# 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
In this case, only the right-hand side of that equation--for example, the value \(IP_{i}\) of the inner product for Eq--needs to be stored. Hence, when a 9-number \(D_{\text{secret}}\) is shared, every participant from \(P_{1}\) to \(P_{5}\) holds 3 inner product values, and \(P_{6}\) holds 4 inner product values._  
**Example 2* (two forbidden combinations). ** Similarly, assume that \(t_{\text{participants}}=3\) and \(n_{\text{participants}}=6\), meaning that any 3 participants can unveil the secret, unless the 3 participants constitute a forbidden combination. As in Example 2 of the polynomial approach, only the two combinations {P\({}_{1}\), \(P_{2}\), \(P_{3}\)} and {P\({}_{4}\), \(P_{5}\), \(P_{6}\)} are forbidden; hence, \(f=2\). Because we have 6 participants, we create 6 shadow files for these 6 participants. Each participant gets a shadow comprising some equations. As in the first step of Example 1*, we use the \((t_{\text{artificial}}=9\), \(n_{\text{artificial}}=16)\) scheme to share the 9-number secret section \(D_{\text{secret}}\) and obtain 16 equations {Eq\({}_{1}\), \(\ldots\), Eq\({}_{16}\)}, where the 9-number secret section \(D_{\text{secret}}\) can be obtained if any 9 equations are used. Because the threshold for \"number of equations\" is 9 and \(P_{1}=\{Eq_{1}\), \(Eq_{2}\), \(Eq_{3}\}\), \(P_{2}=\{Eq_{4}\), \(Eq_{5}\), \(Eq_{6}\}\), and \(P_{3}=\{Eq_{7}\), \(Eq_{8}\), \(\underline{Eq_{3}}\), \(\underline{Eq_{6}}\)}, we can see that {P\({}_{1}\), \(P_{2}\), \(P_{3}\)} 3 people together only have 8 equations {Eq\({}_{1}\), \(\ldots\), Eq\({}_{8}\)}, which is still less than 9, meaning that \(D_{\text{secret}}\) cannot be unveiled. Similarly, {P\({}_{4}\), \(P_{5}\), \(P_{6}\)} together only have 8 equations {Eq\({}_{9}\), \(\ldots\), Eq\({}_{16}\)} because \(P_{4}=\{Eq_{9}\), \(Eq_{10}\), \(Eq_{11}\)}, \(P_{5}=\{Eq_{12}\), \(Eq_{13}\), \(Eq_{14}\}\), and \(P_{6}=\{Eq_{15}\), \(Eq_{16}\), \(\underline{Eq_{11}}\), \(\underline{Eq_{14}}\)}, meaning that \(D_{\text{secret}}\) also cannot be unveiled. As for the other {P\({}_{i}\), \(P_{j}\), \(\overline{P_{k}}\)} combinations, the secret can be revealed because at least 9 equations are available. The proofs of the preceding statements are routine and are thus not presented._  
Type 2: Cross-department support system  
We must demonstrate how the polynomial approach can be extended to the linear-equations approach. Without loss of generality, only demonstrate such an extension for the example where \(t_{artificial}=14\) in the cross-department support algorithm (Algorithm 1) of Section 3.1. Other examples of Type 2 in Section 3.1 can be analogously extended. In Algorithm 3, still use the parameter values used in Figure 2, namely, (\(t_{H}=3\), \(n_{H}=4\)), (\(t_{M}=5\), \(n_{M}=6\)), (\(t_{L}=7\), \(n_{L}=7\)), and \(t_{artificial}=14\). ```
1: Get the input thresholds \(\{t_{H}=3\), \(t_{M}=5\), \(t_{L}=7\}\) of all 3 departments. Also get the input number of participants \(\{n_{H}=4\), \(n_{M}=6\), \(n_{L}=7\}\) of all 3 departments. 2: Use Equations (2)-(4) to find a suitable artificial threshold value \(t_{artificial}=14\). 3: Calculate \(n_{artificial}=\lceil t_{artificial}/n_{H}\rceil\times n_{H}+\lceil t_{artificial}/n_{M} \rceil\times n_{M}+\lceil t_{artificial}/n_{L}\rceil\times n_{L}=\lceil 14/3 \rceil\times 4+\lceil 14/5\rceil\times 6\) + \(\lceil 14/7\rceil\times 7=5\times 4+3\times 6+2\times 7=52\). 4: Create a 52-by-14 matrix \(A\) such that any 14 of the 52 rows are independent. 5: Partition the secret image to non-overlapping segments of 14 pixel values each. Let \(n_{SEG}\) denote the total number of non-overlapping segments. 6: While there are secret segments not shared yet do
7: Grab a \(t_{artificial}\)-values not-yet-shared segment, \(D_{secret}\), of the secret image. 8:for\(k=1\) to \(n_{artificial}\)do
9: Calculate the inner product value \(IP_{k}=Row_{k}\cdot D_{secret}\) for this segment.