2
$\begingroup$

Lets say I have a non-injective function $f()$, adding image for reference. Now lets say I build a quantum circuit to calculate $f^{-1}()$. If the input register has the value $i$, does the output register have all the values $j$ in superpostion, such that $f(j)=i$ ? non-injective function

$\endgroup$
1
  • 1
    $\begingroup$ what would you think $f^{-1}(A)$ would be? $\endgroup$
    – Mark S
    Jul 9 at 2:58
3
$\begingroup$

This is essentially a matter of definition. If you design and build a function which you are claiming to be $f^{-1}$, you first have to define what the output will be if the input is $C$, and what you define will determine how you implement it. Outputting some sort of superposition might seem like a reasonable option, but mathematically the function is unlikely to behave how you would expect it to. But that might all come down to the context you're using it in.

$\endgroup$
1
  • $\begingroup$ The context I'm using it in is to find the different inputs which have the same image under $f$. You can think of $f$ as a function calculating some cube-roots in a finite field where cube-roots are not injective. $\endgroup$ Jul 9 at 14:54
1
$\begingroup$

Any operation on a quantum computer has to be reversible, measurement and reset being an exceptions. In your case the operation is not reversible as for C you have two possible inputs - 3 and 4. To implement your operation on QC you have to distinguish between 3 and 4. The most general approach is to copy input state together with the output. For example if input is $|3\rangle$, then output will be $|3\rangle|C\rangle$ and for input $|4\rangle$ you will have $|4\rangle|C\rangle$. Now, your operation is reversible.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.