Linear Code Problem, Weight & Minimum Distance

Question

Please the highlighted part in the image below. I don't understand why w(c2) must be larger than s(c1, c2) considering s(c1, c2) is counting the position where c1 + c2 = 0, c1 != 0 and c2 != 0 while w(c2) is only counting position where w(c2) != 0.

Should s(c1, c2) be larger than w(c2)?

Thanks for helping out!

Answer 1

$s(c_1, c_2)$ is counting the position where $c_1 + c_2 = 0$, $c_1 \neq 0$ and $c_2 \neq 0$ while $w(c_2)$ is only counting position where $c_2 \neq 0$.

$s(c_1, c_2)$ is counting the positions where all three of the conditions hold at the same time, while $w(c_2)$ is counting the positions where the third condition holds. Since we can't have all three conditions holding without the third condition holding, we must have $s(c_1,c_2) \leq w(c_2)$.

Answer 2

I may have missed something but I cannot fathom why anyone would want to prove this simple fact in this horrible way using that $s(c_1,c_2)$ function. The claim follows more simply as follows.

If $c_1$ and $c_2$ are both non-zero, then $$ w(c_1,c_2,c_1+c_2)\ge w(c_1,c_2,0)=w(c_1)+w(c_2)\ge d_1+d_2\ge2d_1. $$ If $c_1=0$ and $c_2\neq0$, then $$ w(c_1,c_2,c_1+c_2)=w(0,c_2,c_2)=2w(c_2)\ge 2d_2\ge 2d_1. $$ And as the last case if $c_1\neq0$ and $c_2=0$, then $$ w(c_1,c_2,c_1+c_2)=w(c_1,0,c_1)=2w(c_1)\ge 2d_1. $$ In all the cases we saw that the weight of the codeword was at least $2d_1$. By selecting as $c_1$ a word of weight $d_1$ and $c_2=0$ we get a codeword of weight $2d_1$, so that is the minimum weight.