Deutsch–Jozsa algorithm: why is $f$ constant?

Question

I'm trying to understand how the Deutsch–Jozsa algorithm works with the following circuit:

Circuit in Quirk

Since we have the top 2 wires measuring $|0\rangle$ with 100% probability, it means $U_f$ is constant. And that's what I'm having trouble understanding...what exactly is constant?

If I isolate $U_f$ I get this: Oracle function

I understand the concept of balanced and constant functions, and that an n-bit binary string will have $2^n$ mappings, which gives $2^n$ possible functions. So how can I see that $f$ is constant in the isolated circuit above?

Answer 1

The circuit you gave implements an oracle for a function $f(x) = 0$, which is constant.

You can observe that there are no gates leading from the top two wires (inputs $|x\rangle$) to the bottom wire (output $|y\rangle$). Since the oracle is supposed to transform $|x\rangle|y\rangle$ into $|x\rangle|y \oplus f(x)\rangle$, and $|y\rangle$ always remains unaffected, you can see that $f(x) = 0$.