## 3 PPDT Classification Model
### Proposed Protocol
, \tag{32}\]

_and the corresponding classification result matrix_

\[\mathbf{V^{\prime}}= \begin{bmatrix}r_{2}^{\prime}P_{2}+[C_{1},0,0,0]\\ r_{4}^{\prime}P_{4}+[C_{2},0,0,0]\end{bmatrix} = \begin{bmatrix}2r_{2}^{\prime} +C_{1}&r_{2}^{\prime}&-r_{2}^{\prime}&-r_{2}^{\prime}\\ C_{2}&r_{4}^{\prime}&r_{4}^{\prime}&0\end{bmatrix} . \tag{33}\]

_Instead of using the original matrices \(P\) and \(V\), we only need to use the pruned matrices \(P^{\prime}\) and \(V^{\prime}\) for calculation, thus saving computational cost, as shown in Figure 4._

_The following two processes are run on encrypted statuses._

**Comparison on each node:**_To check whether \(X_{i}>t_{i}\) at node \(D_{i}\), let \(\mu\) denote the bit length of the two integers \(X_{i}\) and \(t_{i}\), where \(i=1,2,3\). For ease of explanation, we use \(X_{3},t_{3}\) as an example, with a bit length of \(\mu_{3}=6\)._

* _[PERSON] encrypts each feature of his own data using Equations (6) and (9)._

Figure 4: Image for LNP. \[\llbracket X\rrbracket:=\{\llbracket X_{i}\rrbracket_{1},\llbracket X_{i} \rrbracket_{2}\}_{i=1}^{3}=\{\mathsf{Enc}(pk,\mathsf{poly}_{1}(X_{i})),\mathsf{ Enc}(pk,\mathsf{poly}_{2}(X_{i}))\}_{i=1}^{3},\] (34) _where_ \(X_{3}=32=(100000)_{2}\) _in binary,_ \(\mathsf{poly}_{1}(32)=1,\mathsf{poly}_{2}(32)=x^{6}+x^{12}+x^{18}+x^{24}+x^{30}\)_._
* _Model holder encrypts each threshold of_ \(\mathcal{T}\) _using Equations (_6_) and (_8_)._ \[\llbracket t\rrbracket:=\{\llbracket t_{i}\rrbracket_{1},\llbracket t_{i} \rrbracket_{2}\}_{i=1}^{3}=\{\mathsf{Enc}(pk,\mathsf{poly}_{1}(t_{i})),\mathsf{ Enc}(pk,\mathsf{poly}_{2}(t_{i}))\}_{i=1}^{3},\] (35) _where_ \(t_{3}=25=\)_(_\(11001)_{2}\) _in binary,_ \(\mathsf{poly}_{1}(25)=x+x^{2}+x^{5},\mathsf{poly}_{2}(25)=x^{11}+x^{16}+x^{17}+ x^{22}+x^{23}+x^{28}+x^{29}\)_._
* _Server runs homomorphic operation to obtain polynomials of the comparison result and then adds a random polynomial using Equation (_10_) over a polynomial ring_ \(\mathcal{Z}_{q}\)_, which implies_ \(x^{n}=-1\mod q\)_, as follows._ \[\llbracket d\rrbracket:= \{\llbracket d_{i}\rrbracket\}_{i=1}^{3}\] \[= \{\mathsf{Enc}\Big{(}pk,(\mathsf{poly}_{1}(t_{i})-\mathsf{poly}_{ 1}(X_{i}))(\mathsf{poly}_{2}(t_{i})-\mathsf{poly}_{2}(X_{i})+\mathsf{poly}_{1 \to 3})+\widetilde{\mathsf{poly}(\mathbf{1})}\Big{)}\}_{i=1}^{3},\] (36) \[\llbracket d^{\prime}\rrbracket:= \{\llbracket d_{i}+\gamma\rrbracket\}_{i=1}^{3},\] _where_ \(\gamma\in\mathcal{Z}_{q}\)_._
* _[PERSON] decrypts_ \(\llbracket d^{\prime}\rrbracket\) _and sends_ \(d^{\prime}\) _back to the server. After removing the random polynomial_ \(\gamma\) _from_ \(d^{\prime}\)_, the server can obtain comparison results from_ \(d=\{d_{i}\}_{i=1}^{3}\) _by checking whether_ \(\exists\) _any coefficient of_ \(x^{(i-1)n}=0\)_. If so_ \(X_{i}>t_{i}\)_, set_ \(b_{i}=1\)_; otherwise,_ \(b_{i}=0\)_. For example,_ \(d_{3}=3x^{6}+x^{12}+4x^{18}+4x^{24}+4x^{30}+\mathsf{otherterms}\)_, which is the coefficient of_ \(x^{0}=0\)_, so_ \(b_{3}=1\)_. Similarly,_ \(b_{1}=0,b_{2}=0\)_; therefore,_ \(B=[1,b_{1},b_{2},b_{3}]=[1,0,0,1]\)_._
* _The server runs a multiplication homomorphic operation for Bob using vector_ \(B\) _and the following encrypted path matrix_ \(\llbracket P^{\prime}\rrbracket\) _and classification result matrix_ \(\llbracket V^{\prime}\rrbracket\) _based on Equations (_14_), (_16_) and (_18_)_ \[ \begin{split}\llbracket P^{\prime}\rrbracket&= \mathsf{Enc}\{\mathsf{poly}_{mat}(P^{\prime})\}\\ &=\mathsf{Enc}\Big{(}2r_{2}+r_{2}x-r_{2}x^{2}-r_{2}x^{3}+0 \cdot x^{4}+r_{4}x^{5}+r_{4}x^{6}+0\cdot x^{7}\Big{)},\\ \llbracket V^{\prime}\rrbracket&=\mathsf{Enc} \{\mathsf{poly}_{mat}(V^{\prime})\}\\ &=\mathsf{Enc}\Big{(}(2r_{2}^{2}+C_{1})+r_{2}^{\prime}x-r_{2}^{ \prime}x^{2}-r_{2}^{\prime}x^{3}+C_{2}x^{4}+r_{4}^{\prime}x^{5}+r_{4}^{\prime }x^{6}\Big{)}\end{split} \] (37) _to obtain_ \[