Return to Answer

$$\operatorname{ex}(4k+1,2k,2k) \geq 4k-1$$ if there exists a $(4k-1,2k,k)$-SBIBD. If you don't know what symmetric designs are, see your favorite design theory textbook or, better yet, the original article

Y. M. Chee, C. J. Colbourn, A. C. H. Ling, R. M. Wilson, Covering and packing for pairs, J. Combin. Theory Ser. A 120 (2013) 1440–1449

$$\operatorname{ex}(4k+1,2k,2k) \geq 4k-1$$ if there exists a $(4k-1,2k,k)$-SBIBD. If you don't know what symmetric designs are, you can find the definition in your favorite design theory textbook or, better yet, the original article

Y. M. Chee, C. J. Colbourn, A. C. H. Ling, R. M. Wilson, Covering and packing for pairs, J. Combin. Theory Ser. A 120 (2013) 1440–1449

Edit: Oops, I forgot to mention this, but when a binary code is not equidistant, you may not be able to turn it into a constant-weight one with one less codeword. So the latter half of this post is assuming that the actual bound is probably the same or not far away from the Johnson bound (which is a bound on the number of codewords of a constant-wight code) if $a$ and $b$ are pretty close. Also, I'm not familiar with the case when $b$ is large (compared to $a$). If you impose a condition such as the code must be linear and have a pair of codewords of distance $b = n$, then I think you can find results if you google "linear self-complementary codes."

If you find the case when $a = b = \frac{n}{2}$ interesting, a $q$-ary code $\mathcal{C} \subset \mathbb{F}_q^n$ is said to be equidistant if for any $\boldsymbol{c}, \boldsymbol{c}' \in \mathcal{C}$, $\boldsymbol{c}\not=\boldsymbol{c}'$, of distinct codewords, the Hamming distance $h(\boldsymbol{c}, \boldsymbol{c}')$ is $d$. Your example is the special case when the code is binary and of equidistance $d = \frac{n}{2} (= a = b)$. We focus on the binary case for the moment.

If you find the case when $a = b = \frac{n}{2}$ interesting, a $q$-ary code $\mathcal{C} \subset \mathbb{F}_q^n$ is said to be equidistant if for any pair $\boldsymbol{c}, \boldsymbol{c}' \in \mathcal{C}$, $\boldsymbol{c}\not=\boldsymbol{c}'$, of distinct codewords, the Hamming distance $h(\boldsymbol{c}, \boldsymbol{c}')$ is $d$. Your example is the special case when the code is binary and of equidistance $d = \frac{n}{2} (= a = b)$. We focus on the binary case for the moment.

I think you're assuming $x \not= y$ when you say "for any $x, y \in S$." In any case, your question is more like design theory.

Any equidistant code $\mathcal{C}$ can be transformed into a constant-weight equidistant code $\mathcal{D} \subset \mathbb{F}_2^n$ of size $\vert \mathcal{C} \vert -1$, where $\operatorname{wt}(\boldsymbol{d}) = \operatorname{wt}(\boldsymbol{d}')$ for any $\boldsymbol{d}, \boldsymbol{d}' \in \mathcal{D}$ (see the references given below (or some references therein) for the proof if you need it). Hence, essentially we only need to consider the case when a code is both constant-weight and equidistant. Let $\operatorname{ex}(n,d,w)$ be the number of codewords of a largest binary code of lenght $n$, equidistance $d$, and constant-weight $w$ (i.e., no other constant-weight, equidistant code of the same parameters has more codewords). Then the classical result by Stinson and Rees states that

I think you're assuming $x \not= y$ when you say "for any $x, y \in S$." In any case, your question seems like a mix of coding theory and design theory.

Any equidistant code $\mathcal{C}$ can be transformed into a constant-weight equidistant code $\mathcal{D} \subset \mathbb{F}_2^n$ of size $\vert \mathcal{C} \vert -1$, where $\operatorname{wt}(\boldsymbol{d}) = \operatorname{wt}(\boldsymbol{d}')$ for any $\boldsymbol{d}, \boldsymbol{d}' \in \mathcal{D}$ (see the references given below (or some references therein) for the proof if you need it). Hence, essentially we only need to consider the case when a code is both constant-weight and equidistant. Let $\operatorname{ex}(n,d,w)$ be the number of codewords of a largest binary code of lenght $n$, equidistance $d$, and constant-weight $w$ (i.e., no other constant-weight, equidistant code of the same parameters has more codewords). Then the classical result by Stinson and van Rees states that