## 3 Development of an Imitation-Resistant Zero-Knowledge Authentication Protocol Based on an MPPC
The developed algorithms for the error bursts correction and expansion of the set of bases in the MPPC will be used during the execution of the authentication process for the satellite RSU. Let us consider an execution of the developed ZKAP. The preliminary stage of the protocol is as follows:  
\[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1,\ K=325,\ D(0)=408,\ L(0)=878. \tag{34}\]  
\[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1. \tag{35}\]  
The generation of the session key \(D(n)\) and the parameter \(L(n)\), where \(n=1\), is as follows:  
\[D_{1}(1)=\left|x^{\frac{1}{D(0)+K}}\right|_{m_{1}(x)}=\left|x^{\frac{1}{408+3 25}}\right|_{m_{1}(x)}=\left|x^{\frac{1}{20}}\right|_{m_{2}(x)}=\left|x^{11} \right|_{x^{5}+x^{3}+1}=01111=15. \tag{36}\]  
\[D_{2}(1)=\left|x^{\frac{1}{D(0)+K}}\right|_{m_{2}(x)}=\left|x^{\frac{1}{408+3 25}}\right|_{m_{2}(x)}=\left|x^{\frac{1}{20}}\right|_{m_{2}(x)}=\left|x^{11} \right|_{x^{5}+x^{2}+1}=00111=7. \tag{37}\]  
\[L_{1}(1)=\left|x^{\frac{1}{L(0)+K}}\right|_{m_{1}(x)}=\left|x^{\frac{1}{878+3 25}}\right|_{m_{1}(x)}=\left|x^{\frac{1}{2}}\right|_{m_{1}(x)}=\left|x^{6} \right|_{x^{5}+x^{3}+1}=10010=18. \tag{38}\]  
\[L_{2}(1)=\left|x^{\frac{1}{L(0)+K}}\right|_{m_{2}(x)}=\left|x^{\frac{1}{878+3 25}}\right|_{m_{2}(x)}=\left|x^{\frac{1}{2}}\right|_{m_{2}(x)}=\left|x^{6} \right|_{x^{5}+x^{2}+1}=01010=8. \tag{39}\]  
The representation as a modular code is as follows:  
\[SRSU:\ K=325=0101000101=(10,\,5),\,D(1)=(15,\,7),\,L(1)=(18,\,10). \tag{40}\]  
The computation of the SRSU's true digest in the MPPC is as follows:  
\[SRSU:\ U_{1}^{1}(x)=\left|x^{K_{1}}x^{D_{1}(1)}x^{L_{1}(1)}\right|_{m_{1}(x) }=\left|x^{10}x^{15}x^{18}\right|_{m_{1}(x)}=x^{4}+x^{3}+x^{2}+x=11110=30. \tag{41}\]  
\[SRSU:\ U_{2}^{1}(x)=\left|x^{K_{2}}x^{D_{2}(1)}x^{L_{2}(1)}\right|_{m_{2}(x) }=\left|x^{5}x^{7}x^{10}\right|_{m_{2}(x)}=x^{4}+x^{2}+1=10101=25. \tag{42}\]  
The selection of numbers for the distortion of secret parameters is as follows:  
\[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1,\ K=325,\ D(0)=408,\ L(0)=878. \tag{34}\]0  
The distortion of secret parameters of the protocol is as follows:  
\[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1,\ K=325,\ D(0)=408,\ L(0)=878. \tag{34}\]1  
\[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1,\ K=325,\ D(0)=408,\ L(0)=878. \tag{34}\]2 \[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1,\ K=325,\ D(0)=408,\ L(0)=878. \tag{34}\]3  
The computation of the SRSU's distorted digest in the MPPC is as follows:\[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1,\ K=325,\ D(0)=408,\ L(0)=878. \tag{34}\]4 \[CA\to SRSU:\ m_{1}(x)=x^{5}+x^{3}+1,\ m_{2}(x)=x^{5}+x^{2}+1,\ K=325,\ D(0)=408,\ L(0)=878. \tag{34}\]5  
The selection of the request and its transmission to the SRSU are as follows:  
\[ \begin{array}{l}{{OBU:V(1)=107=1101011=\left(x^{3},\,x^{2},\,x^{4}+x^{3}+x, \,x^{2}+1,\,x^{4}+x^{3}+x^{2}+1,\,x^{4}+1\right)}}\\ {{=(01000,\,00100,\,11010,\,00101,\,11101,\,10001)=(8,\,4,\,27,\,5,\,29,\,17).}} \\ {{OBU\to SRSU:V(1)=(8,\,4,\,27,\,5,\,29,\,17).}} \\ \end{array}  \tag{46}\]  
The translation is performed using a PC-MPPC converter. The error-checking for the request is as follows:  
\[\begin{array}{l}{{|V_{1}(1)B_{1}(x)|_{M_{2}(x)}+|V_{2}(1)B_{2}(x)|_{M_{2}(x) }=\left(x^{3},\,0,\,x^{3}+x^{2},\,x^{3}+1,\,x^{2}+x,\,x^{3}+x^{2}+x\right  
\[U_{6}^{1}(x)=00101+10111=10010=18_{10}. \tag{62}\]  
Then, the SRSU's true digest is of the form \(U^{1}(x)=(30,\,21,\,23,\,9,\,15,\,18)\). The SRSU's distorted digest is \(\hat{U}^{1}(x)=(29,\,19,\,16,\,5,\,31,\,24)\). The responses are  
\[Y^{1}(1) =(26,\,23,\,14,\,7,\,0,\,24). \tag{63}\] \[Y^{2}(1) =(29,\,18,\,2,\,6,\,23,\,14).\] (64) \[Y^{3}(1) =(13,\,12,\,31,\,14,\,5,\,27). \tag{65}\]  
The response signal is transmitted to the OBU. Let us suppose that the error distorts the first residue of the SRSU's true digest. \(\widetilde{U}_{1}(x)=111\bar{0}0=28\) is obtained instead of a residue \(U_{1}(x)=11110\).