Total Memory Optimiser: Proof of Concept and Compromises

Abstract

For most usual optimisation problems, the Nearer is Better assumption is true (in probability). Classical iterative algorithms take this property into account, either explicitly or implicitly, by forgetting some information collected during the process, assuming it is not useful any more. However, when the property is not globally true, i.e. for deceptive problems, it may be necessary to keep all the sampled points and their values, and to exploit this increasing amount of information. Such a basic Total Memory Optimiser is presented here. We experimentally show that this technique can outperform classical methods on small deceptive problems. As it gets very computing time expensive when the dimension of the problem increases, a few compromises are suggested to speed it up.

A Appendix

1.1 A.1 Problem Definitions

Alpine. For dimension D, the search space is $\left[ 0,4D\right] ^{D}$. Function f is defined as:

$$\begin{aligned} f\left( x_{1},\ldots ,x_{D}\right) =\sum _{d=1}^{D}\left| x_{d,\delta }\sin \left( x_{d,\delta }\right) \right| +0.1\left| x_{d,\delta }\right| \end{aligned}$$

(2)

with $x_{d,\delta }=x_{d}-\delta d$. In this case, we have simply chosen $\delta =1$. This parameter serves to ensure that the minimum is not at the centre of the search space or on a diagonal. The problem is multimodal and non-separable.

Deceptive 1 (Flash). The search space is $\left[ 0,1\right]$. Function f is defined as:

$$\begin{aligned} \left\{ \begin{array}{ccl} x\le 2c_{1} &{} \rightarrow &{} f\left( x\right) =c_{2}\\ 2c1<x\le 3c_{1} &{} \rightarrow &{} f\left( x\right) =c_{2}-\frac{c_{2}}{c_{1}}\left( x-2c_{1}\right) \\ 3c_{1}<x\le 4c_{1} &{} \rightarrow &{} f\left( x\right) =\frac{2c_{2}}{c_{1}}\left( x-3c_{1}\right) \\ 4c_{1}<x\le 5c_{1} &{} \rightarrow &{} f\left( x\right) =2c_{2}-\frac{c_{2}}{c_{1}}\left( x-4c_{1}\right) \\ x\ge 5c_{1} &{} \rightarrow &{} f\left( x\right) =c_{2} \end{array}\right) \end{aligned}$$

(3)

with, in this case, $c_{1}=0.1$ and $c_{2}=0.5$. The problem is unimodal, but with plateaus.

Deceptive 2 (Comb). The search space is $\left[ 0,10\right]$. Function f is defined as:

$$\begin{aligned} f(x)=\min \left( c_{2},1+\sin \left( c_{1}x\right) +\frac{x}{c_{1}}\right) \end{aligned}$$

(4)

with, in this case, $c_{1}=10$ and $c_{2}=1$. The problem is multimodal, but with plateaus.

Deceptive 3 (Brush). The search space is $\left[ 0,10\right] ^{2}$. Function f is defined as:

$$\begin{aligned} f\left( x_{1},x_{2}\right) =\min \left( c_{2},\sum _{d=1}^{2}\left| x_{d}\sin \left( x_{d}\right) \right| +\frac{x_{d}}{c1}\right) \end{aligned}$$

(5)

with, in this case, $c_{1}=10$ and $c_{2}=1$. The problem is multimodal and non-separable.

1.2 A.2 When we Know Nothing, the Middle is the Best Choice

On the Search Space. Let $x^{*}$ be the solution point (we do suppose here it is unique). If we sample x, the error is $\left\| x-x^{*}\right\|$. At the very beginning, as we know nothing, the probability distribution of $x^{*}$ is uniform on the search space. Roughly speaking, it can be anywhere with the same probability. So, we have the sample x in order to minimise the risk given by

$$\begin{aligned} r=\intop _{x^{*}\in S}\left\| x-x^{*}\right\| \end{aligned}$$

(6)

Let us solve it for $D=1$, and $S=\left[ x_{min},x_{max}\right]$. We have

$$ \begin{array}{ccl} r &{} = &{} \int _{u=x_{min}}^{x}\left( x-u\right) du+\int _{u=x}^{x_{max}}\left( u-x\right) du\\ &{} = &{} \left[ xu-\frac{u^{2}}{2}\right] _{u=x_{min}}^{x}+\left[ \frac{u^{2}}{2}-xu\right] _{u=x}^{_{x_{max}}}\\ &{} = &{} x^{2}-\left( x_{max}+x_{min}\right) x+\frac{x_{max}^{2}+x_{min}^{2}}{2} \end{array} $$

And the minimum of this parabola is given by

$$ x=\frac{x_{max}+x_{min}}{2} $$

For $D>1$ the proof is technically more complicated (a possible way is to use recurrence and projections), but the result is the same: the less risky first point is the centre of the search space.

On the Value Space. The same reasoning can be applied to the value space, when we do not make any hypothesis like say a positive local NisB correlation, and when we know the lower and upper bounds of the values, respectively $y_{low}$ and $y_{up}$. On any unknown position of the search space the distribution of the possible values on $\left[ y_{low},y_{up}\right]$ is uniform and therefore the less risky is, again, the middle, i.e. $\frac{y_{low}+y_{up}}{2}$.

1.3 A.3 Variability of a Landscape

We use here a specific definition, which is different from the definition of variance in probability theory. Let f be a numerical function on the search space S. What we call variability on a subspace s of S is the quantity

$$\begin{aligned} v=\intop _{s^{4}}\left| \frac{f(x_{2})-f(x_{1})}{\left\| x_{2}-x_{1}\right\| }-\frac{f(x_{3})-f(x_{1})}{\left\| x_{3}-x_{1}\right\| }\right| \end{aligned}$$

(7)

where $\left\{ x_{1},x_{2},x_{3}\right\}$ is an element of $s^{3}=s\otimes s\otimes s$ (Euclidean product), under the constraint $x_{3}=x_{1}+\lambda \left( x_{2}-x_{1}\right)$ or, equivalently, $\left( x_{2}-x_{1}\right) \times \left( x_{3}-x_{2}\right) =0$ (cross product). The definition may seem to be complicated, but it just means that in any direction the slope of the landscape is constantly the same.