Example 1

PFR Proof

The Prover interpolates polynomial $h$ such that $seq_{\mathbb{K}}(h)=\omega^t,\omega^{t+1},..,\omega^{n-1},0,..,0$. Here $t=2$ and $n=5$, therefore $seq_{\mathbb{K}}(h)=\omega^2,\omega^3,\omega^4,0,..,0$. This means that $\{h(k):\hspace{1mm}k\in\mathbb{K}=\{1,43,39,48,73,62,132,65,80\}\}=\{\omega^2,\omega^3,\omega^4,0,..,0\}=\{42,125,135,0,0,0,0,0,0\}$Therefore $h(1)=42$, $h(43)=125$ , $h(39)=135$, $h(48)=h(73)=h(62)=h(132)=h(65)=h(80)=0$. So $h(x)=42L_1(x)+125L_2(x)+135L_3(x)$ where $L_1(x)=\frac{(x-43)(x-39)(x-48)(x-73)(x-62)(x-132)(x-65)(x-80)}{(-42)(-38)(-47)(-72)(-61)(-131)(-64)(-79)}=161x^8+161x^7+161x^6+161x^5+161x^4+161x^3+161x^2+161x+161$,

$L_2(x)=45x^8+125x^7+126x^6+169x^5+27x^4+75x^3+148x^2+29x+161$,

$L_3(x)=125x^8+169x^7+75x^6+29x^5+45x^4+126x^3+27x^2+148x+161$,

Therefore $h(x)=121x^8+133x^7+57x^6+127x^5+103x^4+24x^3+128x^2+140x+114$. The Prover sends $\pi_1=(h_0,h_1,...,h_8)$ where $h_i$ is coefficients of polynomial $h(x)$ to the Verifier.

3-5-1-2- Step 2

The Prover and the Verifier do $Geometric\hspace{1mm}Sequence\hspace{1mm}Test$ on $h$ (to verify correct construction of polynomial $h$). In $Geometric\hspace{1mm} Sequence\hspace{1mm} Test$ want to prove that $Seq_{\mathbb{K}}(h)=a_1,a_1r,..,a_1r^{c_1-1},a_2, a_2r,..., a_2r^{c_2-1},...,a_v,a_vr,...,a_vr^{c_v-1}$.

3-5-1-2- Step 2-A- Geometric Sequence Test

Since $h(\gamma^0)=\omega^2$, $h(\gamma^1)=\omega^3$ ,$h(\gamma^2)=\omega^4$ and $h(\gamma^3)=...=h(\gamma^8)=0$, then Geometric Sequence Test with $a_1=\omega^2$, $a_2=0$, $r=\omega$, $c_1=n-t=3$ and $c_2=m-(n-t)=9-3=6$ run. Let $p_i=\sum_{j<i}c_j$ for $i\in \{1,2,..,v\}$. Here $v$ is the number of $c_i$ that are non-zero. Therefore $v=2$ , $p_1=0$ and $p_2=c_1=3$.

The Prover and the Verifier run $Zero\hspace{1mm}Over\hspace{1mm}\mathbb{K}$ (see the next section) to check that for all $k\in\mathbb{K}$, have $(h(\gamma k)-rh(k))\prod_{i=1}^{v}(k-\gamma^{p_i+c_i-1})=0$. Here, $Zero\hspace{1mm}Over\hspace{1mm}\mathbb{K}$ run to check that for all $k\in\mathbb{K}=\{1,\gamma,\gamma^2,..,\gamma^8\}$, have $(h(\gamma k)-\omega h(k))(k-\gamma^2)(k-\omega^8)=0$. $Note$: It is clear that if the construction of $h$ is correct, it applies this equality, because if $k=1$, $h(\gamma k)=h(\gamma)=\omega h(1)$ then $h(\gamma k)-\omega h(k)=0$. If $k=\gamma$, $h(\gamma k)=h(\gamma^2)=\omega h(\gamma)$, then $h(\gamma k)-\omega h(k)=0$. If $k=\gamma^2$, then $k-\gamma^2=0$. If $k=\omega^3,...,\omega^7$, then $h(\gamma k)=\omega h(k)=0$ and if $k=\gamma^8$, then $k-\gamma^8=0$.

3-5-1-2-Step 2-B- Zero over K

In $Zero\hspace{1mm}Over\hspace{1mm}\mathbb{K}$, want to prove that for all $k\in\mathbb{K}$, $F(k)=0$ where $F(x)=G(x,f_{j_1}(\alpha_1x),f_{j_2}(\alpha_2x),...,f_{j_{t}}(\alpha_{t}x))$. Here $F(x)=(h(\gamma x)-\omega h(x))(x-\gamma^2)(x-\gamma^8)$, therefore since $t=2$, $F(x)=G(x,f_{j_1}(\alpha_1x),f_{j_2}(\alpha_2x))=(h(\gamma x)-\omega h(x))(x-\gamma^2)(x-\gamma^8)$, so $h_1=f_{j_1}=h$, $h_2=f_{j_2}=h$, $\alpha_1=\gamma$ and $\alpha_2=1$.

2-2-1 The Prover selects polynomials $r_i\in \mathbb{F}^{<2}[x]$, $i\in \{1,2,..,t\}$. Then computes masks $m_i(x)=r_i(\alpha_i^{-1}x)Z_{\mathbb{K}}(\alpha_i^{-1}x)$ and computes $h'_i(x)=h_i(x)+m_i(x)$ for $i\in\{1,2,..,t\}$.

Here, the Prover selects $r_1,r_2$. For example, let $r_1(x)=2+x$ and $r_2(x)=2x$. Therefore $m_1(x)=r_1(\alpha_1^{-1}x)Z_{\mathbb{K}}(\alpha_1^{-1}x)$ where $\alpha_1=\gamma=48$ and $Z_{\mathbb{K}}(x)=\prod_{x\in\mathbb{K}}(x-k)=(x-1)(x-43)(x-39)(x-48)(x-73)(x-62)(x-132)(x-65) (x-80)=x^9+180$. Therefore $m_1(x)=r_1(80x)Z_{\mathbb{K}}(80x)=80x^{10}+2x^9+101x+179$ $m_2(x)=r_2(\alpha_2^{-1}x)Z_{\mathbb{K}}(\alpha_2^{-1}x)=r_2(x)Z_{\mathbb{K}}(x)=2x(x^9+180)=2x^{10}+179x$. Then computes $h'_1(x)=h_1(x)+m_1(x)=h(x)+m_1(x)=80x^{10}+2x^9+121x^8+133x^7+57x^6+127x^5+103x^4+$ $24x^3+128x^2+60x+112$ and $h'_2(x)=h_2(x)+m_2(x)=h(x)+m_2(x)=2x^{10}+121x^8+133x^7+$ $57x^6+127x^5+103x^4+24x^3+128x^2+138x+114$.

2-2-2- The Prover computes $F'(x)=G(x,h'_1(\alpha_1x),h'_2(\alpha_2x),...,h'_t(\alpha_tx))$ and $q_1(x)=\frac{F'(x)}{Z_{\mathbb{K}}(x)}$ then the Prover sends $\pi_{PFR_{1+i}}=(m_{i1},...,m_{ideg(m_i)})$ where $m_{ij}$s are coefficients of polynomial $m_i(x)$, $\pi_{PFR_{1+t+i}}=(r_{i1},...,r_{ideg(r_i)})$ where $r_{ij}$s are coefficients of polynomial $r_i(x)$ and $\pi_{PFR_{2t+2}}=(q_{11},...,q_{1deg(q_1)})$ where $q_{1j}$s are coefficients of polynomial $q_1(x)$

Here $F'(x)=G(x,h'_1(43x),h'_2(x))=(h'_1(43x)-\omega h'_2(x))(x-\gamma^2)(x-\gamma^8)=64x^{12}+169x^{11}+168x^{10}$ $+51x^9+117x^3+12x^2+13x+130$ and then $q_1(x)=\frac{F'(x)}{Z_{\mathbb{K}}(x)}=64x^3+169x^2+168x+51$.

The Prover sends $\pi_{PFR_{2}}=(179,101,0,0,0,0,0,0,0,2,80)$, $\pi_{PFR_{3}}=(0,179,0,0,0,0,0,0,0,0,2)$ , $\pi_{PFR_4}=(2,1)$, $\pi_{PFR_5}=(0,2)$ and $\pi_{PFR_6}=(51,168,169,64)$ to the Verifier.

2-2-3- The Verifier derives $h'_i=h_i+m_i$ through additive homomorphism. Then samples $\beta_1$, $\beta_2$ and $c$ of $\mathbb{F}^{*}-\mathbb{K}$ and sends $c$ to the Prover. Suppose $\beta_1=65$, $\beta_2=21$ and $c=171$. The Verifier sends $c=171$ to the Prover.

2-2-4- The Prover computes $q_2=r_1+cr_2+...+c^{t-1}r_t$ and the Verifier derives concrete oracle $q_2$ through additive homomorphism. Here, the Prover computes $q_2(x)=r_1(x)+cr_2(x)=2+x+171(2x)=162x+2$.

2-2-5- Let $M(x)=m_1(\alpha_1 x)+cm_2(\alpha_2 x)+...+c^{t-1}m_t(\alpha_t x)$. The Verifier computes $Z_{\mathbb{K}}(\beta_1)$, $Z_{\mathbb{K}}(\beta_2)$, queries $q_1(\beta_1)$and $q_2(\beta_2)$ and for $i\in \{1,2,..,t\}$, queries $h'_i(\alpha_i\beta_1)$ and $m_i(\alpha_i\beta_2)$to compute $F'(\beta_1)$ and $M(\beta_2)$. The Verifier asserts two identities $M(\beta_2)-q_2(\beta_2)Z_{\mathbb{K}}(\beta_2)=0$ and $F'(\beta_1)-q_1(\beta_1)Z_{\mathbb{K}}(\beta_1)=0$.

Here, $M(x)=m_1(\alpha_1 x)+cm_2(\alpha_2 x)=m_1(\gamma x)+171m_2(x)=162x^{10}+2x^9+19x+179$. The Verifier computes $Z_{\mathbb{K}}(\beta_1)=Z_{\mathbb{K}}(65)=65^9+180=0$ and $Z_{\mathbb{K}}(\beta_2)=Z_{\mathbb{K}}(21)=21^9+180=30$ and queries $q_1(\beta_1)=86$ and $q_2(\beta_2)=146$. Also queries values $h'_1(\alpha_1\beta_1)=h'_1(\gamma\times 65)=h'_1(80)=0$ , $h'_2(\alpha_2\beta_1)=h'_2(65)=0$, $m_1(\alpha_1\beta_2)=m_1(\gamma\times 21)=m_1(179)=147$ and $m_2(\alpha_2\beta_2)=m_2(21)=174$. Then checks$M(\beta_2)-q_2(\beta_2)Z_{\mathbb{K}}(\beta_2)=0$ , that means $36-30\times 146=36-36\equiv 0\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, and $F'(\beta_1)-q_1(\beta_1)Z_{\mathbb{K}}(\beta_1)=0$, that means $0-86\times 0\equiv 0\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$.

3- The Prover and the Verifier run $Subset\hspace{1mm}over\hspace{1mm}\mathbb{K}$ between $row_A$ and $h$ to prove that $row_A(\mathbb{K})\subset h(\mathbb{K})$ where $row_A(\mathbb{K})=\{row_A(k):\hspace{1mm}k\in\mathbb{K}\}$and $h(\mathbb{K})=\{h(k)\hspace{1mm}:\hspace{1mm}k\in\mathbb{K}\}$.

3-5-1-2- Step 3- Subset over $\mathbb{K}$

.................

4- The Prover and the Verifier run $Discrete-log\hspace{1mm} Comparison$ between $row_A$ and $col_A$ as following:

3-5-1-2- Step 4- Discrete-log Comparison

Let parameters $(\Delta \in \mathbb{F}, n = |\mathbb{H}|)$ such that $ord(\Delta) = 2n$ and $\Delta^2=\omega$. Here, $\omega=59$, $\Delta=56$ and $ord(\Delta)=2n=10$.

3-5-1-2- Step 4-Substep A- The Prover interpolates polynomial $s$ so that agrees with $\frac{row_A}{col_A}$ on $\mathbb{K}$. Then send them to the Verifier.

Here, $s(1)=\frac{row_A(1)}{col_A(1)}=\frac{42}{59}\equiv 59\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s(43)=\frac{row_A(43)}{col_A(43)}\equiv125\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s(39)=\frac{row_A(39)}{col_A(39)}=\frac{135}{125}\equiv59\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s(48)=\frac{row_A(48)}{col_A(48)}=\frac{48}{45}\equiv 170\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s(73)=\frac{row_A(73)}{col_A(73)}=\frac{36}{22}\equiv 51\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s(62)=\frac{row_A(62)}{col_A(62)}=\frac{151}{166}\equiv 2\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s(132)=\frac{row_A(132)}{col_A(132)}=\frac{21}{46}\equiv 28\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s(65)=\frac{row_A(65)}{col_A(65)}=\frac{131}{127}\equiv 135\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ and $s(80)=\frac{row_A(80)}{col_A(80)}=\frac{6}{40}\equiv 154\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$. Therefore, $s(x)=59L_1(x)+125L_2(x)+59L_3(x)+170L_4(x)+51L_5(x)+2L_6(x)+28L_7(x)+135L_8(x)+$ $154L_9(x)$.

where $L_1(x)$, $L_2(x)$ and $L_3(x)$ are computed in step $1$ and the rest of $L_i(x)s$ are $L_4(x)=126x^8+75x^7+161x^6+126x^5+75x^4+161x^3+126x^2+75x+161$, $L_5(x)=169x^8+29x^7+126x^6+148x^5+125x^4+75x^3+45x^2+27x+161$, $L_6(x)=27x^8+45x^7+75x^6+125x^5+148x^4+126x^3+29x^2+169x+161$, $L_7(x)=75x^8+126x^7+161x^6+75x^5+126x^4+161x^3+75x^2+126x+161$, $L_8(x)=148x^8+27x^7+126x^6+45x^5+29x^4+75x^3+169x^2+125x+161$and $L_9(x)=29x^8+148x^7+75x^6+27x^5+169x^4+126x^3+125x^2+45x+161$.

Therefore $s(x)=41x^8+101x^7+34x^6+38x^5+x^4+25x^3+152x^2+123x+87$. The Prover sends $s(x)$ to the Verifier.

3-5-1-2- Step 4-Substep B- For $b\in\{row_A, col_A, s\}$, the Prover interpolates polynomial $b'$ so that for all $k\in \mathbb{K}$, $b'(k)=\Delta^{\log_{\omega}^{b(k)}}$.

Here, if $b=row_A$, $row'_A(k)=\Delta^{\log_{\omega}^{row_A(k)}}=56^{\log_{59}^{row_A(k)}}$that means $row'_A(1)=56^{\log_{59}^{42}}=56^2\equiv 59\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $row'_A(43)=56^{\log_{59}^{125}}=56^3\equiv 46\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $row'_A(39)=56^{\log_{59}^{135}}=56^4\equiv 42\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $row'_A(48)=56^{\log_{59}^{48}} \equiv 49\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $row'_A(73)=56^{\log_{59}^{36}} \equiv 114\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $row'_A(62)=56^{\log_{59}^{151}} \equiv \hspace{1mm}(\textrm{mod}\hspace{1mm}181)$.\

Therefore $row'_A(x)=59L_1(x)+46L_2(x)+42L_3(x)+49L_4(x)+114L_5(x)+...$.

if $b=col_A$, $col'_A(k)=\Delta^{\log_{\omega}^{col_A(k)}}=56^{\log_{59}^{col_A(k)}}$that means $col'_A(1)=56^{\log_{59}^{59}}=56^1\equiv 56\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $col'_A(48)=56^{\log_{59}^{1}}=56^5\equiv 180\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ and $col'_A(132)=56^{\log_{59}^{125}}=56^3\equiv 46\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$. Therefore $col'_A(x)=56L_1(x)+180L_2(x)+46L_3(x)=96x^2+47x+94$.

if $b=s$, $s'(k)=\Delta^{\log_{\omega}^{s(k)}}=56^{\log_{59}^{s(k)}}$that means $s'(1)=56^{\log_{59}^{59}}=56^1\equiv 56\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $s'(48)=56^{\log_{59}^{125}}=56^3\equiv 46\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ and $s'(132)=56^{\log_{59}^{59}}=56^1\equiv 56\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$. Therefore $s'(x)=56L_1(x)+46L_2(x)+56L_3(x)=21x^2+103x+113$.

Then the Prover sends $row'_A$, $col'_A$ and $s'$ to the Verifier.

3-5-1-2- Step 4-Substep C- The Prover interpolates polynomial $h$ so that $seq_{\mathbf{K}}(h)=1,\Delta,\Delta^2,..,\Delta^{n-1},0,0,..,0$.

Here $seq_{\mathbf{K}}(h)=1, 56, 59, 46, 42$.

AHP Proof

Since in this example $n_i=1$, therefore input and output are not vector. So, we denote input by $x$ and output by $y$ instead of $X$ and $Y$.

$Proof (\mathbb{F}_{181}, \mathbb{H}=\{1,59,42,125,135\}, \mathbb{K}=\{1,49,48,180,132,133\}, A, B, C, x=4,W=(20,31),y=82)$

1- The Prover puts $x=4$ in $Com_{AHP_X}^1$ . We consider $z=(1,x,w_1,w_2,y)$ where $w_1=5x$, $w_2=w_1+11$ and $y=26w_2$. Therefore $z=(1,x,W,y)=(1,4,20,31,82)$. The Prover calculates $z_A=Az=\begin{bmatrix}0&0&0&0&0\\0&0&0&0&0\\0&1&0&0&0\\1&0&0&0&0\\0&0&0&1&0\\ \end{bmatrix}\begin{bmatrix}1\\4\\20\\31\\82\\\end{bmatrix}=\begin{bmatrix}0\\0\\4\\1\\31\\\end{bmatrix}$, $z_B=Bz=\begin{bmatrix}0&0&0&0&0\\0&0&0&0&0\\5&0&0&0&0\\11&0&1&0&0\\26&0&0&0&0\\\end{bmatrix}\begin{bmatrix}1\\4\\20\\31\\82\end{bmatrix}=\begin{bmatrix}0\\0\\5\\31\\26\\\end{bmatrix}$and $z_C=Cz=\begin{bmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\\end{bmatrix}\begin{bmatrix}1\\4\\20\\31\\82\\\end{bmatrix}=\begin{bmatrix}0\\0\\20\\31\\82\\\end{bmatrix}$

2- The Prover calculates the polynomial $z_A(x)$using indexing $z_A$ by elements of $\mathbb{H}$ that mean $z_A(x)$ is the polynomial where $z_A(1)=0$, $z_A(59)=0$, $z_A(42)=4$, $z_A(125)=1$ and $z_A(135)=31$.

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 $\mathbb{F}-\mathbb{H}$. Here, for simplicity, let $b=2$. The Prover calculates $\hat{z}_A(x)$ such that agree with $z_A(x)$ on $\mathbb{H}$ and also $\hat{z}_A(150)=5$, $\hat{z}_A(80)=47$.

Therefore, we have $\hat{z}_A(x)=4L_3(x)+L_4(x)+31L_5(x)+5L_6(x)+47L_7(x)$ where $L_3(x)=133x^6+155x^5+117x^4+27x^3+48x^2+73x+171$, $L_4(x)=4x^6+123x^5+25x^4+48x^3+27x^2+113x+22$, $L_5(x)=75x^6+115x^5+27x^4+25x^3+117x^2+154x+30$, $L_6(x)=132x^6+119x^5+49x+62$ and $L_7(x)=97x^6+111x^5+84x+70$. So $\hat{z}_A(x)=116x^6+165x^5+63x^4+26x^3+45x^2+141x+168$.

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 $\mathbb{H}$ that mean $\hat{z}_B(1)=0$, $\hat{z}_B(59)=0$, $\hat{z}_B(42)=5$, $\hat{z}_B(125)=31$, $\hat{z}_B(135)=26$ and also $b=2$ random locations $\hat{z}_B(150)=15$ and $\hat{z}_B(80)=170$. So, $\hat{z}_B(x)=5L_3(x)+31L_4(x)+26L_5(x)+15L_6(x)+170L_7(x)=32x^6+178x^5+71x^4+101x^3+137x^2+81x+124$

Similarly, calculates polynomial $\hat{z}_C(x)$ such that $\hat{z}_C(x)\in \mathbb{F}^{<|\mathbb{H}|+b}[x]$ that agree with $z_C(x)$ on $\mathbb{H}$ that mean $\hat{z}_C(1)=0$, $\hat{z}_C(59)=0$, $\hat{z}_C(42)=20$, $\hat{z}_C(125)=31$, $\hat{z}_C(135)=82$ and also $b=2$ random locations $\hat{z}_C(150)=1$ and $\hat{z}_C(80)=100$. So $\hat{z}_C(x)=20L_3(x)+31L_4(x)+82L_5(x)+L_6(x)+100L_7(x)=123x^6+50x^5+80x^4+96x^3+169x^2+157x+49$

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

$\bar{W}:\mathbb{H}[>|x|+1]=\{42,125,135\}\to \mathbb{F}$

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

and $v_{\mathbb{H}[\leq |x|+1]}(h)$ is vanishing polynomial on $\mathbb{H}[\leq |x|+1]=\{1,59\}$, therefore $v_{\mathbb{H}[\leq |x|+1]}(h)=(h-1)(h-59)$. Also $\hat{x}(h)$ is the polynomial such that $\hat{x}(1)=1$ and $\hat{x}(59)=4$, therefore $\hat{x}(h)=128h+54$. Therefore,

$\bar{W}(42)=\frac{W(42)-\hat{x}(42)}{(42-1)(42-59)}=\frac{20-0}{(41)(-17)}\equiv 108\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $\bar{W}(125)=\frac{W(125)-\hat{x}(125)}{(125-1)(125-59)}=\frac{31-126}{(124)(66)}\equiv 160\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$ and $\bar{W}(135)=\frac{W(135)-\hat{x}(135)}{(135-1)(135-59)}=\frac{82-139}{(134)(76)}\equiv 78\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$.

Now, calculates $\hat{W}(x)\in\mathbb{F}^{<n_g+b=3+2}[x]$ so that $\hat{W}(42)=108$, $\hat{W}(125)=160$ , $\hat{W}(135)=78$ and also $b=2$ random locations $\hat{W}(150)=42$ and $\hat{W}(80)=180$. Therefore, $\hat{W}(x)=108L_1(x)+160L_2(x)+78L_3(x)+42L_4(x)+180L_5(x)$ where $L_1(x)=\frac{(x-125)(x-135)(x-150)(x-80)}{(-83)(-93)(-108)(-38)}\equiv 152x^4+92x^3+73x^2+43x+112\hspace{1mm}(\textrm{mod}\hspace{1mm}181)$, $L_2(x)=156x^4+39x^3+84x^2+86x+171$, $L_3(x)=161x^4+157x^3+131x^2+159x+6$, $L_4(x)=60x^4+67x^3+118x^2+50x+20$ and $L_5(x)=14x^4+7x^3+137x^2+24x+54$. Therefore, $\hat{W}(x)=149x^4+97x^3+161x^2+121x+166$.

3- The Prover finds polynomial $h_0(x)$ so that $\hat{z}_A(x)\hat{z}_B(x)-\hat{z}_C(x)=h_0(x)v_{\mathbb{H}}(x)$. Since $\hat{z}_A(x)\hat{z}_B(x)-\hat{z}_C(x)=$ $92x^{12}+45x^{11}+164x^{10}+x^9+20x^8+61x^7+152x^6+49x^5+180x^4+161x^3+28x^2+165x+149$ and $v_{\mathbb{H}}(x)=\prod_{h\in\mathbb{H}}(x-h)=(x-1)(x-59)(x-42)(x-125)(x-135)=x^5+180$, The Prover finds $h_0(x)=92x^7+45x^6+164x^5+x^4+20x^3+153x^2+16x+32$.

4- The Prover samples a fully random $s(x)\in\mathbb{F}^{<2|\mathbb{H}|+b-1=11}[x]$. Consider $s(x)=5x^{10}+101x^8+17x^7+x^5+20x^4+3x+115$. Then, the Prover computes sum $\sigma_1=\sum_{k\in \mathbb{H}}s(k)=$ $s(1)+s(59)+s(42)+s(125)+s(135)=81+47+141+46+109\equiv 62\hspace{1mm}(\hspace{1mm}\textrm{mod}\hspace{1mm}181)$.

5- The Prover sends $Com_{AHP_X}^2=\sum_{i=0}^{deg_{\hat{w}(x)}}\hat{w}_i\hspace{1mm}ck(i)=30$, $Com_{AHP_X}^{3}=\sum_{i=0}^{deg_{\hat{z}_A(x)}}\hat{z}_{A_i}ck(i)=160$, $Com_{AHP_X}^{4}=\sum_{i=0}^{deg_{\hat{z}_B(x)}}\hat{z}_{B_i}ck(i)=69$, $Com_{AHP_X}^{5}=\sum_{i=0}^{deg_{\hat{z}_C(x)}}\hat{z}_{C_i}ck(i)=11$, $Com_{AHP_X}^{6}=\sum_{i=0}^{deg_{h_0(x)}}h_{0_i}ck(i)=18$ and $Com_{AHP_X}^{7}=\sum_{i=0}^{deg_{s(x)}}s_i\hspace{1mm}ck(i)=178$.

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 $\alpha=hash(s(0))$, $\eta_A=hash(s(1))$, $\eta_B=hash(s(2))$, $\eta_C=hash(s(3))$. Let $\alpha=10$, $\eta_A=2$, $\eta_B=30$ and $\eta_C=100$.

7- The Prover finds polynomials $g_1(x)$ and $h_1(x)$ such 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 $r(x,y)=u_{\mathbb{H}}(x,y)=\frac{v_{\mathbb{H}}(x)-v_{\mathbb{H}}(y)}{x-y}$ , $v_{\mathbb{H}}(x)=\prod_{h\in \mathbb{H}}(x-h)=x^{|\mathbb{H}|}-1$. Therefore $r(x,y)=\frac{x^5-y^5}{x-y}$ . Also $r_M(x,y)=\sum_{k\in \mathbb{H}}r(x,k)\hat{M}(k,y)$ for $M\in \{A,B,C\}$.

Now, since $\sum_M\eta_M\hat{z}_M(x)=\eta_A\hat{z}_A(x)+\eta_B\hat{z}_B(x)+\eta_C\hat{z} _C(x)=98x^6+172x^5+120x^4+12x^3+104x^2+131x+87$and $r(\alpha,x)=r(10,x)=\frac{10^5-x^5}{10-x}$, so the second term of the left of equation $(1)$ is $r(\alpha,x)\sum_M\eta_M\hat{z}_M(x)=98x^{10}+66x^9+56x^8+29x^7+32x^6+153x^5+56x^4+136x^3+123x^2+42x+114$

Also, $\hat{z}(x)=\hat{W}(x)v_{\mathbb{H}[\leq |x|+1]}(x)+\hat{x}(x)=(149x^4+97x^3+161x^2+121x+166)(x-1)(x-59)+128x+54=149x^6+26x^5+55x^4+166x^3+52x^2+22x+74$ that agree with $z$ on $\mathbb{H}$. Also, $r_A(10,x)=\sum_{k\in \mathbb{H}}r(10,k)\hat{A}(k,x)$ where $\hat{A}(x,y)$ is a polynomial such that $\hat{A}(42,59)=1$, $\hat{A}(125,1)=1$, $\hat{A}(135,125)=1$ and $\hat{A}(x,y)=0$ for the rest of points in $\mathbb{H}\times\mathbb{H}$. So, $\hat{A}(x,y)$ is a bivariate polynomial that passes from these 25 points. This polynomial can obtain as following: $\hat{A}(x,y)=\sum_{k\in \mathbb{K}}u_{\mathbb{H}}(x,\hat{row}_{AHP_A}(k))u_{\mathbb{H}}(y,\hat{col}_{AHP_A}(k))\hat{val}_{AHP_A}(k)=\sum_{k\in \mathbb{K}}\frac{x^5-\hat{row}_{AHP_A}(k)^5}{x-\hat{row}_{AHP_A}(k)}\frac{y^5-\hat{col}_{AHP_A}(k)^5}{y-\hat{col}_{AHP_A}(k)}\hat{val}_{AHP_A}(k)$ Therefore, $\hat{A}(x,y)=5\frac{x^5-1}{x-42}\frac{y^5-1}{y-59}+5\frac{x^5-1}{x-125}\frac{y^5-1}{y-1}+132\frac{x^5-1}{x-135}\frac{y^5-1}{y-125}$. So $r_A(10,x)=\sum_{k\in\mathbb{H}}r(10,k)\hat{A}(k,x)=70\hat{A}(1,x)+13\hat{A}(59,x)+150\hat{A}(42,x)+26\hat{A}(125,x)+147\hat{A}(135,x)$where $\hat{A}(1,x)=\hat{A}(59,x)=0$, $\hat{A}(42,x)=5\times82\times\frac{x^5-1}{x-59}=48(x^4+59x^3+42x^2+125x+135)=48x^4+117x^3+25x^2+27x+145$, $\hat{A}(125,x)=5\times 29\times\frac{x^5-1}{x-1}=145(x^4+x^3+x^2+x+1)=145x^4+145x^3+145x^2+145x+145$, $\hat{A}(135,x)=132\times 114\times\frac{x^5-1}{x-125}=25(x^4+125x^3+59x^2+135x+42)=25x^4+48x^3+27x^2+117x+145$

Now, calculates $\hat{B}(x,y)$ similarly as following: $\hat{B}(x,y)=\sum_{k\in \mathbb{K}}u_{\mathbb{H}}(x,\hat{row}_{AHP_B}(k))u_{\mathbb{H}}(y,\hat{col}_{AHP_B}(k))\hat{val}_{AHP_B}(k)=\sum_{k\in \mathbb{K}}\frac{x^5-\hat{row}_{AHP_B}(k)^5}{x-\hat{row}_{AHP_B}(k)}\frac{y^5-\hat{col}_{AHP_B}(k)^5}{y-\hat{col}_{AHP_B}(k)}\hat{val}_{AHP_B}(k)$ Therefore, $\hat{B}(x,y)=117\frac{x^5-1}{x-42}\frac{y^5-1}{y-1}+55\frac{x^5-1}{x-125}\frac{y^5-1}{y-1}+29\frac{x^5-1}{x-125}\frac{y^5-1}{y-42}+68\frac{x^5-1}{x-135}\frac{y^5-1}{y-1}$. So, $r_B(10,x)=\sum_{k\in\mathbb{H}}r(10,k)\hat{B}(k,x)=70\hat{B}(1,x)+13\hat{B}(59,x)+150\hat{B}(42,x)+26\hat{B}(125,x)+147\hat{B}(135,x)$ where $\hat{B}(1,x)=\hat{B}(59,x)=0$, $\hat{B}(42,x)=117\times 82\times \frac{x^5-1}{x-1}=1\times(x^4+x^3+x^2+x+1)=x^4+x^3+x^2+x+1$, $\hat{B}(125,x)=55\times 29\times\frac{x^5-1}{x-1}+29\times 29\times\frac{x^5-1}{x-42}=147(x^4+x^3+x^2+x+1)+117(x^4+42x^3+135x^2+59x+125)=83x^4+174x^3+14x^2+172x+111$and $\hat{B}(135,x)=68\times 114\times \frac{x^5-1}{x-1}=150(x^4+x^3+x^2+x+1)=150x^4+150x^3+150x^2+150x+150$

Now, calculates $\hat{C}(x,y)$ similarly as following: $\hat{C}(x,y)=\sum_{k\in \mathbb{K}}u_{\mathbb{H}}(x,\hat{row}_{AHP_C}(k))u_{\mathbb{H}}(y,\hat{col}_{AHP_C}(k))\hat{val}_{AHP_C}(k)=\sum_{k\in \mathbb{K}}\frac{x^5-\hat{row}^5_{AHP_C}(k)}{x-\hat{row}_{AHP_C}(k)}\frac{y^5-\hat{col}^5_{AHP_C}(k)}{y-\hat{col}_{AHP_C}(k)}\hat{val}_{AHP_C}(k)$ Therefore, $\hat{C}(x,y)=114\frac{x^5-1}{x-42}\frac{y^5-1}{y-42}+82\frac{x^5-1}{x-125}\frac{y^5-1}{y-125}+5\frac{x^5-1}{x-135}\frac{y^5-1}{y-135}$. So, $r_C(10,x)=70\hat{C}(1,x)+13\hat{C}(59,x)+150\hat{C}(42,x)+26\hat{C}(125,x)+147\hat{C}(135,x)$ where $\hat{C}(1,x)=\hat{C}(59,x)=0$, $\hat{C}(42,x)=114\times 82\times\frac{x^5-1}{x-42}=117(x^4+42x^3+135x^2+59x+125)=117x^4+27x^3+48x^2+25x+145$, $\hat{C}(125,x)=82\times 29\times\frac{x^5-1}{x-125}=25(x^4+125x^3+59x^2+135x+42)=25x^4+48x^3+27x^2+117x+145$and $\hat{C}(135,x)=5\times 114\times\frac{x^5-1}{x-135}=27(x^4+135x^3+125x^2+42x+59)=27x^4+25x^3+117x^2+48x+145$

Therefore, the third term of the left of equation $(1)$ is

$(\sum_M \eta_M r_M(\alpha,x))\hat{z}(x)=(2r_A(10,x)+30r_B(10,x)+100r_C(10,x))\hat{z}(x)=(23x^4+72x^3+144x^2+10x+19)(149x^6+26x^5+55x^4+166x^3+52x^2+22x+74)=169x^{10}+104x^9+158x^8+161x^7+86x^6+57x^5+85x^4+43x^3+99x^2+72x+139$

Therefore, the left of equation $(1)$ is $s(x)+r(\alpha,x)\sum_M \eta_M \hat{z}M(x)-(\sum_M \eta_M r_M(\alpha,x))\hat{z}(x)=115x^{10}+143x^9+180x^8+66x^7+127x^6+97x^5+172x^4+93x^3+24x^2+154x+90$

Now, the Prover finds polynomials $g_1(x)$ and $h_1(x)$ such that $h_1(x)v_{\mathbb{H}}(x)+xg_1(x)+\frac{\sigma_1}{|\mathbb{H}|}=115x^{10}+143x^9+180x^8+66x^7+127x^6+97x^5+172x^4+93x^3+24x^2+154x+90$

that means, the Prover finds polynomials $g_1(x)$ and $h_1(x)$ such that $h_1(x)(x^5+180)+xg_1(x)+121=115x^{10}+143x^9+180x^8+66x^7+127x^6+97x^5+172x^4+93x^3+24x^2+154x+90$

In this case, polynomials $g_1(x)= 134x^3+92x^2+90x+100$ and $h_1(x)=115x^5+143x^4+180x^3+66x^2+127x+31$ are obtained. The Prover sends , $Com_{AHP_X}^{8}=\sum_{i=0}^{deg_{g_1(x)}}g_{1_i}ck(i)=129$ and $Com_{AHP_X}^{9}=\sum_{i=0}^{deg_{h_1(x)}}h_{1_i}ck(i)=33$ 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 $\beta_1=hash(s(8))$). Let $\beta_1=22$.

9- The Prover calculates $\sigma_2=\sum_{k\in\mathbb{H}}r(\alpha,k)\sum_{M}\eta_M\hat{M}(k,\beta_1)=r(10,1)\sum_{M}\eta_M\hat{M}(1,22)+r(10,59)\sum_{M}\eta_M\hat{M}(59,22)+r(10,42)\sum_{M}\eta_M\hat{M}(42,22)+r(10,125)\sum_{M}\eta_M\hat{M}(125,22)+r(10,135)\sum_{M}\eta_M\hat{M}(135,22)$ that is $\sigma_2=70\times 0+13\times 0+150\times 135+26\times 88+147\times 22=70$.

Then, the Prover finds $g_2(x)$ and $h_2(x)$ such that $r(\alpha,x)\sum_M \eta_M\hat{M}(x,\beta_1)=h_2(x)v_{\mathbb{H}}(x)+xg_2(x)+\frac{\sigma_2}{|\mathbb{H}|}$

where $r(\alpha,x)\sum_M\eta_M \hat{M}(x,\beta_1)=r(10,x)(2\hat{A}(x,22)+30\hat{B}(x,22)+100\hat{C}(x,22))$ where$\hat{A}(x,22)=5\frac{x^5-1}{x-42}\frac{22^5-1}{22-59}+5\frac{x^5-1}{x-125}\frac{22^5-1}{22-1}+132\frac{x^5-1}{x-135}\frac{22^5-1}{22-125}=159\frac{x^5-1}{x-42}+56\frac{x^5-1}{x-125}+114\frac{x^5-1}{x-135}=159(x^4+42x^3+135x^2+59x+125)+56(x^4+125x^3+59x^2+135x+42)+114(x^4+135x^3+125x^2+42x+59)=148x^4+108x^3+104x^2+9x+174$,$\hat{B}(x,22)=117\frac{x^5-1}{x-42}\frac{22^5-1}{22-1}+55\frac{x^5-1}{x-125}\frac{22^5-1}{22-1}+29\frac{x^5-1}{x-125}\frac{22^5-1}{22-42}+68\frac{x^5-1}{x-135}\frac{22^5-1}{22-1}=146x^4+37x^3+95x^2+100x+165$and $\hat{C}(x,22)=114\frac{x^5-1}{x-42}\frac{22^5-1}{22-42}+82\frac{x^5-1}{x-125}\frac{22^5-1}{22-125}+5\frac{x^5-1}{x-135}\frac{22^5-1}{22-135}=6(x^4+42x^3+135x^2+59x+125)+5(x^4+125x^3+59x^2+135x+42)+177(x^4+135x^3+125x^2+42x+59)=7x^4+156x^3+62x^2+137x$

Therefore, $r(\alpha,x)\sum_M \eta_M \hat{M}(x,\beta_1)=\frac{10^5-x^5}{10-x}(127x^4+93x^3+27x^2+66x+49)=(x^4+10x^3+100x^2+95x+45)(127x^4+93x^3+27x^2+66x+49)=127x^8+96x^7+82x^6+162x^5+40x^4+84x^3+77x^2+23x+33$

Hence, the Prover finds $g_2(x)=40x^3+30x^2+173x+105$ and $h_2(x)=127x^3+96x^2+82x+162$ such that $h_2(x)v_{\mathbb{H}}(x)+xg_2(x)+\frac{\sigma_2}{|\mathbb{H}|}=h_2(x)(x^5+180)+xg_2(x)+\frac{70}{5}=127x^8+96x^7+82x^6+162x^5+40x^4+84x^3+77x^2+23x+33$

The Prover sends , $Com_{AHP_X}^{10}=\sum_{i=0}^{deg_{g_2(x)}}g_{2_i}ck(i)=100$ and $Com_{AHP_X}^{11}=\sum_{i=0}^{deg_{h_2(x)}}h_{2_i}ck(i)=179$ 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. For example $\beta_2=80$.

11- The Prover calculates $\sigma_3=\sum_{k\in\mathbb{K}}(\sum_M \eta_M\frac{v_{\mathbb{H}}(\beta_2)v_{\mathbb{H}}(\beta_1)\hat{val'_M}(k)}{(\beta_2-\hat{row'_M}(k))(\beta_1-\hat{col'_M}(k))})=(2\frac{72\times 18\times 5}{38\times 144}+30\frac{72\times 18\times 117}{38\times 21}+100\frac{72\times 18\times 114}{38\times 161})+(2\frac{72\times 18\times 5}{136\times 21}+30\frac{72\times 18\times 55}{136\times 21}+100\frac{72\times 18\times 82}{136\times 78})+(2\frac{72\times 18\times 132}{126\times 78}+30\frac{72\times 18\times 29}{136\times 161}+100\frac{72\times 18\times 5}{126\times 68})+(30\frac{72\times 18\times 68}{126\times 21})=84$ Then, the Prover finds polynomials $g_3(x)$ and $h_3(x)$ such that $h_3(x)v_{\mathbb{K}}(x)=a(x)-b(x)(xg_3(x)+\frac{\sigma_3}{|\mathbb{K}|})$ where $a(x)=\sum_{M\in \{A,B,C\}} \eta_M v_{\mathbb{H}}(\beta_2)v_{\mathbb{H}}(\beta_1)\hat{val}_{AHP_M}(x)\prod_{N\in\{A,B,C\}-\{M\}}(\beta_2-\hat{row}_{AHP_N}(x))(\beta_1-\hat{col}_{AHP_N}(x))$and $b(x)=\prod_{M\in\{A,B,C\}}(\beta_2-\hat{row}_{AHP_M}(x))(\beta_1-\hat{col}_{AHP_M}(x))$.

Therefore, since $v_{\mathbb{H}}(\beta_1)=22^5+180=18$ and $v_{\mathbb{H}}(\beta_2)=80^5+180=72$ ,

$a(x)=2\times 72\times 18\hat{val}_{AHP_A}(x)(80-\hat{row}_{AHP_B}(x))(22-\hat{col}_{AHP_B}(x))(80-\hat{row}_C(x))(22-\hat{col}_{AHP_C}(x))+30\times72\times18\hat{val}_{AHP_B}(x)(80-\hat{row}_{AHP_A}(x))(22-\hat{col}_{AHP_A}(x))(80-\hat{row}_{AHP_C}(x))(22-\hat{col}_{AHP_C}(x))+100\times72\times 18\hat{val}_{AHP_C}(x)(80-\hat{row}_{AHP_A}(x))(22-\hat{col}_{AHP_A}(x))(80-\hat{row}_{AHP_B}(x))(22-\hat{col}_{AHP_B}(x))=63x^{25 }+ 56x^{24} + 45x^{23} + 27x^{22} + 81x^{21} + 80x^{20} + 66x^{19} + 152x^{18} + 97x^{17} + 55x^{16} + 170x^{15} + 142x^{14} + 3x^{13} + 96x^{12} + 11x^{11} + 69x^{10} + 43x^9 + 10x^8 + 75x^7 + 21x^6 + 109x^5 + 31x^4 + 90x^3 + 159x^2 + 119x + 79$

and

$b(x)=(80-\hat{row}_{AHP_A}(x))(22-\hat{col}_{AHP_A}(x))(80-\hat{row}_{AHP_B}(x))(22-\hat{col}_{AHP_B}(x))(80-\hat{row}_{AHP_C}(x))(22-\hat{col}_{AHP_C}(x))=23x^{30} + 176x^{29} + 75x^{28} + 63x^{27} + 174x^{26} + 13x^{25} + 43x^{24} + 169x^{23} + 35x^{22 }+ 66x^{21} + 160x^{20} + 99x^{19} + 150x^{18} + 31x^{17} + 69x^{16} + 83x^{15} + 93x^{14} + 179x^{13} + 30x^{12} + 111x^{11} + 58x^{10} + 131x^9 + 18x^8 + 36x^7 + 56x^6 + 59x^5 + 15x^4 + 171x^3 + 47x^2 + 110x + 142$

Therefore,

$g_3(x)= 110x^4 + 123x^3 + 161x^2 + 111x + 134$

and

$h_3(x)=4x^{29} + 74x^{28} + 65x^{27} + 16x^{26} + 139x^{25} + 135x^{24} + 92x^{23} + 139x^{22} + 60x^{21} + 75x^{20} + 75x^{19} + 99x^{18} + 98x^{17} + 71x^{16} + 15x^{15} + 53x^{14} + 138x^{13} + 128x^{12} + 18x^{11} + 147x^{10} + 111x^9 + 37x^8 + 18x^7 + 97x^6 + 143x^5 + 136x^4 + 53x^3 + 50x^2 + 177x + 99$

The Prover sends , $Com_{AHP_X}^{12}=\sum_{i=0}^{deg_{g_3(x)}}g_{3_i}ck(i)=169$ and $Com_{AHP_X}^{13}=\sum_{i=0}^{deg_{h_3(x)}}h_{3_i}ck(i)=166$ 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 $h_3(x)$.

12- The Prover sends $\pi_{AHP}^1=62$, $\pi_{AHP}^2=(166,121,161,97,149)$, $\pi_{AHP}^3=(168,141,45,26,63,165,116)$, $\pi_{AHP}^4=(124,81,137,101,71,178,32)$, $\pi_{AHP}^5=(49,157,169,96,80,50,123)$, $\pi_{AHP}^6=(32,16,153,20,1,164,45,92)$ and $\pi_{AHP}^7=(115,3,0,0,20,1,0,17,101,0,5)$ to the Verifier that are value of $\sigma_1$, coefficients of polynomials $\hat{W}(x)$, $\hat{z}_A(x)$, $\hat{z}_B(x)$, $\hat{z}_C(x)$,$h_0(x)$ and $s(x)$, respectively.

13- The Prover sends $\pi_{AHP}^8=(100,90,92,134)$ and $\pi_{AHP}^{9}=(31,127,66,180,143,115)$ to the Verifier that are coefficients of polynomials $g_1(x)$ and $h_1(x)$, respectively.

14-The Prover sends $\pi_{AHP}^{10}=70$, $\pi_{AHP}^{11}=(105,173,30,40)$ and $\pi_{AHP}^{12}=(162,82,96,127)$ that are value of $\sigma_2$ and coefficients of polynomials $g_2(x)$ and $h_2(x)$, respectively.

15- The Prover sends $\pi_{AHP}^{13}=84$, $\pi_{AHP}^{14}=(134,111,161,123,110)$ and that are value of $\sigma_3$ and coefficients of polynomials $g_3(x)$ and $h_3(x)$, respectively.

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 $\mathbb{F}$. For example, $\eta_{\hat{w}}=1$, $\eta_{\hat{z}_A}=4$, $\eta_{\hat{z}_B}=10$, $\eta_{\hat{z}_C}=8$, $\eta_{h_0}=32$, $\eta_s=45$, $\eta_{g_1}=92$, $\eta_{h_1}=11$, $\eta_{g_2}=1$, $\eta_{h_2}=5$, $\eta_{g_3}=25$ and $\eta_{h_3}=63$.

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

The Prover obtains

$p(x)=71x^{29} + 137x^{28} + 113x^{27} + 103x^{26} + 69x^{25} + 179x^{24} + 4x^{23} + 69x^{22} + 160x^{21} + 19x^{20} + 19x^{19} + 83x^{18} + 20x^{17} + 129x^{16} + 40x^{15} + 81x^{14} + 6x^{13} + 100x^{12} + 48x^{11} + 74x^{10} + 115x^9 + 179x^8 + 137x^7 + 88x^6 + 126x^5 + 8x^4 + 124x^3 + 37x^2 + 72x + 114$

18- The Prover calculates $p(x)$ in $x=x'$ (value of $x'$ is received from the Verifier), then puts it in $\pi_{AHP}^{16}$ . Therefore $\pi_{AHP}^{16}=p(x')=y'$. For example, if $x'=2$, then $\pi_{AHP}^{16}=p(2)=119$.

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 builds polynomial $q(x)=\frac{p(x)-y'}{x-x'}=\frac{p(x)-119}{x-2}=71x^{28} + 98x^{27} + 128x^{26} + 178x^{25} + 63x^{24} + 124x^{23} + 71x^{22} + 30x^{21} + 39x^{20} + 97x^{19} + 32x^{18} + 147x^{17} + 133x^{16} + 33x^{15} + 106x^{14} + 112x^{13} + 49x^{12} + 17x^{11} + 82x^{10} + 57x^9 + 48x^8 + 94x^7 + 144x^6 + 14x^5 + 154x^4 + 135x^3 + 32x^2 + 101x + 93$ and calculates $\pi_{AHP}^{17}=gq(\tau)$ by using $ck$ as following: $\pi_{AHP}^{17}=\sum_{i=0}^{deg_{q(x)}}q_i\hspace{1mm}ck(i)=149$ where $q_i$ is the coefficient of $x^i$ of $q(x)$.

Proof 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]

    {
    "commitmentId": 64-bit,
    "class": 32-bit Integer,
    "input": 4
    "output": 82    
       
    "P_AHP1": 62
    "P_AHP2": [166,121,161,97,149],
    "P_AHP3": [168,141,45,26,63,165,116],
    "P_AHP4": [124,81,137,101,71,178,32],  
    "P_AHP5": [49,157,169,96,80,50,123], 
    "P_AHP6": [32,16,153,20,1,164,45,92],
    "P_AHP7": [115,3,0,0,20,1,0,17,101,0,5],
    "P_AHP8": [100,90,92,134],
    "P_AHP9": [31,127,66,180,143,115],
    "P_AHP10": 70
    "P_AHP11": [105,173,30,40],
    "P_AHP12": [162,82,96,127],
    "P_AHP13": 84
    "P_AHP14": [134,111,161,123,110],
    "P_AHP15": [99,177,50,53,136,143,97,18,37,111,147,18,128,138,53,15,71,98,99,75,75,60,139,92,135,139,16,65,74,4]
    "P_AHP16": 119
    "P_AHP17": 149
      
    "Com_AHP1_x": 4,
    "Com_AHP2_x": 30,
    "Com_AHP3_x": 160,
    "Com_AHP4_x": 69,
    "Com_AHP5_x": 11,
    "Com_AHP6_x": 18,  
    "Com_AHP7_x": 178,
    "Com_AHP8_x": 129,
    "Com_AHP9_x": 33,
    "Com_AHP10_x": 100,
    "Com_AHP11_x": 179,
    "Com_AHP12_x": 169,
    "Com_AHP13_x": 166
}