1(a2+nˉa2p)andKk(p)=1p−1p−1∑n=1Gk(n). |
In this paper, we will use the analytic methods and properties of the classical Gauss sums and Dirichlet character sums to study the computational problem of Kk(p) for any positive integer k, and give a linear recurrence formulas for Kk(p). That is, we will prove the following result.
Theorem 1. Let p be an odd prime with p≡1mod4, then we have
Kk(p)=(4p+2)⋅Kk−2(p)−8(2α2−p)⋅Kk−3(p)+(16α4−16pα2+4p−1)⋅Kk−4(p), |
for all integer k≥4 with
K0(p)=1,K1(p)=0,K2(p)=2p+1,K3(p)=−3(4α2−2p), |
where
α=α(p)=p−12∑a=1(a+ˉap). |
Applying the properties of the linear recurrence sequence, we may immediately deduce the following corollaries.
Corollary 1. Let p be an odd prime with p≡1mod4. Then we have
1p−1p−1∑n=111+∑p−1a=1(a2+nˉa2p)=16α2p−28α2−8p2+14p16α4−16α2p+4p−1. |
Corollary 2. Let p be an odd prime with p≡1mod4. Then we have
1p−1p−1∑n=1p−1∑m=0(1+p−1∑a=1(a2+nˉa2p))⋅e(nm2p)=−√p. |
Corollary 3. Let p be an odd prime with p≡1mod4. Then we have
1p−1p−1∑n=1p−1∑m=0[1+p−1∑a=1(a2+nˉa2p)]2⋅e(nm2p)=(4α2−2p)⋅√p. |
Corollary 4. Let p be an odd prime with p≡1mod8. Then we have
p−1∑n=1(1+p−1∑a=1(a2+nˉa2p))⋅p−1∑m=0e(nm4p)=√p(−1+B(1))−p, |
where
B(1)=p−1∑m=0e(m4p). |
If we consider such a sequence Fk(p) as follows: Let p be a prime with p≡1mod8, χ4 be any fourth-order character modulo p. For any integer k≥0, we define the Fk(p) as
Fk(p)=p−1∑n=11Gk(n), |
we have
Fk(p)=116α4−16α2p+4p−1Fk−4(p)−(4p+2)16α4−16α2p+4p−1Fk−2(p)+4(4α2−2p)16α4−16α2p+4p−1Fk−1(p). |
Lemma 1. Let p be an odd prime with p≡1mod4. Then for any fourth-order character χ4modp, we have the identity
τ2(χ4)+τ2(¯χ4)=2√p⋅α, |
where
τ(χ4)=p−1∑a=1χ4(a)e(ap) |
denotes the classical Gauss sums, e(y)=e2πiy,i2=−1, and α is the same as in the Theorem 1.
Proof. See Lemma 2.2 in [9].
Lemma 2. Let p be an odd prime. Then for any non-principal character ψ modulo p, we have the identity
τ(ψ2)=ψ2(2)τ(χ2)⋅τ(ψ)⋅τ(ψχ2), |
where χ2=(∗p) denotes the Legendre's symbol modulo p.
Proof. See Lemma 2 in [12].
Lemma 3. Let p be a prime with p≡1mod4, then for any integer n with (n,p)=1 and fourth-order character χ4modp, we have the identity
p−1∑a=1(a2+nˉa2p)=−1−χ2(n)+1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4)). |
Proof. For any integer a with (a,p)=1, we have the identity
1+χ4(a)+χ2(a)+¯χ4(a)=4, |
if a satisfies a≡b4modp for some integer b with (b,p)=1 and
1+χ4(a)+χ2(a)+¯χ4(a)=0, |
otherwise. So from these and the properties of Gauss sums we have
p−1∑a=1(a2+nˉa2p)=p−1∑a=1(a2p)(a4+np)=p−1∑a=1χ2(a4)χ2(a4+n)=p−1∑a=1(1+χ4(a)+χ2(a)+¯χ4(a))⋅χ2(a)⋅χ2(a+n)=p−1∑a=1(1+χ4(na)+χ2(na)+¯χ4(na))⋅χ2(na)⋅χ2(na+n)=p−1∑a=1χ2(a)χ2(a+1)+p−1∑a=1χ4(na)χ2(a)χ2(a+1) | (2) |
+p−1∑a=1χ2(na)χ2(a)χ2(a+1)+p−1∑a=1¯χ4(na)χ2(a)χ2(a+1)=p−1∑a=1χ2(1+ˉa)+p−1∑a=1χ4(na)χ2(a)χ2(a+1)+p−1∑a=1χ2(n)χ2(a+1)+p−1∑a=1¯χ4(na)χ2(a)χ2(a+1). |
Noting that for any non-principal character χ,
p−1∑a=1χ(a)=0 |
and
p−1∑a=1χ(a)χ(a+1)=1τ(ˉχ)p−1∑b=1p−1∑a=1ˉχ(b)χ(a)e(b(a+1)p). |
Then we have
p−1∑a=1χ2(1+ˉa)=−1,p−1∑a=1χ2(a+1)=−1, |
p−1∑a=1χ4(a)χ2(a)χ2(a+1)=1τ(χ2)p−1∑b=1p−1∑a=1χ2(b)χ4(a)χ2(a)e(b(a+1)p)=1τ(χ2)p−1∑b=1¯χ4(b)e(bp)p−1∑a=1χ4(ab)χ2(ab)e(abp) | (3) |
=1τ(χ2)⋅τ(¯χ4)⋅τ(χ4χ2). |
For any non-principal character ψ, from Lemma 2 we have
τ(ψ2)=ψ2(2)τ(χ2)⋅τ(ψ)⋅τ(ψχ2). | (4) |
Taking ψ=χ4, note that
τ(χ2)=√p, τ(χ4)⋅τ(¯χ4)=χ4(−1)⋅p, |
from (3) and (4), we have
p−1∑a=1χ4(a)χ2(a)χ2(a+1)=¯χ42(2)⋅τ(χ24)⋅τ(χ2)⋅τ(¯χ4)τ(χ2)⋅τ(χ4)=χ2(2)⋅τ(χ2)⋅τ2(¯χ4)τ(χ4)⋅τ(¯χ4)=χ2(2)⋅√p⋅τ2(¯χ4)χ4(−1)⋅p | (5) |
=χ2(2)⋅τ2(¯χ4)χ4(−1)⋅√p. |
Similarly, we also have
p−1∑a=1¯χ4(a)χ2(a)χ2(a+1)=χ2(2)⋅τ2(χ4)χ4(−1)⋅√p. | (6) |
Consider the quadratic character modulo p, we have
(2p)=χ2(2)={1,if p≡±1mod8;−1,if p≡±3mod8. | (7) |
And when p≡1mod8, we have χ4(−1)=1; when p≡5mod8, we have χ4(−1)=−1. Combining (2) and (5)–(7) we can deduce that
p−1∑a=1(a2+nˉa2p)=−1−χ2(n)+1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4)). |
This prove Lemma 3.
Lemma 4. Let p be an odd prime with p≡1mod4. Then for any integer k≥4 and n with (n,p)=1, we have the fourth-order linear recurrence formula
Gk(n)=(4p+2)⋅Gk−2(n)+8(p−2α2)⋅Gk−3(n)+[(4α2−2p)2−(2p−1)2]⋅Gk−4(n), |
where
α=α(p)=12p−1∑a=1(a3+ap)=p−12∑a=1(a+ˉap), |
(∗p)=χ2 denotes the Legendre's symbol.
Proof. For p≡1mod4, any integer n with (n,p)=1, and fourth-order character χ4 modulo p, we have the identity
χ44(n)=¯χ44(n)=χ0(n), χ24(n)=χ2(n), |
where χ0 denotes the principal character modulo p.
According to Lemma 3,
p−1∑a=1(a2+nˉa2p)=−1−χ2(n)+1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4)), |
G(n)=1+p−1∑a=1(a2+nˉa2p). |
We have
G(n)=−χ2(n)+1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4)), | (8) |
G2(n)=[−χ2(n)+1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4))]2=1−2χ2(n)⋅1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4))+1p⋅(χ2(n)⋅τ4(¯χ4)+χ2(n)⋅τ4(χ4)+2p2)=1−2χ2(n)⋅1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4))+1p⋅(χ2(n)⋅(τ4(¯χ4)+τ4(χ4))+2p2). |
According to Lemma 1, we have
(τ2(χ4)+τ2(¯χ4))2=τ4(¯χ4)+τ4(χ4)+2p2=4pα2. |
Therefore, we may immediately deduce
G2(n)=1−2(χ2(n)⋅(G(n)+χ2(n))+1p(χ2(n)⋅(τ4(¯χ4)+τ4(χ4))+2p2)=1−2χ2(n)⋅(G(n)+χ2(n)) | (9) |
+1p⋅[χ2(n)((τ2(¯χ4)+τ2(χ4))2−2p2)+2p2]=2p−1−2χ2(n)⋅G(n)+(4α2−2p)⋅χ2(n), |
G3(n)=[−χ2(n)+1√p⋅(χ4(n)⋅τ2(¯χ4)+¯χ4(n)⋅τ2(χ4))]3=(2p−1−2χ2(n)⋅G(n)+(4α2−2p)⋅χ2(n))⋅G(n) | (10) |
=(4α2−2p)χ2(n)⋅G(n)+(2p+3)G(n)−(4p−2)χ2(n)−2(4α2−2p) |
and
[G2(n)−(2p−1)]2=[χ2(n)⋅(4α2−2p)−2χ2(n)⋅G(n)]2, |
which implies that
G4(n)=(4p+2)⋅G2(n)+8(p−2α2)⋅G(n)+[(4α2−2p)2−(2p−1)2]. | (11) |
So for any integer k≥4, from (8)–(11), we have the fourth-order linear recurrence formula
Gk(n)=Gk−4(n)⋅G4(n)=(4p+2)⋅Gk−2(n)+8(p−2α2)⋅Gk−3(n)+[(4α2−2p)2−(2p−1)2]⋅Gk−4(n). |
This proves Lemma 4.
In this section, we will complete the proof of our theorem.
Let p be any prime with p≡1mod4, then we have
K0(p)=1p−1p−1∑n=1G0(n)=p−1p−1=1. | (12) |
K1(p)=1p−1p−1∑n=1G1(n)=1p−1p−1∑n=1(−χ2(n)+1√p⋅(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)))=0, | (13) |
K2(p)=1p−1p−1∑n=1G2(n)=1p−1p−1∑n=1(−χ2(n)+1√p⋅(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)))2=2p+1, | (14) |
K3(p)=1p−1p−1∑n=1G3(n)=1p−1p−1∑n=1(−χ2(n)+1√p⋅(χ4(n)τ2(¯χ4)+¯χ4(n)τ2(χ4)))3=−3(4α2−2p). | (15) |
It is clear that from Lemma 4, if k≥4, we have
Kk(p)=1p−1p−1∑n=1Gk(n)=(4p+2)⋅Kk−2(p)−8(2α2−p)⋅Kk−3(p)+(16α4−16pα2+4p−1)⋅Kk−4(p). | (16) |
Now Theorem 1 follows (12)–(16). Obviously, using Theorem 1 to all negative integers, and that lead to Corollary 1.
This completes the proofs of our all results.
Some notes:
Note 1: In our theorem, know n is an integer, and (n,p)=1. According to the properties of quadratic residual, χ2(n)=±1, χ4(n)=±1.
Note 2: In our theorem, we only discussed the case p≡1mod8. If p≡3mod4, then the result is trivial. In fact, in this case, for any integer n with (n,p)=1, we have the identity
G(n)=1+p−1∑a=1(a2+nˉa2p)=1+p−1∑a=1(a4p)⋅(a4+np)=1+p−1∑a=1(ap)⋅(a+np)=1+p−1∑a=1(a2+nap)=1+p−1∑a=1(1+nˉap)=p−1∑a=0(1+nap)=0. |
Thus, for all prime p with p≡3mod4 and k≥1, we have Kk(p)=0.
The main result of this paper is Theorem 1. It gives an interesting computational formula for Kk(p) with p≡1mod4. That is, for any integer k, we have the identity
Kk(p)=(4p+2)⋅Kk−2(p)−8(2α2−p)⋅Kk−3(p)+(16α4−16pα2+4p−1)⋅Kk−4(p). |
Thus, the problems of calculating a linear recurrence formula of one kind special character sums modulo a prime are given.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors are grateful to the anonymous referee for very helpful and detailed comments.
This work is supported by the N.S.F. (11971381, 12371007) of China and Shaanxi Fundamental Science Research Project for Mathematics and Physics (22JSY007).
The authors declare no conflicts of interest.
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