Example 2

$Setup(1^{\lambda},N)$: This function outputs $pp=PC.Setup(1^{\lambda},d)$.

We consider an input $x$ with size of 32. Hence, $n_i=32$. Considering a program which requires 4 gates for its arithmatization, we have $n_g=4$. In this example, the maximum number of registers which are changed during the execution is $n_r=2$ (Please see Example 2 in the Commitment phase). If the computation is done in $\mathbb{F}$ of order $p=4,767,673$, $|\mathbb{F}|=4,767,673$. Also,$|\mathbb{H}|=n=n_g+n_i+1=37$. Also, $b$ is a random number in {1,...,$|\mathbb{F}|-|\mathbb{H}|$}=$\{1,...,(4,767,673-37)=4,767,636\}$ such as $b=2$. Also, $m=2n_g=8$, $|w|=n_g-n_r=2$, $|\mathbb{K}|=m=8$. Hence:

$d=\{d_{AHP}(N,i,j)\}_{i\in[k_{AHP}]\bigcup\{0\},j\in[s_{AHP}(i)]}=\{8,8,8,8,8,8,8,8,8,4,39,39,39,40,75,36,38,36,36,7,42\}$

Now, we run $KZG.\hspace{1mm}Setup(1^{\lambda},d)$, considering a generator of $\mathbb{F}$, $g=5$, for each element in $d$:

$KZG.Setup(1^{\lambda},138)=(ck,vk)=(\{g\tau^i\}_{i=0}^{137},g\tau)$ that for secret element $\tau=119$ and generator $g=5$ outputs $ck=\{g\tau^i\}_{i=0}^{137}=(5,595,...)$ and $vk=595$.

$KZG.Setup(1^{\lambda},4)=(ck,vk)=(\{g\tau^i\}_{i=0}^{3},g\tau)$ that for secret element $\tau=119$ and generator $g=5$ outputs $ck=\{g\tau^i\}_{i=0}^{3}=(5,595,...)$ and $vk=595$.

$KZG.Setup(1^{\lambda},39)=(ck,vk)=(\{g\tau^i\}_{i=0}^{38},g\tau)$ that for secret element $\tau=119$ and generator $g=5$ outputs $ck=\{g\tau^i\}_{i=0}^{38}=(5,595,...)$ and $vk=595$.

$KZG.Setup(1^{\lambda},40)=(ck,vk)=(\{g\tau^i\}_{i=0}^{39},g\tau)$ that for secret element $\tau=119$ and generator $g=5$ outputs $ck=\{g\tau^i\}_{i=0}^{39}=(5,595,...)$ and $vk=595$.

$KZG.Setup(1^{\lambda},75)=(ck,vk)=(\{g\tau^i\}_{i=0}^{74},g\tau)$ that for secret element $\tau=119$ and generator $g=5$ outputs $ck=\{g\tau^i\}_{i=0}^{74}=(5,595,...)$ and $vk=595$.