Create unique number from 2 numbers

Question

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

• shuhalo: You do not specify what you mean by making a number out of two numbers.
• Henry: Are the pairs ordered? Should 2 and 30 produce the same answer as 30 and 2?
• drizzt: 2and30,30and2 are two different pairs
• Asaf Karagila: I don't get the bounty. Alex Kruckman gave a very good answer whose computational complexity is essentially $O(1)$ if I am not mistaken. Can you please specify what is wrong with the answer given so far?
• Abel: I agree with Asaf. Alex Kruckman's answer mentions a quite fast pairing function. If you are really concerned about the efficiency of the algorithm (for reading and writing), then you should probably look for some other ways of combining two numbers, like a list or an ordered pair, or whatever your language supports. But as far as standard mathematical pairing functions go, Cantor's is reasonably quick. Of course, it would be much easier to answer your question appropriately if we knew what your motivation is.

Answer 1

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},ドル given by: $\pi(a,b) = \frac{1}{2}(a+b)(a+b+1) + b$.

You can find more information here: http://en.wikipedia.org/wiki/pairing_function

• drizzt: thanks, this is good advice. can you suggest me which pairing function has the smallest computational complexity?
• user44190: That's not necessarily the cheapest way, it depends on what you're counting. In most actual computers, there is a multiplication at a reasonable price, while zipping two binary expansions will take many operations. The Cantor pairing function is the cheapest in actual applications, with only three(!) additions, one multiplication, and a single right shift (to get the $\tfrac12$).
• Purple Dragon: Just to make it clear, this formula DOES NOT GENERATE the same output when you change the order of A and B, e.g: f(A,B) != f(B,A).
• Alex Kruckman: @ChristopherDumas No - one reason your suggestion doesn't work is that $\Pi(a,b,c) = \Pi(c,b,a)$. There are many "tripling" functions $\mathbb{N}\times\mathbb{N}\times \mathbb{N}\to \mathbb{N}$. One way to generalize any pairing function $\pi$ to a tripling function $\Pi$ is to define $\Pi(a,b,c) = \pi(\pi(a,b),c)$. When $\pi$ is the Cantor pairing function from my answer: $$\Pi(a,b,c) = \frac{1}{2}\left(\frac{1}{2}(a+b)(a+b+1)+b+c\right)\left(\frac{1}{2}(a+b)(a+b+1)+b + c + 1\right) + c.$$
• Alex Kruckman: @hobwell Yes (0ドル\in \mathbb{N}$). If you visit the wikipedia page I linked to in my answer, you'll find a graphical representation of what this function does, and you can see clearly from that picture that it includes ordered pairs of natural numbers of the form $(0,b)$ and $(a,0)$ in the domain of the function.

Answer 2

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

• Bradley: This method generates exceptionally large numbers fairly quickly. (2^1023 on it's own is 8.9885E+307). If you're using this in excel 2^1023.9 is your upper limit)

Answer 3

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.

Answer 4

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>
   

• Steven Stadnicki: If you look in the comments, the OP specifically notes that for their application, argument order does matter.

Answer 5

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.

• Alex Kruckman: Nice answer. I've rewritten the function in notation which will be more familiar to mathematicians.

Answer 6

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?

• José Carlos Santos: Welcome to MSE. Please use MathJax.
• Ross Millikan: The Cantor pairing function only takes natural numbers for its input. If you want to use rationals or reals you need to work harder.