How would you find the graph of $|||x|-2 | -1 | + |||y|-2 | -1 |=1$without a graphing calculator? I plugged this into a graphing calculator and know what the graph looks like, but how would one tackle this huge problem without a calculator?
 A: I use desmos just to help me explaining the method. We can draw the graph without using a computer.
Note that we can replace $x$ by $-x$ or $y$ by $-y$ in the equation. So the graph is symmetric about both the $x$- and $y$-axis. 

If we can draw the graph of $||x-2|-1|+||y-2|-1|=1$, we can draw the graph of $|||x|-2|-1|+|||y|-2|-1|=1$.

The graph of $||x-2|-1|+||y-2|-1|=1$ can be translated by $(2,2)$ to $||x|-1|+||y|-1|=1$.

If we can draw the graph of $||x|-1|+||y|-1|=1$, we can draw the graph of $||x-2|-1|+||y-2|-1|=1$.

Note that we can replace $x$ by $-x$ or $y$ by $-y$ in the equation $||x|-1|+||y|-1|=1$. So its graph is symmetric about both the $x$- and $y$-axis.

If we can draw the graph of $|x-1|+|y-1|=1$, we can draw the graph of $||x|-1|+||y|-1|=1$. 

The graph of $|x-1|+|y-1|=1$ can be translated by $(1,1)$ to $|x|+|y|=1$.

If we can draw the graph of $|x|+|y|=1$, we can draw the graph of $|x-1|+|y-1|=1$. 

The graph of $|x|+|y|=1$ is the  square with vertices $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$.

The graph of $|x-1|+|y-1|=1$ is

The graph of $||x|-1|+||y|-1|=1$ is 

The graph of $||x-2|-1|+||y-2|-1|=1$ is

The graph of $|||x|-2|-1|+|||y|-2|-1|=1$ is

A: Alright, your relation is $f(x) + f(y) = 1,$ where the graph of $f(x)$ is below. For any $x$ with $-4 \leq x \leq 4$ where $x$ is not an integer, we get $0 < f(x) < 1,$ so that $0 < 1 - f(x) < 1,$ in which case we count exactly $8$ values of $y$ with $f(y) + f(x) = 1.$ Different when $x$ is an integer, when $x$ is an even integer, $5$ values of $y$ (also integers), when $x$ is an odd integer, $4$ values of $y.$ 
On each segment between consecutive integers, we have either $f(x) = n + x$ or $f(x) = m - x.$ In either case, we get eight segments over that, of the form $y = 1-n-x$ or $y = 1 - m + x, $ so slopes also $\pm 1$

