On Determining Deep Holes of Generalized Reed-Solomon Codes ⋆ Qi Cheng1 and Jiyou Li2 and Jincheng Zhuang1 arXiv:1309.3546v2 [cs.IT] 16 Dec 2013 1 School of Computer Science The University of Oklahoma Norman, OK 73019, USA. Email: qcheng@cs.ou.edu, jzhuang@ou.edu 2 Department of Mathematics Shanghai Jiao Tong University Shanghai, PR China Email: lijiyou@sjtu.edu.cn Abstract. For a linear code, deep holes are defined to be vectors that are further away from codewords than all other vectors. The problem of deciding whether a received word is a deep hole for generalized ReedSolomon codes is proved to be co-NP-complete [9][5]. For the extended Reed-Solomon codes RSq (Fq , k), a conjecture was made to classify deep holes in [5]. Since then a lot of effort has been made to prove the conjecture, or its various forms. In this paper, we classify deep holes completely for generalized Reed-Solomon codes RSp (D, k), where p is a prime, |D| > k > p−1 . Our techniques are built on the idea of deep hole 2 trees, and several results concerning the Erdös-Heilbronn conjecture. Keywords: Reed-Solomon code, deep hole, deep hole tree, Erdös-Heilbronn conjecture. 1 Introduction Reed-Solomon codes are of special interest and importance both in theory and practice of error-correcting. Definition 1. Let Fq be a finite field with q elements and characteristic p. Let D = {α1 , . . . , αn } ⊆ Fq be the evaluation set and vi ∈ F∗q , 1 6 i 6 n, be the column multipliers. The set of codewords of the generalized Reed-Solomon code RSq (D, k) of length n and dimension k over Fq is defined as RSq (D, k) = {(v1 f (α1 ), . . . , vn f (αn )) ∈ Fnq | f (x) ∈ Fq [x], deg(f ) 6 k − 1}. ⋆ The research is partially supported by NSF under grants CCF-0830522 and CCF0830524 for Q.C. and J.Z., and by National Science Foundation of China (11001170) and Ky and Yu-Fen Fan Fund Travel Grant from the AMS for J.L. We will write generalized Reed-Solomon codes as GRS codes for short in the sequel. If D = F∗q , it is called primitive. If D = Fq , it is called a singly-extended GRS code. A GRS code is called normalized if its column multipliers are all equal to 1. In this paper, we will work on the normalized GRS without loss of generality. The encoding algorithm of the GRS code can be described by the linear map ϕ : Fkq → Fnq , in which a message (a1 , . . . , ak ) is mapped to a codeword (f (α1 ), . . . , f (αn )), where f (x) = ak xk−1 + ak−1 xk−2 + · · · + a1 ∈ Fq [x]. The Hamming distance between two words is the number of their distinct coordinates. The error distance of a received word u ∈ Fnq to the code is defined as its minimum Hamming distance to codewords. The minimum distance of a code, which is denoted by d, is the smallest distance between any two distinct codewords of the code. The covering radius of a code is the maximum distance from any vector in Fnq to the nearest codeword. A deep hole is a vector achieving the covering radius. A linear code [n, k]q is called maximum distance separable (in shot, MDS) if it attains the Singleton bound, i.e., d = n − k + 1. GRS code is a linear MDS code, and its minimum distance is known to be n − k + 1 and the covering radius is known to be n − k. Thus for the GRS code, u is a deep hole if d(u, RSq (D, k)) = n − k. A linear code can be represented by a generator matrix. In this paper, we assume that the rows of a generator matrix form a basis for the code. 1.1 Related work Efforts have been made to obtain an efficient decoding algorithm for GRS codes. √ Given a received word u ∈ Fnq , if the error distance is smaller than n − nk, then the list decoding algorithm of Sudan [17] and Guruswami-Sudan [8] solves the decoding in polynomial time. However, in general, the maximum likelihood decoding of GRS codes is NP-hard [9]. We would like to determine all the deep holes of the code. To this end, given a received word u = (u1 , u2 , . . . , un ) ∈ Fnq , we consider the following Lagrange interpolating polynomial Q n X j6=i (x − αj ) Q ∈ Fq [x], u(x) = ui j6=i (αi − αj ) i=1 where D = {α1 , . . . , αn } is the evaluation set. The Lagrange interpolating polynomial is the only polynomial in Fq [x] of degree less than n that satisfies u(αi ) = ui , 1 6 i 6 n. In this paper, we say that a function u(x) generates a vector u ∈ Fnq if u = (u(α1 ), u(α2 ), . . . , u(αn )). We have the following conclusions: – If deg(u) 6 k − 1, then u ∈ RSq (D, k) by definition and d(u, RSq (D, k)) = 0. – If deg(u) = k, then it can be shown that u is a deep hole by the following proposition [10], i.e., d(u, RSq (D, k)) = n − k. Proposition 1. ([10]) For k 6 deg(u) 6 n − 1, we have the inequality n − deg(u) 6 d(u, RSq (D, k)) 6 n − k. 2 When the degree of u(x) becomes larger than k, the situation becomes complicated for GRS codes. However, in the case of (singly-)extended GRS codes, the situation seems to be much simpler. Cheng and Murray [5] conjectured in 2007 that the vectors generated by polynomial of degree k are the only possible deep holes. Conjecture 1. ([5]) A word u is a deep hole of RSq (Fq , k) if and only if deg(u) = k. There is an analogous conjecture for deep holes of primitive Reed-Solomon codes by Wu and Hong [20]. Conjecture 2. ([20]) A word u is a deep hole of RSq (Fq∗ , k) if and only if: u(x) = axk + f6k−1 (x), a 6= 0; or u(x) = bxq−2 + f6k−1 (x), b 6= 0; where f6k−1 (x) denotes a polynomial with degree not larger than k − 1. Cheng and Murray [5] got the first result by reducing the problem to the existence of rational points on a hypersurface over Fq . Theorem 1. [5] Let u ∈ Fqq such that 1 6 d := deg(u) − k 6 q − 1 − k. If 13 q > max(k 7+ǫ , d 3 +ǫ ) for some constant ǫ > 0, then u is not a deep hole. Following a similar approach of Cheng-Wan [6], Li and Wan [12] improved the above result with Weil’s character sum estimate. Theorem 2. [12] Let u ∈ Fqq such that 1 6 d := deg(u) − k 6 q − 1 − k. If 2 8 q > max((k + 1)2 , d2+ǫ ), k > ( + 1)d + + 2 ǫ ǫ for some constant ǫ > 0, then u is not a deep hole. Then Liao [13] proved the following result: Theorem 3. [13] Let r > 1 be an integer. For any received word u ∈ Fqq , r 6 d := deg(u) − k 6 q − 1 − k, if   2r + 4 2 k+r q > max(2 +2 + d, d2+ǫ ), k > ( + 1)d + ǫ ǫ 2 for some constant ǫ > 0, then d(u, RSq (Fq , k)) 6 q − k − r, which implies that u is not a deep hole. Antonio Cafure ect. [4] proved the following result with tools of algebraic geometry: 3 Theorem 4. [4] Let u ∈ Fqq such that 1 6 d := deg(u) − k 6 q − 1 − k. If 2 q > max((k + 1)2 , 14d2+ǫ ), k > ( + 1)d, ǫ for some constant ǫ > 0, then u is not a deep hole. Using Weil’s character sum estimate and Li-Wan’s new sieve [11] for distinct coordinates counting, Zhu and Wan [21] showed the following result: Theorem 5. [21] Let r > 1 be an integer. For any received word u ∈ Fqq , r 6 d := deg(u) − k 6 q − 1 − k, there are positive constants c1 and c2 such that if d < c1 q 1/2 , ( d+r + 1) log2 (q) < k < c2 q, 2 then d(u, RSq (Fq , k)) 6 q − k − r. The deep hole problem for Reed-Solomon codes are also closely related to the famous MDS conjecture in coding theory. On one hand, GRS codes are MDS codes. On the other hand, it is known that all long enough MDS codes are essentially GRS codes. Following the notation of [14], let Nmin (k, q) be the minimal integer, if any, such that every [n, k] MDS code over GF (q) with n > Nmin (k, q) is GRS and be q + 2 if no such integer exists. For the case of k = 3, Segre [15] obtained the following result: Theorem 6. [15] If q is odd, every [n, 3] MDS code over GF (q) with q − √ q−7 < n 6 q + 1 is GRS. 4 When q = p is a prime, Voloch [18] obtained the following result: Theorem 7. [18] If p is an odd prime number, every [n, 3] MDS code over p GF (p) with p − + 2 < n 6 p + 1 is GRS. 45 Further, there is a relation for Nmin (k + 1, q) and Nmin (k, q) [14] as follows: Lemma 1. [14] For 3 6 k 6 q − 2, we have Nmin (k + 1, q) 6 Nmin (k, q) + 1. Simeon Ball [2] showed the following result: Theorem 8. [2] Let S be a set of vectors of the vector space Fkq , with the property that every subset of S of size k is a basis. If |S| = q + 1 and k 6 p or 3 6 q − p + 1 6 k 6 q − 2, where p is the characteristic of Fq , then S is equivalent to the following set: {(1, α, α2 , . . . , αk−1 ) | α ∈ Fq } ∪ {(0, . . . , 0, 1)}. 4 1.2 Our result In this paper, we classify the deep holes in many cases. Firstly, we show: Theorem 9. Let p > 2 be a prime number, k > p−1 2 , D = {α1 , α2 , . . . , αn } with k < n 6 p. The only deep holes of RSp (D, k) are generated by functions which are equivalent to the following: f (x) = xk , fδ (x) = 1 , x−δ where δ ∈ Fp \ D. Here two functions f (x) and g(x) are equivalent if and only if there exists a ∈ F∗p and h(x) with degree less than k such that g(x) = af (x) + h(x). Our techniques are built on the idea of deep hole trees, and several results concerning the Erdös-Heilbronn conjecture. We also show the following theorem based on some results of finite geometry. Theorem 10. Given a finite filed Fq with characteristic p > 2, we have – If k + 1 6 p√or 3 6 q − p + 1 6 k + 1 6 q − 2, then Conjecture 1 is true. q+1 , then Conjecture 2 is true. – If 3 6 k < 4 p – If 3 6 k < , where q = p is prime, then Conjecture 2 is true. 45 This paper is organized as follows: Section 2 presents some preliminaries; Section 3 describes the idea of the deep hole tree; Section 4 demonstrates the proof of Theorem 9; Section 5 gives the proof of Theorem 10. 2 2.1 Preliminaries A criterion for deep holes of linear MDS codes By definition, deep holes of a linear code are words that has a maximum distance to the code. In the case of linear MDS codes, there is another way to characterize the deep hole as follows, which connects the concept of deep holes with the MDS codes. The following is well known: Proposition 2. Let Fq be a finite field with characteristic p. Suppose G is a generator matrix for a linear MDS code C = [n, k]q with covering radius ρ = n − k, then u ∈ Fnq is a deep hole of C if and only if   G ′ G = u generates another linear MDS code. A proof is included in Appendix A for the sake of completeness. 5 2.2 Some additive combinatorics results In this section, we introduce some additive combinatorics results that we will use later. The first theorem is about the estimation of the size of restricted sum sets, which is first proved by Dias da Silva and Hamidoune [16]. Then Alon et al. [1] gave a simple proof using the polynomial method. Theorem 11. [16,1] Let F be a field with characteristic p and n be a positive integer. Then for any finite subset S ⊂ F we have |n∧ S| > min{p, n|S| − n2 + 1}, where n∧ S denotes the set of all sums of n distinct elements of S. Brakemeier [3] and Gallardo et al. [7] established the following theorem: Theorem 12. [3,7] Let n be a positive integer and S ⊂ Z/nZ. If |S| > then 2∧ S = Z/nZ, n 2 + 1, where 2∧ S denotes the set of all sums of 2 distinct elements of S. Hence we have the following corollary: Corollary 1. Let Fp be a prime finite field, S ⊂ F∗p . If |S| > element of F∗p is the product of two distinct elements of S. p+1 2 , then each Proof. Let g be a generator of F∗p . Let S ′ = {e|g e ∈ S} ⊂ Z/(p − 1)Z. For any given element α = g a ∈ F∗p , we need to show that there exist two distinct elements b 6= c such that ga = gbgc, where b, c ∈ S ′ . This is equivalent to a = b + c, which follows from Theorem 12. 3 Construction of the deep hole tree Let Fq = {α1 , α2 , · · · , αq = 0}. The polynomials in Fq [x] of degree less than q form a Fq -linear space, with a basis {1, x, . . . , xk−1 , k Y (x − αi ), . . . , i=1 6 q−1 Y i=1 (x − αi )}. Given a polynomial f (x) ∈ Fq [x] with degree q − 1 we have f (x) = l(x) + c1 q−1 k Y Y (x − αi ), (x − αi ) + · · · + cq−k i=1 i=1 where l(x) is of degree less than k, we want to determine when f (x) generates a deep hole. By Proposition 2, f (x) generates a deep hole of RSq (Fq , k) if and only if   G ′ G = u generates an MDS code, where G is the generator matrix of RSq (Fq , k), and u = (f (α1 ), . . . , f (αq )). Observe that the function, which generates a deep hole for RSq (D2 , k), also generates a deep hole for RSq (D1 , k) if D1 ⊂ D2 . Instead of considering the deep holes for RSq (Fq , k) at the first step, we propose to consider a smaller evaluation set at the beginning and make it increase gradually. The advantage of doing so is that we can prune as we increase the evaluation set instead of exhausting search (c1 , . . . , cq−k ). To be more precise, firstly we determine c1 over D1 = {α1 , . . . , αk+1 }, then we determine c2 over D2 = {α1 , . . . , αk+2 } based on the knowledge of c1 , so on and so forth. We present the result as a tree, which we will call a deep hole tree in the sequel. Remark 1. Wu and Hong [19] showed that if D = Fq \ {β1 , . . . , βl } then fβi (x) = 1 x−βi generates a deep hole for RSq (D, k), where 1 6 i 6 l. We can also deduce this from Proposition 2. For convenience, we will call these deep holes and deep holes generated by a function of degree k expected deep holes. Motivated by Remark 1, firstly we construct the expected deep hole tree as follows: – The root node is 1 without loss of generality, i.e., c1 = 1. – There are p−k −1 branches of the tree, each with distinct length in [2, p−k]. And we designate the sequence of nodes in a branch with length l as bl . • If l = p − k, then bp−k = {0, . . . , 0}. • If 2 6 l 6 p− k − 1, then bl = (c1 , . . . , cl ), where f = x−α1 l+1 is equivalent Qk+l−1 Qk to c1 i=1 (x − αi ) + · · · + cl i=1 (x − αi ). Proposition 3. The expected deep hole tree is a part of the full deep hole tree. Proof. This follows from Remark 1. Now we can construct the full deep hole tree based on the expected deep hole tree. – The root node is 1 without loss of generality, i.e., c1 = 1. 7 – The children {ci+1 } of a node ci , 1 6 i 6 q − k − 1 are defined as follows: given the ancestors (c1 , . . . , ci ), for γ ∈ GF (q), if γ is the child of ci in the expected deep hole tree, then keep it; otherwise, if c1 k Y i=1 (x − αi ) + · · · + ci k+i−1 Y i=1 (x − αi ) + γ k+i Y i=1 (x − αi ) satisfies the property of the function which generates a deep hole as in Proposition 2, then γ is a child of ci . That is, we keep the nodes of the expected deep hole tree and add additional ones if necessary. Now we illustrate the procedure to construct the deep hole tree by one example. Example 1. Let p = 7, k = 2. The evaluation set is ordered such that αi = i, 1 6 i 6 7. (1) The expected deep hole tree is as follows: 1 1 4 5 0 4 6 0 6 0 0 The root is corresponding to the evaluation set D1 = {1, 2, 3}. The expected Q2 deep holes are generated by functions equivalent to f = i=1 (x − i). In depth 2, the evaluation set is D2 = {1, 2, 3, 4}. One of the expected deep holes is Q2 Q3 generated by the function f = i=1 (x − i) + i=1 (x − i), which is equivalent 1 . In depth 3, the evaluation set is D3 = {1, 2, 3, 4, 5}. One of the to f = x−5 Q3 Q2 expected deep holes is generated by the function f = i=1 (x − i) + 4 i=1 (x − Q4 1 i) + 4 i=1 (x − i), which is equivalent to f = x−6 . In depth 4, the evaluation set is D4 = {1, 2, 3, 4, 5, 6}. One of the expected deep holes is generated by the Q2 Q3 Q4 Q5 function f = i=1 (x − i) + 5 i=1 (x − i) + 6 i=1 (x − i) + 6 i=1 (x − i), which 1 is equivalent to f = x . In depth 5, the evaluation set is D5 = {1, 2, 3, 4, 5, 6, 7}. Q2 One of the expected deep holes is generated by the function f = i=1 (x − i). (2) The full deep hole tree is as follows: 8 1 1 3 4 6 4 5 1 3 0 6 0 6 0 0 Note that there are more nodes here than the expected ones. For example, in depth 3, there deep hole generated by the function f = Q2 Q3 is an additional Q4 there is an additional deep hole i=1 (x − i) + i=1 (x − i) + 3 Q i=1 (x − i). Also,Q Q generated by the function f = 2i=1 (x − i) + 5 3i=1 (x − i) + 4i=1 (x − i). 4 Proof of Theorem 9 The basic idea of the proof of Theorem 9 is reducing the problem to some additive number theory problems. We first present several lemmas. Lemma 2. In depth d = 2, the nodes are the same in both the expected deep hole tree and full deep hole tree. Proof. We need to show that in depth d = 2, the nodes are the same in both the expected deep hole tree and full deep hole tree. In depth d = 2, the evaluation set is D = {α1 , α2 , . . . , αk+2 }, the candidate generating function is f (x) = k Y i=1 (x − αi ) + c2 k+1 Y i=1 (x − αi ), c2 ∈ F p . Designate the set of nodes in depth 2 of the expected deep hole tree as S. Firstly, we show that if c2 ∈ S, then f (x) generates a deep hole. This follows from Theorem 3. Note that there are p − (k + 1) of them. Now we show that they are all distinct. Considering the following square matrix   1 1 ··· 1   . .. .. ..   .. . . .   k−1 k−1 k−1  , G= α2 · · · αk+2   α1   1 1 1  α1 −δ1 α2 −δ1 · · · αk+2 −δ1  1 1 1 α1 −δ2 α2 −δ2 · · · αk+2 −δ2 9 we compute α1 − δ1 ··· αk+2 − δ1 α1 ∗ (α1 − δ1 ) · · · αk+2 ∗ (αk+2 − δ1 ) .. .. k+2 .. Y . . (αi − δ1 ) det(G) = k−1 . k−1 α ∗ (α − δ ) · · · α ∗ (α 1 1 k+2 − δ1 ) 1 k+2 i=1 1 ··· 1 αk+2 −δ1 α1 −δ1 ··· α1 −δ2 αk+2 −δ2 = α1 α21 .. . αk1 1 α1 −δ1 α1 −δ2 = (−1)k ··· ··· .. . ··· ··· ··· αk+2 α2k+2 .. . αkk+2 1 αk+2 −δ1 αk+2 −δ2 1 α1 α21 .. . ··· ··· ··· .. . 1 αk+2 α2k+2 .. . αk1 ··· ··· αkk+2 αk+2 −δ1 αk+2 −δ2 α1 −δ1 α1 −δ2 = (−1)k 1+ 1 α1 α21 .. . ··· ··· ··· .. . αk1 ··· αkk+2 δ2 −δ1 · · · 1 + αk+2 −δ2 δ2 −δ1 α1 −δ2 = (−1)k (δ2 − δ1 ) 1 αk+2 α2k+2 .. . 1 α1 α21 .. . ··· ··· ··· .. . 1 αk+2 α2k+2 .. . αk1 ··· ··· αkk+2 1 α1 −δ2 6= 0. 1 αk+2 −δ2 The last step follows from the same argument as above. Hence we prove that c2 1 1 are different for f1 = x−δ and f2 = x−δ . 1 2 10 Next, we show that if c2 ∈ / S then f (x) does not generate a deep hole. Consider the following matrix   1 1 ··· 1  α1 α2 · · · αk+2      .. .. .. .. G= . , . . .   k−1 k−1 k−1 α α2 · · · αk+2  1 f (α1 ) f (α2 ) · · · f (αk+2 ) Qk Qk where f (i) = 0, 1 6 i 6 k, f (αk+1 ) = i=1 (αk+1 − αi ), f (αk+2 ) = i=1 (αk+2 − Qk+1 αi ) + c2 i=1 (αk+2 − αi ). Case 1. If k+1 k Y Y (αk+2 − αi ) (αk+2 − αi ) + c2 f (αk+2 ) = i=1 i=1 = k Y i=1 (αk+2 − αi )[1 + c2 (αk+2 − αk+1 )] = 0, 1 i.e., c2 = αk+1 −α , then there are k + 1 columns of G which are linearly k+2 dependent. Thus f (x) does not generate a deep hole in this case. Case 2. Suppose f (αk+2 ) 6= 0. For any k−1 elements {β1 , . . . , βk−1 } ⊂ {α1 , . . . , αk }, consider the submatrix   1 ··· 1 1 1  β1 · · · βk−1 αk+1 αk+2      .. . . .. .. .. ′ G = . . . . . .   k−1 k−1 k−1 k−1  β · · · β α α 1 k−1 k+1 k+2 0 · · · 0 f (αk+1 ) f (αk+2 ) Thus det(G′ ) = 0 is equivalent to f (αk+1 ) k−1 Y i=1 (αk+2 − βi ) = f (αk+2 ) k−1 Y i=1 (αk+1 − βi ), that is, k−1 Y αk+2 − βi f (αk+2 ) = f (αk+1 ) α − βi i=1 k+1 = k−1 Y (1 + i=1 αk+2 − αk+1 ). αk+1 − βi Hence for each subset of {β1 , . . . , βk−1 } ⊂ {α1 , . . . , αk }, there is a unique c2 such that det(G′ ) = 0. 11 In total, there are k + 1 elements of candidate c2 such that the corresponding f (x) does not generate a deep hole. This implies that if c2 ∈ / S then f (x) does not generate a deep hole. In conclusion, in depth d = 2, the nodes in the full deep hole tree are exactly those in the expected deep hole tree. Lemma 3. Let p be an odd prime, k > p−1 2 , d > 2 be a positive integer and Dd = {α1 , . . . , αk+d } ⊂ Fp , δ ∈ Fp \ Dd . For any γ ∈ Fp , there exists a subset {β1 , . . . , βk } ⊂ Dd such that the matrix  1 ··· 1 1  β1 · · · βk δ    .. ..   .. . . A= . , . . .   k−1 k−1 k−1  β1 · · · β δ k 1 1 · · · γ β1 −δ βk −δ  is singular. Proof. Note that det(A) = det(A′ ) + det(A′′ ), where   1 1 ··· 1 1  β1   β1 · · · βk δ    . ..  ′′  ..  . A′ =  .. . . . .. ,A =  . .  k−1   k−1 β1 β1 · · · β k−1 δ k−1  k 1 1 1 · · · 0 β1 −δ βk −δ β1 −δ  ··· 1 · · · βk . .. . .. · · · βkk−1 · · · βk1−δ  0 0   . 0  0 γ Since β1 − δ ··· βk − δ 1 β (β − δ) · · · β (β − δ) δ 1 1 k k k Y .. .. .. ′ . . (βi − δ) det(A ) = . . . . k−1 k−1 i=1 k−1 β1 (β1 − δ) · · · βk (βk − δ) δ 1 ··· 1 0 β1 β12 = ... ··· ··· .. . k β1 · · · 1 ··· = (−1)k βk βk2 .. . 1 2δ .. . βkk kδ k−1 1 0 1 β1 β12 .. . ··· ··· ··· .. . 1 βk βk2 .. . 0 1 2δ .. . β1k · · · βkk kδ k−1 12 1 ··· 1 d dx 1 β1 · · · βk d dx x = (−1)k β 2 · · · β 2 1 k d 2 dx x .. . . .. . . . β1k · · · βkk 1 β1 d β2 1 = (−1)k dx .. . x=δ x=δ x=δ .. . d k dx x ··· ··· ··· .. . 1 βk βk2 .. . x=δ 1 x x2 .. . x=δ β1k · · · βkk xk   k Y Y d  (βj − βi ) (x − βi ) = (−1)k dx i=1 16i<j6k thus ′ det(A ) = Qk (−1)k i=1 (βi = Qk = Y 16i<j6k # " k d Y (x − βi ) (βj − βi ) dx i=1 Y (βj − βi ) − δ) 16i<j6k (−1)k i=1 (βi Y − δ) 16i<j6k (βj − βi ) k X i=1 k Y i=1 (δ − βi ) k X i=1 , x=δ x=δ 1 δ − βi 1 . δ − βi It follows that det(A) = det(A′ ) + det(A′′ ) = Y 16i<j6k (βj − βi ) k X i=1 1 +γ δ − βi Y 16i<j6k (βj − βi ) Hence det(A) = 0 is equivalent to k X i=1 1 + γ = 0. δ − βi 1 { δ−β |i i Designate the set ∈ Dd } as S1 with cardinality k + d. Since k, 2 6 d, from Theorem 11, we conclude that |k ∧ S1 | > min{p, k|S1 | − k 2 + 1} = p, 13 p−1 2 6 which implies that for each γ ∈ Fp , there exists a subset {β1 , . . . , βk } ⊂ Dk such P 1 that ki=1 δ−β + γ = 0. i Lemma 4. Let p be an odd prime, k > p−1 2 , d > 2 be a positive integer and Dd+1 = {α1 , . . . , αk+d+1 = δ} ⊂ Fp . For any δ ′ ∈ Fp , δ ′ ∈ / Dd+1 , γ ∈ Fp , γ 6= 1 , there exists a subset {β , . . . , β } ⊂ D \ {δ} such that the matrix ′ 1 k d+1 δ−δ  is singular. 1 β1 .. . ··· ··· .. . 1 βk .. . 1 δ .. .        B=   k−1   β1 · · · βkk−1 δ k−1  1 1 γ β1 −δ ′ · · · βk −δ ′ Proof. Note that det(B) = det(B ′ ) + det(B ′′ ), where     1 ··· 1 0 1 ··· 1 1   β1 · · · βk  β1 · · · βk 0 δ      ..  ′′  .. . . .. ..   .. . . ′ , B = B = .  .  . . . .  . 0 .    k−1 k−1 k−1  k−1 k−1    β1 · · · βk δ β1 · · · βk 0 1 1 1 1 1 1 β1 −δ ′ · · · βk −δ ′ δ−δ ′ β1 −δ ′ · · · βk −δ ′ γ − δ−δ ′ Since β1 − δ ′ ··· βk − δ ′ δ − δ′ ′ ′ β1 (β1 − δ ) · · · βk (βk − δ ) δ(δ − δ ′ ) k Y . . .. .. .. .. (βi − δ) det(B ′ ) = . . k−1 k−1 i=1 ′ ′ k−1 β1 (β1 − δ ) · · · βk (βk − δ ) δ (δ − δ ′ ) 1 ··· 1 1 β1 β12 = ... ··· ··· .. . k β1 · · · 1 ··· βk βk2 .. . δ δ2 .. . βkk δ k 1 1 1 β1 2 = (−1)k β1 .. . ··· ··· ··· .. . 1 βk βk2 .. . 1 δ δ2 .. . β1k · · · βkk δ k Y = (−1)k 16i<j6k 14 (βj − βi ) k Y i=1 (δ − βi ), we have det(B ′ ) = k Y Y (−1)k (δ − βi ) (β − β ) Qk j i (δ − δ ′ ) i=1 (βi − δ ′ ) 16i<j6j i=1 1 = δ − δ′ Y 16i<j6k k Y βi − δ , (βj − βi ) β − δ′ i=1 i and det(B ′′ ) = (γ − 1 ) δ − δ′ Hence det(B) = 1 δ − δ′ Y 16i<j6k (βj − βi ) Y 16i<j6k (βj − βi ). k Y βi − δ 1 + (γ − ) ′ β − δ δ − δ′ i i=1 16i<j6k k γ(δ − δ ′ ) − 1 1 Y βi − δ + = (βj − βi ) δ − δ ′ i=1 βi − δ ′ δ − δ′ 16i<j6k " # Q k Y βi − δ 16i<j6k (βj − βi ) = + γ(δ − δ ′ ) + 1 . ′ δ − δ′ β − δ i i=1 Y " Y # (βj − βi ) . It follows that det(B) = 0 is equivalent to k Y δ′ − δ (1 + ) = 1 − γ(δ − δ ′ ). ′ β − δ i i=1 If |Dd | = k+2, we consider the “dual” version of the equality. From Corollary δ ′ −δ δ ′ −δ 1, there exist two distinct elements x, y ∈ Dd such that (1 + x−δ ′ )(1 + y−δ ′ ) = θ for any θ ∈ F∗p , hence there exist k distinct elements in Dd such that k Y δ′ − δ (1 + ) = 1 − γ(δ − δ ′ ), αi − δ ′ i=1 1 for any γ 6= δ−δ ′. If |Dd | > k + 2, we select a subset D′ ⊂ Dd such that |D′ | = k + 2, then apply the same argument as above. Lemma 5. Let p be an odd prime, k > p−1 2 , d > 2 be a positive integer and Dd+1 = {α1 , . . . , αk+d+1 = δ} ⊂ Fp . For any γ ∈ Fp , γ 6= δ k , there exists a subset {β1 , . . . , βk } ⊂ Dd+1 \ {δ} such that the matrix   1 ··· 1 1  β1 · · · βk δ     .. . . .. ..  B= .  . . .  k−1  k−1 k−1  β · · · βk δ 1 β1k · · · βkk γ 15 is singular. Proof. Note that det(B) = det(B ′ ) + det(B ′′ ), where  1 β1 .. . ··· ··· .. . 1 βk .. . 1 δ .. .   1 β1 .. . ··· ··· .. . 1 βk .. . 0 0            ′′   B = . ,B =  0   k−1    k−1 k−1  k−1 k−1  β1   · · · βk δ β1 · · · βk 0 1 1 1 1 k · · · δ · · · γ − δk ′ ′ ′ ′ β1 −δ βk −δ β1 −δ βk −δ ′ Since Y det(B ′ ) = 16i<j6k Y det(B ′′ ) = (βj − βi ) 16i<j6k we have Q k Y (δ − βi ), i=1 (βj − βi )(γ − δ k ), k Y 1 (δ − βi ) + γ − δ k . det(B) = 1≤i<j≤k (βj − βi ) i=1 Thus det(B) = 0 is equivalent to k Y i=1 (δ − βi ) = δ k − γ. Consider S = {δ − α|α ∈ D}. If |Dd | = k + 2, we consider the “dual” version of the equality. From Corollary 1, there exist two distinct elements x, y ∈ S such that xy = δ k − γ for any δ k − γ ∈ F∗p , hence there exist k distinct elements in S such that k Y (δ − βi ) = δ k − γ, i=1 k for any γ 6= δ . If |Dd | > k + 2, we select a subset S ′ ⊂ S such that S = k + 2, then apply the same argument as above. Now we prove Theorem 9. Proof. (of Theorem 9) Proceed by induction on the depth of the full deep hole tree. Basis case This follows from Lemma 2. Inductive step We need to show that if the set of nodes of the full deep hole tree coincide with the nodes of the expected deep hole tree in the same depth d > 2, then there are no additional nodes in depth d + 1 except the expected ones. Denote the corresponding evaluation set by Dd = {α1 , . . . , αk+d } in depth 16 d and Dd+1 = {α1 , . . . , αk+d , αk+d+1 = δ} in depth d + 1. In order to show there are no new nodes in depth d + 1, There are three cases to consider. Case 1: We need to show the branch, which is corresponding to the function 1 f = x−δ , will not continue in the depth d+ 1. It suffices to show that there exists a subset {β1 , . . . , βk } ⊂ {α1 , . . . , αk+d } such that for any γ ∈ Fp and matrix  1 ··· 1 1  β1 · · · βk δ    .. ..   .. . . A= .  . . .   k−1 β1 · · · β k−1 δ k−1  k 1 1 γ β1 −δ · · · βk −δ  we have det(A) = 0. This follows from Lemma 3. Case 2: We need to show that the branch, which is corresponding to the function 1 ′ f = x−δ / Dk+1 , has only one child in depth d + 1. It suffices to show ′ , where δ ∈ that there exists a subset {β1 , . . . , βk } ⊂ Dd such that for any δ ′ ∈ / Dd+1 , γ ∈ 1 Fp , γ 6= δ−δ ′ and matrix  1 β1 .. . ··· ··· .. . 1 βk .. . 1 δ .. .        B=    k−1  β1 · · · βkk−1 δ k−1  1 1 γ β1 −δ ′ · · · βk −δ ′ we have det(B) = 0. This follows from Lemma 4. Case 3: We need to show that the branch, which is corresponding to the function f = xk has only one child in each depth. It suffices to show that there exists a subset {β1 , . . . , βk } ⊂ Dd such that for any γ 6= δ k and matrix   1 ··· 1 1  β1 · · · βk δ     .. . . .. ..  B= .  . . .  k−1  k−1 k−1  β · · · β δ 1 k β1k · · · βkk γ we have det(B) = 0. This follows from Lemma 5. From the principle of induction, the theorem is proved. 5 Proof of Theorem 10 Proof. There are 3 cases to prove. 17 Case 1. Let RSq (Fq , k) be an extended GRS code over the finite field Fq whose characteristic p is odd. Let one of its generator matrix be   1 1 ··· 1  α1 α2 · · · αq   2  2 2   G =  α1 α2 · · · αq  ,  .. .. . . .   . . ..  . α1k−1 α2k−1 · · · αqk−1 where α1 , . . . , αq are distinct element of Fq . Suppose a word u ∈ Fqq is a deep hole of RSq (Fq , k). From proposition 2, this is equivalent to the fact that   G G′ = u generates another linear MDS code, where u = (u1 , u2 , . . . , uq ). Thus the set S = {c1 , . . . , cq } ∪ {(0, . . . , 0, 1)}, where ci is the i-th column of G′ for 1 6 i 6 q, has size q + 1 and has the property that every subset of S of size k + 1 is a basis. Since k + 1 6 p or 3 6 q − p + 1 6 k + 1 6 q − 2, by Theorem 8, we deduce that S is equivalent to the set {(1, α, α2 , . . . , αk ) | α ∈ Fq } ∪ {(0, . . . , 0, 1)}. Thus we conclude that u(x) = axk + f6k−1 (x), a 6= 0; where f6k−1 (x) denotes a polynomial with degree not larger than k − 1. Case 2. Firstly, we get an estimation of Nmin (k, q). Combining Theorem 6 and Lemma 1, we conclude that Nmin (k, q) 6 Nmin (3, q) + k − 3 √ q−7 6 ⌈q − ⌉+k−3 4 6 q − 1. Now let G be a generator matrix of RSq (Fq∗ , k) of the following form   1 1 ··· 1  α1 α2 · · · αq−1   2  2 2   G =  α1 α2 · · · αq−1  ,  .. .. . . ..   . . .  . k−1 α1k−1 α2k−1 · · · αq−1 18 where α1 , . . . , αq−1 are distinct element of F∗q . From proposition 2, a word u ∈ Fqq−1 is a deep hole of RSq (Fq∗ , k) if and only if G′ =  G u  generates another linear MDS code C2 , where u = (u1 , u2 , . . . , uq−1 ). Since C2 is of length q − 1, thus the matrix G′ is equivalent to a Vandermonde matrix of rank k. Notice that G is the given Vandermonde matrix of rank k − 1. 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Hence there exist a1 , . . . , ak ∈ Fq , not all zero, such that (u1 , . . . , uk+1 ) = a1 r1,k+1 + · · · + ak rk,k+1 , where ri,k+1 is the vector consisting of the first k + 1 elements of the i-th row of G for 1 6 i 6 k. Let v = a1 r1 + · · · + ak rk ∈ C, where ri is the i-th row of G for 1 6 i 6 k. We have d(u, v) 6 n − (k + 1) < ρ, which is a contradiction with the assumption that u is a deep hole of C. 20 ⇐ Now suppose G′ is a generator matrix for an MDS code, i.e., any k + 1 columns of G′ are linearly independent. We need to show that d(u, C) = n − k. Assume that d(u, C) < n − k. Equivalently, there exist a1 , . . . , ak ∈ Fq such that u and v = a1 r1 + · · · + ak rk have more than k common coordinates, where ri is the i-th row of G for 1 6 i 6 k. Without loss of generality, we assume that the first k + 1 coordinates of u and v are the same. Consider the submatrix consisting of the first k + 1 columns. Since the rank of the matrix is less than k + 1, thus the first k + 1 columns of G′ are linearly dependent, which contradicts the assumption. 21