Elias Bassalygo bound explained

The Elias Bassalygo bound is a mathematical limit used in coding theory for error correction during data transmission or communications.

Definition

Let

C

be a

q

-ary code of length

n

, i.e. a subset of

[q]n

.[1] Let

R

be the rate of

C

,

\delta

the relative distance and

Bq(y,\rhon)=\left\{x\in[q]n:\Delta(x,y)\leqslant\rhon\right\}

be the Hamming ball of radius

\rhon

centered at

y

. Let

Volq(y,\rhon)=|Bq(y,\rhon)|

be the volume of the Hamming ball of radius

\rhon

. It is obvious that the volume of a Hamming Ball is translation-invariant, i.e. indifferent to

y.

In particular,

|Bq(y,\rhon)|=|Bq(0,\rhon)|.

With large enough

n

, the rate

R

and the relative distance

\delta

satisfy the Elias-Bassalygo bound:

R\leqslant1-Hq(Jq(\delta))+o(1),

where

Hq(x)\equivdef-xlogq\left({x\over{q-1}}\right)-(1-x)logq{(1-x)}

is the q-ary entropy function and

Jq(\delta)\equivdef\left(1-{1\overq}\right)\left(1-\sqrt{1-{q\delta\over{q-1}}}\right)

is a function related with Johnson bound.

Proof

To prove the Elias–Bassalygo bound, start with the following Lemma:

Lemma. For

C\subseteq[q]n

and

0\leqslante\leqslantn

, there exists a Hamming ball of radius

e

with at least
|C|Volq(0,e)
qn

codewords in it.

Proof of Lemma. Randomly pick a received word

y\in[q]n

and let

Bq(y,0)

be the Hamming ball centered at

y

with radius

e

. Since

y

is (uniform) randomly selected the expected size of overlapped region

|Bq(y,e)\capC|

is
|C|Volq(y,e)
qn

Since this is the expected value of the size, there must exist at least one

y

such that

|Bq(y,e)\capC|\geqslant{{|C|Volq(y,e)}\over{qn}}={{|C|Volq(0,e)}\over{qn}},

otherwise the expectation must be smaller than this value.

Now we prove the Elias–Bassalygo bound. Define

e=nJq(\delta)-1.

By Lemma, there exists a Hamming ball with

B

codewords such that:

B\geqslant{{|C|Vol(0,e)}\over{qn}}

By the Johnson bound, we have

B\leqslantqdn

. Thus,

|C|\leqslantqnd{{qn}\over{Volq(0,e)}}\leqslant

n(1-Hq(Jq(\delta))+o(1))
q

The second inequality follows from lower bound on the volume of a Hamming ball:

Volq\left(0,\left\lfloor

d-1
2

\right\rfloor\right)\geqslant

H\left
(\delta
2
\right)n-o(n)
q
q

.

Putting in

d=2e+1

and

\delta=\tfrac{d}{n}

gives the second inequality.

Therefore we have

R={logq{|C|}\overn}\leqslant1-Hq(Jq(\delta))+o(1)

See also

Notes and References

  1. Each

    q

    -ary block code of length

    n

    is a subset of the strings of
    n,
    l{A}
    q
    where the alphabet set

    l{A}q

    has

    q

    elements.