# Secret Image Sharing Revisited: Forbidden Type, Support Type, and Their Two Approaches
## 6 Conclusions  
In this paper, two types of secret image sharing, which are extensions of traditional image sharing, are proposed. Type 1 is the forbidden type, and Type 2 is the cross-department support type. Both the polynomial approach and linear-equations approach can be applied for each type. By using the concept of redundant shares when assigning traditional shares to participants, and by mapping between the polynomial approach and linear-equations approach, the two proposed designs to achieve the requirement of the two types are obtained. Notably, hacking becomes more difficult if both approaches are used in the same image. Furthermore, as stated in Remark 3 of Section 3, the value of the threshold \(t_{\it artificial}\) can be randomly chosen; hence, we may also use a predetermined sequence of multiple thresholds such as \(t_{\it artificial}\) = 14, 7, 11, 20, 43, 31,... to share the same secret in order to confuse hackers. Hacking becomes difficult because of this multiplicity of possible combinations. We now compare between approaches and types. First, because Type 1 systems are designed on a case-by-case basis, systems of the forbidden type are harder to design than those of the cross-department support type. Second, the linear-equations approach is harder to apply than the polynomial approach is; although we have Algorithm 2, which is a general procedure to obtain the corresponding instance of the linear-equations approach from an instance of the polynomial approach. This is because Step 1 of Algorithm 2 implies the need to create an \(n_{\it artificial}\)-by-\(t_{\it artificial}\) matrix in the conversion to the linear-equations approach, where \(t_{\it artificial}\) of the \(n_{\it artificial}\) rows are linearly independent. This makes the linear-equations approach slightly harder to apply, but such nontrivial complexity also makes systems designed using the linear-equations approach more difficult to hack. As for storage, if the \(n_{\it artificial}\)-by-\(t_{\it artificial}\) matrix \(A\) needs to be stored, then the required storage space of each participant in the linear-equations approach is approximately \(t_{\it artificial}\) times larger than the corresponding space in the polynomial approach. However, if this matrix \(A\) can be automatically generated by an algorithm or a preassigned method, then matrix \(A\) does not need to be stored, making the storage space of the two approaches approximately equal; we analyze this claim as follows. Each participant \(i\) in the linear-equations approach needs to store a value \(IP_{i}\), which is the inner product of a row of \(A\) and a vector of \(t_{\it artificial}\) pixel values, that is, \(IP_{i}\) = (\(Row_{i}\)) \(\cdot\) (\(D_{\it secret}\)). Because (\(Row_{i}\)) has \(t_{\it artificial}\) elements and \(D_{\it secret}\) also has \(t_{\it artificial}\) pixel values, notice that the inner product value cannot exceed \(t_{\it artificial}\) \(\times\) (\(Max\)\(A\)) \(\times\) 256--where (\(Max\)\(A\)) is the maximal value of the elements of matrix \(A\), 256 is the pixel value range, and \(t_{\it artificial}\) has its value because each inner product is the sum of \(t_{\it artificial}\) integers in which each integer is less than (\(Max\)\(A\)) \(\times\) 256. To share a vector constituted by \(t_{\it artificial}\) pixels, each participant for whom the polynomial approach is used stores a byte, but each participant in the linear-equations approach stores approximately \(log_{256}[t_{\it artificial}\) \(\times\) (\(Max\)\(A\)) \(\times\) 256] = 1 + \(log_{256}\)(\(Max\)  
Figure 7: A blueprint image for a company that designs, builds, and sells houses. \(A)+log_{256}(t_{artificial})\) bytes. Notably, \(1+log_{256}(Max\ A)+log_{256}(t_{artificial})\leq 1+1+1=3\) if the maximal absolute value of elements of matrix \(A\)is \(<256\) and \(t_{artificial}\leq 256\). If \(Max\ A\) becomes 65,535, the size amplification factor between the two approaches is still only \(1+2+1=4\).