A revised monotonicity-based method for computing tight image enclosures of functions

Abstract

The computation of tight interval image enclosures of functions over bounded variable domains is in the heart of interval-based branch and bound optimization (and constraint satisfaction) solvers. Interval arithmetic extends arithmetic operators, such as $+$, −, $*$, $\backslash$, $\sin$, $\cos$, etc., to intervals. In this way, the operators can be used directly for computing image enclosures of real functions over bounded domains (i.e., natural interval evaluations). Importantly, it is widely recognized that when a function f is monotonic w.r.t. some variable(s) in a given domain, we can compute tighter images of f on this domain than by using natural interval evaluations. This work presents a more general monotonicity-based method that may be applied even if the function is non-monotonic w.r.t. its variables. The method combines basic interval-based filtering techniques with a straightforward analysis of function derivatives. First, filtering based on partial derivatives detects sub-intervals in the domain where the function certainly increase or decrease. Then, we can determine in which subdomains within the interval the value should be maximizing (or minimizing) the function. Finally, we use the natural interval evaluation on the subdomains where f is maximized to compute an upper bound of the enclosure. We show that this method is equivalent to computing an enclosure by using the traditional method when f is monotonic. However, it may be more effective when f is not.

Proofs

Proof (Proof of Proposition 1)

Let the box ${{\varvec{x}}}$ consisting of a non-monotonic region ${{\varvec{w}}}$ w.r.t. $x_i$, and a region ${{\varvec{y}}}{\setminus } {{\varvec{w}}}$ that is non-decreasing w.r.t. $x_i$, as illustrated in Fig. 7. The non-decreasing region can be partitioned into a collection of boxes (four boxes in the figure). The minimizers of all these boxes, except for the box to the right of ${{\varvec{w}}}$ (i.e., boxes 1, 2 and 4), reside in $\underline{{{\varvec{y}}}_i}$. The minimizer of the remaining box (i.e., box 3) is in $\overline{{{\varvec{w}}}_i} \subseteq {{{\varvec{w}}}}$.

Fig. 7

A 2D-box ${{\varvec{x}}}$ consists of a region ${{\varvec{w}}}$ that is non-monotonic w.r.t. $x_i$, and a region ${{\varvec{y}}}{\setminus } {{\varvec{w}}}$ that is non-decreasing w.r.t. $x_i$. In this example, it is observed that the minimizer of $f$ within the non-decreasing region must be located at either $\underline{{{\varvec{y}}}}_i$ or $\overline{{{\varvec{w}}}}_i$ (illustrated by the thick vertical lines)

Note that the same argument holds even if more than two dimensions are considered. In such cases, the non-decreasing region would be partitioned into more boxes, and regardless, one box would be to the right side of ${{\varvec{w}}}$. Hence, it can be concluded that the minimizer of the non-decreasing region is either in $\underline{{{\varvec{y}}}_i}$ or in $\overline{{{\varvec{w}}}_i}$, and similarly, the minimizer of the non-increasing region is either in $\overline{{{\varvec{z}}}_i}$ or in $\underline{{{\varvec{w}}}_i}$.

Finally, given that ${{{\varvec{x}}}}=({{\varvec{y}}}{\setminus } {{{\varvec{w}}}}) \cup {{{\varvec{w}}}} \cup ({{\varvec{z}}}{\setminus } {{{\varvec{w}}}})$, the minimizer of f inside ${{\varvec{x}}}$ must be in the non-decreasing region (i.e., in $\underline{{{\varvec{y}}}_i}$ or in $\overline{{{\varvec{w}}}_i}$), or in the non-increasing region (i.e, in $\overline{{{\varvec{z}}}_i}$ or in $\underline{{{\varvec{w}}}_i}$), or in the non-monotonic region (i.e., in ${{\varvec{w}}}$). Hence, it must be in $\underline{{{\varvec{y}}}_{i}} \cup {{{\varvec{w}}}} \cup \overline{{{\varvec{z}}}_{i}}$. $\square$

Proof (Proof of Proposition 2)

Consider the scenario illustrated in Fig. 8. Let the region ${{\varvec{y}}}\setminus {{\varvec{w}}}$ be a non-decreasing region w.r.t. $x_i$. By Proposition 1, the minimum of f is found either in $\underline{{{\varvec{y}}}_{i}}$ or in ${{{\varvec{w}}}}$. The minimum in $\underline{{{\varvec{y}}}_{i}}$ is at the point marked in bold in Fig. 8, with value $v_1=\min \limits _{x \in \underline{{{\varvec{y}}}_{i}}} f(x)$. Since the function f is non-decreasing in the gray region between $\underline{{{\varvec{y}}}_{i}}$ and ${{\varvec{w}}}$ (as shown in Fig. 8), any point in the left bound of ${{\varvec{w}}}$ will have an image larger than or equal to $v_1$.

Fig. 8

Example showing that the minimum value of $f$ within a box ${{\varvec{x}}}$, which contains a non-decreasing region w.r.t. $x_i$ denoted by ${{\varvec{y}}}{\setminus } {{\varvec{w}}}$, should be $v_1 = \min \limits _{x \in \underline{{{\varvec{y}}}_{i}}} f(x)$ or greater than (or equal to) $v_1 + \underline{{{\varvec{g}}}_i({{\varvec{w}}})} \cdot \textsf {w}({{\varvec{w}}}_i)$

Therefore, the image of any point x in the box ${{\varvec{w}}}$ must be greater than or equal to $v_1 + \underline{{{\varvec{g}}}_i({{\varvec{w}}})} \cdot (x_i-\underline{w_i})$, where $\underline{{{\varvec{g}}}_i({{\varvec{w}}})}$ is an underestimation of the minimum value of $\frac{\partial f}{\partial x_i}$ in ${{\varvec{w}}}$. Since $0 \in {{\varvec{g}}}_i({{\varvec{w}}})$ and $\underline{{{\varvec{g}}}_i({{\varvec{w}}})} \le 0$, this expression is minimized when $x_i=\overline{w_i}$, leading to the stated inequality. $\square$

Fig. 9

Example showing that the minimum value of f in a box ${{\varvec{x}}}$, with a horizontal box ${{\varvec{w}}}$ which divides the box in the dimension of a variable $x_j\ne x_i$, should be larger than (or equal to) $\min \limits _{x \in \underline{{{\varvec{y}}}_{i}}} f(x) + \underline{{{\varvec{g}}}_i({{\varvec{w}}})}\cdot \textsf {w}({{\varvec{w}}}_i)+\underline{{{\varvec{g}}}_j({{\varvec{w}}}\cup {{\varvec{z}}})} \cdot ({{\varvec{w}}}_j \cup {{\varvec{z}}}_j)$

Proof (Proof of Proposition 3)

Let ${{\varvec{x}}}$ be a box as illustrated in Fig. 9. As ${{\varvec{y}}}$ is non-decreasing w.r.t. $x_i$, the minimizer of ${{\varvec{y}}}$ is at $\underline{{{\varvec{y}}}_i}$. If we consider the point with image $v_1$ as the minimizer of ${{\varvec{y}}}$, it follows that the image of the point above must be greater than $v_1$. Consequently, the image $v_2$ of the minimizer of the horizontal line $\underline{{{\varvec{w}}}_j}$ must be greater than $v_2+\underline{{{\varvec{g}}}_i({{\varvec{w}}})} \cdot \textsf {w}(\overline{w_i}-\underline{w_i})$, where $\underline{{{\varvec{g}}}_i({{\varvec{w}}})} \le 0$ is an underestimation of the enclosure of $\frac{\partial f}{\partial x_i}$ over the box ${{\varvec{w}}}$. Finally, the image $v_3$ of the minimizer in ${{\varvec{w}}}\cup {{\varvec{z}}}$ must be greater than $v_2+\underline{{{\varvec{g}}}_j({{\varvec{w}}}\cup {{\varvec{z}}})} \cdot (\overline{z_j}-\underline{w_j})$, where $\underline{{{\varvec{g}}}_j({{\varvec{w}}}_j \cup {{\varvec{z}}}_j)}$ is an underestimation of the enclosure of $\frac{\partial f}{\partial x_j}$ over the box ${{\varvec{w}}}\cup {{\varvec{z}}}$. Finally, as $\overline{w_i}-\underline{w_i}=\textsf{w}({{\varvec{w}}})$ and $(\overline{z_j}-\underline{w_j})=\textsf {w}({{\varvec{w}}}\cup {{\varvec{z}}})$, we arrive at the inequality in the proposition.

Similarly, by considering the minimizer of $\overline{{{\varvec{z}}}}$, we can prove that $f(x) \ge \min \limits _{x \in \overline{{{\varvec{z}}}_{i}}} f(x) - \overline{{{\varvec{g}}}_i({{\varvec{w}}})}\cdot \textsf {w}({{\varvec{w}}}_i)-\max {\left( \overline{{{\varvec{g}}}_j({{\varvec{w}}}\cup {{\varvec{y}}})},0\right) }\cdot \textsf {w}({{\varvec{w}}}_j \cup {{\varvec{y}}}_j)$. $\square$