Existentially closed graphs via permutation polynomials over finite fields

Existentially closed graphs via permutation polynomials over finite fields tài liệu, giáo án, bài giảng , luận văn, luận...

Discrete Applied Mathematics ( ) – Contents lists available at ScienceDirect Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam Existentially closed graphs via permutation polynomials over finite fields Nguyen Minh Hai a , Tran Dang Phuc a , Le Anh Vinh b,∗ a Faculty of Mathematics, Mechanics and Informatics, Hanoi University of Science, Vietnam National University, Hanoi, Viet Nam b University of Education, Vietnam National University, Hanoi, Viet Nam article info Article history: Received 28 November 2014 Received in revised form 17 June 2015 Accepted 18 May 2016 Available online xxxx Keywords: Existentially closed graphs Permutation polynomial Distance graphs abstract For a positive integer n, a graph is n-existentially closed or n-e.c if we can extend all n-subsets of vertices in all possible ways It is known that almost all finite graphs are n-e.c Despite this result, until recently, only few explicit examples of n-e.c graphs are known for n > In this paper, we construct explicitly a 4-e.c graph via a linear map over finite fields We also study the colored version of existentially closed graphs and construct explicitly many (3, t )-e.c graphs via permutation polynomials and multiplicative groups over finite fields © 2016 Published by Elsevier B.V Introduction For a positive integer n, a graph is n-existentially closed or n-e.c if we can extend all n-subsets of vertices in all possible ways Precisely, for every pair of subsets A, B of vertex set V of the graph such that A ∩ B = ∅ and |A| + |B| = n, there is a vertex z not in A ∪ B that joined to each vertex of A and no vertex of B From the results of Erdős and Rényi [4], almost all finite graphs are n-e.c Despite this result, until recently, only few explicit examples of n-e.c graphs are known for n > See [2] for a comprehensive survey on the constructions of n-e.c graphs In [13], the third listed author studied a multicolor version of this adjacency property Let n, t be positive integers A t-edge-colored graph G is (n, t )-e.c or (n, t )-existentially closed if for any t disjoint sets of vertices A1 , , At with |A1 | + · · · + |At | = n, there is a vertex x not in A1 ∪ · · · ∪ At such that all edges from this vertex to the set Ai are colored by the ith color Since the complement of a graph can be viewed as a color class, the usual definition of n-e.c graphs is the special case of t = For a positive integer N, the probability space Gt (N , 1t ) consists of all t-colorings of the complete graph of order N such that each edge is colored independently by any color with the probability 1t The third listed author showed [13, Theorem 1.1] that almost all graphs in Gt (N , 1t ) have the property (n, t )-e.c as N → ∞ The proof of this theorem is similar to the proof that almost all finite graphs have n-e.c property (see, for example, [4]) Although this result implies that there are many (n, t )-e.c graphs, it is nontrivial to construct such graphs The third listed author [13, theorem 1.2] constructed explicitly many graphs satisfying this condition Let q be an odd prime power and Fq be the finite field with q elements Let q be a prime power such that t |(q − 1) and ν be a generator of the multiplicative group of the field Fq We identify the color set with the set {0, , t − 1} The graph Pq,t is a graph with vertex set Fq , the edge between two distinct vertices being colored by the ith color if their sum is of the form ν j where j ≡ i mod t One can show that Pq,t is an (n, t )-e.c graph when ∗ Corresponding author E-mail addresses: nguyenminhhai06@gmail.com (N.M Hai), trandangphuc234@gmail.com (T.D Phuc), vinhla@vnu.edu.vn (L.A Vinh) http://dx.doi.org/10.1016/j.dam.2016.05.017 0166-218X/© 2016 Published by Elsevier B.V 2 N.M Hai et al / Discrete Applied Mathematics ( ) – q is large enough More precise, if q is a prime power such that q > 3(t −1)n q1/2 + n2(t −1)n , (1.1) then Pq,t has the (n, t )-e.c property (Note that, from the probabilistic argument, the upper bound for the smallest order of an (n, t )-e.c is better than the bound in (1.1) The probabilistic bound, however, is not explicit.) Note that the main motivation of that work is to construct new classes of n-e.c graphs From any (n, t )-e.c graph, we can obtain an n-e.c graph by dividing the color set into two sets For a positive integer N and < ρ < 1, the probability space G(N , ρ) consists of graphs with vertex set of size N so that two distinct vertices are joined independently with probability ρ It is known that almost all graphs in G(N , ρ) have the n-e.c graphs The above construction supports this statement by constructing explicitly n-e.c graphs with edge density p for any < ρ < For any positive integers n and t, let f (n, t ) be the order of the smallest (n, t )-e.c graph It follows from (1.1) that f (n, t ) ≤ 9(t −1)n + n2(t −1)n In particular, if n = then f (3, t ) = O(93t ), which is of exponential order We recall that the expressions A ≪ B and A = O(B) are each equivalent to the statement that |A| ≤ cB for some constant c > In [15], the second listed author gave new explicit constructions of (3, t )-graphs of polynomial order Let p be a prime such that t |(p − 1), Fp be the finite field of p elements, and ν be a generator of the multiplicative group of the field We identify the color set with the set {0, , t − 1} For any d ≥ 2, the graph Qpd ,t is the complete graph with the vertex set Fdp , the edge between two distinct vertices x, y being colored by the ith color if their distance ∥x − y ∥ = (x1 − y1 )2 + · · · + (xd − yd )2 is of the form ν j where j ≡ i mod t The third listed author [15, Theorem 1.1] showed that Qpd ,t is an (3, t )-e.c graph when p ≥ t and d ≥ As an immediate corollary, f (3, t ) = O(t 30 ), which is of polynomial order 1.1 Permutation polynomials The main purpose of this paper is to give other explicit constructions of (3, t )-graphs via permutation polynomials with two advantages over previous known results First, we can relax the condition t |(p − 1) Second, we can construct explicitly (3, t )-e.c graphs with arbitrarily color density Let p be a prime and Fp be the finite field of p elements Suppose that f (x) is a polynomial over Fp of degree smaller than p A basic question in the theory of finite fields is to estimate the size Vf of the value set {f (a) | a ∈ Fq } Because a polynomial f (x) cannot assume a given value of more than deg(f ) times over a field, one has the trivial bound  p−1  + ≤ Vf ≤ p (1.2) deg(f ) If the lower bound in (1.2) is attained, then f (x) is called a minimal value set polynomial The classification of minimal value set polynomials is the subject of several papers; see [3,5,6,10] The results in these papers assume that p is large compared to the degree of f (x) If the upper bound in (1.2) is attained, then f (x) is called a permutation polynomial The classification of permutation polynomials has received considerable attention See the book of Lidl and Niederreiter [9] and the survey article by Mullen [11] We identify Fp with the set {0, 1, , p − 1} Let A = A1 ∪ · · · ∪ At be a partition of Fp such that each Ai is a block of consecutive numbers in Fp , that is for any ≤ i ≤ t, there exist ti , si such that Ai = {ti + 1, , ti + si } Let f ∈ Fp [x] be a permutation polynomial of degree at least We also need to assume that p is large compared to the degree of f (x) For any ≤ i ≤ t, set Vi = {f (a) : a ∈ Ai } For any d ≥ 2, the graph Gdf,A is the complete graph with the vertex set Fdp ; the edge between two distinct vertices x, y being colored by the ith color if their distance ∥x − y ∥ ∈ Vi We claim that Gdf,A is an (3, t )-e.c graph when d ≥ and |Ai | ≫ deg(f )p5/6 log p for all ≤ i ≤ t Theorem 1.1 Let f be a nonlinear permutation polynomial over Fp and let A = A1 ∪ · · · ∪ At be a partition of Fp such that each Ai is a block of consecutive numbers of cardinality |Ai | ≫ deg(f )p5/6 log p For any d ≥ 5, the graph Gdf,A has (3, t )-e.c property Note that these graphs are just Cayley graphs of Fdp To construct non-Cayley (3, t )-e.c graphs, we need to adjust the definition of Gdf,A slightly using the following notion of mixed distance between two points x, y ∈ Fdp : ∥x − y ∥m = 2x1 y1 + (x2 − y2 )2 + · · · + (xd − yd )2 Theorem 1.2 Let f be a nonlinear permutation polynomial over Fp and let A = A1 ∪ · · · ∪ At be a partition of Fp such that each Ai is a block of consecutive numbers of cardinality |Ai | ≫ deg(f )p5/6 log p For any d ≥ 6, the graph Hfd,A is the complete graph with the vertex set Fdp ; the edge between two distinct vertices x, y being colored by the ith color if their mixed distance ∥x − y ∥m ∈ {f (a) : a ∈ Ai } Then Hfd,A is a non-Cayley (3, t )-e.c graphs The proof of Theorem 1.2 is exactly the same as the proof of Theorem 1.1 and is left to the interested reader N.M Hai et al / Discrete Applied Mathematics ( ) – 1.2 Linear maps We also study the case of linear maps over finite fields Let f (x) = α x + β , where α ∈ F∗p , β ∈ Fp , be a linear map over the finite field Fp We identify Fp with the residue set {0, 1, , p − 1} We partition Fp into three equitable disjoint sets as follows For p = 3k + 1, set A1 = {0, 1, , k}, A2 = {k + 1, k + 2, , 2k}, A3 = {2k + 1, 2k + 2, , 3k}, and for p = 3k + 2, set A1 = {0, 1, , k}, A2 = {k + 1, k + 2, , 2k + 1}, A3 = {2k + 2, 2k + 3, , 3k + 1} For any ≤ i ≤ 3, set Vi = {f (a) : a ∈ Ai } For any d ≥ 2, the graph Gdf,3 is the complete graph with the vertex set Fdp ; the edge between two distinct vertices x, y being colored by the ith color if their distance ∥x − y ∥ ∈ Vi Our next result is the following theorem Theorem 1.3 Let f (x) = α x + β , where α ∈ F∗p , β ∈ Fp , be a linear map over the finite field Fp Suppose that p ≥ 20 is an odd prime and d ≥ be an integer Then the graph Gdf,3 has (3, 3)-e.c property Note that we can partition the finite field Fp into t equitable disjoint sets of consecutive numbers and define the graph Gdf,t similarly The case t = has been studied in [14] More precisely, let A1 = {0, 1, , (p − 1)/2}, the graph Gd2 is defined as follows: the vertices of the graph Gd2 are the points of Fdp ; and there is an edge between two vertices x and y if and only if ∥x − y ∥ = (x1 − y1 )2 + · · · + (xd − yd )2 ∈ A1 The third listed author [14] proved that Gd2 is 3-e.c for p ≥ and d ≥ In this paper, we will show further that Gd2 is indeed a 4-e.c graph More precisely, we have the following theorem Theorem 1.4 Let p ≥ 11 be an odd prime and d ≥ be an integer, then the graph Gd2 is 4-e.c On the other hand, one can prove that the graph Gd2 is not a 5-e.c graph for every odd prime p and integer d Furthermore, in contrary to the case of higher degree polynomials, the graph Gdf,t , however, does not have (3, t )-e.c property when f is a linear map and t ≥ The main reason these constructions fail is that the image set of a linear map of consecutive numbers over Fp is an arithmetic progression This motivates our next construction from the multiplicative group of Fp Let ν be a generator of the multiplicative group of the finite field Fp , then multiplicative group F∗p ≡ {ν, ν , , ν p−1 } Let A = A1 ∪ · · · ∪ At be a partition of {1, , p − 1} such that each Ai is a block of consecutive numbers in Z/(p − 1)Z, that is for any ≤ i ≤ t, there exist ti , si such that Ai = {ti + 1, , ti + si } mod (p − 1) For any ≤ i ≤ t, set Vi = {ν a : a ∈ Ai } d For any d ≥ 2, the graph HA is the graph with the vertex set Fdp ; the edge between two distinct vertices x, y being colored d by the ith color if their distance ∥x − y ∥ ∈ Vi We claim that HA is an (3, t )-e.c graph when d ≥ and |Ai | ≫ p5/6 log p for all ≤ i ≤ t Theorem 1.5 Let A = A1 ∪ · · · ∪ At be a partition of {1, , p − 1} such that each Ai is a block of consecutive numbers of d has (3, t )-e.c property cardinality |Ai | ≫ p5/6 log p For any d ≥ 5, the graph HA d The (3, t )-e.c property of the graphs Gfd,A and HA We first give a proof of Theorem 1.1 in this section Our proof is similar to that of Theorem in [15] It suffices to show that for any three distinct points a = (a1 , , ad ), b = (b1 , , bd ), c = (c1 , , cd ) in Fdp and i, j, k ∈ {1, 2, 3}, there is a point x = (x1 , , xd ) ∈ Fdp , x ̸= a, b, c such that ∥x − a∥ ∈ Vi , ∥x − b∥ ∈ Vj and ∥x − c ∥ ∈ Vk Therefore, we only need to show that there exist u ∈ Vi , v ∈ Vj , and w ∈ Vk such that the following system has at least four solutions (in this case, one of these solutions is different from a, b, and c), ∥x − a∥ = u ∥x − b∥ = v ∥x − c ∥ = w By eliminating x2i ’s (2.1) (2.2) (2.3) from (2.2) and (2.3), we get an equivalent system of equations ∥x − a∥ = u (2.4) x · (b − a) = (u − v + ∥b∥ − ∥a∥)/2 (2.5) x · (c − a) = (u − w + ∥c ∥ − ∥a∥)/2, (2.6) N.M Hai et al / Discrete Applied Mathematics ( ) – where x · y is the usual dot product between two vectors x and y We first show that the system of two equations (2.5) and (2.6) has a solution x0 for some choices of u ∈ Vi , v ∈ Vj , and w ∈ Vk We consider two cases Case Suppose that b − a and c − a are linearly independent For any u ∈ Vi , v ∈ Vj , and w ∈ Vk , it is clear that there is a solution x0 to the system of two equations (2.5) and (2.6) Case Suppose that b − a and c − a are linearly dependent Since b − a ̸= c − a ̸= 0, c − a = l(b − a) for some l ̸= 0, The two equations (2.5) and (2.6) have a common solution if we can choose u ∈ Vi , v ∈ Vj , and w ∈ Vk such that u − w + ∥c ∥ − ∥a∥ = l(u − v + ∥b∥ − ∥a∥), or equivalently, w + (l − 1)u − lv = γ , (2.7) where γ = ∥c ∥ + (l − 1)∥a∥ − l∥b∥ ∈ Fp For any γ ∈ Fq , let Nδ = |{(u, v, w) ∈ Vi × Vj × Vk : (l − 1)u − lv + w = γ }| For any x ∈ Fp , let ep (x) = e2π ix/p From the orthogonality property of the exponential sum [9] p−1  ep (st ) = p if t = 0, and if t ̸= 0, s=0 we have Nγ = p−1   p (u,v,w)∈V ×V ×V s=0 i j k ep (s((l − 1)u − lv + w − γ )), where the inner sum is p if (l − 1)u − lv + w = γ and zero otherwise This implies that Nγ = = |Vi | |Vj | |Vk | p |Vi | |Vj | |Vk | p + + p−1   p (u,v,w)∈V ×V ×V s=1 i j k p−1 1 p s=1  ep (−sγ )  ep (s((l − 1)u − lv + w − γ ))        ep (sw) ep (−slv) ep (s(l − 1)u) u∈Ai v∈Aj (2.8) w∈Ak Note that deg(f ) ≥ It is well-known (see, for example, [1, Lemma 2] and [16]) that        ep (sf (x)) ≪ deg(f )p1/2 log p,   m≤x≤m+h (2.9) for any s ̸= and ≤ m, h ≤ p − Putting (2.8) and (2.9) together, we have Nγ ≥ |Vi | |Vj | |Vk | p   − O (p − 1)p1/2 (log p)3 > for every γ ∈ Fq This implies that Eq (2.7) has a solution (u, v, w) ∈ Vi × Vj × Vk Hence we always can choose u ∈ Vi , v ∈ Vj , and w ∈ Vk such that the two equations (2.5) and (2.6) have a common solution x0 Take a basis of solutions of the system x · (b − a) = x · (c − a) = 0, and the solution x0 Substitute them into (2.4), we get a single quadratic equation of d − variables Since d − 3, this quadratic equation has at least q (≥4) solutions Theorem 1.1 follows immediately Note that, proof of Theorem 1.5 is similar to the proof of Theorem 1.1 One only need to replace the bound (2.9) in the above proof by the following well known inequality, which can be found in [7,8,12] Let ν be a generator of F∗p , for any a ∈ F∗p and any ≤ S , T ≤ p − 1, the bound       u  ep (aν ) ≪ p1/2 log p,  S ≤u≤S +T  holds (2.10) N.M Hai et al / Discrete Applied Mathematics ( ) – The (3, 3)-e.c property of the graph Gfd,3 We give a proof of Theorem 1.3 in this section It suffices to show that for any three distinct points a = (a1 , , ad ), b = (b1 , , bd ), c = (c1 , , cd ) in Fdp and i, j, k ∈ {1, 2, 3}, there is a point x = (x1 , , xd ) ∈ Fdp , x ̸= a, b, c such that ∥x − a∥ ∈ Vi , ∥x − b∥ ∈ Vj and ∥x − c ∥ ∈ Vk Follows the proof of Theorem 1.1 in Section 2, we only need to show that for any ≤ l ≤ p − and δ ∈ Fp , the equation w + (l − 1)u − lv = δ (3.1) has a solution u ∈ Vi , v ∈ Vj , w ∈ Vk Let u1 = (u − β)/α , v1 = (v − β)/α , and w1 = (w − β)/α , then (3.1) becomes w1 + (l − 1)u1 − lv1 = γ /α, where u1 ∈ Ai , v1 ∈ Aj , w1 ∈ Ak Hence, we only need to show that for any ≤ l ≤ p − and for any δ ∈ Fp , the equation w + (l − 1)u − lv = δ has a solution u ∈ Ai , v ∈ Aj , w ∈ Ak Let A0 = {0, 1, , k − 1}, then (3.2) A0 ⊂ A1 , A0 + (k + 1) ≡ {k + 1, , 2k} ⊂ A2 , A0 + (2k + 1) ≡ {2k + 1, , 3k} ⊂ A3 This implies that it suffices to show that for any ≤ l ≤ p − and for any δ ∈ Fp , Eq (3.2) has a solution u, v, w ∈ A0 For any u, w, w , set fl (u, v, w) = w + (l − 1)u − lv For any δ ∈ Fp , we will show an explicit solution (u, v, w) ∈ A30 of fl (u, v, w) = δ First, let u = v = and let w run from to k − 1, we have fl (0, 0, 0) = 0, fl (0, 0, 1) = 1, fl (0, 0, k − 1) = k − Next, let w = and u = v ∈ {1, , k − 1}, we have fl (1, 1, 0) = p − 1, fl (2, 2, 0) = p − 2, fl (k − 1, k − 1, 0) = p − (k − 1) We now show explicit solutions for the remaining cases δ ∈ {k, k + 1, , p − k} We consider four cases Case Suppose that l = Then f2 (u, v, w) = (w + u) − 2v Let v = 0, w = k − 1, and u runs from to k − 1, then we have f2 (1, 0, k − 1) = k, f2 (2, 0, k − 1) = k + 1, f2 (k − 1, 0, k − 1) = 2k − For remaining values of δ ∈ {p − k − 3, p − k − 2, p − k − 1, p − k}, let (u, v, w) ∈ {(k − 5, k − 1, 0), (k − 4, k − 1, 0), (k − 3, k − 1, 0), (k − 2, k − 1, 0)}, respectively This implies that f2 (A30 ) ≡ Fp Case Suppose that ≤ l ≤ k + Then l − ∈ A0 , and we have fl (1, 0, k + − l) = k, fl (2, 0, k + − l) = k + (l − 1), fl (k − 1, 0, k + − l) = k + (k − 2)(l − 1) For any ≤ s ≤ k − 1, let w run from k + − l to k − 1, we have fl (s, 0, k + − l) = k + (s − 1)(l − 1), fl (s, 0, k + − l) = k + (s − 1)(l − 1) + 1, fl (s, 0, k − 1) = k + (s − 1)(l − 1) + (l − 2) 6 N.M Hai et al / Discrete Applied Mathematics ( ) – Hence, we obtained explicit solutions for δ ∈ {k, , k + (k − 2)(l − 1)} On the other hand, l ≥ so k + (k − 2)(l − 1) ≥ k + 2(k − 2) ≥ p − k This implies that fl (A30 ) ≡ Fp for any ≤ l ≤ k + Case Suppose that k + ≤ l ≤ 2k − Set a = (l − 1) − k, then ≤ a ≤ k − We have fl (a + 1, a, 0) = (l − 1)(a + 1) − la = l − a − = k, fl (a + 1, a, 1) = k + 1, fl (a + 1, a, k − 1) = 2k − 1, and fl (0, 1, a − 1) = a − − l = p − k − 2, fl (0, 1, a) = p − k − 1, fl (0, 1, a + 1) = p − k This implies that fl (A30 ) ≡ Fp for any k + ≤ l ≤ 2k − Case Suppose that 2k ≤ l ≤ p − Set l′ = p − l + then ≤ l′ < 2k − and fl (u, v, w) = fl′ (v, u, w) It follows immediately from Case 2.3 that fl (A30 ) = fl′ (A30 ) ≡ Fp Therefore, we showed that for any ≤ l ≤ p − and δ ∈ Fq , Eq (3.1) has a solution u ∈ Vi , v ∈ Vj , w ∈ Vk Theorem 1.3 now follows as in the proof of Theorem 1.1 The 4-e.c property of the graph G2d We give a proof of Theorem 1.4 in this section Let A2 = Fp \ A1 We will show that for every four distinct vertices a, b, c , d of the graph Gd2 , and α, β, γ , δ ∈ {1, 2}, there exists a vertex x ̸= a, b, c , d such that ∥x − a∥ = u ∈ Aα ∥x − b ∥ = v ∈ A β (4.1) ∥x − c ∥ = w ∈ Aγ (4.3) ∥x − d ∥ = t ∈ A δ (4.4) (4.2) By eliminating xi ’s from (4.2), (4.3), and (4.4), we have the following equivalent system of equations ∥x − a∥ = u x · (b − a) = x · (c − a) = x · (d − a) = (4.5) u − v + ∥b∥ − ∥a∥ (4.6) w − v + ∥ c ∥ − ∥ a∥ t − v + ∥d ∥ − ∥a∥ (4.7) (4.8) We first show that the system of three equations (4.6), (4.7), (4.8) has a solution x0 for some choices of u ∈ Aα , v ∈ Aβ , w ∈ Aγ , and t ∈ Aδ We consider three cases Case Suppose that three vectors b − a, c − a, and d − a are linear independent For any u ∈ Aα , v ∈ Aβ , w ∈ Aγ , and t ∈ Aδ , it is clear that there is a solution x0 to the system of three equations (4.6), (4.7), and (4.8) If three vectors b − a, c − a, and d − a are not linear independent, then we have two cases Case Suppose that d − a = t1 (b − a) + t2 (c − a), with (t1 , t2 ) ̸∈ {(0, 0), (0, 1), (1, 0)} and b − a, c − a are linearly independent In this case, the system of equations (4.6), (4.7), (4.8) has a solution if we can choose u ∈ Aα , v ∈ Aβ , w ∈ Aγ , and t ∈ Aδ such that u − t + ∥d ∥ − ∥a∥ = t1 (u − v + ∥b∥ − ∥a∥) + t2 (u − w + ∥c ∥ − ∥a∥), or t = (1 − t1 − t2 )u + t1 v + t2 w + a, where a = ∥d ∥ − ∥a∥ − t1 (∥b∥ − ∥a∥) + t2 (∥c ∥ − ∥a∥) Let A0 = {0, 1, , k − 1}, then A0 ⊂ A1 , A0 + (k + 1) = A2 , hence, it suffices to consider the case (u, v, w, t ) ∈ A40 We have the following subcases (4.9) N.M Hai et al / Discrete Applied Mathematics ( ) – Case 2.1 At least one of − t1 − t2 , t1 , t2 is different from 0, 1, p − (Note that, we identify Fp with the set {0, 1, , p − 1}) Without loss of generality, we can assume that − t1 − t2 ̸= 0, 1, p − as other cases are similar We label Fp around the circle Fix v and w , then Eq (4.9) becomes t = (1 − t1 − t2 )u + b (where b = t1 v + t2 w + a) Going |A0 | = k steps of length (1 − t1 − t2 ) around the circle, we cannot avoid any block of k consecutive points Hence, we can choose u ∈ A0 such that (1 − t1 − t2 )u + b ∈ Aδ This implies that the system of equations (4.6), (4.7), and (4.8) has a solution Case 2.2 Suppose that − t1 − t2 , t1 , t2 ∈ {0, 1, p − 1} Since (t1 , t2 ) ̸∈ {(0, 0), (0, 1), (1, 0)}, we only need to consider three cases: t1 = 1, t2 = 1; t1 = 1, t2 = p − 1; and t1 = p − 1, t2 = If t1 = and t2 = 1, then t = −u + v + w + a Fix u and let v = w , we are back to Case 2.1 We can similarly for t1 = 1, t2 = p − and t1 = p − 1, t2 = Case Suppose that c − a = t1 (b − a) and d − a = t2 (b − a) with t1 ̸= t2 ̸= 0, Similar to Case 2, we will show that for any t1 , t2 , c1 , c2 ∈ Fp , the following system of equations has a solution (u, v, w, t ) ∈ A40 w + (t1 − 1)u − t1 v = c1 t + (t2 − 1)u − t2 v = c2 (4.10) (4.11) Without loss of generality, we assume that t1 < t2 Let k = k − 1, then p = 2k + and A0 = {0, 1, , k′ } Let gt2 (u, v, w, t ) = t + (t2 − 1)u − t2 v and N = {(u, v, w, t ) ∈ A40 : w + (t1 − 1)u − t1 v = c1 } We will show that gt2 (N ) ≡ Fp Note that if (u, v, w, t ) is a solution for the case (t1 , t2 , c1 , c2 ) then (k′ − u, k′ − v, k′ − w, k′ − t ) is a solution for the case (t1 , t2 , p − c1 , p − c2 ) and (v, u, w, t ) is a solution for the case (p − t1 + 1, p − t2 + 1, c1 , c2 ) Therefore, we only need to consider the following sub cases Case 3.1 Suppose that ≤ t1 < t2 ≤ k′ and ≤ c1 ≤ k′ We choose i, m such that k′ − c1 = it1 + m and ≤ i, ≤ m ≤ t1 − Let s be the smallest natural number such that c1 + s = j(t1 − 1) We have j(t1 − 1) ≤ k′ + (t1 − 1) − We start with some specific values of (u, v, w, t ) ∈ N: ′ gt2 (0, 0, c1 , t ) = t , ′ ≤ t ≤ k′ gt2 (1, 1, c1 + 1, 0) = p − 1, gt2 (2, 2, c1 + 2, 0) = p − 2, ··· gt2 (m, m, c1 + m, 0) = p − m, gt2 (m, m + 1, c1 + m + t1 , 0) = p − m − t2 , gt2 (m, m + i, c1 + m + it1 , 0) = p − m − it2 Since t2 ≤ k′ , for any ≤ i′ ≤ i, we start with (u, v, w, t ) = (m, m + i′ , c1 + m + i′ t1 , 0) and let t run from to k′ then gt2 obtains all possible values between −m − i′ t2 and −m − (i′ − 1)t2 Next, we consider the following values of gt2 : gt2 (0, 0, c1 , k′ ) = k′ , gt2 (s + 1, s, c1 + s − (t1 − 1), k′ ) = t2 − + k′ − s, gt2 (s + 2, s, c1 + s − 2(t1 − 1), k′ ) = 2(t2 − 1) + k′ − s, ··· gt2 (s + j, s, c1 + s − j(t1 − 1), k ) = j(t2 − 1) + k′ − s ′ To obtain all values between k′ , t2 − + k′ − s, 2(t2 − 1) + k′ − s, , j(t2 − 1) + k′ − s, we let t run backward from k′ to We, however, need to check that s + j ≤ k′ We have s + j ≤ (t1 − 2) + k′ + t1 − t1 − = (t1 − 1) + k′ − t1 − Let k′ = r (t1 − 1) + s′ with ≤ r , ≤ s′ ≤ t1 − Then (t1 − 1) + k′ − t1 − = (t1 − 1) + r (t1 − 1) + s′ − = (t1 − 1) + r + t1 − s′ − t1 − ≤ r (t1 − 1) + + (s′ − 1) = k′ Therefore, we have a solution (u, v, w, t ) ∈ A40 of the system of equations (4.10) and (4.11) Note that the above listed values of gt2 cover Fp unless p − m − it2 > j(t2 − 1) + k′ − s + Since m = k′ − c1 − it1 , s = c1 − j(t1 − 1), we have 2k′ + > 2k′ + (i + j)(t2 − t1 ) + 1, or ≥ (i + j)(t2 − t1 ) 8 N.M Hai et al / Discrete Applied Mathematics ( ) – We consider two sub cases Subcase 3.1.1 Suppose that i = j = This implies that c1 = 0, s = and k′ = m ≤ t1 − 1, which is a contradiction Subcase 3.1.2 Suppose that i = 1, j = (the case i = 0, j = is similar) This implies that c1 + s = 0, and c1 = If c2 ≤ k′ , then (u, v, w, t ) = (0, 0, 0, c2 ) is a solution of the system of equations (4.10) and (4.11) If c2 ≥ k′ + 3, then we have a solution (u, v, w, t ) = (p − c2 , p − c2 , p − c2 , 0) If c2 = k′ + or k′ + 2, then (t1 , t1 − 1, 0, k′ + − (t2 − t1 )) and (k′ − t1 , k′ − (t1 − 1), k′ , (t2 − t1 ) − 1) are solutions of the system, respectively Case 3.2 Suppose that ≤ t1 < t2 ≤ k′ and c1 = k′ + Let ht2 (u, v, w, t ) = gt2 (u, v, w, t ) − (t2 − 1) Choose i, m such that k′ + (t1 − 1) − c1 = it1 + m and ≤ i, ≤ m ≤ t1 − Let s be the smallest natural number such that c1 − (t1 − 1) + s = j(t1 − 1) Hence, j(t1 − 1) ≤ k′ + (t1 − 1) − We consider the following specific values of (u, v, w, t ) ∈ N ht2 (1, 0, c1 − t1 + 1, t ) = t , ≤ t ≤ k′ , ht2 (2, 1, c1 − t1 + 2, 0) = p − 1, ht2 (3, 2, c1 − t1 + 3, 0) = p − 2, ··· ht2 (m, m − 1, c1 − t1 + m, 0) = p − m + 1, ht2 (m, m, c1 − t1 + m + t1 , 0) = p − m − t2 + 1, ··· ht2 (m, m + i − 1, c1 − t1 + m + it1 , 0) = p − m − it2 + 1, and ht2 (1, 0, c1 − t1 + 1, k′ ) = k′ , ht2 (s + 1, s, c1 − t1 + + s − (t1 − 1), k′ ) = k′ − s, ht2 (s + 2, s, c1 − t1 + + s − 2(t1 − 1), k′ ) = (t2 − 1) + k′ − s, ··· ht2 (s + j, s, c1 − t1 + + s − j(t1 − 1), k ) = (j − 1)(t2 − 1) + k′ − s ′ The remaining of this case can be done similarly as in the proof of Case 3.1 Case 3.3 Suppose that ≤ t1 < t2 ≤ k′ and c1 ≥ k′ + Note that for some c1 , c2 such that the system of equations (4.10) and (4.11) has a solution then the system also has a solution for −c1 , −c2 Since ≤ −c1 ≤ k′ + 1, we come back to Cases 3.1 and 3.2 Case 3.4 Suppose that ≤ t1 ≤ k′ < t2 We can rewrite the system of equations (4.10) and (4.11) as following w + (t1 − 1)u − t1 v = c1 t + (t2′ − 1)v − t2′ u = c2 , (4.12) (4.13) where t2′ = p − t2 + One can show explicitly solutions for the cases t2 = k′ + 1, k′ + 2, k′ + When k′ + ≤ t2 , we can repeat the proof of the above cases Case 3.5 Suppose that k′ < t1 < t2 Similarly, one can show explicitly solutions for the case k′ < t2 ≤ k′ + When k′ + ≤ t2 , we consider (p − t1 + 1, p − t2 + 1) instead of (t1 , t2 ) and go back to the above cases Hence we always can choose u ∈ Aα , v ∈ Aβ , w ∈ Aγ , and t ∈ Aδ such that the system of three equations (4.6), (4.7), (4.8) has a solution x0 Take a basis of solutions of the system x · (b − a) = x · (c − a) = x · (d − a) = 0, and the solution x0 Substitute them into (4.5), we get a single quadratic equation of d − variables Since d − quadratic equation has at least q (≥5) solutions Theorem 1.1 follows immediately 3, this Remarks It is remarkable that the graph Gd2 is not n-e.c for any n ≥ It is clear that we only need to show that for n = Suppose for the contrary that the graph Gd2 is 5-e.c then for every five distinct vertices a, b, b, c , d , e of the graph Gd2 , and α, β, γ , δ, ϵ ∈ {1, 2}, there exists a vertex x ̸= a, b, b, c , d , e such that ∥x − a∥ = u ∈ Aα (5.1) N.M Hai et al / Discrete Applied Mathematics ( ) – ∥x − b∥ = v ∈ Aβ (5.2) ∥x − c ∥ = w ∈ Aγ (5.3) ∥x − d ∥ = t ∈ Aδ ∥x − e∥ = r ∈ Aϵ (5.4) (5.5) By eliminating xi ’s from (5.2), (5.3), (5.4), and (5.5) we have the following equivalent system of equations ∥x − a∥ = u ∈ Aα u − v + ∥b∥ − ∥a∥ x · (b − a) = (5.6) (5.7) x · (c − a) = x · (d − a) = x · (e − a) = u − w + ∥ c ∥ − ∥ a∥ (5.8) u − t + ∥d ∥ − ∥a∥ u − r + ∥e∥ − ∥a∥ (5.9) (5.10) We will consider the case c − a, d − a, and e − a are linearly dependent on b − a and show that the above system of equations has no solution Similarly as in the previous section, it is suffice to consider the case (u, v, w, t , r ) ∈ A50 By the linearly dependent of the above vectors, there exist t1 , t2 , t3 such that c − a = t1 (b − a), d − a = t2 (b − a) v e − a = t3 (b − a) Since a, b, b, c , d , e are distinct, we have t1 , t2 , t3 ̸= 0, If the system of equations (5.7)–(5.10) has a solution, then there exit (u, v, w, t , r ) ∈ A50 such that u − w + ∥c ∥ − ∥a∥ = t1 (u − v + ∥b∥ − ∥a∥) u − r + ∥d ∥ − ∥a∥ = t2 (u − v + ∥b∥ − ∥a∥) u − t + ∥e∥ − ∥a∥ = t3 (u − v + ∥b∥ − ∥a∥), or w + (t1 − 1)u − t1 v = c1 r + (t2 − 1)u − t2 v = c2 t + (t3 − 1)u − t3 v = c3 , (5.11) (5.12) (5.13) where c1 = ∥c ∥ − ∥a∥ − t1 (∥b∥ − ∥a∥), c2 = ∥d ∥ − ∥a∥ − t2 (∥b∥ − ∥a∥), c3 = ∥e∥ − ∥a∥ − t3 (∥b∥ − ∥a∥) We will show a specific case such that the system of equations (5.11), (5.12) v (5.13) has no solution Let t1 = 2, t2 = p−5 3, t3 = 4, c1 = 0, c2 = , c3 = p − 2, we will show that the following system of equations has no solution (u, v, w, t , r ) ∈ A50 w = 2v − u (5.14) t = 3v − 2u + p−5 (5.15) r = 4v − 3u + (p − 2) (5.16) Consider Eq (5.14) Suppose that 2v − u < Since 2v − u ≥ −u ≥ − contradiction to w ∈ A0 Hence, 2v − u ≥ On the other hand, 2v − u ≤ 2v ≤ p − Hence, it follows from w ∈ A0 that p−3 , ≤ 2v − u ≤ p−3 u p−3 we have − p−3 ≤ w < 0, which is a We have the following condition of v and u ≤v≤ u + (5.17) 2 Suppose that 4v − 3u < From (5.16) and (5.17), we have p > r = p − + 4v − 3u ≥ p − + 2u − 3u = p − − u ≥ p − − which is a contradiction to ≤ r ≤ p−3 Hence, 4v − 3u ≥ p−3 > p−3 , 10 N.M Hai et al / Discrete Applied Mathematics ( ) – Since we need r ∈ A0 , it follows that ip ≤ 4v − 3u + p − ≤ ip + p−3 , (5.18) for a positive integer i Suppose that i ≥ From (5.18), we have 4v − 3u + p − ≥ 2p, hence v ≥ Therefore, i = 1, substituting into (5.18), we have p−3 p ≤ 4v − 3u + p − ≤ p + u + p+2 , which is contradiction to (5.17) , which is equivalent to u+ ≤v≤ u+ p+1 (5.19) From (5.15) and (5.19), we have t = 3v − 2u + p−5 ≥ u+ − 2u + p−5 Since we need t ∈ A0 , it follows that 3v − 2u + v≥ Gd2 u+ p+5 = p−5 u + 2p − 4 > p−3 ≥ p, or 3v − 2u ≥ p+5 Hence, From (5.19) and (5.20), we have 32 u + (5.20) p+5 ≤ 34 u + p+1 or p+17 ≤ u, which is a contradiction It concludes that the graph is not n-e.c for any n ≥ Acknowledgment The third author’s research was supported by Vietnam National University Hanoi Project QG.13.02 References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] W.D Banks, A Conflitti, I Shparlinski, Character sums over integers with restricted g-ary digits digits, Illinois J Math 46 (2002) 819–836 A Bonato, The 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Gomez-Calderon, A note on polynomials with minimal value set overfinite fields, Mathematika 35 (1988) 144–148 J Gomez-Calderon, D.J Madden, Polynomials with small value sets overfinite fields, J Number... Niederreiter, Finite Fields, Addison-Wesley, Reading, MA, 1983 W.H Mills, Polynomialswith minimal valuesets, Pacific J Math 14 (1964) 225–241 G.L Mullen, Permutation polynomials over finite fields, ... polynomial order 1.1 Permutation polynomials The main purpose of this paper is to give other explicit constructions of (3, t ) -graphs via permutation polynomials with two advantages over previous known