Home » Data Hacks Ideas » Create Unique Number from 2 Numbers

Create Unique Number from 2 Numbers

Marcus Vance • • 3 min read
Create Unique Number from 2 Numbers
$\begingroup$

is there some way to create unique number from 2 positive integer numbers? Result must be unique even for these pairs: 2 and 30, 1 and 15, 4 and 60. In general, if I take 2 random numbers result must be unique(or with very high probability unique)

Thanks a lot

EDIT: calculation is for computer program,so computational complexity is important

$\endgroup$
8

6 Answers

$\begingroup$

The sort of function you are looking for is often called a pairing function.

A standard example is the Cantor pairing function $\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$, given by: $\pi(a,b) = \frac{1}{2}(a+b)(a+b+1) + b$.

You can find more information here:

$\endgroup$
9
$\begingroup$

if the numbers are $a$ and $b$, take $2^a3^b$. This method works for any number of numbers (just take different primes as the bases), and all the numbers are distinct.

$\endgroup$
1
$\begingroup$

Google pairing function. As I mentioned in the similar question, there are also other pairing functions besides the well-known one due to Cantor. For example, see this "elegant" pairing function, which has the useful property that it orders many expressions by depth.

$\endgroup$
$\begingroup$

For positive integers as arguments and where argument order doesn't matter:

  1. Here's an unordered pairing function:

    $<x, y> = x * y + trunc(\frac{(|x - y| - 1)^2}{4}) = <y, x>$

  2. For x ≠ y, here's a unique unordered pairing function:

    <x, y> = if x < y:
           x * (y - 1) + trunc((y - x - 2)^2 / 4)
         if x > y:
           (x - 1) * y + trunc((x - y - 2)^2 / 4)
       = <y, x>
$\endgroup$
1
$\begingroup$

You could try the function by Matthew Szudzik, which is given as: $$a \geq b ~?~ a*a + a + b : a+b*b$$

In other words: $$(a,b)\mapsto \begin{cases} a^2 + a + b & \text{if } a \geq b\\ a + b^2 & \text{if } a < b \end{cases}$$

I found it here, which is a similar question as this.

$\endgroup$
1
$\begingroup$

Apologies for resurrecting this ancient question, but I've noticed that there are collisions in the results of the Cantor pairing function.

For example, I've noticed that when a and b are 0.0 and 0.0, or 0.6 and 0.0 the result is the same: 0.48.

Is this function expected to work for non-discrete numbers, or does this have something to do with the fact they the numbers are 0.0 and 0.0 in the former case?

$\endgroup$
2
Marcus Vance
Author

Marcus Vance

Marcus Vance is a cybersecurity auditor and technology writer dedicated to educating the public about online safety, data privacy regulations, enterprise security, and emerging cyber threats.