# 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
Then, to satisfy each department's rule for unveiling the secret, we must have:  
\[7\times Q_{L}\geq t_{\textit{artificial}}>6\times Q_{L}, \tag{2}\]  
\[5\times Q_{M}\geq t_{\textit{artificial}}>4\times Q_{M}, \tag{3}\]  
\[3\times Q_{H}\geq t_{\textit{artificial}}>2\times Q_{H}, \tag{4}\]  
Because \(t_{\textit{artificial}}\), \(Q_{H}\), \(Q_{M}\), and \(Q_{L}\) are all positive integers, if \(Q_{L}\) is 1, then Equation (2) implies \(7\geq t_{\textit{artificial}}>6\); hence, \(t_{\textit{artificial}}=7\). Furthermore, the right-hand side of Equation (3) implies \(t_{\textit{artificial}}=7>4\times Q_{M}\), which means the positive integer \(Q_{M}\) is 1, thus contradicting the left-hand side of Equation (3) because \(5\times Q_{M}\geq t_{\textit{artificial}}\) then becomes \(5\times 1\geq 7\). Hence, we cannot use \(Q_{L}=1\). Thus, we try the next smallest value 2 for \(Q_{L}\). Equation (2) then implies \(14\geq t_{\textit{artificial}}>12\). Hence, \(t_{\textit{artificial}}\) is 13 or 14, proven as follows. If \(t_{\textit{artificial}}=13\), Equation (3) becomes \(5Q_{M}\geq 13>4Q_{M}\), which can be satisfied by \(Q_{M}=\lceil 13/5\rceil=3\), and Equation (4) becomes \(3Q_{H}\geq 13>2Q_{H}\), which can be satisfied by \(Q_{H}=\lceil 13/3\rceil=5\). Similarly, if \(t_{\textit{artificial}}=14\), Equation (3) becomes \(5Q_{M}\geq 14>4Q_{M}\), which can be satisfied by \(Q_{M}=\lceil 14/5\rceil=3\), and Equation (4) becomes \(3Q_{H}\geq 14>2Q_{H}\), which can be satisfied by \(Q_{H}=\lceil 14/3\rceil=5\). Likewise, if we try the next smallest value for \(Q_{L}\), 3, then Equation (2) implies \(21\geq t_{\textit{artificial}}>18\). Hence, \(t_{\textit{artificial}}\) is in [19, 20, 21]. We now prove that \(t_{\textit{artificial}}\) can be 19, 20, or 21. If \(t_{\textit{artificial}}=19\), Equation (3) becomes \(5Q_{M}\geq 19>4Q_{M}\), which can be satisfied by \(Q_{M}=\lceil 19/5\rceil=4\), and Equation (4) becomes \(3Q_{H}\geq 19>2Q_{H}\), which can be satisfied by \(Q_{H}=\lceil 19/3\rceil=7\). Analogous steps can be taken to prove that \(t_{\textit{artificial}}\) can be 20 or 21. When we execute the preceding procedure for \(Q_{L}\geq 2\), we obtain valid values for \(t_{\textit{artificial}}\), namely, \(t_{\textit{artificial}}\in\{13,14\}_{\textit{when}\ QL=2\}\), [19, 20, 21]when \(QL=3\), [25, 26, 27, 28]when \(QL=4\), [31, 32, 33, 34, 35]when \(QL=5\), or [37, 38, 39, 40, 41, 42]when \(QL=6\), and so on. We now examine the system further. For example, let the artificial threshold be 14. We can then use (\(t_{\textit{artificial}}\), \(n_{\textit{artificial}}\)) = (14, 52) in the traditional polynomial-based sharing scheme to create 52 shares. Each participant in \(DEP_{H}\), \(DEP_{M}\), and \(DEP_{L}\) gets 5 = \(\lceil 14/3\rceil\), 3 = \(\lceil 14/5\rceil\), and 2 = \(\lceil 14/7\rceil\) shares, respectively. Moreover, for any 2 participants (whether from the same department or from different departments), their set of shares do not intersect. Notably, because no intersection is allowed, any 3 \(DEP_{H}\) participants can hand in \(3\times 5=15>14\) distinct shares; any 5 \(DEP_{M}\) participants can hand in \(5\times 3=15>14\) distinct shares; and any 7 \(DEP_{L}\) participants can hand in \(7\times 2=14\) distinct shares. However, 2 \(DEP_{H}\) participants receive only \(2\times 5=10<14\) distinct shares; 4 \(DEP_{M}\) participants receive only \(4\times 3=12<14\) distinct shares; and 6 \(DEP_{L}\) participants receive only \(6\times 2=12<14\) distinct shares. Hence, the intra-department thresholds to unveil the secret are 3 for \(DEP_{H}\), 5 for \(DEP_{M}\), and 7 for \(DEP_{L}\). We create 52 shares because \(4\times 5+6\times 3+7\times 2=52\). In the deciphering meeting, the secret is revealed once the total number of available shares is equal to or greater than the threshold number 14, regardless of whether the participants are from the same department. Table 2 lists some of the many possible combinations that can be used to reveal the secret.