+
+\begin{enumerate}
+\item It shall integrate the entropy values and, to contrast them, for example, integrate these values into the square.
+\item As this square varies from 0 to 1, in order to denote the imbalance and therefore the inclusion, in order to oppose entropy, the value retained will be the complement to 1 of its square as long as the number of counter-examples is less than half of the observations of a (resp. non b).
+ Beyond these values, as the implications no longer have an inclusive meaning, the criterion will be assigned the value 0.
+\item In order to take into account the two information specific to $a\Rightarrow b$ and $\neg b \Rightarrow \neg a$, the product will report on the simultaneous quality of the values retained.
+The product has the property of cancelling itself as soon as one of its terms is cancelled, i.e. as soon as this quality is erased.
+\item Finally, since the product has a dimension 4 with respect to entropy, its fourth root will be of the same dimension.
+\end{enumerate}
+
+Let $\alpha=\frac{n_a}{n}$ be the frequency of a and $\overline{b}=\frac{n_{\overline{b}}}{n}$ be the frequency of non b.
+Let $t=\frac{n_{a \wedge \overline{b}}}{n}$ be the frequency of counter-examples, the two significant terms of the respective qualities of involvement and its counterpart are:
+
+\begin{eqnarray*}
+ h_1(t) = H(b\mid a) = - (1-\frac{t}{\alpha}) log_2 (1-\frac{t}{\alpha}) - \frac{t}{\alpha} log_2 \frac{t}{\alpha} & \mbox{ if }t \in [0,\frac{\alpha}{2}[\\
+ h_1(t) = 1 & \mbox{ if }t \in [\frac{\alpha}{2},\alpha]\\
+ h_2(t)= H(\overline{a}\mid \overline{b}) = - (1-\frac{t}{\overline{\beta}}) log_2 (1-\frac{t}{\overline{\beta}}) - \frac{t}{\overline{b}} log_2 \frac{t}{\overline{b}} & \mbox{ if }t \in [0,\frac{\overline{\beta}}{2}[\\
+ h_2(t)= 1 & \mbox{ if }t \in [\frac{\overline{\beta}}{2},\overline{\beta}]
+\end{eqnarray*}
+Hence the definition for determining the entropic criterion:
+\definition: The inclusion index of A, support of a, in B, support of b, is the number:
+$$i(a,b) = \left[ (1-h_1^2(t)) (1-h_2^2(t))) \right]^{\frac{1}{4}}$$
+
+which integrates the information provided by the realization of a small number of counter-examples, on the one hand to the rule $a \Rightarrow b$ and, on the other hand, to the rule $\neg b \Rightarrow \neg a$.
+
+\subsection{The implication-inclusion index}
+
+The intensity of implication-inclusion (or entropic intensity), a new measure of inductive quality, is the number:
+
+$$\psi(a,b)= \left[ i(a,b).\varphi(a,b) \right]^{\frac{1}{2}}$$
+which integrates both statistical surprise and inclusive quality.
+
+The function $\psi$ of the variable $t$ admits a representation that has the shape indicated in Figure~\ref{chap2fig4}, for $n_a$ and $n_b$ fixed.
+Note in this figure the difference in the behaviour of the function with respect to the conditional probability $P(B\mid A)$, a fundamental index of other rule measurement models, for example in Agrawal.
+In addition to its linear, and therefore not very nuanced nature, this probability leads to a measure that decreases too quickly from the first counter-examples and then resists too long when they become important.
+
+
+\begin{figure}[htbp]
+ \centering
+\includegraphics[scale=0.5]{chap2fig4.png}
+\caption{Example of implication-inclusion.}
+
+\label{chap2fig4}
+\end{figure}
+
+In Figure~\ref{chap2fig4}, it can be seen that this representation of the continuous function of $t$ reflects the expected properties of the inclusion criterion:
+\begin{itemize}
+\item ``Slow reaction'' to the first counter-examples (noise resistance),
+\item ``acceleration'' of the rejection of inclusion close to the balance i.e. $\frac{n_a}{2n}$,
+\item rejection beyond $\frac{n_a}{2n}$, the intensity of implication $\varphi(a,b)$ did not ensure it.
+\end{itemize}
+
+\noindent Example 1\\
+\begin{tabular}{|c|c|c|c|}\hline
+ & $b$ & $\overline{b}$ & margin\\ \hline
+ $a$ & 200 & 400& 600 \\ \hline
+ $\overline{a}$ & 600 & 2800& 3400 \\ \hline
+ margin & 800 & 3200& 4000 \\ \hline
+\end{tabular}
+\\
+\\
+In Example 1, implication intensity is $\varphi(a,b)=0.9999$ (with $q(a,\overline{b})=-3.65$).
+ The entropic values of the experiment are $h_1=h_2=0$.
+ The value of the moderator coefficient is therefore $i(a,b)=0$.
+ Hence, $\psi(a,b)=0$ whereas $P(B\mid A)=0.33$.
+Thus, the "entropic" functions "moderate" the intensity of implication in this case where inclusion is poor.
+\\
+\\
+\noindent Example 2\\
+ \begin{tabular}{|c|c|c|c|}\hline
+ & $b$ & $\overline{b}$ & margin\\ \hline
+ $a$ & 400 & 200& 600 \\ \hline
+ $\overline{a}$ & 1000 & 2400& 3400 \\ \hline
+ margin & 1400 & 2600& 4000 \\ \hline
+ \end{tabular}
+ \\
+ \\
+ In Example 2, intensity of implication is 1 (for $q(a,\overline{b}) = - 8.43$).
+ The entropic values of the experiment are $h_1 = 0.918$ and $h_2 = 0.391$.
+ The value of the moderator coefficient is therefore $i(a,b) = 0.6035$.
+ As a result $\psi(a,b) = 0.777$ whereas $P(B \mid A) = 0.6666$.
+ \\
+ \\
+{\bf remark}
+ \noindent The correspondence between $\varphi(a,b)$ and $\psi(a,b)$ is not monotonous as shown in the following example:
+
+\begin{tabular}{|c|c|c|c|}\hline
+ & $b$ & $\overline{b}$ & margin\\ \hline
+ $a$ & 40 & 20& 60 \\ \hline
+ $\overline{a}$ & 60 & 280& 340 \\ \hline
+ margin & 100 & 300& 400 \\ \hline
+\end{tabular}
+\\
+Thus, while $\varphi(a,b)$ decreased from the 1st to the 2nd example, $i(a,b)$ increased as well as $\psi(a,b)$. On the other hand, the opposite situation is the most frequent.
+Note that in both cases, the conditional probability does not change.
+\\
+\\
+{\bf remark}
+\noindent We refer to~\cite{Lencaa} for a very detailed comparative study of association indices for binary variables.
+In particular, the intensities of classical and entropic (inclusion) implication presented in this article are compared with other indices according to a "user" entry.
+
+\section{Implication graph}
+\subsection{Problematic}
+
+At the end of the calculations of the intensities of implication in both the classical and entropic models, we have a table $p \times p$ that crosses the $p$ variables with each other, whatever their nature, and whose elements are the values of these intensities of implication, numbers of the interval $[0,~1]$.
+It must be noted that the underlying structure of all these variables is far from explicit and remains largely unimportant.
+The user remains blind to such a square table of size $p^2$.
+It cannot simultaneously embrace the possible multiple sequences of rules that underlie the overall structure of all $p$ variables.
+In order to facilitate a clearer extraction of the rules and to examine their structure, we have associated to this table, and for a given intensity threshold, an oriented graph, weighted by the intensities of implication, without a cycle whose complexity of representation the user can control by setting himself the threshold for taking into account the implicit quality of the rules.
+Each arc in this graph represents a rule: if $n_a < n_b$, the arc $a \rightarrow b$ represents the rule $a \Rightarrow b$ ; if $n_a = n_b$, then the arc $a \leftrightarrow b$ will represent the double rule $a \Leftrightarrow b$, in other words, the equivalence between these two variables.
+By varying the threshold of intensity of implication, it is obvious that the number of arcs varies in the opposite direction: for a threshold set at $0.95$, the number of arcs is less than or equal to those that would constitute the graph at threshold $0.90$. We will discuss this further below.
+
+\subsection{Algorithm}
+
+The relationship defined by statistical implication, if it is reflexive and not symmetrical, is obviously not transitive, as is induction and, on the contrary, deduction.
+However, we want it to model the partial relationship between two variables (the successes in our initial example).
+By convention, if $a \Rightarrow b$ and $b \Rightarrow c$, we will accept the transitive closure $a \Rightarrow c$ only if $\psi(a,c) \geq 0.5$, i.e. if the implicit relationship of $a$ to $c$ is better than neutrality by emphasizing the dependence between $a$ and $c$.
+
+
+{\bf VERIFIER PHI PSI}\\
+\\
+{\bf Proposal:} By convention, if $a \Rightarrow b$ and $b \Rightarrow c$, there is a transitive closure $a \Rightarrow c$ if and only if $\psi(a,c) \geq 0.5$, i.e. if the implicit relationship of $a$ over $c$, which reflects a certain dependence between $a$ and $c$, is better than its refutation.
+Note that for any pair of variables $(x;~ y)$, the arc $x \rightarrow y$ is weighted by the intensity of involvement (x,y).
+\\
+Let us take a formal example by assuming that between the 5 variables $a$, $b$, $c$, $d$, and $e$ exist, at the threshold above $0.5$, the following rules: $c \Rightarrow a$, $c \Rightarrow e$, $c \Rightarrow b$, $d \Rightarrow a$, $d \Rightarrow e$, $a \Rightarrow b$ and $a \Rightarrow e$.
+
+This set of numerical and graphical relationships can then be translated into the following table and graph:
+
+\begin{tabular}{|C{0.5cm}|c|c|c|c|c|}\hline
+\hspace{-0.5cm}\turn{45}{$\Rightarrow$} & $a$ & $b$ & $c$ & $d$ & $e$\\ \hline
+$a$ & & 0.97& & & 0.73 \\ \hline
+$b$ & & & & & \\ \hline
+ $c$ & 0.82 & 0.975& & & 0.82 \\ \hline
+ $d$ & 0.78 & & & & 0.92 \\ \hline
+ $e$ & & & & & \\ \hline
+\end{tabular}
+
+\begin{figure}[htbp]
+ \centering
+\includegraphics[scale=1]{chap2fig5.png}
+\caption{Implication graph corresponding to the previous example.}
+
+\label{chap2fig5}
+\end{figure}
