Area enclosed by the curve $\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$ 
What is the area enclosed by the curve $$\lfloor x + y\rfloor + \lfloor x - y\rfloor = 5$$ $$x\ge y, \forall x, y \ge 0$$
$\lfloor x\rfloor$ stands for the Greatest Integer Function.

I think that the curve will start from $\left(\cfrac{5}{2},\cfrac{5}{2}\right)$ and then be a straight line parallel to the $y$ axis till $\left(\cfrac{7}{2},\cfrac{5}{2}\right)$ (not including this last point). But then I am finding it difficult to manipulate it further.

Blue tiles:
$$
\lfloor{x + y}\rfloor + \lfloor{x - y}\rfloor = 5
$$
Gray triangle
$$
 y > 0, \quad x \le y
$$
Find the area of the blue tiles within the gray triangle.

The black dots are the vertices of the first square:
$$
\left( \begin{array}{c}
  3 \\ 0
\end{array} \right), \quad 
%
\left( \begin{array}{c}
  3.5 \\ 0.5
\end{array} \right), \quad 
%
\left( \begin{array}{c}
  3 \\ 1
\end{array} \right), \quad 
%
\left( \begin{array}{c}
  2.5 \\ 0.5
\end{array} \right)
%
$$
The side length for each square is $s =\frac{1}{\sqrt{2}}$. Therefore each square has area $A=s^{2} = \frac{1}{2}.$ There are three squares within the second region. Therefore the total area is 
$$A_{total} = 3 A = \frac{3}{2}$$
