A bijective mapping from $\mathbb N^k$ to $\mathbb N$?

Question

Having $k$ numbers $N_i\in\mathbb{N}$, I'm looking for a bijective mapping

$f:\mathbb{N}\times\ldots\times\mathbb{N}\rightarrow\mathbb{N}$

So that $f^{-1}\left(N_0\right)=\left(N_1,\ldots,N_k\right)$.

Any ideas?

May I ask what additional structure should be preserved by isomorphism? — Evgeny, Jul 30 '13 at 07:57
Do you mean a bijective mapping? (that is, a one-to-one and onto mapping?) When we say "isomorphic mapping" we mean that in addition to bijective-ness, we need to preserve some structure such as $f(x+y)=f(x)+f(y)$ or perhaps $f(x^2)=f(x)^2$. — Eric Stucky, Jul 30 '13 at 08:07
@EricStucky you're right, I'm looking for a bijective mapping. There's no algebraic structure whatsoever. — Artem Oboturov, Jul 30 '13 at 08:10
If you are looking for a bijective mapping from $\mathbb{N}^k \to \mathbb{N}$, then here is mapping $\mathbb{N}^2 \to \mathbb{N}$ that easily extends to any finite $k$ (just use more colors $\ddot\smile$). Of course, composing with itself many mappings $\mathbb{N}^2 \to \mathbb{N}$ is another solution. — dtldarek, Jul 30 '13 at 08:10
Please edit the title as well :) And consider revising/deleting the third line, because although it's clear what it means, it's not clear why you are taking the time to mention it specifically. — Eric Stucky, Jul 30 '13 at 08:11
@ArtemOboturov I don't care for score very much, maybe there is some yet marvelous answer which we don't know? — dtldarek, Jul 30 '13 at 08:14
@dtldarek : your answer will require $k$ colours to be used. I have an additional requirement : only one colour should be used. Is this enough as a structure? — Artem Oboturov, Jul 30 '13 at 08:16
@dtldarek I didn't notice at first, that you have linked to a question with your proposal, I'm on it now. — Artem Oboturov, Jul 30 '13 at 08:21
@ArtemOboturov That was an informal statement, you don't have to use any colors at all. — dtldarek, Jul 30 '13 at 08:23

score 6 · Answer 1 · edited Apr 13 '17 at 12:19

There is a nice bijection that works like this (here for $k = 3$):

$$ \color{red}{42},\color{blue}{2013},\color{green}{789}, \to \color{blue}{2}\, \color{red}{0}\, \color{green}{7}\, \color{blue}{0}\, \color{red}{0}\, \color{green}{8}\, \color{blue}{1}\, \color{red}{4}\, \color{green}{9}\, \color{blue}{3}\, \color{red}{2}\,,$$

this is an easy extension of this $\mathbb{N}^2 \to \mathbb{N}$ bijection.

Another approach is to use any $f: \mathbb{N}^2 \to \mathbb{N}$ bijection and compose it with itself, e.g.

$$f_k (a_1,a_2,\ldots,a_k) = f(a_1, f(a_2, \ldots f(a_{k-1},a_k)\ldots)).$$

I hope this helps $\ddot\smile$

This is a constructive solution which I could use in a computer program, as I intended to. — Artem Oboturov, Jul 30 '13 at 09:18

score 2 · Answer 2 · answered Jul 30 '13 at 08:36

To give a bijection $A\leftrightarrow\Bbb N$ amounts to define a sequence $a_1,a_2,a_3,...$ which includes all elements of $A$.

When $A=\Bbb N^k$ a standard way to do this is to write $$ \Bbb N^k=\bigcup_{r=0}^\infty A_r,\qquad A_r=\{(n_1,...,n_k)\in\Bbb N^k\,|\,\sum_{j=1}^kn_j=r\} $$ and since every $A_r$ is finite we can list all elements of $A_1$, followed by all elements of $A_2$, followed by all elements of $A_3$, and so on.

This certainly works although it may not be easy to say what is the $n$-th $k$-ple ofthe sequence excplicitly.

P.S.: the above is written under the conventional assumption that $0\in\Bbb N$. If, so to speak, $\Bbb N$ starts with 1 the above works in the same way with the observation that the first non-empty subset $A_r$ is $A_k$.

A bijective mapping from $\mathbb N^k$ to $\mathbb N$?

2 Answers2