Self-Balancing Multimemetic Algorithms in Dynamic Scale-Free Networks

Abstract

We study the behavior and performance of island-based multimemetic algorithms, namely memetic algorithms which explicitly represent and evolve memes alongside solutions, in unstable computational environments whose topology is modeled as scale-free networks, a pattern of connectivity observed in real-world networks, such as peer-to-peer systems. We consider the utilization of self-balancing strategies in order to efficiently adjust population sizes to cope with the phenomenon of churn, as well as the dynamic re-wiring of connections in order to deal with connectivity losses caused by node failures. A broad experimental evaluation on different problems and computational scenarios featuring diverse volatility conditions shows that the combination of these two strategies leads to more robust performances, in particular in situations in which churn rates are large.

A Test Suite

Deb’s 4-bit fully deceptive function (TRAP) is defined as $f_{trap}(s) = 0.6 - 0.2\cdot u(s)$ for $u(s)<4$ and $f_{trap}(s) = 1$ for $u(s)=4$, where $u(s_1\cdots s_i)=\sum _js_j$ is the number of 1 s in binary string $s$. A higher-order problem is built by concatenating $k$ such blocks.

The Hierarchical if-and-only-if (HIFF) function is a recursive epistatic function for binary strings of $2^k$ bits by means of two auxiliary functions $f$ and $t$ defined as

• $f(a,b) = 1$ for $a=b\ne \bullet$ and $f(a,b)=0$ otherwise.

• $t(a,b) = a$ if $a=b$ and $t(a,b)=\bullet$ otherwise.

These two functions are used as follows:

$$\begin{aligned} \mathrm{HIFF}_k(b_1\cdots b_n) = \sum _{i=1}^{n/2} f(b_{2i-1},b_{2i}) + 2 \cdot \mathrm{HIFF}_{k-1}(b'_1,\cdots ,b'_{n/2}) \end{aligned}$$

where $b'_i=t(b_{2i-1}, b_{2i})$ and $\mathrm{HIFF}_0(\cdot ) = 1$.

The basic MMDP block is defined for 6-bit strings as $f_{mmdp}(s) = 1$ for $u(s) \in \{0,6\}$, $f_{mmdp}(s) = 0$ for $u(s) \in \{1,5\}$, $f_{mmdp}(s) = 0.360384$ for $u(s) \in \{2,4\}$ and $f_{mmdp}(s) = 0.640576$ for $u(s) =3$. We concatenate $k$ copies of this basic block to create a harder problem.