> For the complete documentation index, see [llms.txt](https://fidesinnova.gitbook.io/docs/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://fidesinnova.gitbook.io/docs/fides-zero-knowledge-proof-zkp-algorithm/2-commitment-phase/2-3-example-2.md).

# Example 2

Assume the following sample code:\
\
\
$$add\hspace{2mm} R\_1, R\_1, 5$$ => Gate 1\
$$mul\hspace{2mm} R\_2, R\_2, R\_{10}$$ => Gate 2\
$$add\hspace{2mm} R\_2,R\_2,10$$ => Gate 3\
$$mul\hspace{2mm} R\_1,R\_1,R\_{19}$$ => Gate 4

The constraints are as follows considering $$p= 1678321$$:

\
$$R\_1^{(2)}=R\_1^{(1)}+5$$\
$$R\_2^{(2)}=R\_2^{(1)}\times R\_{10}^{(1)}$$\
$$R\_2^{(3)}=R\_2^{(2)}+10$$\
$$R\_1^{(3)}=R\_1^{(2)}\times R\_{19}^{(1)}$$

We continue with $$n\_g=4$$ and $$n\_i=32$$.

The Prover calculates square matrices $$A$$, $$B$$ and $$C$$ of order $$n\_g+n\_i+1=4+32+1=37$$ based on above construction:

$$$
A=\begin{bmatrix}
0&0&0&...&0&0&0&0\\
0&0&0&...&0&0&0&0\\
.&.&.&...&.&.&.&.\\
.&.&.&...&.&.&.&.\\
.&.&.&...&.&.&.&.\\
.&.&.&...&.&.&.&.\\
1&0&0&...&0&0&0&0\\
0&0&1&...&0&0&0&0\\
1&0&0&...&0&0&0&0\\
0&0&0&...&1&0&0&0
\end{bmatrix}
$$,
$$$

B=\begin{bmatrix} 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&...&0&0&0&0\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&...&0&0&0&0\ .&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&...&.&.&.&.\ .&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&...&.&.&.&.\ .&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&...&.&.&.&.\ .&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&.&...&.&.&.&.\ 5&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&...&0&0&0&0\ 0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&..&0&0&0&0\ 10&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&...&0&1&0&0\ 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&...&0&0&0&0 \end{bmatrix}

C=\begin{bmatrix} 0&0&...&0&0&0&0\ 0&0&...&0&0&0&0\ .&.&...&.&.&.&.\ .&.&...&.&.&.&.\ .&.&...&.&.&.&.\ .&.&...&.&.&.&.\ 0&0&...&1&0&0&0\ 0&0&...&0&1&0&0\ 0&0&...&0&0&1&0\ 0&0&...&0&0&0&1 \end{bmatrix}

$$$
At we see, the matrices $$A$$ and $$B$$ are $$33-SLT$$ and matrix $$C$$ is $$33-Diag$$. Also $$Az\hspace{1mm}o\hspace{1mm}Bz=Cz$$.

We consider multiplicative subgroup $$\mathbb{H}$$ of order $$n=n\_g+n\_i+1=37$$ with generator\
 $$\omega=11^{\frac{p-1}{37}}\equiv 1101389 (\textrm{mod}\hspace{1mm}p)$$.  Therefore, $$\mathbb{H}=$$ { $$1,\omega,\omega^2,\omega^3,\omega^4,...,\omega^{36}$$ } = { $$1, 1101389, 876941, 1253722, 373750, 668759, 747302, 1444226, 727349, 1364683, 1298322, 1224122, 644775, 127745, 1610054, 257937,$$\
$$1257144, 1496263, 722913, 1195510, 1536124,1669124, 870923, 784349, 1262357, 1500979, 469942, 1466322, 1364193, 358753, 1173208,$$

$$586481, 1605555, 1190739, 1632256, 159545, 899305$$ }

Note that based on construction for matrices $$A$$, $$B$$ and $$C$$, we can conclude that the most number nonzero entries in $$A$$ and $$C$$ is $$n\_g$$ and in $$B$$ is $$2n\_g$$, therefore we can consider $$m=2n\_g$$.\
Therefore, consider multiplicative subgroup $$\mathbb{K}$$ of order $$m=2n\_g=8$$ with generator $$\gamma=11^{\frac{p-1}{8}}\equiv  216  (\textrm{mod}\hspace{1mm}p)$$. Therefore, $$\mathbb{K}={1,\gamma,\gamma^2,...,\gamma^{7}}={1,216, 46656, 7770, 1678320, 1678105, 1631665, 1670551 }$$.

###  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 $$row\_{AHP\_A}:\mathbb{K}\to\mathbb{H}$$ with $$row\_{AHP\_A}(k=\gamma^i)=\omega^{r\_i}$$ for $$1\leq i\leq ||A||$$ , and otherwise $$row\_{AHP\_A}(k)$$ returns an arbitrary element in $$\mathbb{H}$$. \
So $$row\_{AHP\_A}(k)$$ on $$\mathbb{K}$$ is a polynomial so that $$row\_{AHP\_A}(1)=\omega^{33}$$ , $$row\_{AHP\_A}(\gamma)=\omega^{34}$$, $$row\_{AHP\_A}(\gamma^2)=\omega^{35}$$ , $$row\_{AHP\_A}(\gamma^3)=\omega^{36}$$  and otherwise $$row\_{AHP\_A}(k)$$ returns arbitrary elements of $$\mathbb{H}$$, for example $$row\_{AHP\_A}(\gamma^i)=\omega^i$$ for $$4\leq i\leq 7$$. Therefore, 

$$row\_{AHP\_A}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=739930x^7 + 100821x^6 + 664621x^5 + 147929x^4 + 1066695x^3 + 902750x^2 + 454730x + 469905$$

Also,  $$col\_{AHP\_A}:\mathbb{K}\to\mathbb{H}$$ with $$col\_{AHP\_A}(k=\gamma^i)=\omega^{c\_i}$$ for $$1\leq i\leq ||A||$$ and otherwise $$col\_{AHP\_A}(k)$$ returns an arbitrary element in $$\mathbb{H}$$. \
So, $$col\_{AHP\_A(}k)$$ on $$\mathbb{K}$$ is a polynomial so that $$col\_{AHP\_A}(1)=1$$ , $$col\_{AHP\_A}(\gamma)=\omega^2$$, $$col\_{AHP\_A}(\gamma^2)=1$$ , $$col\_{AHP\_A}(\gamma^3)=\omega^{33}$$  and  otherwise  $$col\_{AHP\_A}(k)$$ returns arbitrary elements of $$\mathbb{H}$$, for example $$col\_{AHP\_A}(\gamma^i)=\omega^i$$ for $$4\leq i\leq 7$$. Therefore, $$col\_{AHP\_A}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=344425x^7 + 1483132x^6 + 490249x^5 + 246919x^4 + 662128x^3 + 101801x^2 + 833805x + 872505$$

and ,  $$val\_{AHP\_A}:\mathbb{K}\to\mathbb{H}$$ with $$val\_{AHP\_A}(k=\gamma^i)=\frac{v\_i}{u\_{\mathbb{H}}(row\_{AHP\_A}(k),row\_{AHP\_A}(k))u\_{\mathbb{H}}(col\_{AHP\_A}(k),col\_{AHP\_A}(k))}$$ for $$1\leq i\leq ||A||=4$$ where $$v\_i$$ is value of  $$i^{th}$$ nonzero entry and otherwise $$val\_{AHP\_A}(k)$$ returns zero. Note that based on definition of $$u\_{\mathbb{H}}(x,y)$$, for each $$x \in \mathbb{H}$$, $$u\_{\mathbb{H}}(x,x)=|\mathbb{H}|x^{|\mathbb{H}|-1}=37x^{36}$$. So $$val\_{AHP\_A}(1)=\frac{1}{(37row\_{AHP\_A}^{36}(1))(37col\_{AHP\_A}^{36}(1))}=\frac{1}{37\times (\omega^{33})^{36}\times37\times1}$$, $$val\_{AHP\_A}(\gamma)=\frac{2}{(37row\_{AHP\_A}^{36}(\gamma))(37col\_{AHP\_A}^{36}(\gamma))}=\frac{2}{37\times (\omega^{34})^{36}\times37\times(\omega^2)^{36}}$$ , $$val\_{AHP\_A}(\gamma^2)=\frac{1}{(37row\_{AHP\_A}^{36}(\gamma^2))(37col\_{AHP\_A}^{36}(\gamma^2))}=\frac{1}{37\times (\omega^{35})^{36}\times37\times1}$$, $$val\_{AHP\_A}(\gamma^3)=\frac{7}{(37row\_{AHP\_A}^{36}(\gamma^3))(37col\_{AHP\_A}^{36}(\gamma^3))}=\frac{7}{37\times (\omega^{36})^{36}\times37\times(\omega^{33})^{36}}$$ ,\
 and $$val\_{AHP\_A}(k)=0$$ for $$k\in \mathbb{K}-{1,\gamma,\gamma^2,\gamma^3}$$. Therefore $$val\_{AHP\_A}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=1349736x^7 + 870027x^6 + 189189x^5 + 1024786x^4 + 201373x^3 + 194896x^2 + 1568354x + 1218943$$

Now, $$\hat{row}*{AHP\_A}$$, $$\hat{col}*{AHP\_A}$$ and $$\hat{val}*{AHP\_A}$$ are extensions of $$row*{AHP\_A}$$, $$col\_{AHP\_A}$$ and $$val\_{AHP\_A}$$ so that are agree on $$\mathbb{K}$$. 

Similarly, $$row\_{AHP\_B}:\mathbb{K}\to\mathbb{H}$$ so that  $$row\_{AHP\_B}(1)=\omega^{33}$$, $$row\_{AHP\_B}(\gamma)=\omega^{33}$$,  $$row\_{AHP\_B}(\gamma^2)=\omega^{34}$$ ,   $$row\_{AHP\_B}(\gamma^3)=\omega^{35}$$, $$row\_{AHP\_B}(\gamma^4)=\omega^{35}$$,  $$row\_{AHP\_B}(\gamma^5)=\omega^{36}$$ and for the rest values of $$\mathbb{K}$$,  $$row\_{AHP\_B}(k)$$ returns arbitrary elements of $$\mathbb{H}$$, for example  $$row\_{AHP\_B}(\gamma^i)=\omega^i$$ for $$6\leq i\leq 7$$.  Therefore $$row\_{AHP\_B}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=745068x^7 + 1550888x^6 + 311204x^5 + 1053454x^4 + 781197x^3 + 709275x^2 + 356449x + 718167$$

Also,  $$col\_{AHP\_B}:\mathbb{K}\to\mathbb{H}$$ so that $$col\_{AHP\_B}(1)=1$$ , $$col\_{AHP\_B}(\gamma)=\omega$$, $$col\_{AHP\_B}(\gamma^2)=\omega^{10}$$,  $$col\_{AHP\_B}(\gamma^3)=1$$ , $$col\_{AHP\_B}(\gamma^4)=\omega^{34}$$, $$col\_{AHP\_B}(\gamma^5)=1$$  and for the rest values of $$\mathbb{K}$$ , $$col\_{AHP\_B}(k)$$ returns arbitrary elements of $$\mathbb{H}$$, for example, $$col\_{AHP\_B}(\gamma^i)=\omega^i$$ for $$6\leq i\leq 7$$.  Therefore $$col\_{AHP\_B}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=757883x^7 + 1334841x^6 + 941775x^5 + 1041045x^4 + 898965x^3 + 1498879x^2 + 781052x + 1137166$$

and ,  $$val\_{AHP\_B}:\mathbb{K}\to\mathbb{H}$$ so that  $$val\_{AHP\_B}(1)=\frac{5}{(37row\_{AHP\_B}^{36}(1))(37col\_{AHP\_B}^{36}(1))}=\frac{5}{37\times (\omega^{33})^{36}\times37\times1}=$$, $$val\_{AHP\_B}(\gamma)=\frac{1}{(37row\_{AHP\_B}^{36}(\gamma))(37col\_{AHP\_B}^{36}(\gamma))}=\frac{1}{37\times (\omega^{33})^{36}\times37\times(\omega)^{36}}$$ , \
$$val\_{AHP\_B}(\gamma^2)=\frac{1}{(37row\_{AHP\_B}^{36}(\gamma^2))(37col\_{AHP\_B}^{36}(\gamma^2))}=\frac{1}{37\times (\omega^{34})^{36}\times37\times1}$$ , $$val\_{AHP\_B}(\gamma^3)=\frac{10}{(37row\_{AHP\_B}^{36}(\gamma^3))(37col\_{AHP\_B}^{36}(\gamma^3))}=\frac{10}{37\times (\omega^{35})^{36}\times37\times1}$$, $$val\_{AHP\_B}(\gamma^4)=\frac{1}{(37row\_{AHP\_B}^{36}(\gamma^4))(37col\_{AHP\_B}^{36}(\gamma^4))}=\frac{1}{37\times (\omega^{35})^{36}\times37\times(\omega^{34})^{36}}$$ , $$val\_{AHP\_B}(\gamma^5)=\frac{1}{(37row\_{AHP\_B}^{36}(\gamma^5))(37col\_{AHP\_B}^{36}(\gamma^5))}=\frac{1}{37\times (\omega^{36})^{36}\times37\times1}$$ , \
and $$val\_{AHP\_B}(k)=0$$ for $$k\in \mathbb{K}-{1,\gamma,\gamma^2,\gamma^3,\gamma^4,\gamma^5}$$. Therefore $$val\_{AHP\_B}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=1592245x^7 + 169676x^6 + 1573426x^5 + 1612428x^4 + 256156x^3 + 1249861x^2 + 995621x + 462292$$

Now, $$\hat{row}*{AHP\_B}$$, $$\hat{col}*{AHP\_B}$$ and $$\hat{val}*{AHP\_B}$$ are extensions of $$row*{AHP\_B}$$, $$col\_{AHP\_B}$$ and $$val\_{AHP\_B}$$ so that are agree on $$\mathbb{K}$$. 

Similarly, $$row\_{AHP\_C}:\mathbb{K}\to\mathbb{H}$$ so that  $$row\_{AHP\_C}(1)=\omega^{33}$$, $$row\_{AHP\_C}(\gamma)=\omega^{34}$$,  $$row\_{AHP\_C}(\gamma^2)=\omega^{35}$$, $$row\_{AHP\_C}(\gamma^3)=\omega^{36}$$ and for the rest values of $$\mathbb{K}$$,  $$row\_{AHP\_C}(k)$$ returns arbitrary elements of $$\mathbb{H}$$, for example  $$row\_{AHP\_C}(\gamma^i)=\omega^{i}$$ for $$4\leq i\leq 7$$. Therefore $$row\_{AHP\_C}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=739930x^7 + 100821x^6 + 664621x^5 + 147929x^4 + 1066695x^3 + 902750x^2 + 454730x + 469905$$

Also,  $$col\_{AHP\_C}:\mathbb{K}\to\mathbb{H}$$ so that $$col\_{AHP\_C}(1)=\omega^{33}$$ , $$col\_{AHP\_C}(\gamma)=\omega^{34}$$,  $$col\_{AHP\_C}(\gamma^2)=\omega^{35}$$, $$col\_{AHP\_C}(\gamma^3)=\omega^{36}$$  and for the rest values of $$\mathbb{K}$$ , $$col\_{AHP\_C}(k)$$ returns arbitrary elements of $$\mathbb{H}$$, for example  $$col\_{AHP\_C}(\gamma^i)=\omega^i$$ for $$4\leq i\leq 7$$. Therefore $$col\_{AHP\_C}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=739930x^7 + 100821x^6 + 664621x^5 + 147929x^4 + 1066695x^3 + 902750x^2 + 454730x + 469905$$

and ,  $$val\_{AHP\_C}:\mathbb{K}\to\mathbb{H}$$ so that \
 $$val\_{AHP\_C}(1)=\frac{1}{(37row\_{AHP\_C}^{36}(1))(37col\_{AHP\_C}^{36}(1))}=\frac{1}{37\times (\omega^{33})^{36}\times37\times(\omega^{33})^{36}}$$, $$val\_{AHP\_C}(\gamma)=\frac{1}{(37row\_{AHP\_C}^{36}(\gamma))(37col\_{AHP\_C}^{36}(\gamma))}=\frac{1}{37\times (\omega^{34})^{36}\times37\times(\omega^{34})^{36}}$$ , $$val\_{AHP\_C}(\gamma^2)=\frac{1}{(37row\_{AHP\_C}^{36}(\gamma^2))(37col\_{AHP\_C}^{36}(\gamma^2))}=\frac{1}{37\times (\omega^{35})^{36}\times37\times(\omega^{35})^{36}}$$ ,$$val\_{AHP\_C}(\gamma^3)=\frac{1}{(37row\_{AHP\_C}^{36}(\gamma^3))(37col\_{AHP\_C}^{36}(\gamma^3))}=\frac{1}{37\times (\omega^{36})^{36}\times37\times(\omega^{36})^{36}}$$ \
 and $$val\_{AHP\_C}(k)=0$$ for $$k\in \mathbb{K}-{1,\gamma,\gamma^2,\gamma^3}$$. Therefore $$val\_{AHP\_C}(x)=\sum\_{i=1}^{8}y\_iL\_i(x)=435354x^7 + 1188860x^6 + 1158417x^5 + 525698x^4 + 1126753x^3 + 467855x^2 + 544916x + 1083027$$

Now, $$\hat{row}*{AHP\_C}$$, $$\hat{col}*{AHP\_C}$$ and $$\hat{val}*{AHP\_C}$$ are extensions of $$row*{AHP\_C}$$, $$col\_{AHP\_C}$$ and $$val\_{AHP\_C}$$ so that are agree on $$\mathbb{K}$$. 

Therefore, 

$$\overrightarrow{O}\_{AHP}=469905,  454730, 902750, 1066695,  147929, 664621, 100821, 739930,872505 ,$$\
$$833805, 101801, 662128,246919, 490249, 1483132, 344425,1218943,1568354, 194896$$\
$$201373, 1024786, 189189,870027, 1349736,718167,356449, 709275, 781197,1053454,$$\
$$311204,1550888,745068, 1137166,  781052,  1498879, 898965, 1041045, 941775, 1334841,$$\
$$757883,462292, 995621, 1249861,256156, 1612428, 1573426, 169676,1592245,469905,$$\
$$454730,902750, 1066695,147929,664621,100821,739930,469905,454730, 902750,$$\
$$1066695, 147929,664621,100821, 739930, 1083027, 544916,467855,1126753,525698,$$\
$$1158417,1188860,435354)$$

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\
$$Com\_{AHP}^0=\sum\_{i=0}^{5}\hat{row}*{AHP*{A\_i}}\hspace{1.1mm}ck(i)=$$\\

and similarly

$$Com\_{AHP}^1=$$, $$Com\_{AHP}^2=$$, $$Com\_{AHP}^3=$$, $$Com\_{AHP}^4=$$, $$Com\_{AHP}^5=$$, $$Com\_{AHP}^6=$$, $$Com\_{AHP}^7=$$ and $$Com\_{AHP}^8=$$.

## 2-3- PFR and AHP Commitment.json File Format

```
{
    "commitmentId": String,
    "deviceType": String,
    "deviceIdType": String,
    "deviceModel": String,
    "manufacturer": String,
    "softwareVersion": String,
    "class":2,
    "m": 8,
    "n": 37,
    "p": 1678321,
    "g": 11,   
        
    // PFR Commitment   
    "row_PFR_A": 64-bit Array,
    "col_PFR_A": 64-bit Array,
    "val_PFR_A": 64-bit Array,
    "row_PFR_B": 64-bit Array,
    "col_PFR_B": 64-bit Array,
    "val_PFR_B": 64-bit Array,
    "row_PFR_C": 64-bit Array,
    "col_PFR_C": 64-bit Array,
    "val_PFR_C": 64-bit Array,
    
    "Com_PFR0": 64-bit Integer,
    "Com_PFR1": 64-bit Integer,
    "Com_PFR2": 64-bit Integer,
    "Com_PFR3": 64-bit Integer,
    "Com_PFR4": 64-bit Integer,
    "Com_PFR5": 64-bit Integer,
    "Com_PFR6": 64-bit Integer,
    "Com_PFR7": 64-bit Integer,
    "Com_PFR8": 64-bit Integer,   

    // AHP Commitment
    "row_AHP_A": [469905, 454730, 902750, 1066695, 147929, 664621, 100821, 739930]
    "col_AHP_A": [ 872505, 833805, 101801, 662128,246919, 490249, 1483132, 344425]
    "val_AHP_A": [1218943,1568354, 194896,201373, 1024786, 189189,870027, 1349736]
    "row_AHP_B": [718167,356449, 709275, 781197,1053454, 311204,1550888,745068]
    "col_AHP_B": [1137166, 781052, 1498879, 898965, 1041045, 941775, 1334841,757883]
    "val_AHP_B": [462292, 995621, 1249861,256156, 1612428, 1573426, 169676,1592245]
    "row_AHP_C": [469905, 454730,902750, 1066695,147929,664621,100821,739930]
    "col_AHP_C": [469905,454730, 902750,1066695, 147929,664621,100821, 739930]
    "val_AHP_C": [1083027, 544916,467855,1126753,525698, 1158417,1188860,435354]
    "Com_AHP0": 64-bit Integer,
    "Com_AHP1": 64-bit Integer,
    "Com_AHP2": 64-bit Integer,
    "Com_AHP3": 64-bit Integer,
    "Com_AHP4": 64-bit Integer,
    "Com_AHP5": 64-bit Integer,
    "Com_AHP6": 64-bit Integer,
    "Com_AHP7": 64-bit Integer,
    "Com_AHP8": 64-bit Integer,    
    
    "curve": String,
    "polynomial_commitment": String
}
```

$$$