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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?

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  • $\begingroup$ This graph can be broken down into some set of line segments. Try to find the vertexes and then connect the dots. To do that, what would it take for to minimize $|||x|-2|-1|$? The smallest it can be is $0.$ For what values of x does it equal $0?$ And then if it equals 0 then $|||y|-2|-1| = 1$ What are the possible values of y? Then this equation is symmetric. i.e. you can swap all of the $x$ and $y$ values, and get the same results. It also has a symmetry that you can swap $x$ for $-x$ and $y$ for $-y.$ $\endgroup$ – Doug M Jan 26 '18 at 18:50
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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 $(0,-1)$.

enter image description here

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

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

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

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

enter image description here

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    $\begingroup$ In case of interest, relating to the appearance of the final graph: there is a certain type of embroidery called cross stitch, often used in decorating pillows or framed and hung on the wall. This site sells patterns and related supplies: shop.subversivecrossstitch.com Here is the woman who runs it talking about how the business began: shop.subversivecrossstitch.com/pages/about-us $\endgroup$ – Will Jagy Jan 26 '18 at 19:17
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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 $\pm 1$

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