I have a C program that does the following calculation:
int b = 16;
int a = max(1, ( (b + 3) / 4 ) ) * max(1, ( (b + 3) / 4 ) ) * 16;
More easily read as: $$ a = \max\left(1, \frac{b + 3}{4}\right)^2 × 16 $$
I would like to know how I could use simple algebra in order to find what is the equivalent algorithm to find the value of b if I am only given the value of a? For example:
int a = 256;
int b = ?
On
No, you cannot. This is because there exists more than one possible value for $b$ that returns the same value of $a$. For example,
$$\max\left(1, \frac{-1+3}{4}\right)\cdot 16=\max\left(1, \frac{-2+3}{4}\right)\cdot 16=16.$$
So, if all you know is that $\max\left(1, \frac{b+3}{4}\right) = 1$, you have no way of knowing what $b$ is, since it could be $-1$ or $-2$ or some other number.
However, there is still something you can extract, if you are given $a$:
If $a=16$:
In this case, you know that $\max\left(1, \frac{b+3}{4}\right)=1$, and this happens if and only if $\frac{b+3}{4}\leq 1$ (since, if it were greater than one, then $1$ would not be the maximum). So, in this case, you can conclude (after multiplying the inequality by $4$) that $b\leq 1$.
If $a\neq 16$:
In this case, you know for a fact that $16\cdot \frac{b+3}{4} = a$, since $a$ can be either $16$ or $16\cdot \frac{b+3}{4}$, and it isn't $16$. Therefore, you have $4b+3=a$ and you can, in this case, calculate exactly what $b$ is: $b=\frac{a-3}{4}$
To say it another way, the range of the function on the RHS is $a \ge 16$ (plot $a$ against $b$ if you don't see it.) There are three cases:
if $a \lt 16$, it is not in the range, so there is no $b$ that can produce that value.
if $a = 16$, then $\max(1, (b+3)/4) = 1$ and any $b$ where $(b+3)/4 \le 1$ will do.
if $a \gt 16$ then there is a unique $b = 4\sqrt{a/16} - 3 = \sqrt{a} - 3$ that produces that $a$ (assuming you want the positive square root only).