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Many gates are not available on a real computer and therefore the circuit must be transpiled into a specific set of gates. I have seen this equation below which is used to to determine the 'cost' of a quantum circuit:

Cost = 10NCNOT + Nother

where:
NCNOT : number of CNOT gates
Nother : number of other gates

I am wondering which other considerations are not captured by this equation? Especially given that we live in the Noisy Intermediate-Scale Quantum (NISQ) technology era.

It has obviously left out the number of qubits in a circuit which contributes massively to noise. What else should it take into account though? very curious

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    $\begingroup$ Are you sure the cost isn't defined as $10 \cdot N_{CNOT} + N_{other}$? $\endgroup$
    – epelaez
    Jun 28, 2021 at 16:54
  • $\begingroup$ Yes, you are right sorry. Typo $\endgroup$
    – John
    Jun 28, 2021 at 16:57
  • $\begingroup$ @epelaaez Do you have any useful sources on the cost equation? I have been trying to find but cannot seem to. $\endgroup$
    – John
    Jun 28, 2021 at 16:59
  • $\begingroup$ I've seen that's the cost function that IBM uses in their challenges, but I haven't seen it stated in some more "formal" place $\endgroup$
    – epelaez
    Jun 28, 2021 at 17:04
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    $\begingroup$ I'd be wary of linear formulas like this. Let's say that all 2 qubit gates have $10\times$ as many errors as single-qubit gates, where we call the single-qubit error gates $\eta$. The total success probability will then be $(1-\eta)^{N_{\mathrm{other}}}(1-10\eta)^{N_{\mathrm{CNOT}}}\approx 1-\eta(N_o+10N_C)+\cdots$. Obviously to first order the error looks like it scales with $N_o+10N_C$, but when you add enough gates to have $\eta(N_o+10N_C)\sim 1$ then all of the other terms become necessary and the relationship breaks down. $\endgroup$ Jun 29, 2021 at 13:01

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