Is it possible to combine two integers in such a way that you can always take them apart later?

Given two integers $n$ and $m$ (assuming for both $0 < n < 1000000$) is there some function $f$ so that if I know $f(n, m) = x$ I can determine $n$ and $m$, given $x$? Order is important, so $f(n, m) \not= f(m,n)$ (unless $n=m$).

Solution 1:

$f(n,m) = 2^n 3^m$.

Alternatively, use the bijection between $\mathbb N \times \mathbb N$ and $\mathbb N$ which is given by $$f(n,m) = \frac{(n+m)(n+m+1)}{2} + m$$

Solution 2:

Since others have answered, here is another idea - less easy to write down explicitly as a mathematical function, but easy to describe and easy to implement if you have a machine which can handle strings. Send the digits of the first number to odd positions and the digits of the other to even positions so (31, 5681) would go to 50,608,311.

Solution 3:

If there are no bounds on your integers, use the Cantor pairing function. It is pleasantly easy to compute, as are its two "inverses."

For the case where your integers are bounded by say $10^6$, you can simply concatenate the decimal expansions, padding with initial $0$'s as appropriate. Or do something similar with binary expansions. Dirt cheap to combine and uncombine, an easy string manipulation even when we allow bounds much larger than $10^6$.

Solution 4:

Define $f(n,m) = n+1000000m$. Then $n = f(n,m) \mod 1000000$, and $m = \frac{f(n,m) - n}{1000000}$.

Solution 5:

f(m,n) = m +(1/n)

If there are numbers to the right of the decimal,
m =number to the left of decimal.
n = the reciprocal of the numbers to the right of the decimal
example
f(m,n) =10.5 then m =10, n=1/(0.5) =2

If there are no numbers to right of the decimal [f(m,n) is itself an integer]
then m = f(m,n) -1,
n =1
example
f(m,n) =20 then m=19, n =1