3- Proof Generation Phase

In this section, we will review the proof generation phase of the protocol. This phase contains two parts; AHP proof and PFR proof. We also provide an example to clarify the method.

3-1- PFR Proof

$Proof (\mathbb{F}, \mathbb{H}, \mathbb{K}, A, B, C)$: This function outputs $\pi_{PFR}=(\pi_{PFR}^1,\pi_{PFR}^2,\pi_{PFR}^3,\pi_{PFR}^4,\pi_{PFR}^5,\pi_{PFR}^6,\pi_{PFR}^7,\pi_{PFR}^8,\pi_{PFR}^9,\pi_{PFR}^{10})$.

Note that in this polynomial oracle proof, the Prover wants to prove three following claims: $1.$ $row_A(x)$ , $col_A(x)$ and $val_A(x)$ is encoding of a $t-SLT$ matrix. $2.$ $row_B(x)$ , $col_B(x)$ and $val_B(x)$ is encoding of a $t-SLT$ matrix. $3.$ $row_C(x)$ , $col_C(x)$ and $val_C(x)$ is encoding of a $t-Diag$ matrix.

The proof of these claims is done in the following steps:

1- To prove strictly lower triangularity of the matrices $A$ and $B$, the Prover must prove that $\log^{row_M(\gamma^i)}_{\omega}> \log^{col_M(\gamma^i)}_{\omega}$ for $i \in \{0,1,..,m-1\}$ and $M\in\{A,B\}$. This does by $Discrete-log\hspace{1mm} comparison\hspace{1mm} protocol$. 2- To prove the first $t$ rows of $A$ and $B$ are all zeros, the Prover must prove that $row_M(\mathbb{K})\subseteq\{\omega^t,\omega^{t+1},...,\omega^{n-1}\}$. This does by $subset\hspace{1mm}over\hspace{1mm} \mathbb{K}\hspace{1mm}protocol$ . 3- To prove the diagonality of the matrix $C$, the Prover must prove that $seq_{\mathbb{K}}(row_C)=seq_{\mathbb{K}}(col_C)$where $seq_{\mathbb{K}}(h)=(h(k):k\in\mathbb{K})$. This does by $Geometric\hspace{1mm} sequence$ and $zero\hspace{1mm}over\hspace{1mm}\mathbb{K}\hspace{1mm}protocols$. 4- To prove the first $t$ rows of $C$ are all zeros, the Prover must prove that there is a vector $v\in(\mathbb{F^*})^{n-t}$ so that $seq_{\mathbb{K}}(val_C)=\vec{v}||\vec{0}$ . This does by $Geometric\hspace{1mm} sequence$ and $zero\hspace{1mm}over\hspace{1mm}\mathbb{K}\hspace{1mm}protocols$.

The steps 3 and 4 result that all the non-zero entries of the matrix $C$ are in the positions $(\omega^t,\omega^t),(\omega^{t+1},\omega^{t+1}),...,(\omega^n,\omega^n)$.

3-2- AHP Proof

$Proof (\mathbb{F}, \mathbb{H}, \mathbb{K}, A, B, C, X,W,Y)$: This function outputs

$\Pi_{AHP}=(Com_{AHP_X},\pi_{AHP})$

where $Com_{AHP_X}=(Com_{AHP_X}^{1},Com_{AHP_X}^2,Com_{AHP_X}^3,Com_{AHP_X}^4,Com_{AHP_X}^5,Com_{AHP_X}^6,Com_{AHP_X}^7,Com_{AHP_X}^8,Com_{AHP_X}^9,$ $Com_{AHP_X}^{10},Com_{AHP_X}^{11},Com_{AHP_X}^{12},Com_{AHP_X}^{13})$

and $\pi_{AHP}=(\pi_{AHP}^{1},\pi_{AHP}^2,\pi_{AHP}^3,\pi_{AHP}^4,\pi_{AHP}^5,\pi_{AHP}^6,\pi_{AHP}^7,\pi_{AHP}^8,\pi_{AHP}^9,\pi_{AHP}^{10},\pi_{AHP}^{11},\pi_{AHP}^{12},\pi_{AHP}^{13},\pi_{AHP}^{14},\pi_{AHP}^{15},$$\pi_{AHP}^{16},\pi_{AHP}^{17})$

as following:

1- The Prover calculates $z_A=Az$, $z_B=Bz$, $z_C=Cz$ where $z=(1,X,W,Y)$, for input $X$ that puts in $Com_{AHP_X}^1$.

2- The Prover calculates polynomial $z_A(x)$using indexing $z_A$ by elements of $\mathbb{H}$. Then, calculates polynomial $\hat{z}_A(x)$ using the polynomial $z_A(x)$such that $\hat{z}_A(x)\in \mathbb{F}^{<|\mathbb{H}|+b}[x]$ that agree with $z_A(x)$ on $\mathbb{H}$. Note that values of up to $b$ locations in this polynomial reveals no information about the witness $w$ provided the locations are in $\mathbb{F}-\mathbb{H}$. Similarly, calculates polynomial $\hat{z}_B(x)$ so that $\hat{z}_B(x)\in \mathbb{F}^{<|\mathbb{H}|+b}[x]$ that agree with $z_B(x)$ on $\mathbb{H}$. Also, calculates polynomial $\hat{z}_C(x)$ so that $\hat{z}_C(x)\in \mathbb{F}^{<|\mathbb{H}|+b}[x]$ that agree with $z_C(x)$ on $\mathbb{H}$.

Then, calculates polynomial $\hat{W}(x)\in \mathbb{F}^{<n_g+b}[x]$ that agree with $\bar{W}(x)$ on $\mathbb{H}[>|X|+1]$ where

$\bar{W}:\mathbb{H}[>|X|+1]\to \mathbb{F}$

$\bar{W}(h)=\frac{W(h)-\hat{X}(h)}{v_{\mathbb{H}[\leq |X|+1]}(h)}$

Note that $\mathbb{H}[>|X|+1]$ includes the members of $\mathbb{H}$ except for the first $|X|+1$ members. Also, $v_{\mathbb{H}[\leq |X|+1]}(h)$ is vanishing polynomial on $\mathbb{H}[\leq |X|+1]$ and $\hat{X}(h)$ is the polynomial obtained using indexing $x$ by elements of $\mathbb{H}[\leq |X|+1]$.

3- The Prover finds polynomial $h_0(x)$ so that $\hat{z}_A(x)\hat{z}_B(x)-\hat{z}_C(x)=h_0(x)v_{\mathbb{H}}(x)$.

4- The Prover samples a fully random $s(x)\in\mathbb{F}^{<2|\mathbb{H}|+b-1}[x]$ and computes sum $\sigma_1=\sum_{k\in \mathbb{H}}s(k)$

5- The Prover sends $Com_{AHP_X}^2=\sum_{i=0}^{deg_{\hat{W}(x)}}\hat{w}_i\hspace{1mm}ck(i)$, $Com_{AHP_X}^{3}=\sum_{i=0}^{deg_{\hat{z}_A(x)}}\hat{z}_{A_i}ck(i)$, $Com_{AHP_X}^{4}=\sum_{i=0}^{deg_{\hat{z}_B(x)}}\hat{z}_{B_i}ck(i)$, $Com_{AHP_X}^{5}=\sum_{i=0}^{deg_{\hat{z}_C(x)}}\hat{z}_{C_i}ck(i)$, $Com_{AHP_X}^{6}=\sum_{i=0}^{deg_{h_0(x)}}h_{0_i}ck(i)$, and $Com_{AHP_X}^{7}=\sum_{i=0}^{deg_{s(x)}}s_i\hspace{1mm}ck(i)$, where $\hat{w}_i$ is coefficient of $x^i$ in polynomial $\hat{W}(x)$, $\hat{z}_{A_i}$ is coefficient of $x^i$ in polynomial $\hat{z}_A(x)$, $\hat{z}_{B_i}$ is coefficient of $x^i$ in polynomial $\hat{z}_B(x)$, $\hat{z}_{C_i}$ is coefficient of $x^i$ in polynomial $\hat{z}_C(x)$, $h_{0_i}$ is coefficient of $x^i$ in polynomial $h_0(x)$, $s_{i}$ is coefficient of $x^i$ in polynomial $s(x)$.

6- The Verifier chooses random numbers $\alpha$, $\eta_A$, $\eta_B$, $\eta_C$ and sends them to the Prover. ( Note that the Prover can choose $\alpha=hash(s(0)+s(1)+1)$, $\eta_A=hash(s(2)+s(3)+2)$, $\eta_B=hash(s(4)+s(5)+3)$, $\eta_C=hash(s(6)+s(7)+4)$.

7- The Prover finds polynomials $g_1(x)$ and $h_1(x)$ so that

$s(x)+r(\alpha,x)\sum_{M}\eta_M\hat{z}_M(x)-(\sum_{M}\eta_Mr_M(\alpha,x))\hat{z}(x)=h_1(x)v_{\mathbb{H}}(x)+xg_1(x)+\frac{\sigma_1}{|\mathbb{H}|}$ $(1)$

where $\hat{z}(x)=\hat{W}(x)v_{\mathbb{H}[\leq |X|+1]}(x)+\hat{X}(x)$ that agree with $z$ on $\mathbb{H}$ and $r(x,y)=u_{\mathbb{H}}(x,y)=\frac{v_{\mathbb{H}}(x)-v_{\mathbb{H}}(y)}{x-y}$ , $v_{\mathbb{H}}(x)=\prod_{h\in \mathbb{H}}(x-h)=x^{|\mathbb{H}|}-1$. Therefore $r(x,y)=\frac{x^{|\mathbb{H}|}-y^{|\mathbb{H}|}}{x-y}$. (Note that $r(x,y)$satisfies two useful algebraic properties. First, the univariate polynomials $(r(x,a))_{a\in \mathbb{H}}$ are linearly independent and $r(x,y)$ is their (unique) low-degree extension. Second, $r(x,y)$ vanishes on the square $\mathbb{H}\times \mathbb{H}$ except for on the diagonal, where it takes on the (non-zero) values $(r(a,a))_{a\in \mathbb{H}}$.) . Also $r_M(x,y)=\sum_{k\in \mathbb{H}}r(x,k)\hat{M}(k,y)$ for $M\in \{A,B,C\}$ where $\hat{A}(x,y)$ is a bivariate polynomial that passes from 25 points where theses points are obtained using indexing rows and columns of $A$ by elements of $\mathbb{H}$. This polynomial can obtain as following: $\hat{A}(x,y)=\sum_{k\in \mathbb{K}}u_{\mathbb{H}}(x,\hat{row'_A}(k))u_{\mathbb{H}}(y,\hat{col'_A}(k))\hat{val'_A}(k)$

, $\hat{B}(x,y)$ similarly as following: $\hat{B}(x,y)=\sum_{k\in \mathbb{K}}u_{\mathbb{H}}(x,\hat{row'_B}(k))u_{\mathbb{H}}(y,\hat{col'_B}(k))\hat{val'_B}(k)$

and $\hat{C}(x,y)$ similarly as following: $\hat{C}(x,y)=\sum_{k\in \mathbb{K}}u_{\mathbb{H}}(x,\hat{row'_C}(k))u_{\mathbb{H}}(y,\hat{col'_C}(k))\hat{val'_C}(k)$

The Prover sends $Com_{AHP_X}^{8}=\sum_{i=0}^{deg_{g_1(x)}}g_{1_i}ck(i)$ and $Com_{AHP_X}^{9}=\sum_{i=0}^{deg_{h_1(x)}}h_{1_i}ck(i)$ to the Verifier where $g_{1_i}$ is coefficient of $x^i$ of polynomial $g_1(x)$ and $h_{1_i}$ is coefficient of $x^i$ of polynomial $h_1(x)$.

8- The Verifier selects $\beta_1\in \mathbb{F}-\mathbb{H}$ and sends it to the Prover. (The Prover can selects $\beta_1=hash(s(8))\in \mathbb{F}-\mathbb{H}$ ).

9- The Prover calculates $\sigma_2=\sum_{k\in\mathbb{H}}r(\alpha,k)\sum_{M}\eta_M\hat{M}(k,\beta_1)$. Then, the Prover finds $g_2(x)$ and $h_2(x)$ so that $r(\alpha,x)\sum_M \eta_M\hat{M}(x,\beta_1)=h_2(x)v_{\mathbb{H}}(x)+xg_2(x)+\frac{\sigma_2}{|\mathbb{H}|}$

The Prover sends $Com_{AHP_X}^{10}=\sum_{i=0}^{deg_{g_2(x)}}g_{2_i}ck(i)$ and $Com_{AHP_X}^{11}=\sum_{i=0}^{deg_{h_2(x)}}h_{2_i}ck(i)$ where $g_{2_i}$ is coefficient of $x^i$ of polynomial $g_2(x)$ and $h_{2_i}$ is coefficient of $x^i$ of polynomial $h_2(x)$.

10- The Verifier selects $\beta_2\in \mathbb{F}-\mathbb{H}$ and sends it to the Prover. ( The Prover can select $\beta_2=hash(s(9))\in \mathbb{F}-\mathbb{H}$ ).

11- The Prover calculates $\sigma_3=\sum_{k\in\mathbb{K}}(\sum_M \eta_M\frac{v_{\mathbb{H}}(\beta_2)v_{\mathbb{H}}(\beta_1)\hat{val'_M}(k)}{(\beta_2-\hat{row'_M}(k))(\beta_1-\hat{col'_M}(k))})$. Then, the Prover finds polynomials $g_3(x)$ and $h_3(x)$ so that $h_3(x)v_{\mathbb{K}}(x)=a(x)-b(x)(xg_3(x)+\frac{\sigma_3}{|\mathbb{K}|})$ where $a(x)=\sum_{M\in \{A,B,C\}} \eta_M v_{\mathbb{H}}(\beta_2)v_{\mathbb{H}}(\beta_1)\hat{val'_M}(x)\prod_{N\in\{A,B,C\}-\{M\}}(\beta_2-\hat{row'_N}(x))(\beta_1-\hat{col'_N}(x))$and $b(x)=\prod_{M\in\{A,B,C\}}(\beta_2-\hat{row'_M}(x))(\beta_1-\hat{col'_M}(x))$.

The Prover sends $Com_{AHP_X}^{12}=\sum_{i=0}^{deg_{g_3(x)}}g_{3_i}ck(i)$ and $Com_{AHP_X}^{13}=\sum_{i=0}^{deg_{h_3(x)}}h_{3_i}ck(i)$ where $g_{3_i}$ is coefficient of $x^i$ of polynomial $g_3(x)$ and $h_{3_i}$ is coefficient of $x^i$ of polynomial $h_3(x)$.

and

12- The Prover sends $\pi_{AHP}^1=\sigma_1$, $\pi_{AHP}^2=(\hat{w}_0,\hat{w}_1,\hat{w}_3,...,\hat{w}_{|W|+b-1})$, $\pi_{AHP}^3=(\hat{z}_{A_0},\hat{z}_{A_1},...,\hat{z}_{A_{|H|+b-1}})$, $\pi_{AHP}^4=(\hat{z}_{B_0},\hat{z}_{B_1},...,\hat{z}_{B_{|H|+b-1}})$, $\pi_{AHP}^5=(\hat{z}_{C_0},\hat{z}_{C_1},...,\hat{z}_{C_{|H|+b-1}})$, $\pi_{AHP}^6=(h_{0_0},h_{0_1},...,h_{0_{|H|+2b-2}})$ and $\pi_{AHP}^7=(s_0,s_1,...,s_{2|H|+b-2})$ 13- The Prover sends $\pi_{AHP}^8=(g_{1_0},...,g_{1_{|H|-2}})$ and $\pi_{AHP}^{9}=(h_{1_0},...,h_{1_{|H|+b-2}})$ .

14-The Prover sends $\pi_{AHP}^{10}=\sigma_2$, $\pi_{AHP}^{11}=(g_{2_0},...,g_{2_{|H|-2}})$ and $\pi_{AHP}^{12}=(h_{2_0},...,h_{2_{|H|-2}})$.

15- The Prover sends $\pi_{AHP}^{13}=\sigma_3$, $\pi_{AHP}^{14}=(g_{3_0},...,g_{3_{|K|-2}})$ and $\pi_{AHP}^{15}=(h_{3_0},...,h_{3_{6|K|-6}})$ .

16- The Prover chooses random values $\eta_{\hat{w}}$, $\eta_{\hat{z}_A}$, $\eta_{\hat{z}_B}$, $\eta_{\hat{z}_C}$, $\eta_{h_0}$, $\eta_s$, $\eta_{g_1}$, $\eta_{h_1}$, $\eta_{g_2}$, $\eta_{h_2}$, $\eta_{g_3}$ and $\eta_{h_3}$ of $\mathbb{F}$ as following: $\eta_{\hat{w}}=hash(s(10))$, $\eta_{\hat{z}_A}=hash(s(11))$, $\eta_{\hat{z}_B}=hash(s(12))$, $\eta_{\hat{z}_C}=hash(s(13))$, $\eta_{h_0}=hash(s(14))$, $\eta_{s}=hash(s(15))$, $\eta_{g_1}=hash(s(16))$, $\eta_{h_1}=hash(s(17))$, $\eta_{g_2}=hash(s(18))$, $\eta_{h_2}=hash(s(19))$, $\eta_{g_3}=hash(s(20))$, $\eta_{h_3}=hash(s(21))$.

17- The Prover builds the linear combination$p(x)=\eta_{\hat{W}}\hat{W}(x)+\eta_{\hat{z}_A}\hat{z}_A(x)+\eta_{\hat{z}_B}\hat{z}_B(x)+\eta_{\hat{z}_C}\hat{z}_C(x)+\eta_{h_0}h_0(x)+\eta_ss(x)+\eta_{g_1}g_1(x)$ $+\eta_{h_1}h_1(x)+\eta_{g_2}g_2(x)+\eta_{h_2}h_2(x)+\eta_{g_3}g_3(x)+\eta_{h_3}h_3(x)$.

18- The Prover calculates $p(x)$ in $x=x'$ (value of $x'$ is received from the Verifier. Also, can select as $x'=hash(s(22)))$, then puts it in $\pi_{AHP}^{16}$ . Therefore $\pi_{AHP}^{16}=p(x')=y'$.

19- The Prover computes $\pi_{AHP}^{17}=PC.Eval(ck,p(x),d_p,r_p,x')$ where $d_p$ is degree bound of $p(x)$ and $r_p$ is a random value. For example, if the polynomial commitment scheme $KZG$ is used, then the Prover calculates polynomial $q(x)=\frac{p(x)-y'}{x-x'}$ and $\pi_{AHP}^{17}=g\hspace{1mm}q(\tau)$ by using $ck$ as following: $\pi_{AHP}^{17}=\sum_{i=0}^{deg_{q(x)}}q_i\hspace{1mm}ck(i)$, where $q_i$ is the coefficient of $x^i$ of $q(x)$.

3-3- Proof Structure

Proof set is

$\Pi_{AHP}=(Com_{AHP_X},\pi_{AHP})$

where $Com_{AHP_X}=(Com_{AHP_X}^{1},Com_{AHP_X}^2,Com_{AHP_X}^3,Com_{AHP_X}^4,Com_{AHP_X}^5,Com_{AHP_X}^6,Com_{AHP_X}^7,Com_{AHP_X}^8,Com_{AHP_X}^9,$ $Com_{AHP_X}^{10},Com_{AHP_X}^{11},Com_{AHP_X}^{12},Com_{AHP_X}^{13})$

$Com_{AHP_X}^1=X$, $Com_{AHP_X}^2=\sum_{i=0}^{deg_{\hat{W}(x)}}\hat{w}_i\hspace{1mm}ck(i)$, $Com_{AHP_X}^{3}=\sum_{i=0}^{deg_{\hat{z}_A(x)}}\hat{z}_{A_i}ck(i)$, $Com_{AHP_X}^{4}=\sum_{i=0}^{deg_{\hat{z}_B(x)}}\hat{z}_{B_i}ck(i)$, $Com_{AHP_X}^{5}=\sum_{i=0}^{deg_{\hat{z}_C(x)}}\hat{z}_{C_i}ck(i)$, $Com_{AHP_X}^{6}=\sum_{i=0}^{deg_{h_0(x)}}h_{0_i}ck(i)$, $Com_{AHP_X}^{7}=\sum_{i=0}^{deg_{s(x)}}s_i\hspace{1mm}ck(i)$, $Com_{AHP_X}^{8}=\sum_{i=0}^{deg_{g_1(x)}}g_{1_i}ck(i)$, $Com_{AHP_X}^{9}=\sum_{i=0}^{deg_{h_1(x)}}h_{1_i}ck(i)$, $Com_{AHP_X}^{10}=\sum_{i=0}^{deg_{g_2(x)}}g_{2_i}ck(i)$, $Com_{AHP_X}^{11}=\sum_{i=0}^{deg_{h_2(x)}}h_{2_i}ck(i)$, $Com_{AHP_X}^{12}=\sum_{i=0}^{deg_{g_3(x)}}g_{3_i}ck(i)$, $Com_{AHP_X}^{13}=\sum_{i=0}^{deg_{h_3(x)}}h_{3_i}ck(i)$

and $\pi_{AHP}=(\pi_{AHP}^{1},\pi_{AHP}^2,\pi_{AHP}^3,\pi_{AHP}^4,\pi_{AHP}^5,\pi_{AHP}^6,\pi_{AHP}^7,\pi_{AHP}^8,\pi_{AHP}^9,\pi_{AHP}^{10},\pi_{AHP}^{11},\pi_{AHP}^{12},\pi_{AHP}^{13},\pi_{AHP}^{14},\pi_{AHP}^{15},$$\pi_{AHP}^{16},\pi_{AHP}^{17})$

$\pi_{AHP}^1=\sigma_1$, $\pi_{AHP}^2=(\hat{w}_0,\hat{w}_1,\hat{w}_3,...,\hat{w}_{n_g+b-1})$, $\pi_{AHP}^3=(\hat{z}_{A_0},\hat{z}_{A_1},...,\hat{z}_{A_{|H|+b-1}})$, $\pi_{AHP}^4=(\hat{z}_{B_0},\hat{z}_{B_1},...,\hat{z}_{B_{|H|+b-1}})$, $\pi_{AHP}^5=(\hat{z}_{C_0},\hat{z}_{C_1},...,\hat{z}_{C_{|H|+b-1}})$, $\pi_{AHP}^6=(h_{0_0},h_{0_1},...,h_{0_{|H|+2b-2}})$, $\pi_{AHP}^7=(s_0,s_1,...,s_{2|H|+b-2})$, $\pi_{AHP}^8=(g_{1_0},...,g_{1_{|H|-2}})$, $\pi_{AHP}^{9}=(h_{1_0},...,h_{1_{|H|+b-2}})$, $\pi_{AHP}^{10}=\sigma_2$, $\pi_{AHP}^{11}=(g_{2_0},...,g_{2_{|H|-2}})$, $\pi_{AHP}^{12}=(h_{2_0},...,h_{2_{|H|-2}})$, $\pi_{AHP}^{13}=\sigma_3$, $\pi_{AHP}^{14}=(g_{3_0},...,g_{3_{|K|-2}})$, $\pi_{AHP}^{15}=(h_{3_0},...,h_{3_{6|K|-6}})$, $\pi_{AHP}^{16}=y'$, $\pi_{AHP}^{17}=\sum_{i=0}^{deg_{q(x)}}q_i\hspace{1mm}ck(i)$.

where $\hat{w}_i$ is coefficient of $x^i$ in polynomial $\hat{W}(x)$, $\hat{z}_{A_i}$ is coefficient of $x^i$ in polynomial $\hat{z}_A(x)$, $\hat{z}_{B_i}$ is coefficient of $x^i$ in polynomial $\hat{z}_B(x)$, $\hat{z}_{C_i}$ is coefficient of $x^i$ in polynomial $\hat{z}_C(x)$, $h_{0_i}$ is coefficient of $x^i$ in polynomial $h_0(x)$, $s_{i}$ is coefficient of $x^i$ in polynomial $s(x)$, $g_{1_i}$ is coefficient of $x^i$ of polynomial $g_1(x)$ and $h_{1_i}$ is coefficient of $x^i$ of polynomial $h_1(x)$, $g_{2_i}$ is coefficient of $x^i$ of polynomial $g_2(x)$ and $h_{2_i}$ is coefficient of $x^i$ of polynomial $h_2(x)$, $g_{3_i}$ is coefficient of $x^i$ of polynomial $g_3(x)$ and $h_{3_i}$ is coefficient of $x^i$ of polynomial $h_3(x)$.

• Size of AHP proof: $|\Pi_{AHP}|=10|\mathbb{H}|+7|\mathbb{K}|+|W|+8b+6$.

• CommitmentID is explained on the Commitment Phase page.

• DeviceEncodedID = Base64<MAC>

• Input and Output are the device input and output, respectively.

3-4- Proof JSON file format

{
    "CommitmentID": 64-bit,
    "DeviceEncodedID":   
    "Input": String,
    "Output": String,
    "P1AHP": 64-bit Integer,
    "P2AHP": 64-bit Array,
    "P3AHP": 64-bit Array,
    "P4AHP": 64-bit Array,
    "P5AHP": 64-bit Array,
    "P6AHP": 64-bit Array,
    "P7AHP": 64-bit Array,
    "P8AHP": 64-bit Array,
    "P9AHP": 64-bit Array],
    "P10AHP": 64-bit Integer,
    "P11AHP": 64-bit Array,
    "P12AHP": 64-bit Array,
    "P13AHP": 64-bit Integer,
    "P14AHP": 64-bit Array,
    "P15AHP": 64-bit Array,
    "P16AHP": 64-bit Array,    
    "Protocol": String
}