The Elias-Bassalygo bound is a mathematical limit used in coding theory for error correction during data transmission or communications. The properties of the Elias-Bassalygo bound are defined, below, using mathematical expressions.
Let C be a q -ary code of length n , i.e. a subset of [ q ] n . Let R be the rate of C , δ the relative distance and
B q ( y , ρ n ) = { x ∈ [ q ] n : Δ ( x , y ) ⩽ ρ n } be the Hamming ball of radius ρ n centered at y . Let Vol q ( y , ρ n ) = | B q ( y , ρ n ) | be the volume of the Hamming ball of radius ρ n . It is obvious that the volume of a Hamming Ball is translation-invariant, i.e. indifferent to y . In particular, | B q ( y , ρ n ) | = | B q ( 0 , ρ n ) | .
With large enough n , the rate R and the relative distance δ satisfy the Elias-Bassalygo bound:
R ⩽ 1 − H q ( J q ( δ ) ) + o ( 1 ) , where
H q ( x ) ≡ def − x log q ( x q − 1 ) − ( 1 − x ) log q ( 1 − x ) is the q-ary entropy function and
J q ( δ ) ≡ def ( 1 − 1 q ) ( 1 − 1 − q δ q − 1 ) is a function related with Johnson bound.
To prove the Elias–Bassalygo bound, start with the following Lemma:
Lemma. For
C ⊆ [ q ] n and
0 ⩽ e ⩽ n , there exists a Hamming ball of radius
e with at least
| C | Vol q ( 0 , e ) q n codewords in it.
Proof of Lemma. Randomly pick a received word
y ∈ [ q ] n and let
B q ( y , 0 ) be the Hamming ball centered at
y with radius
e . Since
y is (uniform) randomly selected the expected size of overlapped region
| B q ( y , e ) ∩ C | is
| C | Vol q ( y , e ) q n Since this is the expected value of the size, there must exist at least one
y such that
| B q ( y , e ) ∩ C | ⩾ | C | Vol q ( y , e ) q n = | C | Vol q ( 0 , e ) q n , otherwise the expectation must be smaller than this value.
Now we prove the Elias–Bassalygo bound. Define e = n J q ( δ ) − 1. By Lemma, there exists a Hamming ball with B codewords such that:
B ⩾ | C | Vol ( 0 , e ) q n By the Johnson bound, we have B ⩽ q d n . Thus,
| C | ⩽ q n d ⋅ q n Vol q ( 0 , e ) ⩽ q n ( 1 − H q ( J q ( δ ) ) + o ( 1 ) ) The second inequality follows from lower bound on the volume of a Hamming ball:
Vol q ( 0 , ⌊ d − 1 2 ⌋ ) ⩽ q H q ( δ 2 ) n − o ( n ) . Putting in d = 2 e + 1 and δ = d n gives the second inequality.
Therefore we have
R = log q | C | n ⩽ 1 − H q ( J q ( δ ) ) + o ( 1 )