I have a complicated system of equations and I would like your help to provide conditions such that it has a unique solution.
In alternative, any direction on how to rewrite the system in a nicer way would be really useful and I will reward it.
The system is the following: $$ \begin{aligned} d_k(Z_1,Z_2,..., Z_{11})=& \sum_{R_1=0}^1\sum_{R_2=0}^1 ... \sum_{R_{11}=0}^1 \sum_{X=1}^4 \Big[ c_k(R_1,R_2,..., R_{11}, X ) \\ \\ &\times \text{ } \Pi_{h=1}^{10} g_h(R_{12-h} , R_{12-h-1}, Z_{12-h-1}, Z_{12-h}, Z_{12-h+1},..., Z_{11}, X) \\ \\ & \times b(R_1, Z_1, Z_2,..., Z_{11}, X)\\ \\ & \times a(X, Z_1, Z_2, ..., Z_{11})\Big]\\ \\ & \text{ for each $Z_1\in \{0,1\}^2$, $Z_2\in \{0,1\}^2$, ..., $Z_{11}\in \{0,1\}^2$}\\ \\ & \text{ for each $k=1,..., 2^{11}-2$} \end{aligned} $$ [When I write $Z_1\in \{0,1\}^2$, I mean that $Z_1$ can take 4 values: $(0,0), (0,1), (1,1), (1,0)$]
The terms that we know are the functions on the left: $$ \begin{aligned} & d_k(Z_1,Z_2,..., Z_{11}) \\ & \text{ for each $Z_1\in \{0,1\}^2$, $Z_2\in \{0,1\}^2$, ..., $Z_{11}\in \{0,1\}^2$}\\ & \text{ for each $k=1,..., 2^{11}-2$} \end{aligned} $$
The unknowns are the functions on the right:
- (A): $\hspace{1cm}c_k(R_1,R_2,..., R_{11}, X ) $ $$ \text{ for each $R_1\in \{0,1\}$, ..., $R_{11}\in \{0,1\}$, $X\in \{1,2,3,4\}$, $k=1,..., 2^{11}-2$} $$
- (B): $\hspace{1cm} g_h(R_{12-h} , R_{12-h-1}, Z_{12-h-1}, Z_{12-h}, Z_{12-h+1},..., Z_{11}, X)$ $$\text{ for each $R_{12-h}\in \{0,1\}$, $R_{12-h-1}\in \{0,1\}$,$Z_{12-h-1}\in \{0,1\}^2$, ..., $Z_{11}\in \{0,1\}^2$, $X\in \{1,2,3,4\}$, $h=1,..., 10$} $$
- (C) $\hspace{1cm} b(R_1, Z_1, Z_2,..., Z_{11}, X)$ $$ \text{ for each $R_{1}\in \{0,1\}$, $Z_{1}\in \{0,1\}^2$,..., $Z_{11}\in \{0,1\}^2$, $X\in \{1,2,3,4\}$} $$
- (D) $\hspace{1cm} a(X, Z_1, Z_2, ..., Z_{11})$ $$ \text{ for each $X\in \{1,2,3,4\}$, $Z_{1}\in \{0,1\}^2$,..., $Z_{11}\in \{0,1\}^2$} $$
There are some constraints which reduce the number of unknowns:
- $g_h(1 , R_{12-h-1}, Z_{12-h-1}, Z_{12-h}, Z_{12-h+1},..., Z_{11}, X)=1-g_h(0 , R_{12-h-1}, Z_{12-h-1}, Z_{12-h}, Z_{12-h+1},..., Z_{11}, X)$
- $b(1, Z_1, Z_2,..., Z_{11}, X)=1-b(0, Z_1, Z_2,..., Z_{11}, X)$
- $ \sum_{X=1}^3 a(X, Z_1, Z_2, ..., Z_{11})= 1-a(4, Z_1, Z_2, ..., Z_{11})$
Therefore, from counting the cardinality of the codomains of the functions $\{d_k(\cdot)\}_{k=1}^{2^{11}-2}$, we have that $$ \text{Number of equations:} \quad 4^{11}*(2^{11}-2) $$
From counting the cardinality of the codomains of the unknown functions and keeping in mind the constraints, we have that $$ \text{Number of unknowns:} \quad \underbrace{(2^{11}*4)*(2^{11}-2)}_{(A)}+\underbrace{[2*4*\sum_{n=2}^{11}4^n]}_{(B)}+\underbrace{(1*4*4^{11})}_{(C)}+\underbrace{(3*4^{11})}_{(D)} $$
We have less unknowns than equations (I made the computation in Matlab).