1.7 Proof of theorems in Chapter 1 43

The notations are the same as in [11].

• Check node symmetry: For any sequence (b 1 , . . .,b dc−1) in GF(q), we have

Ψ (l)

c (m +b 1

1 , . . .,m +b dc−1

d c−1

) = Ψ (l)

c (m 1 , . . .,m dc−1) +b 1+···+b dc−1

• Variable node symmetry: We also have, for any b ∈ GF(q):

Ψ (l)

v (m+b 0 ,m+b 1 , . . .,m+b d v−1 ) = Ψ(l) v (m 1, . . .,m dc−1) +b

With same notation as in [11], we define y = z +x , where x is a vector of size q, denoting

an arbitrary codeword over GF(q). y **and** z are sets of vectors, **and** each element y t

corresponds to y t = z +xt

t .

Still with same notations as in [11], we easily prove that:

m (0)

ij (y) = m(0) ij (z)+x i

; .

We also prove that, since x is a codeword, then ∑ k:∃e=(v k ,c j ) x k = 0. Hence, as in [11],

we conclude that

m (l+1)

ji

thanks to the check node symmetry, **and**

(y) = m (l+1)

ji (z) +x i

m (l+1)

ij

thanks to the variable node symmetry.

(y) = m (l+1)

ij (z) +x i

Lemma 3.If the channel is symmetric, then, under the all-zero codeword assumption,

the initial message density P 0 in LDR form is symmetric:

P 0 (W = w) = e w i

P 0 (W = w +i )

Proof: Let us define y by y = LDR −1 (w). If we call x noisy the noisy observation

of the sent symbol value, by following the notation of [10], we have w = L(x noisy ).

( )

Hence, the i th component of y is y i = P(x noisy ∈ L −1 (w)|x = i), **and** w i = log

y0

y i

=

)

log

also.

(

P(xnoisy ∈L −1 (w)|x=0)

P(x noisy ∈L −1 (w)|x=i)

Given the symmetry of the channel, let us prove that P 0 (W = w) satisfies equation

□