Getting 5 or 7 and returning the opposite? [closed]

Question

I get 5 or 7 and if i get 5 i need to return 7 if i get 7 i need to return 5.

i need to do this in 1 mathematical formula.

I have those:

12 - x

35 / x

There are more solutions ?

Example in code:

public static int Transform(int x)
{
    return (12-x);
    //return (35/x);
}

Why are you unsatisfied with the simple ones you already have? — J. M. ain't a mathematician, May 11, 2011 at 15:57
If $h$ is an invertible function we can define $f(x) = h^{-1}(h(5) + h(7) - h(x))$. This $f$ has the property that you want. This covers the two methods you have (with $h(x)=x$ and $h(x)=\log(x)$, respectively, and quite a few more. — Thomas Andrews, May 11, 2011 at 16:08

Michael Burge · Accepted Answer · 2011-05-11 16:08:42Z

5

return x ^ 2;

Are you sure you aren't missing any constraints?

answered May 11, 2011 at 16:08

Michael Burge

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2

$\begingroup$ This doesn't return 5 when x is 7...nor does it return 7 when x is 5 $\endgroup$
– Nicolas Villanueva
May 11, 2011 at 16:12
2

$\begingroup$ He was using Java or C++ in his original question; ^ is a bitwise XOR. A test confirms that x^2 swaps both 5 and 7 in Visual C++ 10.0. $\endgroup$
– Michael Burge
May 11, 2011 at 16:20
2

$\begingroup$ A way to verify this answer: $2 = 5\ \text{XOR}\ 7$. (So basically return x ^ (5 ^ 7);) $\endgroup$
– Aryabhata
May 11, 2011 at 16:25
$\begingroup$ Woops, my bad. I was naively thinking exponentiation. $\endgroup$
– Nicolas Villanueva
May 11, 2011 at 16:30
2

$\begingroup$ yeah, I read it as exponentiation, too. You might want to make it clear in the answer that it is a bitwise operator. (Mathematicians tend to not think of bitwise operators as "mathematical formulas" as a rule.) $\endgroup$
– Thomas Andrews
May 11, 2011 at 16:39

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