Example 1

Assume the following sample code: mul R1, R1, 5 => Gate 1, p=181 add R1, R1, 11 => Gate 2, p=181 mul R1, R1, 26 => Gate 3, p=181

The constraints are as follows considering $p=181$:

$R_1^{(2)}-5R_1^{(1)}=0$ $R_1^{(3)}-R_1^{(2)}-11=0$ $R_1^{(4)}-26R_1^{(3)}=0$

To keep the example simple and understandable, we continue with $n_g=3$ and $n_i=1$.

The Prover calculates square matrices $A$, $B$ and $C$ of order $n_g+n_i+1=3+1+1=5$ based on above construction:

$A=\begin{bmatrix} 0&0&0&0&0\\ 0&0&0&0&0\\ 0&1&0&0&0\\ 1&0&0&0&0\\ 0&0&0&1&0\\ \end{bmatrix}$, $B=\begin{bmatrix}0&0&0&0&0\\0&0&0&0&0\\5&0&0&0&0\\11&0&1&0&0\\26&0&0&0&0\\\end{bmatrix}$and $C=\begin{bmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\\end{bmatrix}$.

At we see, the matrices $A$ and $B$ are $2-SLT$ and matrix $C$ is $2-Diag$. Also $Az\hspace{1mm}o\hspace{1mm}Bz=Cz$.

We consider multiplicative subgroup $\mathbb{H}$ of order $n=n_g+n_i+1=5$ with generator $\omega=2^{36}\equiv 59 (\textrm{mod}\hspace{1mm}181)$. Note that if $g$ is a generator of field $\mathbb{F}$ of order $p$, then, $g^{\frac{p-1}{n}}$is a generator of a multiplicative subgroup of it of order $n$. Therefore, $\mathbb{H}=\{1,\omega,\omega^2,\omega^3,\omega^4\}=\{1,59,42,125,135\}$.

Also, consider multiplicative subgroup $\mathbb{K}$ of order $m=2n_g=6$ where $t=n_i+1=2$ with generator $\gamma=g^{\frac{p-1}{m}}=2^{30}\equiv49(\textrm{mod}\hspace{1mm}181)$. Therefore,$\mathbb{K}=\{1,\gamma,\gamma^2,...,\gamma^5\}=\{1,49,48,180,132,133\}$.

PFR Commitment

$Commit(ck=(2,66,83,91,96,24,2,66,83),m_f=(A,B,C),s\in R)$:

1- The Prover Selects $s=(s_1,...,s_{s_{AHP}(0)})$ of random space $R$.

2- The Prover calculates $\overrightarrow{O}=Enc(m_f=(A,B,C))$ as encoded index as following:.

First, calculates the polynomial $PFRrow_A(x)$. Since matrix $A$ has three non-zero entries that the first of them is in the second row, so $c_0=2$ and $PFRrow_A(\gamma^0)=\omega^2$ , note that rows numbered of zero, the second is in the third row, so $c_1=3$ and $PFRrow_A(\gamma^1)=\omega^3$ and the third is in the fourth row, so $c_2=4$ and $PFRrow_A(\gamma^2)=\omega^4$. Note that to construct all three polynomials, we read the entries in the matrix in row or column order. Here they are read in row order.

According to the values ​​defined for $\gamma$ and $\omega$, we have $PFRrow_A(1)=42$, $PFRrow_A(43)=125$ and $PFRrow_A(39)=135$. Therefore based on Lagrange polynomials, have $PFRrow_A(x)=\sum_{i=1}^{3}y_iL_i(x)$, so $PFRrow_A(x)=42L_1(x)+125L_2(x)+135L_3(x)$, where $L_1(x)=\frac{(x-43)(x-39)}{(1-43)(1-39)}=\frac{(x-43)(x-39)}{(-42)(-38)}$. Now, since $-42\equiv139\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $-38\equiv143\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $-43\equiv138\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$and $-39\equiv142\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, therefore $L_1(x)=\frac{(x+138)(x+142)}{(139)(143)}$and since $(139)(143)\equiv 148\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$and $148^{-1}\equiv 170\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, have $L_1(x)=170(x+138)(x+142)=170x^2+178x+15$. We get in a similar way that $L_2(x)=167x^2+17x+178$ and $L_3(x)=25x^2+167x+170$. Therefore $PFRrow_A(x)=77x^2+109x+37$.

Now, calculates $PFRcol_A(x)$. Since matrix $A$ has three non-zero entries that the first of them is in the first column, so $r_0=1$ and $PFRcol_A(\gamma^0)=\omega^1$ , note that columns numbered of zero, the second is in the zero column, so $r_1=0$ and $PFRcol_A(\gamma^1)=\omega^0$ and the third is in the third column, so $r_2=3$ and $PFRcol_A(\gamma^2)=\omega^3$. Therefore $PFRcol_A(1)=59$, $PFRcol_A(43)=1$ and $PFRcol_A(39)=125$. So $PFRcol_A(x)=59L_1(x)+L_2(x)+125L_3(x)=109x^2+81x+50$.

Now, calculates $PFRval_A(x)$. Since matrix $A$ has three non-zero entries such that the value of all of them is one, therefore $v_0=v_1=v_2=1$. So, $PFRval_A(x)=L_1(x)+L_2(x)+L_3(x)=1$.

Therefore, the matrix $A$ is encoded by $PFRrow_A(x)=77x^2+109x+37$, $PFRcol_A(x)=109x^2+81x+50$ and $PFRval_A(x)=1$.

Now, encodes the matrix $B$. First, calculates the polynomial $row_B(x)$. Since matrix $B$ has four non-zero entries such that the first of them is in the second row, so $c_0=2$ and $row_B(\gamma^0)=\omega^2$ . The second and the third are in the third row, so $c_1=c_2=3$ and $PFRrow_B(\gamma^1)=PFRrow_B(\gamma^2)=\omega^3$ and the fourth is in the fourth row, so $c_3=4$ and $PFRrow_B(\gamma^3)=\omega^4$. Therefore, $PFRrow_B(1)=42$, $PFRrow_B(43)=125$, $PFRrow_B(39)=125$ and $PFRrow_B(48)=135$. So $PFRrow_B(x)=42L_1(x)+125L_2(x)+125L_3(x)+135L_4(x)$ where $L_1(x)=\frac{(x-43)(x-39)(x-48)}{(1-43)(1-39)(1-48)}=\frac{(x+138)(x+142)(x+133)}{(139)(143)(134)}=\frac{(x+138)(x+142)(x+133)}{103}$. Since $103^{-1}\equiv58\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, so $L_1(x)=58(x+138)(x+142)(x+133)=58x^3+62x^2+116x+127$. We get in a similar way that $L_2(x)=39x^3+7x^2+19x+116$, $L_3(x)=138x^3+155x^2+7x+62$ and $L_4(x)=127x^3+138x^2+39x+58$. Therefore $PFRrow_B(x)=76x^3+35x^2+174x+119$.

Now, calculates $PFRcol_B(x)$. Since matrix $B$ has four non-zero entries that the first, second and fourth of them are in zero column, so $r_0=r_1=r_3=0$ and $PFRcol_B(\gamma^0)=PFRcol_B(\gamma^1)=PFRcol_B(\gamma^3)=\omega^0$ and the third is in the second column, so $r_2=2$ and $PFRcol_B(\gamma^2)=\omega^2$ . Therefore $PFRcol_B(1)=1$, $PFRcol_B(43)=1$, $PFRcol_B(39)=42$and $PFRcol_B(48)=1$. So $PFRcol_B(x)=L_1(x)+L_2(x)+42L_3(x)+L_4(x)=47x^3+20x^2+106x+9$.

Now, calculates $PFRval_B(x)$. Since matrix $B$ has four non-zero entries that value of them are $5$, $11$, $1$ and $26$, respectively. Therefore $v_0=v_1=v_2=1$$PFRval_B(\gamma^0)=5$, $PFRval_B(\gamma^1)=11$, $PFRval_B(\gamma^2)=1$and $PFRval_B(\gamma^3)=26$. So, $PFRval_B(1)=5$, $PFRval_B(43)=11$, $PFRval_B(39)=1$and $PFRval_B(48)=26$. Therefore, $PFRval_B(x)=5L_1(x)+11L_2(x)+L_3(x)+26L_4(x)=177x^3+148x^2+42$.

Therefore, the matrix $B$ is encoded by $PFRrow_B(x)=76x^3+35x^2+174x+119$, $PFRcol_B(x)=47x^3+20x^2+106x+9$ and $PFRval_B(x)=177x^3+148x^2+42$.

Now, calculates polynomial $PFRrow_C(x)$. In a similar way, Since matrix $C$ has three non-zero entries that are in the third, fourth and fifth rows, therefore $PFRrow_C(\gamma^0)=\omega^2$ , $PFRrow_C(\gamma^1)=\omega^3$ and $PFRrow_C(\gamma^2)=\omega^4$. Then $PFRrow_C(1)=42$ , $PFRrow_C(43)=125$ and $PFRrow_C(39)=135$. So$PFRrow_C(x)=42L_1(x)+125L_2(x)+135L_3(x)$, therefore $PFRrow_C(x)=77x^2+109x+37$.

Now, since $C$ is a diagonal matrix, polynomials $PFRrow_C(x)$ and $PFRcol_C(x)$are equal. Also, since the matrix $C$ has three non-zero entries such that the value of all of them is one, therefore $v_0=v_1=v_2=1$. So, $PFRval_C(x)=L_1(x)+L_2(x)+L_3(x)=1$.

Therefore, the matrix $C$ is encoded by $PFRrow_C(x)=77x^2+109x+37$, $PFRcol_C(x)=77x^2+109x+37$ and $PFRval_C(x)=1$. Therefore, encoding of $m_f=(A,B,C)$ calculates as following:$\overrightarrow{O}_{PFR}=(37,109,77,0,0,0,0,0,0,50,81,109,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,119,174,35,76,0,0,$ $0,0,0,9,106,20,47,0,0,0,0,0,42,0,148,177,0,0,0,0,0,37,109,77,0,0,0,0,0,0,37,109,$$77,0,0,0,0,0,0,37,109,77,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0)$

3- The Prover calculates commitment by using of KZG commitment scheme as following:

$Com_{T}=\sum_{i=0}^{deg_T}a_ig\tau^i=\sum_{i=0}^{deg_T}a_ick(i)$ where $a_i$ is coefficient of $x^i$ in polynomial $T(x)$.

Therefore

$PFR\hspace{1mm}Com^0=\sum_{i=0}^{2}PFRrow_{A_i}(g\tau^i)=\sum_{i=0}^{2}PFRrow_{A_i}ck(i)=37ck(0)+109ck(1)+77ck(2)=37\times2+109\times 57+77\times 86\equiv\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$

$PFR\hspace{1mm}Com^1=$

$PFR\hspace{1mm}Com^2=$

and similarly

$PFR\hspace{1mm}Com^3=$, $PFR\hspace{1mm}Com^4=$, $PFR\hspace{1mm}Com^5=$, $PFR\hspace{1mm}Com^6=$, $PFR\hspace{1mm}Com^7=$ and $PFR\hspace{1mm}Com^8=$.

4- The Prover sends $PFR\hspace{1mm}Com=$ to the Verifier.

AHP Commitment

$Commit(ck,m_f=(A,B,C),s\in R)$:

1 - The Prover Selects $s=(s_1,...,s_{s_{AHP}(0)})$ of random space $R$. 2- The Prover calculates $\overrightarrow{O}_{AHP}=Enc(m_f=(A,B,C))$ as encoded index as following:

The polynomial $AHProw_A:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ with $AHProw_A(k=\gamma^i)=\omega^{r_i}$ for $1\leq i\leq ||A||$ , and otherwise $AHProw_A(k)$ returns an arbitrary element in $\mathbb{H}$. So $AHProw_A(k)$ on $\mathbb{K}$ is a polynomial so that $AHProw_A(1)=42$ , $AHProw_A(49)=125$, $AHProw_A(48)=135$ and otherwise $AHProw_A(k)$ returns arbitrary elements of $\mathbb{H}$, for example $AHProw_A(180)=125$, $AHProw_A(132)=135$, $AHProw_A(133)=1$ Therefore, $AHProw_A(x)=\sum_{i=1}^{6}y_iL_i(x)=162x^5+62x^4+161x^3+169x^2+88x+124$.

Also, $AHPcol_A:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ with $AHPcol_A(k=\gamma^i)=\omega^{c_i}$ for $1\leq i\leq ||A||$ and otherwise $AHPcol_A(k)$ returns an arbitrary element in $\mathbb{H}$. So $AHPcol_A(k)$ on $\mathbb{K}$ is a polynomial so that $AHPcol_A(1)=59$ , $AHPcol_A(49)=1$, $AHPcol_A(48)=125$ and otherwise $AHPcol_A(k)$ returns arbitrary elements of $\mathbb{H}$, for example $AHPcol_A(180)=125$, $AHPcol_A(132)=135$ and $AHPcol_A(133)=1$. Therefore, $AHPcol_A(x)=\sum_{i=1}^{6}y_iL_i(x)=128x^5+150x^4+32x^3+109x^2+169x+14$

and , $AHPval_A:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ with $AHPval_A(k=\gamma^i)=\frac{v_i}{u_{\mathbb{H}}(AHProw_A(k),AHProw_A(k))u_{\mathbb{H}}(AHPcol_A(k),AHPcol_A(k))}$ for $1\leq i\leq ||A||$ where $v_i$ is value of $i^{th}$ nonzero entry and otherwise $AHPval_A(k)$ returns zero. Note that based on definition of $u_{\mathbb{H}}(x,y)$, for each $x \in \mathbb{H}$, $u_{\mathbb{H}}(x,x)=|\mathbb{H}|x^{|\mathbb{H}|-1}=5x^4$. So $AHPval_A(1)=\frac{1}{(5AHProw^4_A(1))(5AHPcol^4_A(1))}=\frac{1}{5\times 42^4\times5\times59^4}=\frac{1}{145}\equiv 5\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $AHPval_A(49)=\frac{1}{(5AHProw^4_A(49))(5AHPcol^4_A(49))}=\frac{1}{5\times 125^4\times5\times1^4}=\frac{1}{145}\equiv 5\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ , $AHPval_A(48)=\frac{1}{(5AHProw^4_A(48))(5AHPcol^4_A(48))}=\frac{1}{5\times 135^4\times5\times 125^4}=\frac{1}{48}\equiv 132\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ and $AHPval_A(k)=0$ for $k\in \mathbb{K}-\{1,49,48\}$. Therefore $AHPval_A(x)=\sum_{i=1}^{6}y_iL_i(x)=72x^5+79x^4+22x^3+111x^2+180x+84$.

Now, $\hat{AHProw_A}$, $\hat{AHPcol_A}$ and $\hat{AHPval_A}$ are extensions of $AHProw_A$, $AHPcol_A$ and $AHPval_A$ so that are agree on $\mathbb{K}$.

Similarly, $AHProw_B:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ so that $AHProw_B(1)=42$, $AHProw_B(49)=125$, $AHProw_B(48)=125$, $AHProw_B(180)=135$ and for the rest values of $\mathbb{K}$, $AHProw_B(k)$ returns arbitrary elements of $\mathbb{H}$, for example $AHProw_B(132)=135$ and $AHProw_B(133)=1$. Therefore $AHProw_B(x)=\sum_{i=1}^{6}y_iL_i(x)=20x^5+85x^4+37x^3+151x^2+168x+124$.

Also, $AHPcol_B:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ so that $AHPcol_B(1)=1$ , $AHPcol_B(49)=1$, $AHPcol_B(48)=42$, $AHPcol_B(180)=1$ and for the rest values of $\mathbb{K}$ , $AHPcol_B(k)$ returns arbitrary elements of $\mathbb{H}$, for example $AHPcol_B(132)=135$ and $AHPcol_B(133)=1$. Therefore $AHPcol_B(x)=\sum_{i=1}^{6}y_iL_i(x)=18x^5+164x^4+180x^3+18x^2+164x$

and , $AHPval_B:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ so that $AHPval_B(1)=\frac{5}{(5AHProw^4_B(1))(5AHPcol^4_B(1))}=\frac{5}{5\times 42^4\times5\times1^4}=\frac{5}{48}\equiv 117\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $AHPval_B(49)=\frac{11}{(5AHProw^4_B(49))(5AHPcol^4_B(49))}=\frac{11}{5\times 125^4\times5\times1^4}=\frac{11}{145}\equiv 55\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ , $AHPval_B(48)=\frac{1}{(5AHProw^4_B(48))(5AHPcol^4_B(48))}=\frac{1}{5\times 125^4\times5\times 42^4}=\frac{1}{25}\equiv 29\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ , $AHPval_B(180)=\frac{26}{(5AHProw^4_B(180))(5AHPcol^4_B(180))}=\frac{26}{5\times 135^4\times5\times 1^4}=\frac{26}{27}\equiv 68\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ and $AHPval_B(k)=0$ for $k\in \mathbb{K}-\{1,49,48,180\}$. Therefore $AHPval_B(x)=\sum_{i=1}^{6}y_iL_i(x)=86x^5+53x^4+34x^3+55x^2+176x+75$.

Now, $\hat{AHProw_B}$, $\hat{AHPcol_B}$ and $\hat{AHPval_B}$ are extensions of $AHProw_B$, $AHPcol_B$ and $AHPval_B$ so that are agree on $\mathbb{K}$.

Similarly, $AHProw_C:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ so that $AHProw_C(1)=42$, $AHProw_C(49)=125$, $AHProw_C(48)=135$ and for the rest values of $\mathbb{K}$, $AHProw_C(k)$ returns arbitrary elements of $\mathbb{H}$, for example $AHProw_C(180)=125$, $AHProw_C(132)=135$ and $AHProw_C(133)=1$. Therefore $AHProw_C(x)=\sum_{i=1}^{6}y_iL_i(x)=162x^5+62x^4+161x^3+169x^2+88x+124$.

Also, $AHPcol_C:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ so that $AHPcol_C(1)=42$ , $AHPcol_C(49)=125$, $AHPcol_C(48)=135$, $AHPcol_C(180)=125$ and for the rest values of $\mathbb{K}$ , $AHPcol_C(k)$ returns arbitrary elements of $\mathbb{H}$, for example $AHPcol_C(132)=135$ and $AHPcol_C(133)=1$. Therefore $AHPcol_C(x)=\sum_{i=1}^{6}y_iL_i(x)=162x^5+62x^4+161x^3+169x^2+88x+124$.

and , $AHPval_C:\mathbb{K}=\{1,49,48,180,132,133\}\to\mathbb{H}=\{1,59,42,125,135\}$ so that $AHPval_C(1)=\frac{1}{(5AHProw^4_C(1))(5AHPcol^4_C(1))}=\frac{1}{5\times 42^4\times5\times42^4}=\frac{1}{27}\equiv 114\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $AHPval_C(49)=\frac{1}{(5AHProw^4_C(49))(5AHPcol^4_C(49))}=\frac{1}{5\times 125^4\times5\times125^4}=\frac{1}{117}\equiv 82\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ , $AHPval_C(48)=\frac{1}{(5AHProw^4_C(48))(5AHPcol^4_C(48))}=\frac{1}{5\times 135^4\times5\times135 ^4}=\frac{1}{145}\equiv 5\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ and $AHPval_C(k)=0$ for $k\in \mathbb{K}-\{1,49,48\}$. Therefore $AHPval_C(x)=\sum_{i=1}^{6}y_iL_i(x)=65x^5+61x^4+157x^3+53x^2+16x+124$.

Now, $\hat{AHProw_C}$, $\hat{AHPcol_C}$ and $\hat{AHPval_C}$ are extensions of $AHProw_C$, $AHPcol_C$ and $AHPval_C$ so that are agree on $\mathbb{K}$.

Therefore, $\overrightarrow{O}_{AHP}=(124,88,169,161,62,162,14,169,109,32,150,128,84,180,111,22,79,$ $72,124,168,151,37,85,20,0,164,18,180,164,18,75,176,55,34,53,86,124,88,169,$ $161,62,162,124,88,169,161,62,162,124,16,53,157,61,65)$.

3- The Prover calculates commitment by using of $KZG$ commitment scheme as following:

$Com_{T}=\sum_{i=0}^{deg_T}a_ig\tau^i=\sum_{i=0}^{deg_T}a_ick(i)$ where $a_i$ is coefficient of $x^i$ in polynomial $T(x)$.

4- The Prover sends $AHP\hspace{1mm}Com^0=\sum_{i=0}^{5}\hat{AHProw}_{A_i}\hspace{1.1mm}ck(i)=124ck(0)+88ck(1)+169ck(2)+161ck(3)+62ck(4)+$ $162ck(5) \equiv 166\hspace{1mm}(\textrm{mod}\hspace{1mm} 181)$

and similarly

$AHP\hspace{1mm}Com^1=36$, $AHP\hspace{1mm}Com^2=108$, $AHP\hspace{1mm}Com^3=58$, $AHP\hspace{1mm}Com^4=73$, $AHP\hspace{1mm}Com^5=157$, $AHP\hspace{1mm}Com^6=166$, $AHP\hspace{1mm}Com^7=166$ and $AHP\hspace{1mm}Com^8=36$.

AHP Commitment JSON File Example 1

IoT_Manufacturer_Name = "zkIoT" IoT_Device_Name = "MultiSensor" Device_Hardware_Version = "1.0" Firmware_Version = "1.0" Device Picture= <> Lines = [200, 350, 4000-4010]

{
    "commitment_id": String,
    "iot_developer_name": String,
    "iot_device_name": String,
    "device_hardware_version": String,
    "firmware_version": String,
    "class": 32-bit Integer,
    "m": 6,
    "n": 5,
    "p": 181,
    "g": 2,   
        
    // PFR Commitment   
    "PFRrow_A": 64-bit Array,
    "PFRcol_A": 64-bit Array,
    "PFRval_A": 64-bit Array,
    "PFRrow_B": 64-bit Array,
    "PFRcol_B": 64-bit Array,
    "PFRval_B": 64-bit Array,
    "PFRrow_C": 64-bit Array,
    "PFRcol_C": 64-bit Array,
    "PFRval_C": 64-bit Array,
    
    "PFR Com0": 64-bit Integer,
    "PFR Com1": 64-bit Integer,
    "PFR Com2": 64-bit Integer,
    "PFR Com3": 64-bit Integer,
    "PFR Com4": 64-bit Integer,
    "PFR Com5": 64-bit Integer,
    "PFR Com6": 64-bit Integer,
    "PFR Com7": 64-bit Integer,
    "PFRCom8": 64-bit Integer,   

    // AHP Commitment       
    "AHProw_A":[124,88,169,161,62,162],
    "AHPcol_A":[14,169,109,32,150,128],
    "AHPval_A":[84,180,111,22,79,72],
    "AHProw_B":[124,168,151,37,85,20],
    "AHPcol_B":[0,164,18,180,164,18],
    "AHPval_B":[75,176,55,34,53,86],
    "AHProw_C":[124,88,169,161,62,162],
    "AHPcol_C":[124,88,169,161,62,162],
    "AHPval_C":[124,16,53,157,61,65],
    
    "AHP Com0":166
    "AHP Com1":36 
    "AHP Com2":108
    "AHP Com3":58
    "AHP Com4":73
    "AHP Com5":157
    "AHP Com6":166
    "AHP Com7":166
    "AHP Com8":36  
    
    "Curve": "bn128",
    "PolynomialCommitment": "KZG"
}