Explicit bijection between $\mathbb N$ and $\mathbb N \times \mathbb N$

Question

quadratic scheme

We can consider the quadratic scheme above for a possible explicit bijection between $\mathbb N$ and $\mathbb N \times \mathbb N$.

The part $\mathbb N \times \mathbb N \to \mathbb N$ is easy via $(m,n) \mapsto m + \frac{(n + m)\cdot(n + m + 1)}{2}$ for $m$ going down (as row index) and $n$ going right (as column index).

Consider $(0,0) \mapsto 0$ and $(2,1) \mapsto 8$ and $(3,2) \mapsto 18$ as examples.

What ist the other map $\mathbb N \to \mathbb N \times \mathbb N$?

You may want to check http://en.wikipedia.org/wiki/Pairing_function#Inverting_the_Cantor_pairing_function — Andreas Caranti, Mar 07 '15 at 14:05

score 4 · Accepted Answer · answered Mar 07 '15 at 14:08

4

For $s \in \mathbb N$ define $r := \left\lfloor \frac{-1 + \sqrt{1 + 8 \cdot s}}{2}\right\rfloor$.

Then, $m := s - \frac{r \cdot (r + 1)}{2}$ and $n := r - m$ and so $s \mapsto (m, n)$.

answered Mar 07 '15 at 14:08

Curiosity

716

Did you find out or look it up? – Achilles Mar 11 '15 at 07:19
3

@Achilles: I found out by myself in 2010. – Curiosity Mar 11 '15 at 07:20

Explicit bijection between $\mathbb N$ and $\mathbb N \times \mathbb N$

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