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What is the explicit form of the inverse of the function $f:\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow\mathbb{Z}^+$ where $$f(i,j)=\frac{(i+j-2)(i+j-1)}{2}+i?$$

Henry B.
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  • Does you $\mathbb Z^+$ include $0$? – Mariano Suárez-Álvarez Mar 03 '11 at 03:01
  • No. Zero is not included. – Henry B. Mar 03 '11 at 03:13
  • What is your motivation for inverting this function? – Abel Mar 03 '11 at 03:59
  • @Abel: It's just curiosity. I realize that the function, I mentioned, shows that the number of entries in an infinite grid are countable. Naturally,I wanted to see the inverse. However, when I tried to construct it, I realized it was more complicated than I thought. I wanted to see how to work this example. – Henry B. Mar 03 '11 at 04:06
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    @Henry: In that case, the key phrase is "pairing function". Your $f$ is a specific example of one, particularly, a "displacement" of Cantor's pairing function. Notice that you can apply it repeatedly in order to demonstrate the isomorphism between $\mathbb{N}^n$ and $\mathbb{N}$, for a natural $n$. In my opinion, it is conceptually easier to use Gödel numbering, based on the fundamental theorem of arithmetic, to "encode" an $n$-tuple of natural numbers into one. Lastly, keep in mind that one explicitly writes these functions in order to constructively demonstrate an isomorphism, which in ... – Abel Mar 03 '11 at 04:43
  • ... the end does not rely on the details of the construction. That is to say, we don't usually care whether these pairing functions have nice inverses in the sense of having a closed-form expression, but only about the existence of the function and its inverse. – Abel Mar 03 '11 at 04:45

2 Answers2

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Let $i+j-2 = n$.

We have $f = 1 + 2 + 3 + \cdots + n + i$ with $1 \leq i \leq n+1$. Note that the constraint $1 \leq i \leq n+1$ forces $n$ to be the maximum possible $n$ such that the sum is strictly less than $f$.

Hence given $f$, find the maximum $n_{max}$ such that $$1 + 2 + 3 + \cdots + n_{max} < f \leq 1 + 2 + 3 + \cdots + n_{max} + (n_{max} + 1)$$ and now set $i = f - \frac{n_{max}(n_{max}+1)}{2}$ and $j = n_{max} + 2 - i$.

$n_{max}$ is given by $\left \lceil \frac{-1 + \sqrt{1 + 8f}}{2} - 1 \right \rceil$ which is obtained by solving $f = \frac{n(n+1)}{2}$ and taking the ceil of the positive root minus one. (since we want the sum to strictly smaller than $f$ as we need $i$ to be positive)

Hence, $$ \begin{align} n_{max} & = & \left \lceil \frac{-3 + \sqrt{1 + 8f}}{2} \right \rceil\\\ i & = & f - \frac{n_{max}(n_{max}+1)}{2}\\\ j & = & n_{max} + 2 - i \end{align} $$

  • It looks like the final formula for the maximum n is in conflict with the one in the body of the text, the ceiling function has become a floor function. – Henry B. Mar 03 '11 at 06:00
  • @Henry: edit done –  Mar 03 '11 at 06:33
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Since your function seems to be Cantor's pairing function $$ p(x,y) = \frac{(x+y)(x+y+1)}{2} + y $$ applied to $x= j-2, y = i$, and since the inverse of the pairing function is $$ p^{-1}(z) = \left( \frac{ \left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor^2 + 3\left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor}{2}-z \, , \, z-\frac{\left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor^2 + \left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor}{2}\right), $$ the inverse of your function is: $$ f^{-1}(z) = \left( z-\frac{\left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor^2 + \left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor}{2}\, , \, 2+ \frac{ \left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor^2 + 3\left\lfloor \frac{\sqrt{8z+1}-1}{2} \right\rfloor}{2}-z \right), $$ which can be a bit ugly. What is your motivation for inverting this function?

Saaqib Mahmood
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Abel
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    A minor correction though. The OP wants $i,j \in \mathbb{Z}^+$, so you need to make slight modifications to the answer. For instance, $z=1$ should give $(1,1)$ but your solution gives $(0,3)$ –  Mar 03 '11 at 03:41
  • @Sivaram: You're probably right. I'm pretty sure that after some modification (probably replacing all instances of $\lfloor \frac{\sqrt{8z+1}-1}{2}\rfloor$ with $\lfloor \frac{\sqrt{8z+1}-3}{2}\rfloor$), one would obtain the desired inverse, ending up with your solution. – Abel Mar 03 '11 at 04:06