# Secret Image Sharing Revisited: Forbidden Type, Support Type, and Their Two Approaches
## 3 Proposed Types and Approaches
### Using the Polynomial Approach to Design Types 1 and 2  
Type 1 (Sharing with forbidden combinations):  
Without loss of generality, consider 6 participants {\(P_{1}\),..., \(P_{6}\)} in the company, of which, any 3 can unveil the secret, unless the 3 participants constitute a forbidden combination. Examples 1 and 2 illustrate the steps in creating a sharing scheme with one and two forbidden combinations, respectively, and the Appendix A at the end of paper illustrates the cases when the number of forbidden combinations is 3 or 4. In general, the design of Type 1 is case by case, and the design of Type 2 is easier. **Example 1 (one forbidden combination)**. : _A sharing scheme with one forbidden combination is easy to design. Without loss of generality, let the only forbidden combination be the combination {\(P_{4}\), \(P_{5}\), \(P_{6}\)}. Becausethere are only 6 participants, we shall create 6 \"shadows\" for the 6 participants per the following steps 1 and 2. This results in the creation of a (\(t_{\text{participants}}=3\), \(n_{\text{participants}}=6,f=1\)) threshold scheme, where {P_4, _P5, _P6_} is the only forbidden combination_. Step 1: Use (\(t_{\text{artificial}}\), \(n_{\text{artificial}}\)) = (9, 17) in polynomial-based sharing to share the secret and obtain \(n_{\text{artificial}}=17\) shares {\(S_{1}\),..., \(S_{17}\)} such that any \(t_{\text{artificial}}=9\) of them can recover the secret. Step 2: Distribute these 17 shares to 6 participants. The distributions of the 17 shares are \(P_{1}=\{S_{1}\), \(S_{2}\), \(S_{3}\}\), \(P_{2}=\{S_{4}\), \(S_{5}\), \(S_{6}\}\), \(P_{3}=\{S_{7}\), \(S_{8}\), \(S_{9}\}\), \(P_{4}=\{S_{10}\), \(S_{11}\), \(S_{12}\}\), \(P_{5}=\{S_{13}\), \(S_{14}\), \(S_{15}\}\), and \(P_{6}=\{S_{16}\), \(S_{17}\), \(S_{12}\), \(S_{15}\}\). Note that we deliberately let the final two components of the participant \(P_{6}\) be the already-used shares \(S_{12}\) and \(S_{15}\), which had already appeared in \(P_{4}\) and \(P_{5}\). By doing this, the total number of different shares in the combination {\(P_{4}\), \(P_{5}\), \(P_{6}\)} is only \(3+3+2=8\), which is less than the threshold of 9, meaning that the secret cannot be revealed. Conversely, any other combination {\(P_{i}\), \(P_{j}\), \(P_{k}\)} yields at least 9 distinct shares, and hence, can reveal the secret. Notably, for Participants \(P_{1}\) to \(P_{5}\), every participant holds a shadow (the data held by a participant), which is constituted by 3 shares; and each share is 9 times smaller in size than the original secret in the (9, 17) threshold scheme. Therefore, in our (\(t_{\text{participants}}=3\), \(n_{\text{participants}}=6,f=1\)) scheme, the data size of each participant \(P_{1}\) to \(P_{5}\) is \(3\times 1/9=1/3\) of the size of the original secret. By contrast, because \(P_{6}\) holds 4 shares, the data size is \(4\times 1/9=4/9\) of the size of the original secret. **Example 2 (two forbidden combinations). **_Figure 1 illustrates the design._  
Still assume \(t_{\text{participants}}=3\), and \(n_{\text{participants}}=6\). Hence, any 3 participants can unveil the secret, unless the 3 participants constitute a forbidden combination. In Example 2, only two combinations {\(P_{1}\), \(P_{2}\), \(P_{3}\)} and {\(P_{4}\), \(P_{5}\), \(P_{6}\)} are forbidden, hence, \(f=2\). These combinations may be forbidden because, for example, the groups {\(P_{1}\), \(P_{2}\), \(P_{3}\)} and {\(P_{4}\), \(P_{5}\), \(P_{6}\)} do not trust each other. Because we have 6 participants, we now create 6 shadows for these 6 participants, where each participant gets a shadow comprising some shares. As with the first step of Example 1, we still use \(t_{\text{artificial}}=9\) in the traditional polynomial sharing scheme to share the secret and obtain many shares {\(S_{1}\), \(S_{2}\),..., \(S_{13}\),...}, allowing the secret to be unveiled if any 9 of the shares are used.