Bijective Function from N to N x N

Question

I have an idea but I don't know how to formalize my idea in a function. That's what n should give me back as (x,y):

n = 0 -> (0,0)
n = 1 -> (1,0)
n = 2 -> (0,1)
n = 3 -> (2,0)
n = 4 -> (1,1)
n = 5 -> (0,2)
n = 6 -> (3,0)
n = 7 -> (2,1)
n = 8 -> (1,2)
n = 9 -> (0,3)
n = 10 -> (4,0)

How do I formulate a function for this one? I also need one for N -> N x {1,...,n) but I think it's the same function, is it?

score 10 · Accepted Answer · answered May 15 '18 at 00:18

10

Each $n \in \mathbb N$ can be uniquely written as $n=m(m+1)/2 + j$ with $0 \le j \le m$, namely $m = \left\lfloor \dfrac{\sqrt{8n+1}-1}{2} \right\rfloor$. Then take $f(n) = (m-j,j)$.

It's simpler, though, to write the inverse function from $\mathbb N \times \mathbb N$ to $\mathbb N$:

$$ (i,j) \mapsto \frac{(i+j)(i+j+1)}{2}+j $$

answered May 15 '18 at 00:18

Robert Israel

448,999

That looks very irritating for me. If I try it with n = 1, then I get m = 1 and j = 0, means f(n) = (1,0) which is true. But if I try it with n = 2, I need to take the root of 17. But that wouldn't result in a natural number for m. Where is my error? – CoreNoob May 15 '18 at 00:33
@corenoob Note the floor operation – Bram28 May 15 '18 at 00:35
1

@Bram28. Thanks. I'm such an idiot.
How do you guys come up with such functions? I'm a c.s. student in the 2nd semester but I wouldn't be able to come up with the function above.
– CoreNoob May 15 '18 at 00:39
@CoreNoob The bijection you indicated is pretty standard, even as the actual function is a pain to figure out. My guess is Robert had this function laying around in his notes somewhere ... though it's possible he figured it out on the spot! – Bram28 May 15 '18 at 00:43

score 2 · Answer 2 · answered May 15 '18 at 00:38

2

There is an easier bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$. If $n$ $\in$ $\mathbb{N}$, there is a unique pair $(x_n,y_n)$ $\in$ $\mathbb{N}\times\mathbb{N} $ such that $n = 2^{x_n-1}\cdot (2y_n - 1)$ (using fundamental theorem of arithmetic), defining the map

\begin{align*}F:&\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N}\\ &n\rightarrow (x_n,y_n) \end{align*} you get a bijection.

answered May 15 '18 at 00:38

Matheus Manzatto

3,231

Thanks for this one. But that is N x N -> N and not N -> N x N, right? f(n) would be N -> N x N. – CoreNoob May 15 '18 at 00:51
So you can use the inverse of $F$, $F^{-1} (n,m) = 2^{n-1} \cdot (2m - 1)$ – Matheus Manzatto May 15 '18 at 00:54
If you are concerned about the domain of $F$, is correct, this map associates every natural $n$ $\in$ $\mathbb{N}$ to a pair $(x_n,y_n)$ $\in$ $\mathbb{N}\times \mathbb{N}$ – Matheus Manzatto May 15 '18 at 00:57
What is the inverse function of f(n,m) = 2^(n-1) * (2m-1) ? – CoreNoob May 15 '18 at 01:07
The inverse of $f $ is the function $F $ that I explain in my answer. Just observe that $F (f (n,m)) =(n,m) $ – Matheus Manzatto May 15 '18 at 01:09
Sry, I don't get it. I can't see a function in your answer where I can put in a n and get back a pair (x,y). I just see the inverse of it: Put in x,y and get n. – CoreNoob May 15 '18 at 01:26
Let us continue this discussion in chat. – CoreNoob May 15 '18 at 01:27
I figured out how I can put in n and get (x,y). You can write f(n) = f(2^v * u) = (v+1, (u+1)/2) where v+1 = x and (u+1)/2 = y. – CoreNoob May 15 '18 at 02:20
I think there is a mistake in the formula: $0$ cannot be written as $2^a(2b-1)$. I think the mapping $2^{x_n}(2y_n+1)-1$ works though ... – rambi Jan 10 '24 at 15:40
If I remember this correctly, I am not considering $0 \in \mathbb N$ (which goes a bit against the def proposed by the OP...) – Matheus Manzatto Jan 10 '24 at 17:57

Bijective Function from N to N x N

2 Answers2

Linked