Setup(1λ,N): This function outputs pp=PC.Setup(1λ,d).
We consider an input x with size of 1. Hence, ni=1. Considering a program which requires three gates for its arithmatization, we have ng=3. In this example, the maximum number of registers which are changed during the execution is nr=1. If the computation is done in F of order p=181, ∣F∣=181. Also,∣H∣=n=ng+ni+1=5. Also, b is a random number in {1,...,∣F∣−∣H∣}={1,...,176} such as b=2. Also, m=2ng=6, ∣w∣=ng−nr=2, ∣K∣=m=6. Hence:
d={dAHP(N,i,j)}i∈[kAHP]⋃{0},j∈[sAHP(i)]={6,6,6,6,6,6,6,6,6,4,7,7,7,8,11,4,6,4,4,5,30}
Now, we run KZG.Setup(1λ,d), considering a generator of F, g=2, for each element in d:
KZG.Setup(1λ,6)=(ck,vk)=({gτi}i=05,gτ) that for secret element τ=119 and generator g=2 outputs ck={gτi}i=05=(2,57,86,98,78,51) and vk=57.
KZG.Setup(1λ,4)=(ck,vk)=({gτi}i=03,gτ) that for secret element τ=119 and generator g=2 outputs ck={gτi}i=03=(2,57,86,98) and vk=57.
KZG.Setup(1λ,7)=(ck,vk)=({gτi}i=06,gτ) that for secret element τ=119 and generator g=2 outputs ck={gτi}i=06=(2,57,86,98,78,51,96) and vk=57.
KZG.Setup(1λ,8)=(ck,vk)=({gτi}i=07,gτ) that for secret element τ=119 and generator g=2 outputs ck={gτi}i=07=(2,57,86,98,78,51,96,21) and vk=57.
KZG.Setup(1λ,11)=(ck,vk)=({gτi}i=010,gτ) that for secret element τ=119 and generator g=2 outputs ck={gτi}i=010=(2,57,86,98,78,51,96,21,146,179,124) and vk=57.
KZG.Setup(1λ,6)=(ck,vk)=({gτi}i=05,gτ) that for secret element τ=119 and generator g=2 outputs ck={gτi}i=05=(2,57,86,98,78,51) and vk=57.
KZG.Setup(1λ,30)=(ck,vk)=({gτi}i=029,gτ) that for secret element τ=119 and generator g=2 outputs ck={gτi}i=029=(2,57,86,98,78,51,96,21,146,179,124,95,83,103,130,85,160,35, 2,57,86,98,78,51,96,21,146,179,124,95) and vk=57.