# Secret Image Sharing Revisited: Forbidden Type, Support Type, and Their Two Approaches
## Appendix A  
This appendix details the steps for designing (3, \(n_{participants},f\)) forbidden-type sharing. That is, the secret can be unveiled if any 3 of the \(n_{participants}\) participants {\(P_{1}\), \(P_{2}\),..., \(P_{nparticipants}\)} are present, unless these 3 participants constitute a forbidden combination. With \(f\) being the number of banned combinations, in this appendix, we assume that each of the \(f\) banned combinations has the 2 troulemakers \(P_{1}\) and \(P_{2}\). In other words, forbidden combinations are of the form {\(P_{1}\), \(P_{2}\), \(P_{k}\)} for some \(k\in\{3\), \(4\),..., \(n_{participants}\)}. Case 1: 1 \(\leq f\leq 3\). Step 1: Use (\(t_{artificial}\), \(n_{artificial}\)) in traditional sharing to create \(n_{artificial}\) shares where \(t_{artificial}=9\) and \(n_{artificial}=[3+3+2\times f+3\times(n_{participants}-2-f)]\), where \(n_{participants}\) is the total number of participants. As we will demonstrate below, the distribution of nonredundant shares to participants follow the sequence 3, 3, 2,..., 2, 3, 3, 3. In this sequence, 2 appears \(f\) times. Step 2: Distribute the first \(3+3=6\) shares to Participants \(P_{1}\) and \(P_{2}\). That is, \(P_{1}=\{S_{1}\), \(S_{2}\), \(S_{3}\)} and \(P_{2}=\{S_{4}\), \(S_{5}\), \(S_{6}\)}. Step 3: The forbidden combinations are of the form {\(P_{1}\), \(P_{2}\), \(P_{k}\)} for, say, \(k\in\{3\), \(4\), \(5\)} if \(f=3\). We then let the contents of Participants 3 to 5 be as follows. \(P_{3}=\{\underline{S_{1}},\underline{S_{4}},\underline{S_{7}},\underline{S_{8 }}\}\), where the underlined \(\underline{S_{1}}\) and \(\underline{S_{4}}\) also appear elsewhere. \(P_{4}=\{\underline{S_{2}},\underline{S_{5}},\underline{S_{9}},\underline{S_{ 10}}\}\), where the underlined \(\underline{S_{2}}\) and\(\underline{S_{5}}\) also appear elsewhere. \(P_{5}=\{\underline{S_{3}},\underline{S_{6}},\underline{S_{11}},\underline{S_{ 12}}\}\), where the underlined \(\underline{S_{3}}\) and\(\underline{S_{6}}\) also appear elsewhere. \(P_{6}=\{\text{Any three $S_{i}$ which never appear in other participants. }\}\)  
\(P_{7}=\{\text{Any three $S_{i}$ which never appear in other participants. }\}\)  
...  
Each of the three combinations {\(P_{1}\), \(P_{2}\), \(P_{3}\)} or {\(P_{1}\), \(P_{2}\), \(P_{4}\)} or {\(P_{1}\), \(P_{2}\), \(P_{5}\)} only has \(3+3+(4-2)=8\) shares, which is insufficient to unveil the secret under a system whose threshold is \(t_{artificial}=9\). All other combinations of any other 3 participants have at least \(3+3+3=9=t_{artificial}\) shares. It is sufficient for unveiling the secret. Step 3\({}^{\prime}\): In Step 3, if we also apply a nonredundancy condition to \(P_{5}\)--that is, we also let \(P_{5}\) be of the form {Any three \(S_{i}\) which never appear in other participants}--to obtain say, \(P_{5}=\{S_{11}\), \(S_{12}\), \(S_{13}\}\), \(P_{6}\) = {\(S_{14}\), \(S_{15}\), \(S_{16}\)},..., then there are only 2 forbidden combinations, namely, {\(P_{1}\), \(P_{2}\), \(P_{3}\)} and {\(P_{1}\), \(P_{2}\), \(P_{4}\)}. Again, each of these two forbidden combinations will only extract \(3+3+2=8<t_{artificial}=9\) shares, meaning that the secret cannot be unveiled. This solves the case \(f=2\). Step 3\({}^{\prime\prime}\): Likewise, if we also apply a nonredundancy condition to \(P_{4}\)--that is, we also let \(P_{4}\) be also of the form {any three \(S_{i}\) which never appear in other participants}--to obtain, say, \(P_{4}=\{S_{9}\), \(S_{10}\), \(S_{11}\)}, \(P_{5}=\{S_{12}\), \(S_{13}\), \(S_{14}\}\), \(P_{6}=\{S_{15}\), \(S_{16}\), \(S_{17}\},\ldots\), then, the combination {\(P_{1}\), \(P_{2}\), \(P_{3}\)} is still forbidden because these participants only have a total of 8 shares. However, every other combination of 3 participants will yield at least \(3+3+3=9\) nonredundant shares, meaning that the secret can be unveiled. As a result, the only forbidden combination is {\(P_{1}\), \(P_{2}\), \(P_{3}\)}. This solves the case \(f=1\). Case 2: \(f=4\).