Can somebody prove my finding which I am proposing here as a conjecture regarding the binary BCH code?

Question

I am a PhD student and I work in the area of applied mathematics. I am studying the different kind of uniform column weight binary matrices with prescribed inner products between its columns. I am particularly interested in binary matrices arising from narrow-sense primitive binary BCH codes. So, in the first part, I will describe what had already been proven and can be mathematically proved regarding the BCH code I am dealing with. And what I want to be proved is presented here as a Conjecture.

Let "$[495]$-BCH code" denotes the narrow-sense primitive binary BCH code of length $1023$ with designed minimum distance $495$. Let "$[496]$-BCH code" denotes the even parity sub-code of the $[495]$-BCH code consisting of all even parity code vector in it. Let the symbol $a_j$ denote the number of code vectors of weight $j$ in the given even parity BCH code. Please note that the statements written in "italic" font do already have a mathematical proof.

Now, from T.Kasami's work, the weight distribution of the $[496]$-BCH code is as follows:

$$a_0 := 1$$ $$a_{496}:= 16368$$ $$a_{512}:=1023$$ $$a_{528}:= 15376.$$

Now, let $\mathcal{H} := \{\mathbf{b}_1, \mathbf{b}_2, \mathbf{b}_3,...,\mathbf{b}_{16368}\}$ denote the set of all the code vector of weight $496$, and let $\mathcal{G}:= \{\mathbf{c}_1 , \mathbf{c}_2 , \mathbf{c}_3 ,...., \mathbf{c}_{1023} \} $ denote the set of all the code vector of weight $512$, and let $\mathcal{T}:= \{\mathbf{d}_1 , \mathbf{d}_2 , \mathbf{d}_3 ,..., \mathbf{d}_{15376}\} $ denote the set of all the code vector of weight $528$.

Due to the above given weight distribution it follows that, if $\mathbf{b}_i, \mathbf{b}_j \in \mathcal{H}$ are two distinct code vectors (i.e. $ i \neq j $), then the inner product between them defined over field of real numbers can only take one of these three values: $248, 240, 232$, and this can be easily proved. Note that $\langle \mathbf{b}_i , \mathbf{b}_j \rangle = 248 \implies \mathbf{b}_i \oplus \mathbf{b}_j \in \mathcal{H}$, $\langle \mathbf{b}_i , \mathbf{b}_j \rangle = 240 \implies \mathbf{b}_i \oplus \mathbf{b}_j \in \mathcal{G}$, $\langle \mathbf{b}_i , \mathbf{b}_j \rangle = 232 \implies \mathbf{b}_i \oplus \mathbf{b}_j \in \mathcal{T}$, where $\oplus$ is the $\mod 2$ addition (element-wise). The above facts are the result of some simple calculations. Now, define the matrix $B$ as $$B:= \big[\mathbf{b}_1 \big| \mathbf{b}_2 \big| \mathbf{b}_3 \big|,...., \big| \mathbf{b}_{16368}\big],$$ basically the matrix $B$ is consist of all the code-vector of weight $496$. Now, simple calculation can show us that each row of $B$ contains exactly $7936$ number of $1's$, and I have proof this also.

Now, I begin explaining my finding regarding the above BCH code of which I don't have a mathematical proof:

Conjecture 1: Let $j\in \{1,2,3,...,16368\}$ be arbitrary. Define the set $S$ as $$ S:= \{ \mathbf{b}_j \oplus \mathbf{b}_i : \forall i \in [16368]\setminus \{j\} \},$$ where $[16368] := \{1,2,3,...,16368\}$ Then the set $S$ can be partitioned into three pair-wise disjoint subsets $\mathcal{O}, \mathcal{P}, \mathcal{Q}$ such that $\big|\mathcal{O}\big| = 8400,\ \big|\mathcal{P}\big| = 527,\ \big|\mathcal{Q}\big| = 7440$, in addition to that, $\mathcal{O}\subset \mathcal{H},\ \mathcal{P}\subset \mathcal{G}$, and $\mathcal{Q}\subset \mathcal{T}$.

I am have tried a lot but can't find a direct way of proving Conjecture 1.

Answer 1

Work in progress:

Perhaps looking at the (nonzero) cycles of this $[1023,15,496]$ cyclic code will help. The code's parity-check polynomial has two factors: a primitive polynomial $h_{10}(x)$ of degree $10$ and a primitive polynomial $h_5(x)$ of degree $5$. The $2^{15}-1 = 32,767$ nonzero codewords can be partitioned into $32$ cycles of $1023$ codewords each, and one cycle of $31$ codewords.

The single cycle of $31$ codewords is just the nonzero codewords of the $[31,5,16]$ shortened and expurgated first-order Reed-Muller code (of length $31$ and parity-check polynomial $h_5(x)$) that have been repeated $33$ times to make the length $31\times 33=1023$ and weight $16\times 33 = 528$. Let $\mathbf v$ denote a cycle representative and $\mathsf T$ denote the "cyclic-shift-by one place" operator. Then, the $31$ codewords are $\mathbf v$, $\mathsf T\mathbf v$, $\mathsf T^2\mathbf v$, $\mathsf T^3\mathbf v\cdots$, $\mathsf T^{30}\mathbf v$.

The 32 cycles of 1023 codewords are comprised of

• $1$ cycle of weight $512$
  Th cycle is comprised of the nonzero codewords of the $[1023,10,512]$ shortened and expurgated first-order Reed-Muller code of length $1023$ whose parity-check polynomial is $m_{10}(x)$. Let $\mathbf u$ denote a cycle representative. Then the codewords comprising this cycle are $\mathbf u$, $\mathsf T\mathbf u$, $\mathsf T^2\mathbf u$, $\mathsf T^3\mathbf u, \cdots$, $\mathsf T^{1022}\mathbf u$.

• $16$ cycles of weight $496$
  Each cycle representative is of the form $\mathbf u + \mathsf T^c\mathbf v$ for some $c \in C$ where $C \subset \{0,1,2, \cdots, 30\}$ has $16$ elements. The codewords comprising this cycle are $\mathbf u + \mathsf T^c\mathbf v,$ $\mathsf T(\mathbf u + \mathsf T^c\mathbf v) = \mathsf T\mathbf u + \mathsf T^{c+1}\mathbf v,$ $\mathsf T^2\mathbf u + \mathsf T^{c+2}\mathbf v$, $\mathsf T^3\mathbf u + \mathsf T^{c+3}\mathbf v, \cdots$, $\mathsf T^{1022}\mathbf u + \mathsf T^{c+1022}\mathbf v$. Note that since $\mathbf v$ is of period 31, the exponent of $\mathsf T$ can be reduced modulo $31$, that is, the $j$-th codeword in the cycle can be expressed as $\mathsf T^j\mathbf u + \mathsf T^{(c+j)\bmod 31}\mathbf v$.

• $15$ cycles of weight $528$
  Each cycle representative is of the form $\mathbf u + \mathsf T^{c^\prime}\mathbf v$ for some $c^\prime \in C^\prime$ where $C^\prime = \{0,1,2, \cdots, 30\} -C$ has $15$ elements. Similar remarks apply with respect to the codewords comprising each cycle.

Let's check the statements numerically: $16\times 1023 = 16368 = a_{496}$, $15\times 1023 +33 = 15378 = a_{528}$ and of course $1\times 1023 = 1023 = a_{512}$.

The set of integers $\{0,1,2, \cdots ,30\}$ can be partitioned into $6$ cyclotomic cosets of size $5$ and $\left\{0\right\}$. Since $|C| = 16, |C^\prime| = 15$, I conjecture that $C$ is the union of three cyclotomic cosets plus $\left\{0\right\}$ while $C^\prime$ is the union of the other three cyclotomic cosets.

Let us now consider the $16368$ codewords of weight 496 but instead of labeling them with integers $1$ through $16368$ as the OP has done, we arrange them into a $16\times 1023$ array with the codeword in the $i$-th row and $j$-th column being $\mathbf b_{i,j} =\mathsf T^j\mathbf u + \mathsf T^{(c_i+j)\bmod 31}\mathbf v$. Note that each of $\mathbf v$, $\mathsf T\mathbf v$, $\mathsf T^2\mathbf v$, $\mathsf T^3\mathbf v\cdots$, $\mathsf T^{30}\mathbf v$ occurs in exactly 33 sums in each row.

After these preliminaries, lets's take up the OP's question: Add $\mathbf b_{r,s}$ to all the other codewords in the $16\times 1023$ array. This results in $16367$ codewords of the $[1023,15,496]$ code of which $527$ have weight $512$, $8400$ have weight $496$ and $7440$ have weight $528$. Why?

Well, the $527$ codewords of weight $512$ are easily explained. $\mathbf b_{r,s}$ is of the form $\mathsf T^s\mathbf + \mathsf T^\ell \mathbf v$ for some $\ell\in \{0,1,\cdots,30\}$ and so when we add $\mathbf b_{r,s}$ to a $\mathbf b_{r,s^\prime}$ (same row), in $32$ instances, the two $\mathsf T^\ell \mathbf v$'s cancel out leaving us with the sum of two cyclic shifts of $\mathbf u$ which is just another shift of $\mathbf u$, i.e. a codeword of weight $512$. In all other rows, this cancellation occurs in $33$ places, giving us that the modified array has $32+ 15\times 33 =527$ codewords of weight $512$.

Answer 2

I suspect that Dilip's approach may be more likely to bear fruit, but I am adding the following more algberaic point of view to the mix anyway. I will study Dilip's description later in the fond hope that one of us can take it further :-)

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Let $F=\Bbb{F}_{2^n}=GF(2^n)$ to be the finite fiels with $q=2^n$ elements, and let $E=\Bbb{F}_{2^m}=GF(Q), m=2n, Q=q^2,$ be its unique quadratic extension. The case $n=5$ is relevant here, but I want to take a look at the more general $n$ for the time being. Let $$tr^n_1:F\to \Bbb{F}_2, x\mapsto \sum_{i=0}^{n-1}x^{2^i}$$ be the (absolute) trace function of $F$, $$ tr^m_1:E\to \Bbb{F}_2, x\mapsto \sum_{i=0}^{m-1}x^{2^i} $$ the absolute trace of $E$, and $tr^m_n:E\to F, x\mapsto x+x^q$ the relative trace. It is well known that $tr^m_1=tr^n_1\circ tr^m_n$ (called transitivity of the trace). It is also well known that the words of the Kasami code can be thought of as the following functions $$ c(\alpha,\beta):E^*\to \Bbb{F}_2, x\mapsto tr^m_1(\alpha x)+tr^n_1(\beta x^{q+1}),\tag{1} $$ where $\alpha$ ranges over $E$ (so $2^m$ choices) and $\beta$ ranges over $F$ ($2^n$ choices). Altogether $2^{m+n}=2^{3n}$ codewords (when $n=5$ we get a code of $2^{15}$ words).

It is easy to see that the choice $\beta=0$ gives exactly the $2^m$ words of weight $Q/2=2^{m-1}$. Those are the words from the (shortened) first order Reed-Muller code $R(1,m)^*$ (aka the simplex code). The rest of them have differing weights, deviating from the expected $Q/2$ by $q/2$ into either direction. I want to characterize the pairs $(\alpha,\beta)$ that yield the words of low weight $=(Q-q)/2$.

To that end I use the direct sum decomposition (note to the experts: Niho's method for studying certain correlation functions is not unlike this, but that method won't work as cleanly as it does here). $$ E^*=F^*\times S,\tag{2} $$ where $$S=\{\alpha\in E\mid \alpha^{q+1}=1$$ is cyclic of order $q+1$, and is the $E/F$ analogue of the unit circle of the complex plane $\Bbb{C}/\Bbb{R}$. Observe that $F^*$ is cyclic of order $q-1$, and as $q$ is even, $\gcd(q-1,q+1)=1$, whence their product is direct, and all of $E^*$.

For all $y\in F$ we have $y^q=y$, and for all $\eta\in S$ we have $\eta^q=\eta^{-1}$. Due to the decomposition $(2)$ we can write $x=y\eta$ in $(1)$, and let $y$ range over $F^*$, and $\eta$ range over the unit circle $S$. We end up with $$ c(\alpha,\beta)=tr^n_1(y[\alpha\eta+\alpha^q\eta^{-1}])+tr^n_1(\beta y^2) =tr^n_1(y[\sqrt{\beta}+\alpha\eta+\alpha^q\eta^{-1}]),\tag{3} $$ where I used (in addition to the transitivity of the trace and the earlier results the fact that $tr^n_1(y^2)=tr^n_1(y)$ for all $y\in F$.

Let's look at $(3)$ by fixing a value of $\eta$, but $y^*$ still ranging over $F^*$. That subword (with $q-1$ bits) is a word of the Reed-Muller code $R(1,n)^*$. It is the all zeros word, iff and only if the quantity in square brackets is equal to zero, and a word of weight $q/2$ otherwise. Working under the earlier assumption that $\beta\neq0$, we see (taking Kasami's results into account) that the bracket quantity vanishes either for no choice of $\eta$ or for exactly two choice of $\eta$, and the latter case is of interest to us as they yield the low weight words.

Next I rewrite the square bracket coefficient using the variable $s=\alpha^{-(q-1)/2}\eta$. Observe that $\alpha^{-(q-1)/2}\in S$ for all $\alpha\neq0$. Rewriting then gives that $$[\sqrt{\beta}+\alpha\eta+\alpha^q\eta^{-1}]=0$$ if and only if $$ s+\frac1s=\sqrt{\beta/\alpha^{q+1}}. $$ Here $\sqrt{\beta/\alpha^{q+1}}\in F^*$. The quadratic equation $$ x+\frac1x=\sqrt{\beta/\alpha^{q+1}}\tag{4} $$ thus has its coefficients in $F$. By the well known solvability criterion the roots of $(4)$ are in $F$ if and only if $tr^n_1(\alpha^{q+1}/\beta)=0$. So if $tr^n_1(\alpha^{q+1}/\beta)=1$, then the solutions of $(4)$ will be in the quadratic extension field $E$. Furthermore, the roots of $(4)$ are always each others reciprocals, and in the latter case also $F$-conjugates. These force the solutions into the subgroup $S$. Observe that the solution $x=1$ will not pop out as $\beta\neq0$.

Conclusion. The word $c(\alpha,\beta)$ is of the minimal weight $(Q-q)/2$ if and only if $\beta\neq0$ and $$tr^n_1(\alpha^{q+1}/\beta)=1.$$

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That's where I am at the moment. A possible continuation is then to study how often the two conditions in my Conclusion are satisfied for the sum of a pairs of $(\alpha,\beta)$ meeting them (one of them fixed). Hopefully more to follow :-)