Maximum of the difference What is the maximum value of
$f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{2012})$
where $x_{1}, x_{2}, … , x_{2012}$ are distinct integers in the set ${1, 2, 3, …, 2012}$ and $f$ is the absolute value function?

I am solving with respect to the general case that is $$\max_{\substack{
   {x_i\in\{1,2,...,n\}\\ x_i\ne x_j,\ i\ne j}}} f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{n}), \quad f(x)=|x|$$
Since we are looking for a maximum solution, then it is sufficient to distribute the numbers $$1,2,3,...,n$$ in an order so that we get a maximum value for any value of $n$ and that order is what I explained below.
Consider the case where $\forall k=1,2,\ ...,\ n-1$ we have $x_k=n-k$ and $x_n=n$ i.e $x_1=n-1$, $x_2=n-2$, $x_3=n-3$ $ , ..., $ $x_k=n-k$ $, ...,$ $ x_{n-1}=1$  but $x_n=n$.
So that for $n=8$ for example, we compute the value of $f(f(f(f(f(f(f(7-6)-5)-4)-3)-2)-1)-8)$
Let $f_1=x_1$, $f_2=(x_{1} – x_{2})$, $f_3=(f(x_{1} – x_{2}) – x_{3})$, $...,$ $f_{n }=f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{n})$
Using the following notation where $0\le n_4,k_4\le3$ $$\quad\quad \quad\quad\quad n\equiv n_4 \mod 4 \quad\quad\quad\text { and }$$ $$k \equiv k_4 \mod 4$$ we have the following pattern depending on the value of $n$ (it's preferable to work out some cases to verify what I wrote below).
$$\boxed{\begin{array}{ll} f_1=n-1\\
f_2=1 \\
f_3=n- 4 \\
f_4=0 \end{array}} $$ $$ \boxed{f_{n-k}=\left\{\begin{array}{ll}1 \quad \quad \quad \quad \quad \quad\text { if } (n_4,k_4)=\{(0,2),(1,3),(2,0),(3,1)\} \\0  \quad \quad \quad \quad \quad \quad\text { if } n_4=k_4\\f_{n-k-4}+4 \end{array}\right.}$$ $$ \boxed{\begin{array}{ll}f_{n- 4 }=\left\{\begin{array}{ll}0 \quad\text   { if } n_4= 0 \quad      \\4   \quad\text   { if } n_4= 1 \quad     \\1 \quad\text   { if } n_4= 2 \quad      \\3 \quad\text   { if } n_4= 3 \quad      \end{array}\right.\\f_{n-3}=\left\{\begin{array}{ll}3 \quad\text   { if } n_4= 0 \quad      \\1 \quad\text   { if } n_4= 1 \quad     \\2 \quad\text   { if } n_4= 2 \quad     \\0 \quad\text   { if } n_4= 3 \quad      \end{array}\right.\\ f_{n-2}=\left\{\begin{array}{ll}1 \quad\text   { if } n_4= 0,1       \\0 \quad\text   { if } n_4= 2 \quad     \\2 \quad\text   { if } n_4= 3 \quad      \end{array}\right.\\ f_{n-1}=\left\{\begin{array}{ll}0 \quad\text   { if } n_4= 0,1       \\1 \quad\text   { if } n_4= 2,3       \end{array}\right.\end{array}}$$
And lastly $$f_n=|f_{n-1}-x_n| = |f_{n-1}-n|=\left\{\begin{array}{ll}n&\text   { if } n_4= 0,1       \\n-1&\text   { if } n_4= 2,3       \end{array}\right.$$
$$2012\equiv 0 \mod 4 \quad \implies \quad f_{2012}=2012$$
