Communication - Chapter 5 Flashcards
How to computer A(n,d)=M
- A(n,d)≤M: show every (n,d)-code has size ≤M
2. A(n,d)≥M: there exists an (n,d)-code of size M
How to view {0,1}ⁿ as a vector space
vector space of dimension n over the binary field 𝔽₂ denoted 𝔽₂ⁿ. Where addition of any two vectors is pointwise modulo 2.
Addition of binary strings.
Pointwise modulo 2. i.e., 0+0 = 0 1 + 1 = 0 1 + 0 = 1 0 + 1 = 0 NO CARRYING
sum code informal
just add x to each element of C via vector addition
How to use proposition 5.5 to show A(n,d)≤m.
An optimal (n,d)-code contains the zero vector.
- For a contradiction suppose there exists an optimal (n,d)-code of size |C|≥m
- By proposition 5.5 we may assume 00…0ϵC
- since an (n,d)-code we have minimum hamming distance d. Use this to construct the remaining codewords.
- If any of these codewords have hamming distance
Can you use punctured codes to show A(n,2)=2ⁿ⁻¹
yes
If a(x,y) denoted the number of positions where x and y are both 1 then how can we use weights to calculate hamming distance?
dₕ(x,y) = w(x) + w(y) - 2a(x,y)
punctured code informal
a punctured code is made from truncating the last ℓ bits from each codeword
parity check informal
the parity check code is the code where each codeword has a 0 added to the end if the weight of the codeword is even and a one added if the weight of the codeword is odd.
parity check add a 0 if…
w(c) is even
parity check add a 1 if…
w(c) is odd
condition for A(n,d)=A(n+1,d+1)
d must be odd
How to use punctured codes to show A(n,2)≤2ⁿ⁻ˡ
- let C be an arbitrary (n,d)-code consider c*(l)
- As dₕ(C)≥d => dₕ(C(l))≥d-l and |C|=|C(l)|
- Note C*(l) has length n-l so it contains at most 2ⁿ⁻ˡ codewords.
- Thus |C|=|C*(l)|≤2ⁿ⁻ˡ
If d is odd…
A(n,d)=A(n+1,d+1)
Let C be an optimal (n,d)-code then
|C|=A(n,d)