Confusion with the number of CNOTs in a circuit

Question

I am a bit puzzled on the following circuit. According to this Quantum Computing SE thread it holds that $$ e^{i(Z\otimes Z)t} = {\rm CNOT} (I\otimes e^{iZt}){\rm CNOT} \qquad (1) $$

As a result we have the following circuit (C1):

Furthermore, for $e^{iZ\otimes Z\otimes Z t}$ we obtain the following circuit (C2):

I was reading arxiv:2003.13599 and in Figure 3 we see three unitaries corresponding to $I\otimes Z \otimes I$, $I \otimes Z \otimes Z$ and $Z \otimes Z \otimes I$ seen below (C3):

Which is very confusing. Specifically, let's focus for example on the middle part of Diagram (a) corresponding to $I\otimes Z \otimes Z$. The $I$ term (in the first qubit) can be ignored. Then, if we ignore the first wire we essentially should have the term $Z\otimes Z$ from Eq. (1) but nevertheless we see two pairs of CNOT gates just like in (C2) that corresponds to $e^{iZ\otimes Z \otimes Z t}$.

Why is this the case? I suspect this is somehow related to the extra wire we ( C2 has three wires but for some reason C3 has four).

Answer 1

I think the key fact you're missing is that $Z_2 \otimes Z_3 \otimes Z_4 = Z_2 \otimes Z_3$ when you know qubit 4 is in the $|0\rangle$ state; in the +1 eigenstate of $Z$.

I'm not sure why that paper is using six CNOT gates instead of four. The ancilla isn't helping. Maybe it's just supposed to be an example for a more general case where it is helpful.