# Secret Image Sharing Revisited: Forbidden Type, Support Type, and Their Two Approaches
## 3 Proposed Types and Approaches
### Using the Linear-Equations Approach to Design Types 1 and 2
Example 1* is derived from Example 1 of Section 3.1, where Example 1* steps are such that Step 1* is derived from Step 1 of Section 3.1, Step 2* is derived from Step 2 of Section 3.1, and so on. Type 1: Sharing with forbidden combinations  
Because the linear-equations approach can be mapped from the polynomial approach, we can, without loss of generality, simply use the examples in Section 3.1 to detail such a mapping. Hence, as in Section 3.1, we still consider 6 participants (\(P_{1}\),..., \(P_{6}\)), where any 3 of the 6 participants can unveilthe secret, unless the 3 participants form a forbidden combination. Examples 1* and 2* illustrate the steps required to create the sharing scheme with one and two forbidden combinations, respectively. Every other example of Section 3.1 also has a linear-equations counterpart in this section, Section 3.2. **Example 1*** (one forbidden combination). : _As in Example 1 of Section 3.1, [P\({}_{4}\), P\({}_{5}\), P\({}_{6}\)] is the only forbidden combination. Because there are 6 participants, we create 6 shadows for the 6 participants per the following steps 1* and 2*. Notably, this is a \((t_{\text{participants}}=3\), \(n_{\text{participants}}=6,f=1)\) threshold scheme, and [P\({}_{4}\), P\({}_{5}\), P\({}_{6}\)] is the only forbidden combination._  
Step 1*: As in Step 1 of Example 1 for the polynomial-based approach, \(t_{\text{artificial}}=9\) and \(n_{\text{artificial}}=17\). First, create a matrix with 17 rows where each row is 9-dimensional and any \(t_{\text{artificial}}\) of the \(n_{\text{artificial}}\) rows are independent. Subsequently, grab the next \(t_{\text{artificial}}=9\) not-shared-yet numbers from the secret, where these secret numbers are termed \(D_{\text{secret}}\). Then, for \(i=1\) to \(i=n_{\text{artificial}}\) (1 to 17 in this case), let \(Eq_{i}\) be the equation \((Row_{i})\cdot(D_{\text{secret}})=IP_{i}\), where \(IP_{i}\) is the value of the inner product of the two vectors \(Row_{i}\) and \(D_{\text{secret}}\). Notably, because any \(t_{\text{artificial}}=9\) of the \(n_{\text{artificial}}\) rows are independent, any \(t_{\text{artificial}}=9\) of the \(n_{\text{artificial}}\) equations can uniquely solve for \(D_{\text{secret}}\), which has \(t_{\text{artificial}}=9\) secret numbers. Step 2*: Distribute these 17 equations to 6 participants. Recall that in Example 1's polynomial-based approach (Section 3.1), the distribution of shares were \(P_{1}=\{S_{1}\), \(S_{2}\), \(S_{3}\)}, \(P_{2}=\{S_{4}\), \(S_{5}\), \(S_{6}\}\), \(P_{3}=\{S_{7}\), \(S_{8}\), \(S_{9}\}\), \(P_{4}=\{S_{10}\), \(S_{11}\), \(S_{12}\}\), \(P_{5}=\{S_{13}\), \(S_{14}\), \(S_{15}\}\), and \(P_{6}=\{S_{16}\), \(S_{17}\), \(S_{\underline{12}}\), \(\underline{S_{15}}\}\). Thus, for this corresponding instance of the linear-equations approach, we let the distribution of equations for the 6 participants be \(P_{1}=\{Eq_{1}\), \(Eq_{2}\), \(Eq_{3}\}\), \(P_{2}=\{Eq_{4}\), \(Eq_{5}\), \(Eq_{6}\}\), \(P_{3}=\{Eq_{7}\), \(Eq_{8}\), \(Eq_{9}\}\), \(P_{4}=\{Eq_{10}\), \(Eq_{11}\), \(Eq_{12}\}\), \(P_{5}=\{Eq_{13}\), \(Eq_{14}\), \(Eq_{15}\}\), and \(P_{6}=\{Eq_{16}\), \(Eq_{17}\), \(\underline{Eq_{12}}\), \(\underline{Eq_{15}}\}\). Similarly, the final two components of the participant \(P_{6}\) already appeared in \(P_{4}\) and \(P_{5}\). Thus, the total number of different equations in the combination \(\{P_{4}\), \(P_{5}\), \(P_{6}\}\) is only \(3+3+2=8\), which is less than the threshold \(t_{\text{artificial}}=9\), meaning that the secret \(D_{\text{secret}}\) cannot be revealed. Conversely, any other combination \(\{P_{i}\), \(P_{j}\), \(P_{k}\}\) yields 9 independent equations, meaning that the 9 secret numbers in \(D_{\text{secret}}\) can be revealed. **Remark 3**. : _Assume that the rule to create the independent 17-by-9 matrix \(A\) in Step 1* is from an algorithm--for example, each element \(a_{i,j}\) is (i)\({}^{i-1}\)--then there is no need to store the matrix.