solve for $f(x)$ in $4\lfloor\frac{x}{3}\rfloor - f(x) - 3\lfloor\frac{x-3f(x)}{9}\rfloor=x$

Question

Given that $x$ is an integer in the interval $[0,8]$ and $f(x)$ is also an integer in the interval $[0,2]$, I have the following equality:

$$4\lfloor\frac{x}{3}\rfloor - f(x) - 3\lfloor\frac{x-3f(x)}{9}\rfloor=x$$

Is it possible to solve for $f(x)$ in this case?

Given the intervals of $x$ and $f(x)$, $\frac{x-3f(x)}{9}$ must be in the interval $[-1,1)$, so its floor must be either -1 or 0. I would write it as this:

$$4\lfloor\frac{x}{3}\rfloor-f(x)+3\big(x<3f(x)\big)=x$$

But I suspect that makes simplifying further impossible. (should probably mention I come from a programming background, the above is probably wrong mathematically)

Edit: The formula is from the following table:

x    f(x)
0     0
1     2
2     1
3     1
4     0
5     2
6     2
7     1
8     0

Answer 1

Solution 1: Lagrange Interpolation:

Given any $n$ data set, there exists a polynomial of degree $n-1$ that fits them.
I did this on Mathematica:
Expand[InterpolatingPolynomial[{{0,0},{1,2},{2,1},{3,1},{4,0},{5,2},{6,2},{7,1},{8,0}},$x$]]

Where the braces represent $(x,f(x))$ values. Or you may think of them as $(x,y)$ coordinates.
This gives me the following function:

$f(x)=\dfrac{3662 x}{105}-\dfrac{10707 x^2}{140}+\dfrac{3247 x^3}{48}-\dfrac{4941 x^4}{160}+\dfrac{947 x^5}{120}-\dfrac{91 x^6}{80}+\dfrac{29 x^7}{336}-\dfrac{3 x^8}{1120}$ $=\dfrac{-x (-117184+256968 x-227290 x^2+103761 x^3-26516 x^4+3822 x^5-290 x^6+9 x^7)}{3360}$

The problem with this is clearly the fact that the starting polynomial is ugly.
We can make this simpler given your range though.

Find out what is the factorization of the denominator: $3360=2^5\cdot 3\cdot 5\cdot 7$.
Choose the smallest prime not in this factorization and bigger than the max $f(x)$. i.e. $p=11$.
We can form a new polynomial under modulo $p$:
$f(x)=9 x+5 x^2+6 x^3+7 x^4+10 x^5+10 x^6+3 x^7+7 x^8$ where each coefficient is $<p$.

The draw back is you need to do a mod $p$ after computing $f(x)$.
The conversion is simple, just perform modulo $p$ on all coefficients (division is modulo inverse).
$p$ is chosen to not divide denominator so that inverse exists.

Solution 2: Solving equation by cases:

Since the greatest denominator in the floor function is 9, it can be solved in 9 cases.
We start by considering $x=9m,9m+1,\dots,9m+8$.
This works because it allows you to remove the floor function.

First, a property of the floor function: $\lfloor x+n\rfloor = \lfloor x\rfloor + n$, where $x$ is real and $n$ an integer.
This means that whenever we have an integer within the floor function, we may take it out.

Let $x=9m+a$, where $0\leq a<9$.
$4\lfloor \frac{9m+a}{3}\rfloor-f(x)-3\lfloor \frac{9m+a-3f(x)}{9}\rfloor=9m+a$
$4\lfloor 3m+\frac{a}{3}\rfloor-f(x)-3\lfloor m+\frac{a}{9}-\frac{f(x)}{3}\rfloor=9m+a$
$12m+4\lfloor\frac{a}{3}\rfloor-f(x)-3m-3\lfloor \frac{a}{9}-\frac{f(x)}{3}\rfloor=9m+a$
$4\lfloor\frac{a}{3}\rfloor-f(x)-3\lfloor \frac{a}{9}-\frac{f(x)}{3}\rfloor=a$

There are 2 approaches here, by restricting $a$ or $f(x)$.
It is faster to consider $f(x)=3n+b$, where $0\leq b<3$.

$4\lfloor\frac{a}{3}\rfloor-(3n+b)-3\lfloor \frac{a}{9}-\frac{3n+b}{3}\rfloor=a$
$4\lfloor\frac{a}{3}\rfloor-3n-b-3\lfloor \frac{a}{9}-n-\frac{b}{3}\rfloor=a$
$4\lfloor\frac{a}{3}\rfloor-3n-b+3n-3\lfloor \frac{a}{9}-\frac{b}{3}\rfloor=a$
$4\lfloor\frac{a}{3}\rfloor-b-3\lfloor \frac{a-3b}{9}\rfloor=a$

Up till this point, this shows that indeed we only need to consider $x\in [0,8]$ and $f(x)\in [0,2]$.
For any solution set $(x,f(x))$, we may add $9m$ to $x$ and $3n$ to $f(x)$.

$\underline{\text{Case } b=0}$:
$4\lfloor\frac{a}{3}\rfloor-3\lfloor \frac{a}{9}\rfloor=a$
$4\lfloor\frac{a}{3}\rfloor=a$
$a=\lbrace 0,4,8\rbrace$
Hence we have $(9m+0,3n+0),(9m+4,3n+0),(9m+8,3n+0)$

$\underline{\text{Case } b=1}$:
$4\lfloor\frac{a}{3}\rfloor-1-3\lfloor \frac{a-3}{9}\rfloor=a$
If $0\leq a<3$:
$4\lfloor\frac{a}{3}\rfloor-1-3(-1)=a$
$4\lfloor\frac{a}{3}\rfloor+2=a$
$a=2\implies$ solution $(9m+2,3n+1)$
If $(3\leq a<9)$:
$4\lfloor\frac{a}{3}\rfloor-1=a$
$a=\lbrace 3,7\rbrace$
Solution: $(9m+3,3n+1),(9m+7,3n+1)$

$\underline{\text{Case } b=2}$:
$4\lfloor\frac{a}{3}\rfloor-2-3\lfloor \frac{a-6}{9}\rfloor=a$
If $0\leq a<6$:
$4\lfloor\frac{a}{3}\rfloor-2-3(-1)=a$
$4\lfloor\frac{a}{3}\rfloor+1=a$
$a=\lbrace 1,5\rbrace$
Solution: $(9m+1,3n+2),(9m+5,3n+2)$
If $6\leq a<9$:
$4\lfloor\frac{a}{3}\rfloor-2=a\implies a=6$
Solution: $(9m+6,3n+2)$

This describes the solution set of the equation entirely.

Solution 3: Karnaugh Map:
We view the input $x$ as $x_3x_2x_1x_0$, where $x_3$ is the most significant bit.
We view the output $f(x)$ as $f(x)_1f(x)_0$, where $f(x)_1$ is the most significant bit.
$$\begin{array}{|c|c|c|c|c|c|}\hline & & x_1x_0 & x_1x_0 & x_1x_0 & x_1x_0 \\\hline & f(x)_0 & 00 & 01 & 11 & 10 \\\hline x_3x_2 & 00 & 0 & 0 & 1 & 1 \\\hline x_3x_2 & 01 & 0 & 0 & 1 & 0 \\\hline x_3x_2 & 11 & X & X & X & X \\\hline x_3x_2 & 10 & 0 & X & X & X \\\hline \end{array} $$

$f(x)_0 = \bar{x_3}x_2x_1x_0+\bar{x_3}\bar{x_2}x_1=\bar{x_3}x_1(x_2x_0+\bar{x_2})$
C code: y0 = (~x3) & x1 & ((x2 & x0) | (~x2));

$$\begin{array}{|c|c|c|c|c|c|}\hline & & x_1x_0 & x_1x_0 & x_1x_0 & x_1x_0 \\\hline & f(x)_1 & 00 & 01 & 11 & 10 \\\hline x_3x_2 & 00 & 0 & 1 & 0 & 0 \\\hline x_3x_2 & 01 & 0 & 1 & 0 & 1 \\\hline x_3x_2 & 11 & X & X & X & X \\\hline x_3x_2 & 10 & 0 & X & X & X \\\hline \end{array} $$

$f(x)_1=\bar{x_3}x_2x_1\bar{x_0}+\bar{x_3}\bar{x_1}x_0=\bar{x_3}(x_2x_1\bar{x_0}+\bar{x_1}x_0)$
C code: y1 = (~x3) & ((x2 & x1 & (~x0)) | ((~x1) & x0));
y = (y1<<1) | y0;