# On Using Linear Diophantine Equations for in-Parallel Hiding of Decision Tree Rules

^{*}

## Abstract

**:**

## 1. Introduction

## 2. Materials and Methods

#### 2.1. Adding Instances to Preserve the Class Balance Using Linear Diophantine Equations: A Proof of Concept and an Indicative Example

**Lemma**

**1.**

**Definition**

**1.**

**Theorem**

**1.**

**Theorem**

**2.**

#### 2.2. Fully Specifying Instances

**Lemma**

**2.**

#### 2.3. Hiding in Parallel: Grouping of Hiding Requests

**Lemma**

**3.**

## 3. Results

## 4. Brief Discussion and Conclusions

## Author Contributions

## Conflicts of Interest

## Appendix A

**Lemma**

**1.**

**Proof.**

**Lemma**

**2.**

**Proof.**

**Lemma**

**3.**

**Proof.**

Parallel | Serially |
---|---|

$\frac{a+{p}_{{X}_{I}}}{b}\left(I\right)$ | $\frac{a+{p}_{1}+{p}_{2}}{b}\left(1\right)$ |

$\frac{b}{a+{p}_{1}+{p}_{2}}\left(2\right)$ | |

$\frac{b}{a+{p}_{{X}_{II}}}\left(II\right)$ | $\frac{a+{p}_{1}}{b+{n}_{2}}\left(3\right)$ |

$\frac{b+{n}_{2}}{a+{p}_{1}}\left(4\right)$ | |

$\frac{a}{b+{n}_{{X}_{III}}}\left(III\right)$ | $\frac{a}{b+{n}_{1}+{n}_{2}}\left(5\right)$ |

$\frac{b+{n}_{1}+{n}_{2}}{a}\left(6\right)$ | |

$\frac{b+{n}_{{X}_{IV}}}{a}\left(IV\right)$ | $\frac{a+{p}_{2}}{b+{n}_{1}}\left(7\right)$ |

$\frac{b+{n}_{1}}{a+{p}_{2}}\left(8\right)$ |

- Case (I-1):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{a+{p}_{1}+{p}_{2}}{b}\iff {p}_{{X}_{I}}={p}_{1}+{p}_{2}Q.E.D.$$
- Case (I-2):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{b}{a+{p}_{1}+{p}_{2}}$$The reason that we selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ is that ${p}_{{X}_{I}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${p}_{{X}_{I}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{II}}\ge {p}_{{X}_{I}}\text{}(*).$ If we had selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$, we would have case (II-2).$$\mathrm{i}.\mathrm{e}.\frac{b}{a+{p}_{{X}_{II}}}=\frac{b}{a+{p}_{1}+{p}_{2}}\iff {p}_{{X}_{II}}={p}_{1}+{p}_{2}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{p}_{2}\ge {p}_{{X}_{I}}$ Q.E.D.
- Case (I-3):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{a+{p}_{1}}{b+{n}_{2}}\iff $$$$\iff ab+a{n}_{2}+b{p}_{{X}_{I}}+{n}_{2}{p}_{{X}_{I}}=ab+b{p}_{1}\iff $$$$\iff a{n}_{2}+b\left({p}_{{X}_{I}}-{p}_{1}\right)+{p}_{{X}_{I}}{n}_{2}=0$$If ${p}_{{X}_{I}}\ge {p}_{1}$, then this case is impossible because all terms in the left-hand side in the above equation are positive.If ${p}_{{X}_{I}}<{p}_{1}$ then we have ${p}_{1}+{n}_{2}>{p}_{{X}_{I}}$ Q.E.D.
- Case (I-4):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{b+{n}_{2}}{a+{p}_{1}}$$The reason that we selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ is that ${p}_{{X}_{I}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${p}_{{X}_{I}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{II}}\ge {p}_{{X}_{I}}\text{}(*).$ If we had selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$, we would have case (II-4), for which it had been proved that$${p}_{1}+{n}_{2}>{p}_{{X}_{II}}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{n}_{2}>{p}_{{X}_{I}}$ Q.E.D.
- Case (I-5):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{a}{b+{n}_{1}+{n}_{2}}\iff $$$$\iff ab+a({n}_{1}+{n}_{2})+b{p}_{{X}_{I}}+{p}_{{X}_{I}}({n}_{1}+{n}_{2})=ab\iff $$$$\iff a({n}_{1}+{n}_{2})+b{p}_{{X}_{I}}+{p}_{{X}_{I}}({n}_{1}+{n}_{2})=0$$This case is impossible because all the terms in the left-hand side in the above equation are positive.
- Case (I-6):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{b+{n}_{1}+{n}_{2}}{a}$$The reason that we selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ is that ${p}_{{X}_{I}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${p}_{{X}_{I}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${n}_{{X}_{IV}}\ge {p}_{{X}_{I}}\text{}(*).$ If we had selected option $\left(IV\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b+{n}_{{X}_{IV}}}{a}$ we would have case (IV-6),$$\mathrm{i}.\mathrm{e}.\frac{b+{n}_{{X}_{IV}}}{a}\text{}=\frac{b+{n}_{1}+{n}_{2}}{a}\iff {n}_{{X}_{IV}}={n}_{1}+{n}_{2}\left(**\right)$$From (*), (**) above, we have ${n}_{1}+{n}_{2}>{p}_{{X}_{I}}$ Q.E.D.
- Case (I-7):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{a+{p}_{2}}{b+{n}_{1}}\iff $$$$\iff ab+a{n}_{1}+b{p}_{{X}_{I}}+{n}_{1}{p}_{{X}_{I}}=ab+b{p}_{2}\iff $$$$\iff a{n}_{1}+b\left({p}_{{X}_{I}}-{p}_{2}\right)+{n}_{1}{p}_{{X}_{I}}=0$$If ${p}_{{X}_{I}}\ge {p}_{2}$ then this case is impossible, since the left-hand side in the above equation is positive.If ${p}_{{X}_{I}}<{p}_{2}$ then we have ${p}_{2}+{n}_{1}>{p}_{{X}_{I}}$ Q.E.D.
- Case (I-8):$$\frac{a+{p}_{{X}_{I}}}{b}=\frac{b+{n}_{1}}{a+{p}_{2}}$$Therefore, ${p}_{{X}_{I}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{II}}\ge {p}_{{X}_{I}}\text{}(*).$ If we had selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$ we would have case (II-8), for which it had been proved that$${n}_{1}+{p}_{2}>{p}_{{X}_{II}}\left(**\right)$$From (*), (**) above, we have ${n}_{1}+{p}_{2}>{p}_{{X}_{I}}$ Q.E.D.
- Case (II-1):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{a+{p}_{1}+{p}_{2}}{b}$$The reason that we selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$ is that ${p}_{{X}_{II}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${p}_{{X}_{II}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{I}}\ge {p}_{{X}_{II}}\text{}(*).$ If we had selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ we would have the case (I-1), and for that it had been proven that$${p}_{1}+{p}_{2}={p}_{{X}_{I}}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{p}_{2}\ge {p}_{{X}_{II}}$ Q.E.D.
- Case (II-2):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{b}{a+{p}_{1}+{p}_{2}}\iff {p}_{{X}_{II}}={p}_{1}+{p}_{2}Q.E.D.$$
- Case (II-3):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{a+{p}_{1}}{b+{n}_{2}}$$The reason that we selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$ is that ${p}_{{X}_{II}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${p}_{{X}_{II}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{I}}\ge {p}_{{X}_{II}}\text{}(*).$ If we had selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ we would have the case (I-3), and for that it had been proved that$${p}_{1}+{n}_{2}\ge {p}_{{X}_{I}}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{n}_{2}\ge {p}_{{X}_{II}}$ Q.E.D.
- Case (II-4):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{b+{n}_{2}}{a+{p}_{1}}\iff $$$$\iff ab+b{p}_{1}=ab+a{n}_{2}+b{p}_{{X}_{II}}+{n}_{2}{p}_{{X}_{II}}\iff $$$$\iff a{n}_{2}+{n}_{2}{p}_{{X}_{II}}+b\left({p}_{{X}_{II}}-{p}_{1}\right)=0$$If ${p}_{{X}_{II}}\ge {p}_{1}$ then this case is impossible, since the left-hand side in the above equation is positive.If ${p}_{{X}_{II}}<{p}_{1}$ then we have ${p}_{1}+{n}_{2}>{p}_{{X}_{II}}$ Q.E.D.
- Case (II-5):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{a}{b+{n}_{1}+{n}_{2}}$$The reason that we selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$ is that ${p}_{{X}_{II}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${p}_{{X}_{II}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${n}_{{X}_{III}}\ge {p}_{{X}_{II}}\text{}(*).$ If we had selected option $\left(III\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a}{b+{n}_{{X}_{III}}}$ we would have case (III-5), and for that it had been proved that$${n}_{1}+{n}_{2}={n}_{{X}_{III}}\left(**\right)$$From (*), (**) above, we have ${n}_{1}+{n}_{2}\ge {p}_{{X}_{II}}$ Q.E.D.
- Case (II-6):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{b+{n}_{1}+{n}_{2}}{a}\iff $$$$\iff ab=ab+a({n}_{1}+{n}_{2})+b{p}_{{X}_{II}}+{p}_{{X}_{II}}({n}_{1}+{n}_{2})\iff $$$$\iff a({n}_{1}+{n}_{2})+b{p}_{{X}_{II}}+{p}_{{X}_{II}}({n}_{1}+{n}_{2})=0$$This case is impossible because all the terms in the left-hand side in the above equation are positive.
- Case (II-7):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{a+{p}_{2}}{b+{n}_{1}}$$Therefore, ${p}_{{X}_{II}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{I}}\ge {p}_{{X}_{II}}\text{}(*).$ If we had selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ we would have the case (I-7), and for that it had been proved that$${p}_{2}+{n}_{1}>{p}_{{X}_{I}}\left(**\right)$$From (*), (**) above, we have ${p}_{2}+{n}_{1}\ge {p}_{{X}_{II}}$ Q.E.D.
- Case (II-8):$$\frac{b}{a+{p}_{{X}_{II}}}=\frac{b+{n}_{1}}{a+{p}_{2}}\iff $$$$\iff ab+b{p}_{2}=ab+a{n}_{1}+b{p}_{{X}_{II}}+{n}_{1}{p}_{{X}_{II}}\iff $$$$\iff a{n}_{1}+{n}_{1}{p}_{{X}_{II}}+b\left({p}_{{X}_{II}}-{p}_{2}\right)=0$$If ${p}_{{X}_{II}}\ge {p}_{2}$ then this case is impossible, since the left-hand side in the above equation is positive.If ${p}_{{X}_{II}}<{p}_{2}$ then we have ${p}_{2}+{n}_{1}>{p}_{{X}_{II}}$ Q.E.D.
- Case (III-1):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{a+{p}_{1}+{p}_{2}}{b}\iff $$$$\iff ab=ab+b({p}_{1}+{p}_{2})+a{n}_{{X}_{III}}+{n}_{{X}_{III}}({p}_{1}+{p}_{2})\iff $$$$\iff b({p}_{1}+{p}_{2})+a{n}_{{X}_{III}}+{n}_{{X}_{III}}({p}_{1}+{p}_{2})=0$$This case is impossible because all the terms in the left-hand side in the above equation are positive.
- Case (III-2):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{b}{a+{p}_{1}+{p}_{2}}$$The reason that we selected option $\left(III\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a}{b+{n}_{{X}_{III}}}$ is that ${n}_{{X}_{III}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${n}_{{X}_{III}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{II}}\ge {n}_{{X}_{III}}\text{}(*).$ If we had selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$ we would have case (II-2), and for that it had been proved that$${p}_{{X}_{II}}={p}_{1}+{p}_{2}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{p}_{2}\ge {p}_{{X}_{II}}$ Q.E.D.
- Case (III-3):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{a+{p}_{1}}{b+{n}_{2}}\iff $$$$\iff ab+a{n}_{2}=ab+b{p}_{1}+a{n}_{{X}_{III}}+{p}_{1}{n}_{{X}_{III}}\iff $$$$\iff b{p}_{1}+{p}_{1}{n}_{{X}_{III}}+a\left({n}_{{X}_{III}}-{n}_{2}\right)=0$$If ${n}_{{X}_{III}}\ge {n}_{2}$ then this case is impossible, since the left-hand side in the above equation is positive.If ${n}_{{X}_{III}}<{n}_{2}$ then we have ${n}_{2}+{p}_{1}>{n}_{{X}_{III}}$ Q.E.D.
- Case (III-4):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{b+{n}_{2}}{a+{p}_{1}}$$The reason that we selected option $\left(III\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a}{b+{n}_{{X}_{III}}}$ is that ${n}_{{X}_{III}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${n}_{{X}_{III}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{I}}\ge {n}_{{X}_{III}}\text{}(*).$ If we had selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ we would have case (I-4), and for that it had been proved that$${p}_{1}+{n}_{2}>{p}_{{X}_{I}}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{n}_{2}\ge {n}_{{X}_{III}}$ Q.E.D.
- Case (III-5):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{a}{b+{n}_{1}+{n}_{2}}\iff {n}_{{X}_{III}}={n}_{1}+{n}_{2}Q.E.D.$$
- Case (III-6):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{b+{n}_{1}+{n}_{2}}{a}$$The reason that we selected option $\left(III\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a}{b+{n}_{{X}_{III}}}$ is that ${n}_{{X}_{III}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${n}_{{X}_{III}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$ meaning that ${n}_{{X}_{IV}}\ge {n}_{{X}_{III}}\text{}(*).$ If we had selected option $\left(IV\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b+{n}_{{X}_{IV}}}{a}$ we would have the case (IV-6), and for that it had been proved that$${n}_{1}+{n}_{2}={n}_{{X}_{IV}}\left(**\right)$$From (*), (**) above, we have ${n}_{1}+{n}_{2}\ge {n}_{{X}_{III}}$ Q.E.D.
- Case (III-7):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{a+{p}_{2}}{b+{n}_{1}}\iff $$$$\iff ab+a{n}_{1}=ab+b{p}_{2}+a{n}_{{X}_{III}}+{p}_{2}{n}_{{X}_{III}}\iff $$$$\iff b{p}_{2}+{p}_{2}{n}_{{X}_{III}}+a\left({n}_{{X}_{III}}-{n}_{1}\right)=0$$If ${n}_{{X}_{III}}\ge {n}_{1}$ then this case is impossible, since the left-hand side in the above equation is positive.If ${n}_{{X}_{III}}<{n}_{1}$ then we have ${n}_{1}+{p}_{2}>{n}_{{X}_{III}}$ Q.E.D.
- Case (III-8):$$\frac{a}{b+{n}_{{X}_{III}}}=\frac{b+{n}_{1}}{a+{p}_{2}}$$The reason that we selected option $\left(III\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a}{b+{n}_{{X}_{III}}}$, is that ${n}_{{X}_{III}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${n}_{{X}_{III}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$ meaning that ${p}_{{X}_{II}}\ge {n}_{{X}_{III}}\text{}(*).$ If we had selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$ we would have case (II-8), and for that it had been proved that$${p}_{2}+{n}_{1}>{p}_{{X}_{II}}\left(**\right)$$From (*), (**) above, we have ${p}_{2}+{n}_{1}>{n}_{{X}_{III}}$ Q.E.D.
- Case (IV-1):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{a+{p}_{1}+{p}_{2}}{b}$$The reason that we selected option $\left(IV\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b+{n}_{{X}_{IV}}}{a}$, is that ${n}_{{X}_{IV}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${n}_{{X}_{IV}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$ meaning that ${p}_{{X}_{II}}\ge {n}_{{X}_{IV}}\text{}(*).$ If we had selected option $\left(II\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b}{a+{p}_{{X}_{II}}}$ we would have case (II-1), and for that it had been proved that$${p}_{1}+{p}_{2}\ge {p}_{{X}_{II}}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{p}_{2}\ge {n}_{{X}_{III}}$ Q.E.D.
- Case (IV-2):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{b}{a+{p}_{1}+{p}_{2}}\iff $$$$\iff ab=ab+b({p}_{1}+{p}_{2})+a{n}_{{X}_{IV}}+{n}_{{X}_{IV}}({p}_{1}+{p}_{2})\iff $$$$\iff b({p}_{1}+{p}_{2})+a{n}_{{X}_{IV}}+{n}_{{X}_{IV}}({p}_{1}+{p}_{2})=0$$
- Case (IV-3):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{a+{p}_{1}}{b+{n}_{2}}$$The reason that we selected option $\left(IV\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b+{n}_{{X}_{IV}}}{a}$ is that ${n}_{{X}_{IV}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore,${n}_{{X}_{IV}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{I}}\ge {n}_{{X}_{IV}}\text{}(*).$ If we had selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ we would have case (I-3), and for that it had been proved that$${p}_{1}+{n}_{2}>{p}_{{X}_{I}}\left(**\right)$$From (*), (**) above, we have ${p}_{1}+{n}_{2}\ge {n}_{{X}_{IV}}$ Q.E.D.
- Case (IV-4):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{b+{n}_{2}}{a+{p}_{1}}\iff $$$$\iff ab+a{n}_{2}=ab+b{p}_{1}+a{n}_{{X}_{IV}}+{p}_{1}{n}_{{X}_{IV}}\iff $$$$\iff b{p}_{1}+{p}_{1}{n}_{{X}_{IV}}+a\left({n}_{{X}_{IV}}-{n}_{2}\right)=0$$If ${n}_{{X}_{IV}}\ge {n}_{2}$ then this case is impossible, since the left-hand side in the above equation is positive.If ${n}_{{X}_{IV}}<{n}_{2}$ then we have ${n}_{2}+{p}_{1}>{n}_{{X}_{IV}}$ Q.E.D.
- Case (IV-5):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{a}{b+{n}_{1}+{n}_{2}}$$The reason that we selected option $\left(IV\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b+{n}_{{X}_{IV}}}{a}$ is that ${n}_{{X}_{IV}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${n}_{{X}_{IV}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$ meaning that ${n}_{{X}_{III}}\ge {n}_{{X}_{IV}}\text{}(*).$ If we had selected option $\left(III\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ we would have case (III-5), and for that it had been proved that$${n}_{{X}_{III}}={n}_{1}+{n}_{2}\left(**\right)$$From (*), (**) above, we have ${n}_{1}+{n}_{2}\ge {n}_{{X}_{IV}}$ Q.E.D.
- Case (IV-6):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{b+{n}_{1}+{n}_{2}}{a}\iff {n}_{{X}_{IV}}={n}_{1}+{n}_{2}Q.E.D.$$
- Case (IV-7):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{a+{p}_{2}}{b+{n}_{1}}$$The reason that we selected option $\left(IV\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{b+{n}_{{X}_{IV}}}{a}$ is that ${n}_{{X}_{IV}}$ is the minimum number of instances to add in order to maintain the ratio in the parent node.Therefore, ${n}_{{X}_{IV}}=min\left\{{p}_{{X}_{I}},{p}_{{X}_{II}},{n}_{{X}_{III}},{n}_{{X}_{IV}}\right\}$, meaning that ${p}_{{X}_{I}}\ge {n}_{{X}_{IV}}\text{}(*).$ If we had selected option $\left(I\right),\text{}\mathrm{i}.\mathrm{e}.,\text{}\frac{a+{p}_{{X}_{I}}}{b}$ we would have case (I-7), and for that it had been proven that$${p}_{2}+{n}_{1}>{p}_{{X}_{I}}\left(**\right)$$From (*), (**) above, we have ${p}_{2}+{n}_{1}>{p}_{{X}_{I}}\ge {n}_{{X}_{IV}}$ Q.E.D.
- Case (IV-8):$$\frac{b+{n}_{{X}_{IV}}}{a}=\frac{b+{n}_{1}}{a+{p}_{2}}\iff $$$$\iff ab+a{n}_{1}=ab+b{p}_{2}+a{n}_{{X}_{IV}}+{p}_{2}{n}_{{X}_{IV}}\iff $$$$\iff b{p}_{2}+{p}_{2}{n}_{{X}_{IV}}+a\left({n}_{{X}_{IV}}-{n}_{1}\right)=0$$If ${n}_{{X}_{IV}}\ge {n}_{1}$ then this case is impossible, since the left-hand side in the above equation is positive.If ${n}_{{X}_{IV}}<{n}_{1}$ then we have ${n}_{1}+{p}_{2}>{n}_{{X}_{IV}}$ Q.E.D.

## Appendix B

process (node X) begin if (X.affected != DO_NOTHING) then // X has been selected for hiding if X.is-leaf() then X.parent.affected = MAKE_LEAF else // X is an internal node if (X.parent != null) then // set X’s parent to be affected X.parent.affected = ADJUST_RATIO if (X.affected == MAKE_LEAF) then // if X is the parent of a leaf make-leaf (X) elseif (X.affected == ADJUST_RATIO) then // if X is a node on the path adjust-ratio (X) end-if end-if end-if end-if end make-leaf (node X) begin // calculate required local changes $\mathtt{compute}\text{}(\pm {p}_{{\rm X}}^{*},\mp {n}_{{\rm X}}^{*})$ // modify local instance population $\mathtt{X}.\mathtt{add}\text{}(\pm {p}_{{\rm X}}^{*},\mp {n}_{{\rm X}}^{*})$ if (X.parent != null) then // propagate change to parent $\mathtt{X}.\mathtt{parent}.\mathtt{must}-\mathtt{add}\text{}(\pm {p}_{{\rm X}}^{*},\mp {n}_{{\rm X}}^{*})$ end-if // turn X into a leaf if you can if (X.left.is-leaf() && X.right.is-leaf()) then X.left = null X.right = null end-if end adjust-ratio (node Y) begin // calculate ratio to be preserved $\mathtt{compute}\text{}{r}_{Y}={p}_{Y}:{n}_{Y}$ // absorb changes from children $\mathtt{Y}.\mathtt{add}\text{}(\pm {p}_{Y.left}^{*}\pm {p}_{Y.right}^{*},\mp {n}_{Y.left}^{*}\mp {n}_{Y.right}^{*})$ if (Y.parent != null) then // propagate changes to parent $\mathtt{Y}.\mathtt{parent}.\mathtt{must}-\mathtt{add}\text{}(\pm {p}_{Y.left}^{*}\pm {p}_{Y.right}^{*},\mp {n}_{Y.left}^{*}\mp {n}_{Y.right}^{*})$ end-if // calculate LDE pairs $\mathtt{calculate}\text{}\mathtt{Diophantine}\text{}\mathtt{to}\text{}\mathtt{add}\text{}\left({p}_{Y}^{m},{n}_{Y}^{m}\right),\text{}m\in \mathbb{N},m=1,\dots $ // select minimum pair to accommodate nodes below $\mathtt{select}\text{}\mathtt{from}\text{}\left({p}_{Y}^{m},{n}_{Y}^{m}\right)\text{}\mathtt{according}\text{}\mathtt{to}\text{}\left({p}_{Y.left}^{m.left},{n}_{Y.left}^{m.left}\right)+\left({p}_{Y.right}^{m.right},{n}_{Y.right}^{m.right}\right)$ $\mathtt{add}\text{}\left({p}_{Y}^{m.Y},{n}_{Y}^{m.Y}\right)\text{}\mathtt{to}\text{}\mathtt{Y}.\mathtt{instances}$ end

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**Figure 1.**A binary decision tree, before (left) and after (right) hiding and the associated rule sets.

**Figure 5.**Bottom-up propagation of instances (−10p,+10n) from the left (+5p,−5n) and right side of the tree.

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**MDPI and ACS Style**

Feretzakis, G.; Kalles, D.; Verykios, V.S. On Using Linear Diophantine Equations for in-Parallel Hiding of Decision Tree Rules. *Entropy* **2019**, *21*, 66.
https://doi.org/10.3390/e21010066

**AMA Style**

Feretzakis G, Kalles D, Verykios VS. On Using Linear Diophantine Equations for in-Parallel Hiding of Decision Tree Rules. *Entropy*. 2019; 21(1):66.
https://doi.org/10.3390/e21010066

**Chicago/Turabian Style**

Feretzakis, Georgios, Dimitris Kalles, and Vassilios S. Verykios. 2019. "On Using Linear Diophantine Equations for in-Parallel Hiding of Decision Tree Rules" *Entropy* 21, no. 1: 66.
https://doi.org/10.3390/e21010066