Hamming distances from a function to all codewords of a Generalized Reed-Muller code of order one

Abstract

For any finite field ${\mathbb {F}}_q$ with q elements, we study the set ${\mathscr {F}}_{(q,m)}$ of functions from ${\mathbb {F}}_q^m$ into ${\mathbb {F}}_q$ from geometric, analytic and algorithmic points of view. We determine a linear system of $q^{m+1}$ equations and $q^{m+1}$ unknowns, which has for unique solution the Hamming distances of a function in ${\mathscr {F}}_{(q,m)}$ to all the affine functions. Moreover, we introduce a Fourier-like transform which allows us to compute all these distances at a cost $O(mq^m)$ and which would be useful for further problems.

Appendices

Appendix 1: Fast computation algorithm of $T_{(q,m)}$

The ${\mathbb {C}}{\mathbb {F}}_q$-matrix of $T_{(q,m)}$

1.1 Presentation of the matrix

From now on we suppose that we have chosen an order on ${\mathbb {F}}_q^m$, so that the writing of vectors and matrices indexed by this set is well defined.

If $\phi \in {\mathscr {G}}_{(q,m)}$ then it is determined by the ${\mathbb {C}}{\mathbb {F}}_q$-vector $\left( \phi (u)\right) _{u \in {\mathbb {F}}_q^m}$. Let us consider the ${\mathbb {C}}{\mathbb {F}}_q$-matrix

$$\begin{aligned} {\mathscr {M}}_{(q,m)}= {(Z^{-<u,v>})}_{\begin{array}[t]{c} {\scriptstyle u \in {\mathbb {F}}_q^m} \\ {\scriptstyle v \in {\mathbb {F}}_q^m} \end{array}} . \end{aligned}$$

(13)

Then the function $T_{(q,m)}(\phi )$ is determined by the ${\mathbb {C}}{\mathbb {F}}_q$-vector

$$\begin{aligned} \left( T_{(q,m)}(\phi )(u)\right) _{u \in {\mathbb {F}}_q^m}= {\mathscr {M}}_{(q,m)}.\left( \phi (u)\right) _{u \in {\mathbb {F}}_q^m}. \end{aligned}$$

The problem is to find a fast algorithm to compute this product, namely the product of a ${\mathbb {C}}{\mathbb {F}}_q$-vector by the ${\mathbb {C}}{\mathbb {F}}_q$-matrix ${\mathscr {M}}_{(q,m)}$.

Remark 6

Do not confuse the matrix ${\mathscr {T}}{(q,m)}$ with ${\mathscr {M}}_{(q,m)}$. The first one is the matrix of the transform $T_{(q,m)}$ when the set of functions ${\mathscr {G}}_{(q,m)}$ is considered as a vector space over ${\mathbb {C}}$. The second one is the matrix of the transform $T_{(q,m)}$ when the set of functions ${\mathscr {G}}_{(q,m)}$ is considered as a module over the ring ${\mathbb {C}}{\mathbb {F}}_q$. However, it is not difficult to pass from one to the other.

1.2 The tensor structure of the matrix ${\mathscr {M}}_{(q,m)}$

Proposition 2

The matrix ${\mathscr {M}}_{(q,m)}$ can be written as

$$\begin{aligned} {\mathscr {M}}_{(q,m)}=\underbrace{{\mathscr {M}}_{(q,1)} \otimes {\mathscr {M}}_{(q,1)} \otimes \cdots \otimes {\mathscr {M}}_{(q,1)}}_{m~terms}, \end{aligned}$$

where $\otimes$ denotes the Kronecker product.

Proof

The proof is done by induction on m. The proposition is true for $m=1$. Let us suppose that it is true for a value $m\ge 1$. Then

$$\begin{aligned} {(Z^{-<u,v>})}_{\begin{array}[t]{c} {\scriptstyle u \in {\mathbb {F}}_q^m} \\ {\scriptstyle v \in {\mathbb {F}}_q^m} \end{array}} \otimes {\mathscr {M}}_{(q,1)}= {(Z^{-<u,v>-ab})}_{\begin{array}[t]{c} {\scriptstyle u \in {\mathbb {F}}_q^m, a\in {\mathbb {F}}_q} \\ {\scriptstyle v \in {\mathbb {F}}_q^m, b\in {\mathbb {F}}_q } \end{array}} = {(Z^{-<u,v>})}_{\begin{array}[t]{c} {\scriptstyle u \in {\mathbb {F}}_q^{m+1}} \\ {\scriptstyle v \in {\mathbb {F}}_q^{m+1}} \end{array}} \end{aligned}$$

which proves that the proposition is true for $m+1$. $\square$

Proposition 3

The matrix ${\mathscr {M}}_{(q,m)}$ can be written as

$$\begin{aligned} {\mathscr {M}}_{(q,m)}=M_{q^m}^{(1)}M_{q^m}^{(2)}\ldots M_{q^m}^{(m)} \end{aligned}$$

where

$$\begin{aligned} M_{q^m}^{(i)}=I_{q^{m-i}}\otimes {\mathscr {M}}_{(q,1)} \otimes I_{q^{i-1}} \end{aligned}$$

and $I_s$ is the identity matrix of size $s \times s$.

Proof

The proof is done by induction on m. For $m=1$ we have

$$\begin{aligned} M_q^{(1)}=I_1 \otimes {\mathscr {M}}_{(q,1)} \otimes I_1={\mathscr {M}}_{(q,1)}, \end{aligned}$$

then the proposition is true. Let us suppose that the proposition is true for a value $m\ge 1$. By definition, for $1 \le i \le m$ we have the following formulas

$$\begin{aligned} M_{q^{m+1}}^{(i)}= & {} I_{q^{m+1-i}} \otimes {\mathscr {M}}_{(q,1)} \otimes I_{q^{i-1}} \\ M_{q^{m+1}}^{(i)}= & {} I_q \otimes I_{q^{m-i}} \otimes {\mathscr {M}}_{(q,1)} \otimes I_{q^{i-1}} \\ M_{q^{m+1}}^{(i)}= & {} I_q \otimes M_{q^{m}}^{(i)} \end{aligned}$$

and we have also

$$\begin{aligned} M_{q^{m+1}}^{(m+1)}={\mathscr {M}}_{(q,1)} \otimes I_{q^m}. \end{aligned}$$

Then

$$\begin{aligned} M_{q^m}^{(1)}M_{q^m}^{(2)}\ldots M_{q^m}^{(m+1)}= & {} (I_q \otimes M_{q^{m}}^{(1)}) \ldots (I_q \otimes M_{q^{m}}^{(m)}) ({\mathscr {M}}_{(q,1)} \otimes I_{q^m}), \\ M_{q^m}^{(1)}M_{q^m}^{(2)}\ldots M_{q^m}^{(m+1)}= & {} {\mathscr {M}}_{(q,1)} \otimes (M_{q^m}^{(1)}M_{q^m}^{(2)}\ldots M_{q^m}^{(m)}), \end{aligned}$$

and by induction hypothesis

$$\begin{aligned} M_{q^m}^{(1)}M_{q^m}^{(2)}\ldots M_{q^m}^{(m+1)}= & {} {\mathscr {M}}_{(q,1)} \otimes {\mathscr {M}}_{(q,m)}, \\ M_{q^m}^{(1)}M_{q^m}^{(2)}\ldots M_{q^m}^{(m+1)}= & {} {\mathscr {M}}_{(q,{m+1})}, \end{aligned}$$

which proves that the proposition is true for $m+1$. $\square$

Let us remark that, using a similar proof, we can show that the matrix can also be written as

$$\begin{aligned} {\mathscr {M}}_{(q,m)}=M_{q^m}^{(m)}M_{q^m}^{(m-1)}\cdots M_{q^m}^{(1)}. \end{aligned}$$

The multiplication of a vector in $\left( {\mathbb {C}}{\mathbb {F}}_q\right) ^{q^m}$ by a matrix $M_{q^m}^{(i)}$ is quite simple. Any row of the matrix has q non-zero elements and these elements are of the form $Z^s$. The multiplication of an element of ${\mathbb {C}}{\mathbb {F}}_q$ by an element $Z^s$ for some s is a shift. Hence, for any matrix $M_{q^m}^{(i)}$ and for any row of this matrix we have to do q shifts and $q-1$ additions in ${\mathbb {C}}{\mathbb {F}}_q$, namely $q(q-1)$ additions in ${\mathbb {C}}$. Consequently, as we have to do a product involving m matrices $M_{q^m}^{(i)}$ of $q^m$ rows each, the following proposition occurs

Proposition 4

The multiplication of a vector in $\left( {\mathbb {C}}{\mathbb {F}}_q\right) ^{q^m}$ by the matrix ${\mathscr {M}}_{(q,m)}$ needs $q(q-1)mq^m$ additions in ${\mathbb {C}}$ and $mq^{m+1}$ shifts in ${\mathbb {C}}{\mathbb {F}}_q$.

Appendix 2: Practice of the transform

Let us remark that the multiplication by a matrix of the form $M_{q^i}^{(i)}$ has a particular importance. Indeed, to multiply a vector $\omega =(\xi _u)_{u \in {\mathbb {F}}_q^m}$ on length $q^m$ by ${\mathscr {M}}_{(q,m)}$ it is sufficient to compute $M_{q^m}^{(m)}.\omega$, then to split the result in q vectors of length $q^{m-1}$ (by taking in the order q blocks of $q^{(m-1)}$ components), to transform these q vectors by $T_{(q,{m-1})}$ and finally to reconstruct the result vector by concatenation of the q resulting blocks. By induction we see that we only need multiplications by matrices of the form $M_{q^i}^{(i)}$.

The multiplication of a vector $\omega =(\xi _u)_{u \in {\mathbb {F}}_q^i}$ of length $q^i$ (where $i>0$) can be realized in the following way: from $\omega$ we construct the matrix $\varOmega$ having q rows and $q^{i-1}$ columns, where the q rows are constituted by the q blocks of $q^{i-1}$ components of $\omega$ taken in the order. The result is obtained by computing the product ${\mathscr {M}}_{(q,1)} . \varOmega$, then by reconstructing, by concatenation of the q rows of this product, the result vector.

Appendix 3: A toy example

Let us take the simple example where $q=3$ and $m=2$. Let us define the function $f \in \{{\mathbb {F}}_3^2 \rightarrow {\mathbb {F}}_3\}$ and the function $F=Z^f$ by the following array

u	(0,0)	(1,0)	(2,0)	(0,1)	(1,1)	(2,1)	(0,2)	(1,2)	(2,2)
f(u)	0	2	1	1	2	0	2	0	1
F(u)	$\hbox {Z}^0$	$\hbox {Z}^2$	$\hbox {Z}^1$	$\hbox {Z}^1$	$\hbox {Z}^2$	$\hbox {Z}^0$	$\hbox {Z}^2$	$\hbox {Z}^0$	$\hbox {Z}^1$

Now there is two steps. The first one is to multiply by $M_{3^2}^2$. This is done using the matrix $\varOmega$ introduced in the previous section. Let us remark that the matrix ${\mathscr {M}}_{(3,1)}$ is

$$\begin{aligned} \begin{pmatrix} Z^0 &{} \quad Z^0 &{} \quad Z^0 \\ Z^0 &{} \quad Z^2 &{} \quad Z^1 \\ Z^0 &{} \quad Z^1 &{} \quad Z^2 \end{pmatrix}. \end{aligned}$$

Let us compute for this first step the product by ${\mathscr {M}}_{(3,1)}$ of the matrix obtained by splitting F(u) into three rows.

$$\begin{aligned} \begin{pmatrix} Z^0 &{} \quad Z^0 &{} \quad Z^0 \\ Z^0 &{} \quad Z^2 &{} \quad Z^1 \\ Z^0 &{} \quad Z^1 &{} \quad Z^2 \end{pmatrix} \begin{pmatrix} Z^0 &{} \quad Z^2 &{} \quad Z^1 \\ Z^1 &{} \quad Z^2 &{} \quad Z^0 \\ Z^2 &{} \quad Z^0 &{} \quad Z^1 \end{pmatrix} = \begin{pmatrix} Z^0+Z^1+Z^2 &{} \quad Z^0+ 2 Z^2 &{} \quad Z^0 + 2Z^1 \\ 3 Z^0 &{}\quad 2 Z^1+Z^2 &{} \quad Z^1+ 2 Z^2 \\ Z^0+Z^1+Z^2 &{} \quad Z^0+2Z^2 &{} \quad Z^0+2Z^1 \\ \end{pmatrix}. \end{aligned}$$

The second step computes the product by $M_{3}^1$ of the vectors given by the three rows of the previous result matrix. This can be done by computing the three following products: the first one is

$$\begin{aligned} \begin{pmatrix} Z^0 &{} \quad Z^0 &{} \quad Z^0 \\ Z^0 &{} \quad Z^2 &{} \quad Z^1 \\ Z^0 &{} \quad Z^1 &{} \quad Z^2 \end{pmatrix} \begin{pmatrix} Z^0+Z^1+Z^2 \\ Z^0+ 2 Z^2 \\ Z^0 + 2Z^1 \end{pmatrix} = \begin{pmatrix} 3Z^0+3Z^1+3Z^2 \\ Z^0+4Z^1+4Z^2 \\ 5Z^0+2Z^1+2Z^2 \end{pmatrix} \end{aligned}$$

the second one is

$$\begin{aligned} \begin{pmatrix} Z^0 &{} \quad Z^0 &{} \quad Z^0 \\ Z^0 &{} \quad Z^2 &{} \quad Z^1 \\ Z^0 &{} \quad Z^1 &{} \quad Z^2 \end{pmatrix} \begin{pmatrix} 3Z^0 \\ 2Z^1+Z^2 \\ Z^1+2Z^2 \end{pmatrix} = \begin{pmatrix} 3Z^0 + 3 Z^1 +3 Z^2\\ 7Z^0+Z^1+Z^2\\ 5Z^0 + 2Z^1+2Z^2 \end{pmatrix}. \end{aligned}$$

the third one is the same that the first one.

Now we have computed all the $N_{v,t}(f)$

$$\begin{aligned} N_{(0,0),0}(f)= & {} 3,N_{(0,0),1}(f)=3,N_{(0,0),2}(f)=3, \\ N_{(1,0),0}(f)= & {} 1,N_{(1,0),1}(f)=4,N_{(1,0),2}(f)=4, \\ N_{(2,0),0}(f)= & {} 5,N_{(2,0),1}(f)=2,N_{(2,0),2}(f)=2, \\ N_{(0,1),0}(f)= & {} 3,N_{(0,1),1}(f)=3,N_{(0,1),2}(f)=3, \\ N_{(1,1),0}(f)= & {} 7,N_{(1,1),1}(f)=1,N_{(1,1),2}(f)=1, \\ N_{(2,1),0}(f)= & {} 5,N_{(2,1),1}(f)=2,N_{(2,1),2}(f)=2, \\ N_{(0,2),0}(f)= & {} 3,N_{(0,2),1}(f)=3,N_{(0,2),2}(f)=3, \\ N_{(1,2),0}(f)= & {} 1,N_{(1,2),1}(f)=4,N_{(1,2),2}(f)=4, \\ N_{(2,2),0}(f)= & {} 5,N_{(2,2),1}(f)=2,N_{(2,2),2}(f)=2. \end{aligned}$$

The distance from f to the code $RM_{(3,2)}^1$ is $q^m - \sup _{(v,t)\in {{\mathbb {F}}_q^m}\times {{\mathbb {F}}_q}}N_{v,t}(f)=2.$ From Formulas (10) and (11) it follows that the covering radius of the code $RM_{(3,2)}^1$ is 5.

Remark 7

Let us consider the finite field ${\mathbb {F}}_9$ as ${\mathbb {F}}_3[X]/(X^2+X+2)$. Let $\alpha$ be the class of X. Let us identify ${\mathbb {F}}_3^2$ to ${\mathbb {F}}_9$ by setting $(u_1,u_2)=u_1+\alpha u_2$. Then the previous function f is defined by $f(0)=0$ and $f(u)=trace(u^{-1})$ if $u\ne 0$. In contrary to the binary case (cf. [3]), the distance of the function f to the affine functions is not maximal.