Mining skypatterns in fuzzy tensors

Abstract

Many data mining tasks rely on pattern mining. To identify the patterns of interest in a dataset, an analyst may define several measures that score, in different ways, the relevance of a pattern. Until recently, most algorithms have only handled constraints in an efficient way, i.e., every measure had to be associated with a user-defined threshold, which can be tricky to determine. Skypatterns were introduced to allow analysts to simply define the measures of interest, and to get as a result a set of globally optimal and semantically relevant patterns. Skypatterns are Pareto-optimal patterns: no other pattern scores better on one of the chosen measures and scores at least as well on every remaining measure. This article tackles the search of the skypatterns in a more general context than the 0/1 (aka Boolean) matrix: the fuzzy tensor. The proposed solution supports a large class of measures. After explaining why and how their common mathematical property enables a safe pruning of the search space, an algorithm is presented. It builds upon multidupehack, a generalist pattern mining framework, which is now able to efficiently list skypatterns in addition to enforcing constraints on them. Experiments on two real-world fuzzy tensors illustrate the versatility of the proposal. Other experiments show it is typically more than one order of magnitude faster than the state-of-the-art algorithms, which can only mine 0/1 matrices.

A Piecewise (Anti-)Monotonicity of the Slope Measure

To simplify the proof that the slope is piecewise (anti-)monotone, all the outputs of the x and y data-access functions, i.e., the abscissas and the ordinates of the points, are supposed positive. If it is not the case, $\min _{t \in \prod _{i \in I} X_i} x(t)$ is subtracted from every abscissa and $\min _{t \in \prod _{i \in I} X_i} y(t)$ is subtracted from every ordinate, what moves all the points to the positive quadrant of the Cartesian coordinate system. The slope of the fitting line being invariant under translation, $x \ge 0$ and $y \ge 0$ are assumed without loss of generality.

A rewriting $m'_{\text {slope}}$ of the slope $m_{\text {slope}}$ maps $({L}, {U}) \in \left( \prod _{i = 1}^n 2^{D_i}\right) ^2$ to:

• case 1. if denom$({U}, {L}) > 0$ then

  • (a) $\displaystyle \frac{\text {num}({L}, {U})}{\text {denom}({U}, {L})}$ if num$({L}, {U}) > 0$

  • (a) $\displaystyle \frac{\text {num}({L}, {U})}{\text {denom}({L}, {U})}$ otherwise

• case 2. if denom$({L}, {U}) < 0$ then

  • (a) $\displaystyle \frac{\text {num}({U}, {L})}{\text {denom}({L}, {U})}$ if num$({U}, {L}) < 0$

  • (b) $\displaystyle \frac{\text {num}({U}, {L})}{\text {denom}({U}, {L})}$ otherwise

• case 3. otherwise $+\infty$

where $\forall (X^1, X^2) = (X_1^1, \dots , X_n^1, X_1^2, \dots , X_n^2) \in \left( \prod _{i = 1}^n 2^{D_i}\right) ^2$:

• num$(X^1, X^2) = \displaystyle \sum _{t \in \prod _{i \in I} X_i^2} x(t) \sum _{t \in \prod _{i \in I} X_i^2} y(t) - \left| \prod _{i \in I} X_i^1\right| \sum _{t \in \prod _{i \in I} X_i^1} x(t)y(t)$;

• denom$(X^1, X^2) = \displaystyle \left( \sum _{t \in \prod _{i \in I} X_i^2} x(t)\right) ^2 - \left| \prod _{i \in I} X_i^1\right| \sum _{t \in \prod _{i \in I} X_i^1} x(t)^2$.

The equality $m'_{\text {slope}}(X, X) = m_{\text {slope}}(X)$, for any pattern $X \in \prod _{i = 1}^n 2^{D_i}$, derives from the equality $\frac{\text {num}(X, X)}{\text {denom}(X, X)} = m_{\text {slope}}(X)$, for cases 1 and 2 in the definition of $m'_{\text {slope}}$, and from the nullity of denom(X, X) in case 3.

The rewriting $m'_{\text {slope}}$ actually proves that $m_{\text {slope}}$ is piecewise (anti-)monotone. To show it, following Definition 8, let us take $U \in \prod _{i = 1}^n 2^{D_i}$, $X \in \prod _{i = 1}^n 2^{U_i}$ and $L \in \prod _{i = 1}^n 2^{X_i}$. L being a sub-pattern of X, its subsets of the dimensions with indexes in I are subsets of those of X, i.e., $\forall i \in I$, $L_i \subseteq X_i$. That implies $\prod _{i \in I} L_i \subseteq \prod _{i \in I} X_i$, which in turn implies both $\left| \prod _{i \in I} L_i\right| \le \left| \prod _{i \in I} X_i\right|$ and $\sum _{t \in \prod _{i \in I} L_i} x(t)^2 \le \sum _{t \in \prod _{i \in I} X_i} x(t)^2$. As a consequence, the (positive) quantity subtracted in the expression of denom is smaller if L, rather than X, is input as the first argument. U being a super-pattern of X, the first sum, in the expression of denom, involves more terms when U, rather than X, is input as the second argument. Because $x \ge 0$, that sum is greater and so is its square. Combining the results on both parts in the expression of denom, $\hbox {denom}(X, X) \le$$\hbox {denom}(L, U)$ stands. It entails $\hbox {denom}(X, X) > 0 \Rightarrow$$\hbox {denom}(L, U) > 0$, i.e., if (X, X) triggers case 1 of $m'_{\text {slope}}$ then (L, U) cannot trigger case 2.

The same steps as in the previous paragraph, but considering X or its super-pattern U as the first input of denom, X or its sub-pattern L as the second input of denom, prove $\hbox {denom}(U, L) \le$$\hbox {denom}(X, X)$. That inequality entails $\hbox {denom}(X, X) < 0 \Rightarrow$$\hbox {denom}(U, L) < 0$, i.e., if (X, X) triggers case 2 of $m'_{\text {slope}}$ then (L, U) cannot trigger case 1. Also, $\hbox {denom}(X, X) = 0$ implies both $\hbox {denom}(U, L) \le 0$ and $\hbox {denom}(L, U) \ge 0$, i.e., if (X, X) triggers case 3 then (L, U) triggers neither case 1 nor case 2. Given all the impossibilities proven so far, if (X, X) triggers case $k \in \{1, 2, 3\}$ in the definition of $m'_{\text {slope}}$ then (L, U) triggers either case k or case 3.

If (L, U) triggers case 3, $m_{\text {slope}}(X) = m'_{\text {slope}}(X, X) \le m'_{\text {slope}}(L, U) = +\infty$. It remains to prove $m_{\text {slope}}(X) \le m'_{\text {slope}}(L, U)$ when (X, X) and (L, U) both trigger case 1 or when they both trigger case 2. An analysis of the expression of num, which is analog to the earlier analysis of denom and uses both $x \ge 0$ and $y \ge 0$, proves $\hbox {num}(U, L) \le$$\hbox {num}(X, X) \le$$\hbox {num}(L, U)$ and, in sequence, the impossibility for (L, U) to trigger a sub-case (b) if (X, X) triggers the related sub-case (a). If, on the contrary, (X, X) triggers a sub-case (b) and (L, U) triggers the related sub-case (a) then $m(X) = m'_{\text {slope}}(X, X) \le m'_{\text {slope}}(L, U)$. Indeed, given the tests in $m'_{\text {slope}}$ and the inequations $\hbox {denom}(U, L) \le$$\hbox {denom}(X, X) \le$$\hbox {denom}(L, U)$ that were proven above, the sub-cases (a) always provide positive outputs, whereas the sub-cases (b) always provide negative (hence smaller) outputs.

Finally, when (X, X) and (L, U) trigger, in the definition of $m'_{\text {slope}}$, not only a same case but also a same sub-case, $m_{\text {slope}}(X) \le m'_{\text {slope}}(L, U)$ still stands. Indeed, the inequality $\hbox {num}(U, L) \le$$\hbox {num}(X, X) \le$$\hbox {num}(L, U)$ and the inequality $\hbox {denom}(U, L) \le$$\hbox {denom}(X, X) \le$$\hbox {denom}(L, U)$ together entail:

• $m_{\text {slope}}(X) = \frac{\text {num}(X, X)}{\text {denom}(X, X)} \le \frac{\text {num}(L, U)}{\text {denom}(U, L)}$ if the two numerators and the two denominators are positive, i.e., in case 1a;

• $m_{\text {slope}}(X) = \frac{\text {num}(X, X)}{\text {denom}(X, X)} \le \frac{\text {num}(L, U)}{\text {denom}(L, U)}$ if the two numerators are negative and the two denominators are positive, i.e., in case 1b;

• $m_{\text {slope}}(X) = \frac{\text {num}(X, X)}{\text {denom}(X, X)} \le \frac{\text {num}(U, L)}{\text {denom}(L, U)}$ if the two numerators and the two denominators are negative, i.e., in case 2a;

• $m_{\text {slope}}(X) = \frac{\text {num}(X, X)}{\text {denom}(X, X)} \le \frac{\text {num}(U, L)}{\text {denom}(U, L)}$ if the two numerators are positive and the two denominators are negative, i.e., in case 2b.