# Bar product

(Redirected from Bar product (coding theory))

In information theory, the bar product of two linear codes C2 ⊆ C1 is defined as

${\displaystyle C_{1}\mid C_{2}=\{(c_{1}\mid c_{1}+c_{2}):c_{1}\in C_{1},c_{2}\in C_{2}\},}$

where (a | b) denotes the concatenation of a and b. If the code words in C1 are of length n, then the code words in C1 | C2 are of length 2n.

The bar product is an especially convenient way of expressing the Reed–Muller RM (dr) code in terms of the Reed–Muller codes RM (d − 1, r) and RM (d − 1, r − 1).

The bar product is also referred to as the | u | u+v | construction[1] or (u | u + v) construction.[2]

## Properties

### Rank

The rank of the bar product is the sum of the two ranks:

${\displaystyle \operatorname {rank} (C_{1}\mid C_{2})=\operatorname {rank} (C_{1})+\operatorname {rank} (C_{2})\,}$

#### Proof

Let ${\displaystyle \{x_{1},\ldots ,x_{k}\}}$ be a basis for ${\displaystyle C_{1}}$ and let ${\displaystyle \{y_{1},\ldots ,y_{l}\}}$ be a basis for ${\displaystyle C_{2}}$. Then the set

${\displaystyle \{(x_{i}\mid x_{i})\mid 1\leq i\leq k\}\cup \{(0\mid y_{j})\mid 1\leq j\leq l\}}$

is a basis for the bar product ${\displaystyle C_{1}\mid C_{2}}$.

### Hamming weight

The Hamming weight w of the bar product is the lesser of (a) twice the weight of C1, and (b) the weight of C2:

${\displaystyle w(C_{1}\mid C_{2})=\min\{2w(C_{1}),w(C_{2})\}.\,}$

#### Proof

For all ${\displaystyle c_{1}\in C_{1}}$,

${\displaystyle (c_{1}\mid c_{1}+0)\in C_{1}\mid C_{2}}$

which has weight ${\displaystyle 2w(c_{1})}$. Equally

${\displaystyle (0\mid c_{2})\in C_{1}\mid C_{2}}$

for all ${\displaystyle c_{2}\in C_{2}}$ and has weight ${\displaystyle w(c_{2})}$. So minimising over ${\displaystyle c_{1}\in C_{1},c_{2}\in C_{2}}$ we have

${\displaystyle w(C_{1}\mid C_{2})\leq \min\{2w(C_{1}),w(C_{2})\}}$

Now let ${\displaystyle c_{1}\in C_{1}}$ and ${\displaystyle c_{2}\in C_{2}}$, not both zero. If ${\displaystyle c_{2}\not =0}$ then:

{\displaystyle {\begin{aligned}w(c_{1}\mid c_{1}+c_{2})&=w(c_{1})+w(c_{1}+c_{2})\\&\geq w(c_{1}+c_{1}+c_{2})\\&=w(c_{2})\\&\geq w(C_{2})\end{aligned}}}

If ${\displaystyle c_{2}=0}$ then

{\displaystyle {\begin{aligned}w(c_{1}\mid c_{1}+c_{2})&=2w(c_{1})\\&\geq 2w(C_{1})\end{aligned}}}

so

${\displaystyle w(C_{1}\mid C_{2})\geq \min\{2w(C_{1}),w(C_{2})\}}$