math proof for METHOD 2 in http://www.geeksforgeeks.org/two-elements-whose-sum-is-closest-to-zero/

math-proof.tex

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The invariant is that $min\_sum$ is the min-abs-sum of pairs whose one element
is among $a[0]...a[l-1]$ or $a[r+1]...a[n-1]$.

\textbf{Initialization}: Before the first iteration, $l = 0$ and $r = n- 1$.
$a[0]...a[-1]$ or $a[n]...a[n-1]$ is empty. It is trivial to see that
the invariant is true.

\textbf{Maintenance}: If $sum > 0$, the min-abs-sum which involves $a[r]$ and
does not involve elements among $a[0]...a[l-1]$ or $a[r+1]...a[n-1]$ is $sum$
since $sum \ge a[i] + a[r] (l < i < r)$. So picking one of  $min\_sum$ and $sum$
with the less absulate value as $min\_sum$ ensures that $min\_sum$ is the
min-abs-sum of pairs whose one element is among $a[0]...a[l-1]$ or
$a[r]...a[r]$.  Before the next iteration, $r$ is increased by $1$. So the
invariant holds.  The case for $sum < 0$ is symmetric.

\textbf{Termination}: Before the iteration for $l = r$, $min\_sum$ is the
min-abs-sum of pair whose one element is among $a[0]...[l-1]$ or 
$a[l]...a[n-1]$ which is $a[0]...a[n]$.


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