In chapter 10.6.1 in Nielsen and Chuang, the section on concatenated codes and the threshold theorem (pages 480-481) states:

The size of the simulating circuit goes as $d^k$ times the size of the original circuit, where $d$ is a constant representing the maximum number of operations used in a fault-tolerant procedure to do an encoded gate and error correction.

It is further explained that if we wish to achieve a final accuracy of $\epsilon$, we are required to concatenate the simulating circuit $k$ times such that:

$$ \frac{(cp)^{2^k}}{c} \leq \frac{\epsilon}{p(n)} \tag{10.113} $$

where $p(n)$ is the number of gates in the simulating circuit and $\frac{(cp)^{2^k}}{c}$ is the failure probability for a procedure at the highest level.

After this, the size of the circuit is to achieve this accuracy is then given as:

$$ d^k = \left(\frac{\log{(p(n)/c\epsilon)}}{\log(1/pc)}\right)^{\log{d}} \tag{10.114} $$

My question is, how is the size of the circuit derived in Eq(10.114)? It is closely related to Eq(10.113) after some rearranging, but I cannot see exactly how to get Eq(10.114).

  • $\begingroup$ All the logarithms are base 2. And $(2^k)^{\log d}$ is ... $\endgroup$
    – ChrisD
    Nov 21, 2022 at 4:38

1 Answer 1


From Eq(10.113) we can work out the size a simulating circuit must be to achieve the desired accuracy of $\epsilon/p(n)$. Rearranging Eq(10.113), we see $1/(cp)^{2^k} = p(n)/c\epsilon$. Taking the logarithm (base 2, as per @ChrisD) of both sides and rearranging \begin{equation} 2^k = \frac{\log(p(n)/c\epsilon)}{\log(1/cp)} \end{equation} Raising both sides to the power $\log(d)$ and noting $(2^k)^{\log{d}} = d^k$ (as per @ChrisD, again), we can express the size of our circuit as \begin{equation} d^k = \left(\frac{\log\left(p(n)/c\epsilon\right)}{\log\left(1/pc\right)}\right)^{\log{d}} \end{equation} The expression for $d^k$ suggests that the size of the simulating circuit scales as \begin{equation} O\left(\text{poly}(\log p(n)/\epsilon)\right) \end{equation} i.e. the size of the circuit grows no faster than some polynomial function of $\log p(n)/\epsilon$.


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