# Inverse of a bijective function involving cases

In continutation to a question that i asked earlier and got answered here :Discretizing a mathematical equation This is a bijective mapping from the set of ordered tuples $$(x,y,z)$$ where each $$x,y,z\in \{1,2,\dots,n\}$$ to itself. \begin{align} x'&=\text{mod}(2(x-1),n)+1+\mathbf{1}[\text{mod}(z,4)= 2\text{ or }\text{mod}(z,4)=0] \\ y'&=\text{mod}(2(y-1),n)+1+\mathbf{1}[\text{mod}(z,4)= 3\text{ or }\text{mod}(z,4)=0] \\ z'&=\lceil z/4\rceil + (n/4)\Big(1.\mathbf{1}[n/2 where $$\def\1{{\bf 1}}\1[S]$$ be a function which is equal to $$1$$ if the statement $$S$$ is true, and zero otherwise.

Since it is a bijective mapping I want to find it's inverse. I have deduced the following results but still not been able to get the inverse. So here s my work

$$1.$$ Given $$(x',y',z')$$ if I see that if $$x'$$ and $$y'$$ both are odd then $$z\equiv 1$$ mod $$4$$.

$$2.$$ Observing the value of $$z'$$ I deduce that $$\begin{cases} z'\leq n/4, &\implies x\leq n/2~\text{and}~y\leq n/2\\ n/4< z'\leq n/2, &\implies x>n/2 ~\text{and}~y\leq n/2~\\ n/2 n/2~\\ z'>3n/4, &\implies x> n/2~\text{and}~y>n/2~ \end{cases}$$

$$3.$$ So if $$z'$$ belongs to the first case then $$z=4\cdot(z')+1$$

$$4.$$ if $$z'$$ belongs to the second case then $$z=4\cdot(z'-n/4-1)+1$$

$$5.$$ if $$z'$$ belongs to the third case then $$z=4\cdot(z'-n/2-1)+1$$

$$6.$$ if $$z'$$ belongs to the fourth case then $$z=4\cdot(z'-3n/4-1)+1$$

The similar cases if $$x'$$ is odd and $$y'$$ is even $$\implies z\equiv 3~\text{mod}~4$$

If $$x'$$ is even and $$y'$$ is odd $$\implies z\equiv 2~\text{mod}~4$$

If $$x'$$ is even and $$y'$$ is even $$\implies z\equiv 0~\text{mod}~4$$

But i doubt that this is correct. and also how to deduce expressions for $$x,y,z$$. Can somebody check?

• Statement 1 isn't true. Take, for example, n=7, x=1, y=5 and z=3 – André Porto May 24 at 18:47
• i forgot to mention $n\equiv 0\mod~4$ – Upstart May 24 at 18:58
• Well, it seems to be a hypothesis to construct the bijection. – André Porto May 24 at 19:13
• Yeah, indeed, because of the n/4 on the z' expression. – André Porto May 24 at 19:14
• yes indeed it is a hypothesis – Upstart May 24 at 19:16

Let us construct the inverse function. First, fix the parameter $$q=\lfloor (z'-1)/(n/4)\rfloor$$.

Analyzing the terms $$\Big(1.\mathbf{1}[n/2 on the definition of $$z'$$, we conclude that:

(i) $$q=0 \Rightarrow$$ $$x, y\leq n/2$$;

(ii) $$q=1 \Rightarrow$$ $$x> n/2$$ and $$y\leq n/2$$;

(iii) $$q=2 \Rightarrow$$ $$x\leq n/2$$ and $$y> n/2$$;

(iv) $$q=3 \Rightarrow$$ $$x,y>n/2$$.

Recovering $$x$$:

Analyzing the definition of $$x'$$, we conclude that

• If $$x'$$ is odd then $$x'= mod(2(x-1),n) + 1$$, and since $$1\leq x\leq n$$, there are only two possibilities: $$x'= 2(x-1) + 1\mbox{ or } x'= 2(x-1) - n + 1,$$ the first one when $$x\leq n/2$$ and the second one when $$x> n/2$$. Both these cases are expressed, relying on the value of $$q$$, by: $$x'=2(x-1) - n\mathbf{1}[q=1\mbox{ or } 3] + 1,$$ which conclusion is made by using the facts (i)-(iv). Therefore, obtaining $$x$$ in terms of $$x'$$ above, we get: $$x=\frac{x'+1+n\mathbf{1}[q=1\mbox{ or } 3 \mbox{ or } 4]}{2}$$
• If $$x'$$ is even, then $$x'= mod(2(x-1),n) + 2$$, and proceeding analogously we get that $$x=\frac{x'+n\mathbf{1}[q=1\mbox{ or } 3]}{2}.$$

The difference between the two cases above can be expressed by a $$\mathbf1$$ function as follows: $$x=\frac{x'+n\mathbf{1}[q=1\mbox{ or } 3]+\mathbf{1}[\mbox{x' is odd}]}{2}$$

Finally, substituing the value of $$q$$ above, we get $$x=\frac{x'-n\mathbf{1}\Big[\lfloor(z'-1)/(n/4)\rfloor=1\mbox{ or } 3\Big]+\mathbf{1}[\mbox{x' is odd}]}{2}.$$

Recovering $$y$$

To recover $$y$$ the argument is analogous to the one concerning $$x$$. We find that $$y=\frac{y'+n\mathbf{1}\Big[\lfloor (z'-1)/(n/4)\rfloor\geq2\Big]+\mathbf{1}[\mbox{y' is odd}]}{2}.$$

Recovering z:

As you said, we can determine $$mod(z,4)$$ by the following:

• Both $$x'$$ and $$y'$$ are even $$\Rightarrow$$ $$mod(z,4)=0$$;
• Both $$x'$$ and $$y'$$ are odd $$\Rightarrow$$ $$mod(z,4)=1$$;
• $$x'$$ is even and $$y'$$ is odd $$\Rightarrow$$ $$mod(z,4)=2$$;
• $$x'$$ is odd and $$y'$$ is even $$\Rightarrow$$ $$mod(z,4)=3$$.

Define $$Mod(a,b)$$ to be the usual $$mod(a,b)$$ function, except that it returns $$b$$ instead of $$0$$.

Write $$z= 4\cdot s + mod(z,4)$$, for some $$0\leq s \leq n/4$$. There are two cases:

• Case 1. $$mod(z,4)=0$$. Then, by the definition of $$z'$$, $$z'=s + (n/4)\Big(\mathbf{1}[n/2 Also, since in this case $$z=4s$$ and $$z\neq0$$, we have $$1\leq s \leq n/4$$, and therefore, $$s=Mod(z',n/4)$$. Consequently, $$z=4Mod(z',n/4).$$

• Case 2. $$mod(z,4)>0$$. Then, by the definition of $$z'$$, $$z'=s + 1 + (n/4)\Big(\mathbf{1}[n/2 Also, in this case, since $$z=4s+mod(z,4)>4s$$ and $$z\leq n$$, then certainly, $$0\leq s \leq n/4-1\ \Rightarrow\ 1\leq s + 1 \leq n/4,$$ and therefore, $$s + 1 = Mod(z',n/4)$$. Consequently, $$z=4(Mod(z',n/4)-1) + mod(z,4).$$

The differences between the cases above may be expressed by a $$\mathbf 1$$ function as follows: $$z=4\Big(Mod(z',n/4)-\mathbf{1}[mod(z,4)\neq 0]\Big) + mod(z,4).$$

• i guess you didn't edit $q$ to $s$ but thanx for a detailed explaination, i will check it – Upstart May 25 at 6:53
• Yeah, sorry about that – André Porto May 25 at 10:38
• There are a few errors i guess. 1). $x'=2(x-1)+1$ or $x'=2(x-1)-n$ not $+n$ as you have written, because that part is extra so we fold it back in. – Upstart May 26 at 20:43
• i figured out the problems in your equations for $x$ and $y$. Consider the odd case for $x$ . If $(z' \leq n/4~\text{or}~ n/2< z'\leq 3n/4)$ Then the case $x= 2(x-1) +1$ will be implemented. and if $( n/4< z'\leq n/2 ~\text{or}~z'>3n/4)$ then $2(x-1) +n +1$ will be implemented – Upstart May 27 at 8:10
• Yeah, you are right. We shall fix those mistakes then. You were right about the change of the sign of $n$ (oops). But I didn't understand the other mistake. I guess it ha to do with the values $z'=n/4$, $z'=n/2$, and so on. It's a mistake with the conclusions on the value $q$. One way to fix it is to change the definition of $q$ to $q=\lfloor (z'-1)/(n/4)\rfloor$. – André Porto May 27 at 13:33