I've tried to figure out what exactly the paper is saying. They start by noting that this circuit: │0〉-H─•───√X-[=1!]─ │0〉─H─┼─•─√X─[=1!]─ │0〉─H─┼

I've tried to figure out what exactly the paper is saying. They start by noting that this circuit: │0〉-H─•───√X-[=1!]─ │0〉─H─┼─•─√X─[=1!]─ │0〉─H─┼─┼─√X─[=1!]─ │0〉───X─X────────── (will now =1 when measured) Which puts three qubits each into the state │0〉+ │1〉, stashes the parity of two of those qubits' values onto a scratch line, then performs a post-selection along the Y axis of all three qubits. The interesting thing is that the stashed parity ends up always equal to 1, indicating that the two qubits had opposite value. The authors then start talking about this meaning that "all the qubits are unequal" because no matter which parity you stash, you end up discovering the two qubits were not equal. I think that's a bit of a stretch, framing-wise. It's the wrong way to think about it. For example, it stops working if you stash two of the parities instead of just one (you'll start getting "they were equal" results half of the time). Instead of talking about the qubits all being unequal, I prefer to think of this in terms of winning a game against a referee. You get three coins and can choose if each is set to heads or set to tails. Then a referee secretly picks two of the coins at random, measures their parity, and otherwise does nothing except pass the coins back to you. Then you do whatever you want to the coins, but have to choose whether or not to "keep" this run or "discard" it. You win if, whenever you say "keep", the referee wrote down "the coins differed". (Also you can't just never say "keep".) Classically, there's no strategy that wins the game 100% of the time. But if you're using qubits instead of coins, quantum 2-level systems instead of classical 2-level systems, you can win 100% of the time (by using the circuit described above). In fact, the strategy works even if you have a million coins/qubits instead of just three!