Consider an additive map $\varphi\colon\mathbb{F}_2^m\to\mathbb{F}_2^n$ and its image $C$. (Interpret $C$ as a code)
Now, optimal decoding maps $x\in\mathbb{F}_2^n$ to the nearest $c\in C$ in terms of the hamming-distance.
Consider the factorgroup $\mathbb{F}_2^n/C$ and use as representatives of the cosets the elements of the coset of minimum hemming-weight: $$\begin{array}{rclrrcr} C &= & \{&0,& c_1,&\dots,&c_{2^m-1}\}\\ b_1+C &=& \{&b_1,& b_1+c_1,&\dots,&b_1+c_{2^m-1}\}\\ b_2+C &=& \{&b_2,& b_2+c_1,&\dots,&b_2+c_{2^m-1}\}\\ & \dots&&&&&\\ b_{2^{n-m}-1}+C &=& \{&b_{2^{n-m}-1},& b_{2^{n-m}-1}+c_1,&\dots,&b_{2^{n-m}-1}+c_{2^m-1}\}\end{array}$$
Now, for $x=b_j+c_i\in\mathbb{F}_2^n$ the nearest code-word is $c_j$, the top-element of x's columns.
I have problems understanding the proof for this theorem, that goes as follows:
Suppose $c_{j_0}$ was sent and $x=c_{j_0}+e$ arrived. Consider the coset $e+C$, and represent it as above: $e+C=b_{j_0}+C$ for some $j_0$. Now, as $x=b_{j_0}+c_{k_0}$ for some $k_0$, $c_{k_0}$ ist the nearest code-word.
Why? The steps are too big for me.
Thanks alot in advance
The key is that we have built this standard array using
(which in general is not required, BTW).
That means that we have chosen $b_i$ (coset leader) as the element with minimum weight inside the coset: $w(b_j) \le w(b_j +C_i)$
Now we receive $x = C + e $ for some unknown $C$ that we will try to guess (lets call this guess $\hat C$). And we find that $x$ is in the position $(i,j)$ in our standard array, so $x = C_i+ b_j$.
To guess that $\hat C =C_t $ for some $t$ implies to guess an error
$$\hat e= x + \hat C=C_i+b_j + C_t$$
If we choose $i=t$, we get $\hat e= b_j$
Elsewhere , because $C_i + C_t = C_s$, for some $s\ne 0$, we get $\hat e = b_j +C_s$. This corresponds to another element of the same coset.
But we have stated above that $w(b_j) \le w(b_j +C_i)$. Hence the first alternative gives us the guessed error with minimum weight.