New bounds on $\bar{2}$-separable codes of length 2

Abstract

Let $\mathbb{C }$ be a code of length $n$ over an alphabet of $q$ letters. The descendant code $\mathsf{desc}(\mathbb C _0)$ of $\mathbb C _0 = \{\mathbf{c}^1, \mathbf{c}^2, \ldots , \mathbf{c}^t\} \subseteq \mathbb{C }$ is defined to be the set of words $\mathbf{x} = (x_1, x_2, \ldots ,x_n)$ such that $x_i \in \{c^1_i, c^2_i, \ldots , c^t_i\}$ for all $i=1, \ldots , n$. $\mathbb{C }$ is a $\overline{t}$-separable code if for any two distinct $\mathbb{C }_1, \mathbb{C }_2 \subseteq \mathbb{C }$ such that $|\mathbb{C }_1| \le t$, $|\mathbb{C }_2| \le t$, we always have $\mathsf{desc}(\mathbb{C }_1) \ne \mathsf{desc}(\mathbb{C }_2)$. The study of separable codes is motivated by questions about multimedia fingerprinting for protecting copyrighted multimedia data. Let $M(\overline{t},n,q)$ be the maximal possible size of such a separable code. In this paper, we provide an improved upper bound for $M(\overline{2},2,q)$ by a graph theoretical approach, and a new lower bound for $M(\overline{2},2,q)$ by deleting suitable points and lines from a projective plane, which coincides with the improved upper bound in some places. This corresponds to the bounds of maximum size of bipartite graphs with girth $6$ and a construction of such maximal bipartite graphs.

1 Introduction

Let $n,M$ and $q$ be positive integers, and $Q$ an alphabet with $|Q|=q$. A set $\mathbb{C } = \{\mathbf{c}^1,\mathbf{c}^2,\ldots , \mathbf{c}^M\} \subseteq Q^n$ is called an $(n,M,q)$ code and each $\mathbf{c}^i$ is called a codeword. Without loss of generality, we may assume $Q=\{0,1,\ldots ,q-1\}$.

For any subset of codewords $\mathbb{C }_0\subseteq \mathbb{C }$, we define the set of $i$th coordinates of $\mathbb{C }_0$ as

$$\begin{aligned} \mathbb{C }_0(i)=\{c_i \in Q | \mathbf{c} = (c_1,c_2, \ldots , c_n) \in \mathbb{C }_0\}, \quad 1 \le i \le n, \end{aligned}$$

and the descendant code of $\mathbb{C }_0$ as

$$\begin{aligned} \mathsf{desc}(\mathbb{C }_0)=\{\mathbf{x}=(x_1,x_2,\ldots ,x_n) \in Q^n \ | \ x_i \in \mathbb{C }_0(i), \quad 1 \le i \le n\}, \end{aligned}$$

that is,

$$\begin{aligned} \mathsf{desc}(\mathbb{C }_0)=\mathbb{C }_0(1) \times \mathbb{C }_0(2) \times \cdots \times \mathbb{C }_0(n). \end{aligned}$$

Definition 1.1

Suppose $\mathbb{C }$ is an $(n,M,q)$ code and $t \ge 2$ is an integer. $\mathbb{C }$ is a $\overline{t}$-separable code, or $\overline{t}$-$\text{ SC }(n,M,q)$ in short, if for any $\mathbb{C }_1, \mathbb{C }_2 \subseteq \mathbb{C }$ such that $|\mathbb{C }_1| \le t,\, |\mathbb{C }_2| \le t$ and $\mathbb{C }_1 \ne \mathbb{C }_2$, we always have $\mathsf{desc}(\mathbb{C }_1) \ne \mathsf{desc}(\mathbb{C }_2)$, that is, there is at least one coordinate $i$, $1 \le i \le n$, such that $\mathbb{C }_1(i) \ne \mathbb{C }_2(i)$.

Let $M(\overline{t},n,q) = \text{ max } \{M \ | \ \text{ there } \text{ exists } \text{ a } \overline{t} \text{-SC }(n,M,q)\}$. A $\overline{t}$-SC$(n,M,q)$ is said to be optimal if $M = M(\overline{t},n,q)$, and asymptotically optimal if $\displaystyle \lim _{q \rightarrow \infty } \frac{M}{M(\overline{t},n,q)}=1$.

The study of separable codes is motivated by questions about multimedia fingerprinting which can effectively trace and even identify the sources of pirate copies of copyrighted multimedia data, see, e.g., [6, 19]. It is not difficult to see [6] that identifiable parent property codes [15, 24], frameproof codes [2, 4], perfect hash families [20, 25] and some other structures in digital fingerprinting all imply separable codes.

In multimedia fingerprinting, $\mathsf{desc}(\mathbb{C }_0)$ consists of all the $n$-tuples that could be produced by a coalition holding the codewords in $\mathbb{C }_0$, where the length $n$ corresponds to the number of orthogonal basis signals in the multimedia content. Since the size $M$ of $\overline{t}$-SC$(n,M,q)$ corresponds to the number of fingerprints assigned to authorized users, we should try to construct separable codes with size $M$ as large as possible, given length $n$. Cheng and Miao [6] showed that long-length separable codes can be constructed by concatenating short-length separable codes. This stimulates the investigation of separable codes with length $n=2$.

In [7] an upper bound on $M(\overline{2},2,q)$ was derived, and two infinite series of optimal $\overline{2}$-$(2,M,q)$-SCs were constructed.

Theorem 1.2

[7] For any positive integer $q$, $M(\overline{2},2,q) \le qk+t$, where $k=\lfloor \frac{1+\sqrt{4q-3}}{2} \rfloor$, and

$$\begin{aligned} \displaystyle t = \left\{ \begin{array}{l@{\quad }l} \lfloor \frac{q(q-1-k^2+k)}{2k}\rfloor , &{} \quad { if~~} k^2-k+1 \le q\le k^2;\\ \lfloor \frac{qk}{(k+1)^2-q}\rfloor , &{} \quad { if~~} k^2+1\le q \le k^2+k. \end{array} \right. \end{aligned}$$

Furthermore, $M(\overline{2},2,q) = qk+t$ if $q = k^2-k+1$ for any prime power $k-1\ge 2$ and $q = k^2+k$ for any prime power $k\ge 2$.

In this paper, by using graph theoretical terminologies, we obtain a tighter upper bound on $M(\overline{2},2,q)$. By using projective geometrical terminologies, we also obtain a lower bound on $M(\overline{2},2,q)$, parts of which agree with the new derived upper bound. In other words, we construct several new infinite series of optimal $\overline{2}$-$(2,M,q)$-SCs.

2 Related combinatorial objects

In order to investigate separable codes, in this section, we describe several related combinatorial structures.

For any $(2,M,q)$ code $\mathbb{C }$ defined on $Q=\{0,1,\ldots ,q-1\}$, we define $A_i$ for $i \in Q$ as $A_i = \{x_2 \ | \ (x_1, x_2) \in \mathbb{C }, \ x_1=i\}$. Obviously, $A_i \subseteq Q$ holds for any $i \in Q$, and $|A_0|+ |A_1| + \cdots + |A_{q-1}|=M$.

Definition 2.1

Let $K$ be a subset of non-negative integers, and $v,b$ be two positive integers. A generalized $(v,b,K,1)$ packing is a pair $(X,\mathbb{B })$ where $X$ is a set of $v$ elements and $\mathbb{B }$ is a set of $b$ subsets of $X$ called blocks that satisfy

• (1) $|B| \in K$ for any $B\in \mathbb{B }$;

• (2) every pair of distinct elements of $X$ occurs in at most one block of $\mathbb{B }$.

Cheng et al. [7] showed a relationship between separable codes and generalized packings.

Lemma 2.2

[7] There exists a $\overline{2}$-SC$(2,M,q)$ defined on $Q$ if and only if there exists a generalized $(q,q,K,1)$ packing $(Q, \{A_0, A_1, \ldots , A_{q-1}\})$, with $K=\{|A_0|, |A_1|,\ldots , |A_{q-1}|\}$, and $M = |A_0| + |A_1| + \cdots + |A_{q-1}|$.

A generalized $(q,q,\{k\},1)$ packing can be constructed by developing a near difference set. A $(q,k,1)$ near difference set defined on an additively written group $G$ of order $|G| = q$ is a $k$-subset $D$ of $G$ such that the differences $\{x-y \ | \ x,y \in D, x \ne y\}$ contains $k(k-1)$ distinct elements of $G$.

Lemma 2.3

For any integer $k\ge 2$, let $q \ge k^2-k+1$. If there exists a $(q,k,1)$ near difference set, then there exists a generalized $(q, q, \{k\},1)$ packing.

Proof

Let $D$ be a $(q,k,1)$ near difference set defined on an additively written group $G$. For any $g \in G$, define $D+g = \{x+g \ | \ x \in D\}$ and $\mathbb B = \{D+g \ | \ g \in G\}$. Then $(G, \mathbb B )$ is the desired generalized $(q, q, \{k\},1)$ packing.$\square$

Near difference sets are not easy to construct. However, a $(k^2+k+1,k,1)$ near difference set always exists [23] for any prime power $k$. This Singer difference set generates a generalized $(k^2+k+1, k^2+k+1, \{k\},1)$ packing, which corresponds to an optimal $\overline{2}-\text{ SC }(2,(k+1)(k^2+k+1),k^2+k+1)$ described in Theorem 1.2.

Given a generalized $(v,b,K,1)$ packing $(Q,\mathbb B )$, we can define its associated element-block graph as the bipartite graph $G_{Q,\mathbb B }$ with vertex partition $Q$ and $\mathbb B$ such that $x\in Q$ is adjacent to $B \in \mathbb B$ if and only if $x\in B$. It is clear that the corresponding element-block graph of a generalized $(v,b,K,1)$ packing $(Q,\mathbb B )$ is a $C_4$-free bipartite subgraph of $K_{v,b}$, because any pair of distinct elements of $Q$ can occur in at most one block of $\mathbb B$. In other words, the girth of this bipartite graph is at least $6$, where the girth of a graph is the length of a shortest cycle contained in the graph.

Zarankiewicz numbers [27] involve bounds on the maximum number of edges in a bipartite graph without a particular subgraph. We denote by $z(m,n;s,t),\, m\le n$ and $s\le t$, the maximum number of edges in a subgraph of $K_{m,n}$ that does not contain a subgraph isomorphic to $K_{s,t}$. In particular, when $m=n$ and $s=t$, simply put $z(n;t)=z(n,n;t,t)$. It is clear that $z(q;2)$, which is the maximum size of a $C_4$-free bipartite subgraph of $K_{q,q}$, is equals to $M(\overline{2},2,q)$ by Lemma 2.2. Meanwhile, García-Vázquez et al. [11] stated that any $C_4$-free bipartite subgraph of $K_{q,q}$ with size $z(q;2)$ must have girth $6$. Therefore, our problem is equivalent to finding the maximum size of bipartite graphs with girth $6$, where the size of a graph refers to the number of edges it contains, and constructing such maximal bipartite graphs.

We can see our problem in one more way. Given a generalized $(q,q,K,1)$ packing $(Q,\mathbb B )$, if we define two elements of $Q$ are adjacent in $B \in \mathbb B$ if they occur in the same block $B$, then each block can be seen as a clique of order $|B|$ belonging to $K$. Since each pair of distinct elements of $Q$ occurs in a block of $\mathbb B$ at most once, this generalized $(q,q,K,1)$ packing can be viewed as a packing of $K_q$ by $q$ cliques of orders belonging to $K$. Therefore, in order to evaluate $z(q;2)=M(\overline{2},2,q)$, it is sufficient to pack $K_q$ by $q$ cliques so that the sum of order of the $q$ cliques is maximum.

It is well known [3] that $z(q;2) \le (q+q \sqrt{4q-3})/2$ and the equality holds when $q=k^2+k+1$ for any prime power $k$. Goddard et al. [12] found the exact values of $z(q;2)$ for $q \le 10$. Very recently, Damásdi et al. [8] also found the exact values of $z(q;2)$ for $q = k^2+k-2, k^2+k-1$ with $k$ being a prime power, among others. Theorem 1.2 is an improvement of the results made by Cheng et al. [7]. It is also known [3] that if $q$ is sufficiently large then

$$\begin{aligned} q^{3/2} - q^{4/3} < z(q;2) \le (q+q \sqrt{4q-3})/2; \end{aligned}$$

In particular, $\displaystyle \lim _{q \rightarrow \infty } \frac{z(q;2)}{q^{3/2}}=1$. For the up-to-date information on Zarankiewicz numbers, the reader is referred to [8].

3 Upper bound

Bipartite graphs with high girth and their related graphs have been extensively investigated, see, e.g., [1, 8–11, 13, 16–18, 21, 22, 26]. We start this section with the following proposition.

Proposition 3.1

[5] Suppose $(X,\mathbb B )$ is a generalized $(v,b,\{k,k+1\},1)$ packing, for some $k$, with $\mathbb B = \{B_1, B_2, \ldots , B_b\}$. If $\displaystyle {\left( \begin{array}{l}v\\ 2\end{array}\right) } - \sum _{i=1}^b {\left( \begin{array}{c}|B_i|\\ 2\end{array}\right) } < k$, then $G_{X,\mathbb B }$, the element-block graph of $(X,\mathbb B )$, is a $C_4$-free subgraph of $K_{v,b}$ with maximum size.

If $K_q$ can be packed by $q$ cliques $K_{x_1}, K_{x_2}, \ldots , K_{x_q}$ with leave $L$, where $x_i\le x_j$ for $1 \le i < j \le q$, then we say $K_q$ admits a feasible $(x_1,x_2,\ldots , x_q)$ packing with leave $L$. For convenience, we replace $(x_1,x_2,\ldots , x_q)$ packing by $(k^{q-t},(k+1)^t)$ packing when

$$\begin{aligned} k = x_1=\cdots =x_{q-t} \mathrm{~and~} x_{q-t+1}=\cdots =x_q=k+1 \end{aligned}$$

for some $k\in \mathbb N$ and $1\le t\le q$. For any $(k^{q-t},(k+1)^t)$ packing $\mathcal P$ of $K_q$, we have

$$\begin{aligned} q{\left( \begin{array}{l}k\\ 2\end{array}\right) } \le (q-t) {\left( \begin{array}{l}k\\ 2 \end{array}\right) } + t{\left( \begin{array}{c}k+1\\ 2\end{array}\right) } \le {\left( \begin{array}{l}q\\ 2 \end{array}\right) }, \end{aligned}$$

which implies $k\le \frac{1+\sqrt{4q-3}}{2}$. In order to maximize $\sum _{i=1}^q x_i$ which subjects to an $(x_1,x_2,\ldots , x_q)$ packing, Proposition 3.1 promises to consider a feasible $(k^{q-t},(k+1)^t)$ packing with $k = \lfloor \frac{1+\sqrt{4q-3}}{2} \rfloor$ and $|L|<k$. Therefore, our objective is to find the maximum index $t$. Note that $k = \lfloor \frac{1+\sqrt{4q-3}}{2} \rfloor$ implies $k^2-k+1\le q< k^2+k+1$. In this section, we investigate $z(q;2)$ by fixing the index $k$ and then classifying $q$, from $k^2-k+1$ to $k^2+k$, into several cases. The following Theorem 3.2 is contained in Theorem 1.2.

Theorem 3.2

[3, 7] For any prime power $k-1 \ge 2,\, z(k^2-k+1;2)=k^3-k^2+k$. For any prime power $k \ge 2,\, z(k^2+k;2) = k^3+2k^2$.

Theorem 3.3

For any $k^2+1\le q\le k^2+k-2$ and $k\ge 2$, we have

$$\begin{aligned} z(q;2)\le qk+ \left\lfloor \frac{(k-1)q}{(k+1)^2-(q+1)} \right\rfloor . \end{aligned}$$

Proof

Let $q=k^2+k-s,\, s=2,3,\ldots ,k-1$. Assume $\mathcal P$ is a $(k^{q-t},(k+1)^t)$ packing of $K_q$, where $0\le t\le q-1$. We claim by contradiction that $t\le \lfloor \frac{(k-1)q}{(k+1)^2-(q+1)} \rfloor$. That is, suppose $t > \lfloor \frac{(k-1)q}{(k+1)^2-(q+1)} \rfloor$.

For $i\ge 0$, let $r_i$ be the number of vertices that is contained in exactly $i$ cliques of order $k+1$ in $\mathcal P$. Since the degree of each vertex is $k^2+k-s-1$, trivially $r_i=0$ for all $i> k$. We now claim $r_k=0$. Suppose not, that is, there exists a vertex $v$ contained in exactly $k$ cliques of order $k+1$, say $C_1, C_2, \ldots , C_k$. Let $A=\displaystyle \{v\} \cup \bigcup _{i=1}^k V(C_i)$ and $B=V(K_q)\setminus A$, where $V(G)$ is the vertex set of graph $G$. Since there is no other subgraph isomorphic to $K_{k+1}$ out of $A$ except $C_1, C_2, \ldots , C_k$, each of the remaining cliques of order $k+1$ must contain at least one vertex in $B$. That is, each of such cliques needs at least $k$ edges between $A$ and $B$. Therefore, we have at most $k+ \lfloor \frac{(k^2+1)(k-s-1)}{k} \rfloor$ cliques of order $k+1$. Thus,

$$\begin{aligned} k(k-s) \ge k+ \left\lfloor \frac{(k^2+1)(k-s-1)}{k} \right\rfloor \ge t > \left\lfloor \frac{(k-1)q}{(k+1)^2-(q+1)} \right\rfloor . \end{aligned}$$

This implies $ks^2-ks-k+s<0$, so $ks(s-1) \le k-s < k$, that is, $s(s-1) <1$, which contradicts $2\le s\le k-1$. So $r_k=0$.

Consider the number of ordered pairs $(v,C)$, where $v$ is a vertex in the clique $C$ of order $k+1$ in $\mathcal P$. Under our assumption, there are exactly $t$ cliques of order $k+1$, then

$$\begin{aligned} t(k+1) = (k-1)r_{k-1} + (k-2)r_{k-2} + \cdots + (k-s)r_{k-s} + \cdots +r_1. \end{aligned}$$

(1)

This implies that

$$\begin{aligned} t(k+1) \le (k-1)(r_{k-1}+ \cdots + r_{k-s+1}) + (k-s)(q- (r_{k-1}+ \cdots + r_{k-s+1})), \end{aligned}$$

so

$$\begin{aligned} \frac{t(k+1)-q(k-s)}{s-1}\le r_{k-1}+ \cdots + r_{k-s+1}. \end{aligned}$$

(2)

Now, we drop all the $t$ cliques of order $k+1$ from $K_q$. Denote by $G$ the remaining subgraph. We again consider the number of ordered pairs $(v^{\prime },C^{\prime })$, where $v^{\prime }$ is a vertex in the clique $C^{\prime }$ of order $k$ in $\mathcal P$. On one hand there are exactly $q-t$ cliques of order $k$, and on the other hand there are exactly $r_{i}$ vertices of degree $q-1-ki$, for $i=0,1, \ldots , k-1$. Since the vertex of degree $q-1-ki$ can be contained in at most $\frac{q-1-ki}{k-1}$ cliques of order $k$, we have

$$\begin{aligned} (q-t)k \le r_{k-1}+2r_{k-2}+ \cdots + (s-1)r_{k-s+1} + (s+1)r_{k-s}+ \cdots + (k+1)r_0. \end{aligned}$$

(3)

Combining (1) and (3) we have

$$\begin{aligned} t(k+1)+(q-t)k \le k(r_{k-1}+ \cdots + r_{k-s+1}) + (k+1)(q-(r_{k-1}+ \cdots + r_{k-s+1})), \end{aligned}$$

and thus

$$\begin{aligned} r_{k-1}+ \cdots + r_{k-s+1} \le q-t. \end{aligned}$$

(4)

Finally, (2) and (4) imply that $t\le \frac{q(k-1)}{k+s}=\frac{(k-1)q}{(k+1)^2-(q+1)}$, a contradiction to the hypothesis. Thus we complete the proof.$\square$

Theorem 3.4

For any $q = k^2$ with $k\ge 2$, we have

$$\begin{aligned} z(q;2)\le qk+ \lfloor \frac{(3k^2+k-1)-\sqrt{5k^4+6k^3-k^2-2k+1}}{2} \rfloor . \end{aligned}$$

Proof

Assume $\mathcal P$ is a $(k^{q-t},(k+1)^t)$ packing of $K_q$. For $i\ge 0$, let $r_i$ be the number of vertices that is contained in exactly $i$ cliques of order $k+1$ in $\mathcal P$. Since $q=k^2$, we have $r_i=0$ for all $i\ge k$. Similar to the Proof of Theorem , we first consider the number of ordered pairs $(v,C)$, where $v$ is a vertex in the clique $C$ of order $k+1$ in $\mathcal P$. Then after dropping those cliques of order $k+1$, we consider the number of ordered pairs $(v^{\prime },C^{\prime })$, where $v^{\prime }$ is a vertex in the clique $C^{\prime }$ of order $k$ in $\mathcal P$. Note that in the remaining graph after dropping $t$ cliques of order $k+1$, there are exactly $r_i$ vertices of degree $k^2-ik-1$. Then we have

$$\begin{aligned} \displaystyle \left\{ \begin{array}{ll} t(k+1) &{} = ~(k-1)r_{k-1}+ \cdots + 2r_2+r_1\\ (q-t)k &{} \le ~r_{k-1}+ \cdots + (k-2)r_2+(k-1)r_1+(k+1)r_0 \end{array} \right. \end{aligned}$$

which implies $t\le r_0$. This concludes that the $t$ cliques of order $k+1$ are out of at most $k^2-t$ vertices somewhere in $\mathcal P$. We immediately have

$$\begin{aligned} t {\left( \begin{array}{c}{k+1}\\ 2\end{array}\right) } \le {\left( \begin{array}{c}k^2-t \\ 2\end{array}\right) }. \end{aligned}$$

That is,

$$\begin{aligned} t^2+(1-k-3k^2)t+(k^4-k^2)\ge 0. \end{aligned}$$

Since $t \le k^2$, the above inequality is true only when $t\le \frac{(3k^2+k-1)-\sqrt{5k^4+6k^3-k^2-2k+1}}{2}$. Hence we complete the proof.$\square$

Theorem 3.5

For any $k^2-k+2\le q\le k^2-1$ and $k\ge 2$, we have $z(q;2)\le qk$.

Proof

Let $q=k^2-s$, where $s=1,2,\ldots ,k-2$. Assume $\mathcal P$ is a $(k^{q-t},(k+1)^t)$ packing of $K_q$. Suppose $t\ge 1$. Define $G$ to be the graph by dropping one of the cliques of order $k+1$, say $\widehat{K}$, from $K_q$. Let $A\subseteq V(G)$ be the collection of vertices whose degree is equal to $q-1-k$, and $B=V(G)\setminus A$. Note that $|A|=k+1$ and $|B|=q-k-1$. Now, consider the number of ordered pairs $(v,C)$, where $v$ is a vertex in the clique $C$ in $\mathcal P$ different from $\widehat{K}$. Notice that for each $v\in A,\, \mathsf{deg}_G(v)=k^2-s-1-k=(k-1)^2+(k-s-2)$, then $v$ is contained in at most $k-1$ cliques different from $\widehat{K}$. Similarly, each vertex in $B$ can be contained in at most $k$ cliques. By counting the number of pairs $(v,C)$, we have

$$\begin{aligned} (t-1)(k+1)+(q-t)k \le (k+1)(k-1)+(q-k-1)k. \end{aligned}$$

This implies that $t \le 0$, a contradiction occurs. Thus the result follows.$\square$

4 Lower bound

Now we derive a lower bound on $z(q;2)= M(\overline{2},2,q)$ via projective planes. A projective plane consists of a set of lines, a set of points, and a relation between points and lines called incidence, having the following properties:

• (1) Given any two distinct points, there is exactly one line incident with both of them.

• (2) Given any two distinct lines, there is exactly one point incident with both of them.

• (3) There are four points such that no line is incident with more than two of them.

Clearly, a projective plane of order $k$ is a generalized $(k^2+k+1,k^2+k+1,\{k+1\},1)$ packing $(X, \mathbb{B })$ in which every pair of distinct elements of $X$ occurs in exactly one block of $\mathbb{B }$. It is well-known [14] that a projective plane of order $k$ always exists for any prime power $k$.

Theorem 4.1

For any prime power $k \ge 2$, let $k^2-1 \le q \le k^2+k-1$. Then there exists a generalized $(q,q,\{k,k+1\},1)$ packing, $(X^{\prime },\mathbb B^{\prime } )$, with $|X^{\prime }| = |\mathbb B^{\prime } | = q$ such that exactly $k^3-k^2-k-qk+2q+1$ blocks out of $\mathbb B^{\prime }$ are of size $k$. That is,

$$\begin{aligned} z(q;2) \ge 2qk-k^3+k^2+k-q-1. \end{aligned}$$

Proof

We start from a projective plane of order $k$, $(X,\mathbb B )$. Note that $|X|=|\mathbb B |=k^2+k+1$, and for any $B \in \mathbb B ,\, |B|=k+1$. Pick an arbitrary point $a \in X$ and an arbitrary line $B^* = \{ x_1,x_2,\ldots , x_{k+1} \}\in \mathbb B$ which does not contain the point $a$. For each $i=1,\ldots , k+1$, let $B_i \in \mathbb{B }$ be the line containing the points $a$ and $x_i$. Let $2 \le s \le k+2$. Dropping $s$ lines $B^*, B_1, \ldots , B_{s-1}$ and $s$ points $a, x_1,\ldots , x_{s-1}$ from $(X,\mathbb B )$, we obtain a generalized $(q,q,\{k,k+1\},1)$ packing, $(X^{\prime },\mathbb B^{\prime } )$, with $q=k^2+k+1-s,\, X^{\prime }= X \setminus \{a,x_1,\ldots ,x_{s-1}\},\, \mathbb B^{\prime } = \mathbb{B } \setminus \{ B^*, B_1, \ldots , B_{s-1}\}$, having ${\Delta } = (s-1)(k-1)+(k+1-s+1)=k^3-k^2-k-qk+2q+1$ blocks of size $k$ and $k^2+k+1-{\Delta }-s$ blocks of size $k+1$. Therefore, $z(q;2) \ge k{\Delta }+(k+1)(k^2+k+1-\Delta -s) = 2qk-k^3+k^2+k-q-1$.$\square$

Applying Theorems 1.2, and 3.5, we immediately have the following result.

Corollary 4.2

For any prime power $k \ge 2,\, z(k^2-1;2)= k^3-k,\, z(k^2+k-2;2)= k^3+2k^2-4k+1,\, z(k^2+k-1;2)= k^3+2k^2-2k$.

We remark that Damaásdi et al. [8] obtained independently the same results for $q=k^2+k-2, k^2+k-1$ in Corollary 4.2. It is easy to verify that the corresponding $\overline{2}$-$\text{ SC }(2,M,q)$s constructed in Theorem  are asymptotically optimal for all $k^2-1 \le q \le k^2+k-1$ with prime power $k$. The lower bound described in Theorem  is better than $q^{3/2} - q^{4/3}$ described in [3] for any prime power $k$.

5 Summary

The main results in the previous sections can be summarized in the following theorem.

Theorem 5.1

For any positive integer $q,\, M(\overline{2},2,q) \le qk+t$, where $k=\lfloor \frac{1+\sqrt{4q-3}}{2} \rfloor$, and

$$\begin{aligned} \displaystyle t = \left\{ \begin{array}{l@{\quad }l} 0 &{} { if~~} k^2-k+1\le q\le k^2-1;\\ \left\lfloor \frac{(3k^2+k-1)-\sqrt{5k^4+6k^3-k^2-2k+1}}{2} \right\rfloor &{} { if~~} q = k^2;\\ \left\lfloor \frac{(k-1)q}{(k+1)^2-(q+1)}\right\rfloor &{} { if~~} k^2+1\le q\le k^2+k-2;\\ k^2-k &{} { if~~} q = k^2+k-1;\\ k^2 &{} { if~~} q = k^2+k. \end{array} \right. \end{aligned}$$

Furthermore, $M(\overline{2},2,q) = qk+t$ if $q=k^2-k+1$ for any prime power $k-1 \ge 2$, and $q=k^2-1, k^2+k-2,k^2+k-1, k^2+k$ for any prime power $k \ge 2$.

The above Figs. 1 and 2 illustrate our improvement on the upper bound of $M(\overline{2},2,q)$. Figure 1 depicts the known upper bound given in [7] and the new upper bound given in Theorem 5.1 when $k=12$, while Fig. 2 depicts those upper bounds when $q=k^2$. It can be seen that our new upper bound is much tighter than the known upper bound.

Fig. 1

Bounds for $k=12$

Fig. 2

Bounds for $q=k^{2}$