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. $$ \begin{align} x'&=\text{mod}(2(x-1),n)+1+\mathbf{1}[\text{mod}(z,4)= 2\text{ or }\text{mod}(z,4)=0] \\ y'&=\text{mod}(2(y-1),n)+1+\mathbf{1}[\text{mod}(z,4)= 3\text{ or }\text{mod}(z,4)=0] \\ z'&=\lceil z/4\rceil + (n/4)\Big(1.\mathbf{1}[n/2<x] + 2\cdot \mathbf{1}[n/2<y]\Big) \end{align} $$ 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 $\begin{cases} z'\leq n/4, &\implies x\leq n/2~\text{and}~y\leq n/2\\ n/4< z'\leq n/2, &\implies x>n/2 ~\text{and}~y\leq n/2~\\ n/2 <z'\leq 3n/4, &\implies x\leq n/2~\text{and}~y>n/2~\\ z'>3n/4, &\implies x> n/2~\text{and}~y>n/2~ \end{cases}$

$3.$ 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?

  • $\begingroup$ Statement 1 isn't true. Take, for example, n=7, x=1, y=5 and z=3 $\endgroup$ – André Porto May 24 at 18:47
  • $\begingroup$ i forgot to mention $n\equiv 0\mod~4$ $\endgroup$ – Upstart May 24 at 18:58
  • $\begingroup$ Well, it seems to be a hypothesis to construct the bijection. $\endgroup$ – André Porto May 24 at 19:13
  • $\begingroup$ Yeah, indeed, because of the n/4 on the z' expression. $\endgroup$ – André Porto May 24 at 19:14
  • $\begingroup$ yes indeed it is a hypothesis $\endgroup$ – 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 $\Big(1.\mathbf{1}[n/2<x] + 2\cdot \mathbf{1}[n/2<y]\Big)$ 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 $z'$, $$ z'=s + (n/4)\Big(\mathbf{1}[n/2<x] + 2\cdot \mathbf{1}[n/2<y]\Big). $$ 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 $z'$, $$ z'=s + 1 + (n/4)\Big(\mathbf{1}[n/2<x] + 2\cdot \mathbf{1}[n/2<y]\Big). $$ 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: $$ z=4\Big(Mod(z',n/4)-\mathbf{1}[mod(z,4)\neq 0]\Big) + mod(z,4). $$

  • $\begingroup$ i guess you didn't edit $q$ to $s$ but thanx for a detailed explaination, i will check it $\endgroup$ – Upstart May 25 at 6:53
  • $\begingroup$ Yeah, sorry about that $\endgroup$ – André Porto May 25 at 10:38
  • $\begingroup$ 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. $\endgroup$ – Upstart May 26 at 20:43
  • $\begingroup$ 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 $\endgroup$ – Upstart May 27 at 8:10
  • $\begingroup$ 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 $q=\lfloor (z'-1)/(n/4)\rfloor$. $\endgroup$ – André Porto May 27 at 13:33

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