# Integrating Classical Preprocessing into an Optical Encryption Scheme
## 3 Results
### A Hybrid Construction
#### 3.2.1 Description and Design Rationale  
Suppose party \(A\) wants to send an \(s\)-bit message to party \(B\). We would like to invoke a linear AONT to transform the ciphertext such that if some blocks are missing, the entropy about other blocks is not reduced. We are particularly interested in the situation \(q=2\) and \(l=1\), with field elements representing bits. We try to \"hide\" individual input bits of an AONT where almost the complete output of the AONT is potentially available to an adversary. Unfortunately, the conditions of Theorem 1 cannot be satisfied when \(\ell=1\), \(q=2\) and \(s\geq 2\), since the matrix in which all entries equal \(1\) is not invertible. To fix this issue, consider an \(s\) by \(s\) matrix \(M\) with entries in \(\mathbb{F}_{2}\) with exactly \(s-1\) entries equal to zero. From [21] (Lemma 7), \(M\) is invertible over \(\mathbb{F}_{2}\) if and only if the zero entries occur in \(s-1\) different rows and in \(s-1\) different columns. We will consider such matrices \(M\) with the zero-entries being arranged as follows:  
\[M= \begin{bmatrix}1&1&1&1&\ldots&1\\ 1&0&1&1&\ldots&1\\ 1&1&0&1&\ldots&1\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 1&1&1&1&\ldots&0\end{bmatrix} \]  
So the first row of \(M\) contains all entries qual to \(1\). This means that each \(x_{j}\) value of the input depends on the value \(y_{1}\) of the output. Thus, if \(y_{1}\) is unknown, any value of the input is completely undetermined. Therefore, if we consider this restriction, the above linear transformation \(M\) behaves as an AONT. As with Definition 1, we can define a restricted AONT. Definition 7: Let \(X_{1},\ldots,X_{s},Y_{1},\ldots,Y_{s}\) be random variables taking on values in the finite set \(X\). These \(2\)s random variables define a \(\{1\}\)-restricted AONT provided that the following conditions are satisfied:  
1. \(H(Y_{1},\ldots,Y_{s}|X_{1},\ldots,X_{s})=0\)
2. \(H(X_{1},\ldots,X_{s}|Y_{1},\ldots,Y_{s})=0\)
3. For all \(i\) such that \(1\leq i\leq s\), \(H(X_{i}|Y_{2},\ldots,Y_{s})=H(X_{i})\). We can generalize Definition 7 to an Y-restricted AONT. Definition 8: Let \(X_{1},\ldots,X_{s},Y_{1},\ldots,Y_{s}\) be random variables taking on values in the finite set \(X\). These \(2\)s random variables define an \(\mathsf{Y}\)-restricted AONT provided that the following conditions are satisfied:  
1. \(H(Y_{1},\ldots,Y_{s}|X_{1},\ldots,X_{s})=0\)
2. \(H(X_{1},\ldots,X_{s}|Y_{1},\ldots,Y_{s})=0\)
3. Let \(\mathcal{Y}=\{Y_{v}:v\in\mathrm{Y}\}\) represent the collection of hidden bits. For all \(i\) such that \(1\leq i\leq s\), it holds that \[H(X_{i}|\{Y_{1},\ldots,Y_{s}\}\backslash\mathcal{Y})=H(X_{i}).\] (7)  
For convenience, we will simply speak of an \(\ell\)-_restricted AONT_ instead of an \(\{1,\ldots,\ell\}\)-restricted AONT. Proposition 1: Let \(M\in\mathrm{GL}_{s}(\mathbb{F}_{2})\). Let \(\mathbf{x}=(x_{1}\,x_{2}\ldots\,x_{s})\) and \(\mathbf{y}=(y_{1}\,y_{2}\ldots\,y_{s})\) where \(\mathbf{x}=\mathbf{y}M\). If \(\ell\) coordinates of \(\mathbf{y}\) are unknown, then there are \(2^{\ell}\) possible preimages \(\mathbf{x}\). Proof: Suppose one hides \(l\) values of \(\mathbf{y}\). Since each bit has two possibilities, this leads to \(2^{l}\) possible choices for \(\mathbf{y}\). Since \(M\) is a bijection, it implies that there are also \(2^{\ell}\) candidates for the correct preimage \(\mathbf{x}\). **Remark 3**. : _While being strong, one should note that the guarantees of our restricted AONT are also limited: if available, two \(y_{i}\)-values can be combined to obtain the corresponding sum of \(x_{i}\)-values. Specifically, one can express \(\mathbf{y}\) in terms of \(\mathbf{x}\) as \(\mathbf{y}=\mathbf{x}M^{-1}\) where_  
\[M^{-1}= \begin{bmatrix}0&1&1&\ldots&1\\ 1&1&0&\ldots&0\\ 1&0&1&\ldots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&0&0&\ldots&1\end{bmatrix} \]  
_if \(s\) is even. If \(s\) is odd, the form of \(M^{-1}\) is similar except that \(m_{11}=1\). It can be observed that \(y_{i}=x_{1}+x_{i}\), for all \(i=2,\ldots,s\).