THE RESULTS ON OPTIMAL VALUES OF SOME COMBINATORIAL BATCH CODES

. An ( n,N,k,m )-combinatorial batch code (CBC) was deﬁned by Paterson, Stinson and Wei as a purely combinatorial version of batch codes which were ﬁrst proposed by Ishai, Kushilevitz, Ostrovsky and Sahai. It is a system consisting of m subsets of an n -element set such that any k distinct elements can be retrieved by reading at most one (or in general, t ) elements from each subset and the number of total elements in m subsets is N . For given parameters n,k,m , the goal is to determine the minimum N , denoted by N ( n,k,m ). So far, for k ≥ 5, m + 3 ≤ n < (cid:0) mk − 2 (cid:1) , precise values of N ( n,k,m ) have not been established except for some special parameters. In this paper, we present a lower bound on N ( n,k,k + 1), which is tight for some n and k . Based on this lower bound, the monotonicity of optimal values of CBC and several constructions, we obtain N ( m + 4 , 5 ,m ), N ( m + 4 , 6 ,m ) and N ( m + 3 , 7 ,m ) in diﬀerent ways.


Introduction
Batch codes were introduced by Ishai, Kushilevitz, Ostrovsky and Sahai [7]. A batch code specifies a method to distribute a database of n items among m servers in such a way that any k items can be retrieved by reading at most t items from each server, and the total number of items stored in m servers is N . It can be employed in a distributed database scenario for load balancing. The goal is to minimize the total storage, which is interesting and practically important.
Batch codes can be defined as follows: [7] Section 1.1) An (n, N, k, m, t) batch code over an alphabet encodes a string x ∈ n into an m-tuple of strings y 1 , y 2 , . . . , y m ∈ * (also referred to as servers) of total length N , such that for each k-tuple (batch) of distinct indices i 1 , i 2 , . . . , i k ∈ {1, 2, . . . , n}, the entries x i1 , x i2 , . . . ,x i k from x can be decoded by reading at most t symbols from each server.
General batch codes allow encoding/decoding during the storage/retrival process. A simplified version, called replication-based batch code [7], is where each server stores a subset of items and decoding simply means reading items from servers. Paterson, Stinson and Wei [9] call this construct a combinatorial batch code. They introduce the study of this family of batch codes.
Combinatorial batch codes can be defined as follows: In this definition, the points (elements of X) and blocks (subsets in B) correspond to items and servers, respectively, and t serves to balance the load among servers. We will only consider the case t = 1 in this paper. Such a batch code permits at most one item to be retrieved from each server. We follow the convention as given in [9], and denote (n, N, k, m, 1)-CBC by (n, N, k, m)-CBC or (n, k, m)-CBC.
Recall that N is the total number of items stored in m servers. To save the total storage, we want N to be as small as possible. An (n, N, k, m)-CBC is optimal [9] if N ≤ N for all (n, N , k, m)-CBC. We denote the minimum of N by N (n, k, m). Given n, k, m, the objective is to determine N (n, k, m) and to give an explicit construction of an optimal CBC with corresponding parameters.
Precise values of N (n, k, m) have been given for the majority of parameters. Some known results are summarized as follows: (i) ([9] Theorem 2.9) If n ≥ (k − 1) m k−1 , then Therefore, the problem of determining N (n, k, m) has been totally solved when n ≥ m k−2 . For small n, the case of n = m is trivial and N (n, k, n) = n ([9] Theorem 2.1). In [9], Paterson et al. obtain an optimal solution for the special case m = k. They also give a construction of (n, N, k, m)-CBC when n is not too much bigger than m, and prove the construction is optimal in the case n = m + 1.
Brualdi et al. [2,3] explore the connection between combinatorial batch codes and transversal matroids. By applying various properties of the latter and analyzing the case of optimality, they determine the function N (m + 2, k, m). And Bujtás et al. [5] obtain N (m + 2, k, m) by a purely combinatorial proof and several explicit constructions.
(iv) ([2] Theorem 4 and [5] Theorem 2) For any two integers m and k, if k ≤ m ≤ k + √ k, then Besides, N (n, k, m) with fixed k is also interesting. The case of k = 1 is trivial and N (n, 1, m) = n. From (i) and (ii), we can easily obtain N (n, 2, m) and N (n, 3, m). The value of N (n, 4, m) is presented in [4], which turns out to be more complex than that of N (n, 3, m).
Apart from the above results, Silberstein and Gál [11] provide an optimal (q 2 + q − 1, q 3 − q, q 2 − q − 1, q 2 − q)-CBC based on resolvable transversal designs, i.e. (2) where q ≥ 3 is a prime power. Chen et al. [6] show the monotonicity of optimal values of CBC. By using the monotonicity and some constructions, Jia, Zhang and Yuan [8] obtain the following results.  In this paper, we obtain the following conclusions in different ways. 2) Let (X, B) be an optimal (n, N, k, m)-CBC. If 2 ≤ k ≤ m < n, then there exists at least one point in X whose degree is greater than 1. Furthermore, the blocks containing a point of degree greater than 1 have length greater than 1.
Next we introduce some known results on monotonicity of optimal values of combinatorial batch codes.
By adding a new point and a singleton containing it to an optimal (n, N, k, m)-CBC, we obtain an (n + 1, N + 1, k, m + 1)-CBC. Therefore, the following lemma holds.
By repeatedly applying Lemma 1.4, we have where p is a non-negative integer. In fact, add p new points and p singletons to an optimal (n, N, k, m)-CBC such that each new point belongs to a unique singleton.
Then the system derived is an (n + p, N + p, k, m + p)-CBC. Lemma 1.3 implies that the number of blocks in an optimal (n + 1, N, k, m)-CBC will not be decreased if we delete a point of degree d (d ≥ 2) from the point set and the blocks containing it. Furthermore, the system derived is an (n, N − d, k, m)-CBC. Hence we have To complete the proof of the main theorems in Section 5, we need to improve the bound in Lemma 1.6.
If there exists a point of degree one which belongs to a block of length s, then Proof. Let x ∈ X be a point of degree one, x ∈ B ∈ B and |B| = s. If s = 1, then from Lemma 1.6, the proof is completed. Next we suppose that s ≥ 2. Notice that if B contains another point y of degree one, then the two points x, y only can be retrieved from the unique block B. Since k ≥ 2, it follows that each point in B except for x has degree at least two, which implies that replacing B with {x} will not decrease the cardinality of B∈B B which equals |X|.
Define (X, B ) as the set system obtained from (X, B) by replacing B with B = {x}. We claim that (X, B ) is an (n, N 1 − s + 1, k − 1, m)-CBC. In fact, for any (k − 1)-subset X 1 of X, if x ∈ X 1 , then X 1 can be retrieved in B by the same way as that in B. If x does not belong to X 1 , then {x} ∪ X 1 is a k-subset of X and can be retrieved in B in such a way that x is obtained from B and X 1 is obtained from B\{B}. Hence X 1 can be retrieved in B . This paper is organized as follows. In Section 2, we provide a lower bound on N (n, k, k + 1), which is tight for some n and k. By employing this lower bound, some results on the monotonicity of optimal values and several constructions, we determine N (m + 4, 5, m) and N (m + 3, 7, m) in Section 3 and 5, respectively. In Section 4, we study the condition for which the conclusion in Lemma 1.4 is an equality if n = m + 4, k = 6. Furthermore, we obtain N (m + 4, 6, m).

2.
A lower bound on N (n, k, k + 1) To complete the results in the following sections, we provide a lower bound on N (n, k, k + 1) in this section.
Theorem 2.1. Let n, k be integers and n > k ≥ 2. If n − k is odd, then If n − k is even and n − k ≥ 4, then Proof. Let (X, B) be an (n, N, k, k + 1)-CBC. If there exist two blocks in B with union of size at most n − k, then there exist at least k points contained in neither of the two blocks which only can be retrieved from the remaining k − 1 blocks. This contradicts the fact that the combinatorial batch codes with t = 1 permit at most one point to be retrieved from each block. Therefore the union of any two blocks has length at least n − k + 1 in an (n, N, k, k + 1)-CBC, which implies that the number of blocks of size less than n−k+1 2 is at most 1. Suppose that the minimum size of blocks in B is x. If x > n−k+1 2 , then , then there exists exactly one block in B of size x and the others have size at least n − k + 1 − x. It follows that , then except for the blocks of size x, the others have size at least n − k + 1 − x = x, so the inequality (6) also holds. Hence we obtain a lower bound on N (n, k, k + 1) when x = n−k+1 N (n, k, k + 1) ≥ (n − k + 3)(k + 1) 2 − 2n n − k + 1 .
If n − k + 1 is odd and B contains a block B 0 of length n−k 2 ≥ 2, then the other blocks have length at least n−k+2 2 , and each of the blocks of size n−k+2 2 is disjoint from B 0 . Since k ≥ 2, there exists at least one block intersecting B 0 and of size greater than n−k+2 2 . Otherwise the points of B 0 only can be retrieved from the unique block B 0 . Therefore, we have Note that some direct calculations show that both the lower bounds in (7) and (8) are not greater than that in (6) when x < n−k+1 2 .
Notice that the bound in Theorem 2.1 is tight for some n and k. For example, from Theorem 2.1, we have where the equality holds according to Equation (1). Also the bound for n = q 2 +q−1, k = q 2 − q − 1 is tight using Equation (2), where q ≥ 3 is a prime power. However for some n, k, the bound in Theorem 2.1 is definitely not tight. (ii) B consists of eight blocks of size 6. For (i), since the union of any two blocks has length at least 11, each pair of blocks of size 6 and 5 are disjoint. Hence the 5 points contained in the block of size 5 only can be retrieved from at most 2 blocks in B, which leads to a contradiction.
For (ii), note that the intersection of any two blocks has length at most 1. Hence the union of the eight blocks in B has cardinality at least 6 × 8 − 8 2 = 20, which contradicts the fact that n = 17. Therefore N (17, 7, 8) > 48.
In what follows we determine N (m + 4, 5, m) for m = 6 and m = 7.  We can infer that the union of any three blocks in an (11,5,7)-CBC has length at least 7 since if there exist three blocks with union of size at most 6, then the remaining at least 5 points will be retrieved from the remaining 4 blocks. Hence at most two pairs belong to B.
If B contains exactly one pair, then B consists of one pair and six triplets, say, B 1 = {u, v}. Since the two points u, v should be retrieved from at least two blocks in B, there is an another block containing u or v, say B 2 . If |B 1 ∩ B 2 | = 2, then for any other block B ∈ B we have |B 1 ∪ B 2 ∪ B| < 7. Thus |B 2 \ B 1 | = 2. Likewise, since the three points of B 2 should be retrieved from at least three blocks, at least one point of B 2 \ B 1 belongs to another block B 3 . Then we have |B 1 ∪ B 2 ∪ B 3 | < 7, which leads to a contradiction.
If two pairs belong to B, then B consists of two pairs, four triplets and one block of size 4. Let |B 1 | = |B 2 | = 2, |B 3 | = 4. Recall that the union of any three blocks has length at least 7, which suggests that B 1 ∩ B 2 = ∅ and each triplet is disjoint from B 1 ∪ B 2 . Hence we can only obtain the 4 points contained in B 1 ∪ B 2 from the 3 blocks B 1 , B 2 and B 4 . This yields a contradiction.
Because of Lemma 1.5, if there exists a point of degree at least 3 in X, then N (m + 4, 6, m + 1) ≤ m + 16 − 3 = m + 13, which is contrary to (4). Therefore, all points in X have degree either one or two. Since |X| = m + 5 and N = m + 16, it follows that the number of points of degree one is m − 6.
Since the two points of a pair should be retrieved from two blocks, at least one point in a pair has degree two. Suppose that both the points in a pair B 1 ∈ B have degree two. Let B 1 = {x, y}. If both x and y belong to another block B 2 , then B 1 = B 2 (i.e. |B 2 | = 2) since otherwise B remains an (m + 5, 6, m + 1)-CBC when B 2 is replaced with {y}, contradicting minimality. Next we apply some local modifications to (X, B). If x, y ∈ B 2 and |B 2 | > 2, then we delete x from X, and replace B 1 and B 2 with B 1 = {y} and B 2 = B 2 \ {x, y}, respectively. If x ∈ B 2 and y ∈ B 3 , then B 2 = B 3 since both x and y have degree 2. We delete x from X and replace B 1 , B 2 and B 3 with B 1 = {y}, B 2 = B 2 \{x} and B 3 = B 3 \{y}, respectively. Recall that B contains no singleton, which implies that neither B 2 nor B 3 is empty.
Let Y = X \ {x}. We can claim that the set system (Y, B ) derived by applying the modifications above satisfies the definition of CBC with k = 6. In fact, for any 6-element subset X of Y , if X does not contain y, then it can be retrieved by the same method as employed in the original CBC. If X contains y, then the point y can be obtained from B 1 , and the remaining points in X can be retrieved in the same way as shown in the original CBC. Therefore (Y, B ) is an (m+4, m+16−3, 6, m+1)-CBC, which contradicts the fact that N (m + 4, 6, m + 1) = m + 14.
Thus only one point has degree two in each pair of (X, B), and the other has degree one. Recall that the number of pairs is at least 2m − 13, so is the number of points of degree one. It follows that m − 6 ≥ 2m − 13, which implies that m ≤ 7. Therefore if m ≥ 8, then N (m + 5, 6, m + 1) = N (m + 4, 6, m) + 1.
Assume that N (m + 3, 7, m) = m + 13. Then all points in the optimal (m + 3, N, 7, m)-CBC have degree at most 2. Let (X 1 , B 1 ) be an optimal (m + 3, m + 13, 7, m)-CBC. It is obvious that the number of points of degree one in X 1 is m − 7. If each of the points of degree one belongs to a singleton, then omitting these points and the singletons containing them, we can get a (10, 20, 7, 7)-CBC. This contradicts the fact that N (10, 7, 7) = 28. Hence at least one point of degree one belongs to a block of size greater than one. From Theorem 1.7, we have N (m + 3, 6, m) ≤ N (m + 3, 7, m) − 1 = m + 12, but N (m + 3, 6, m) = m + 13. This leads to a contradiction. Hence for m ≥ 9, N (m + 3, 7, m) = m + 14.