Problem: Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]

Solution: We can easily see that only $x, y \in \left[0,1 \right]$ may satisfy both functions.
We call function $y = 4g \left( f \left( \sin \left( 2 \pi x \right) \right) \right)$ as Function 1 and function $x = 4g \left( f \left( \cos \left( 3 \pi y \right) \right) \right)$ as Function 2.
For Function 1, in each interval $\left[ \frac{i}{4} , \frac{i+1}{4} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ , Function 1's value oscillates between 0 and 1. It attains 1 at $x = \frac{i}{4}$ $\frac{i+1}{4}$ and another point between these two.
Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1. 
So the graph of this function in this interval consists of 4 monotonic pieces.
For Function 2, in each interval $\left[ \frac{i}{6} , \frac{i+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 5 \right\}$ , Function 2's value oscillates between 0 and 1. It attains 1 at $y = \frac{i}{6}$ $\frac{i+1}{6}$ and another point between these two.
Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1.
So the graph of this function in this interval consists of 4 monotonic curves.
Consider any region $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ and $j \in \left\{0, 1, \cdots , 5 \right\}$ but $\left( i, j \right) \neq \left( 3, 5 \right)$ .
Both functions have four monotonic pieces.
Because Function 1's each monotonic piece can take any value in $\left[ \frac{j}{6} , \frac{j+1}{6} \right]$ and Function 2' each monotonic piece can take any value in $\left[ \frac{i}{4} , \frac{i+1}{4} \right]$ , Function 1's each monotonic piece intersects with Function 2's each monotonic piece. 
Therefore, in the interval $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ , the number of intersecting points is $4 \cdot 4 = 16$
Next, we prove that if an intersecting point is on a line $x = \frac{i}{4}$ for $i \in \left\{ 0, 1, \cdots, 4 \right\}$ , then this point must be $\left( 1, 1 \right)$
For $x = \frac{i}{4}$ , Function 1 attains value 1. 
For Function 2, if $y = 1$ , then $x = 1$ . 
Therefore, the intersecting point is $\left( 1, 1 \right)$
Similarly, we can prove that if an intersecting point is on a line $y = \frac{i}{6}$ for $i \in \left\{ 0, 1, \cdots, 6 \right\}$ , then this point must be $\left( 1, 1 \right)$
Therefore, in each region $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ and $j \in \left\{0, 1, \cdots , 5 \right\}$ but $\left( i, j \right) \neq \left( 3, 5 \right)$ , all 16 intersecting points are interior.
That is, no two regions share any common intersecting point.
Next, we study region $\left[ \frac{3}{4} , 1 \right] \times \left[ \frac{5}{6} , 1 \right]$ .
Consider any pair of monotonic pieces, where one is from Function 1 and one is from Function 2, except the pair of two monotonic pieces from two functions that attain $\left( 1 , 1 \right)$ .
Two pieces in each pair intersects at an interior point on the region.
So the number of intersecting points is $4 \cdot 4 - 1 = 15$
Finally, we compute the number intersection points of two functions' monotonic pieces that both attain $\left( 1, 1 \right)$
One trivial intersection point is $\left( 1, 1 \right)$ .
Now, we study whether they intersect at another point.
Define $x = 1 - x'$ and $y = 1 - y'$ .
Thus, for positive and sufficiently small $x'$ and $y'$ , Function 1 is reduced to \[ y' = 4 \sin 2 \pi x' \hspace{1cm} (1) \] and Function 2 is reduced to \[ x' = 4 \left( 1 - \cos 3 \pi y' \right) . \hspace{1cm} (2) \]
Now, we study whether there is a non-zero solution.
Because we consider sufficiently small $x'$ and $y'$ , to get an intuition and quick estimate, we do approximations of the above equations.
Equation (1) is approximated as \[ y' = 4 \cdot 2 \pi x' \] and Equation (2) is approximated as \[ x' = 2 \left( 3 \pi y' \right)^2  \]
To solve these equations, we get $x' = \frac{1}{8^2 \cdot 18 \pi^4}$ and $y' = \frac{1}{8 \cdot 18 \pi^3}$ .
Therefore, two functions' two monotonic pieces that attain $\left( 1, 1 \right)$ have two intersecting points.
Putting all analysis above, the total number of intersecting points is $16 \cdot 4 \cdot 6 + 1 = \boxed{385}$

Answer:  385