Optimization using lion algorithm: a biological inspiration from lion’s social behavior

Abstract

Nature-inspired optimization algorithms, especially evolutionary computation-based and swarm intelligence-based algorithms are being used to solve a variety of optimization problems. Motivated by the obligation of having optimization algorithms, a novel optimization algorithm based on a lion’s unique social behavior had been presented in our previous work. Territorial defense and territorial takeover were the two most popular lion’s social behaviors. This paper takes the algorithm forward on rigorous and diverse performance tests to demonstrate the versatility of the algorithm. Four different test suites are presented in this paper. The first two test suites are benchmark optimization problems. The first suite had comparison with published results of evolutionary and few renowned optimization algorithms, while the second suite leads to a comparative study with state-of-the-art optimization algorithms. The test suite 3 takes the large-scale optimization problems, whereas test suite 4 considers benchmark engineering problems. The performance statistics demonstrate that the lion algorithm is equivalent to certain optimization algorithms, while outperforming majority of the optimization algorithms. The results also demonstrate the trade-off maintainability of the lion algorithm over the traditional algorithms.

Appendix

Definition 1

An ${X^{cub}}$ of an ${X^{male}}$ and an ${X^{female}}$ is the sum of Hadamard product of crossover mask and ${X^{male}}$ and Hadamard product of complement of the same crossover mask and ${X^{female}}$, provided the crossover mask is essentially to be a binary vector with ${C_r} \cdot L$ number of binary ones.

Assumption 1

Within a pride, male lions are always stronger than female lions, applicable to cubs also.

Lemma 1

If the product of ${C_r}$ and $L$ is equal to zero, then ${X^{cubs}}$ are equal to ${X^{female}}$.

Proof of Lemma 1

According to Definition 1, $B$, which is a vector with $L$ elements, has ${C_r} \cdot L$number of ones and$L\left( {1 - {C_r}} \right)$ number of zeros. If ${C_r} \cdot L=0$, then $B$ has no ones and $L$ number of zeros, which means $B$ is a vector of zeros. By applying this $B$ in the mathematical representation of crossover operation given in [11], the first term of RHS becomes $B$ and the second term of RHS becomes ${X^{female}}$ as $\bar {B}=1 - B$. Hence, ${X^{cubs}}$ becomes ${X^{female}}$ when ${C_r} \cdot L=0$.

Lemma 2

If the product of ${C_r}$ and $L$ is equal to one, then ${X^{cubs}}$ are equal to ${X^{male}}$.

Proof of Lemma 2

Similar to the proof of Lemma 1, $B$ is a vector of ones, when ${C_r} \cdot L=L$ and $\bar {B}$ becomes a vector of zeros. This in turn, makes the first RHS term of Eq. (11) as ${X^{male}}$ and the second RHS term as a vector of zeros. Hence, ${X^{cubs}}$ become ${X^{male}}$ when ${C_r} \cdot L=L$.

Lemma 3

$E_{1}^{{nomad}}$ is greater than $E_{2}^{{nomad}}$, if ${d_1}$ and $f\left( {X_{1}^{{nomad}}} \right)$ are greater than ${d_2}$ and $f\left( {X_{2}^{{nomad}}} \right)$, respectively.

Proof of Lemma 3

According to the lemma,

$$\text{max} \left( {{d_1},{d_2}} \right)={d_1}$$

(16)

$$\text{max} \left( {f\left( {X_{1}^{{nomad}}} \right),f\left( {X_{2}^{{nomad}}} \right)} \right)=f\left( {X_{1}^{{nomad}}} \right)$$

(17)

Hence, Eqs. (24) and (25) (from Theorem 2) takes the form,

$$E_{1}^{{nomad}}=\exp \left( 1 \right)=2.71$$

(18)

$$E_{2}^{{nomad}}=\exp \left( {\frac{{{d_2}}}{{{d_1}}}} \right)\frac{{f\left( {X_{1}^{{nomad}}} \right)}}{{f\left( {X_{2}^{{nomad}}} \right)}}.$$

(19)

$E_{2}^{{nomad}}$ can be greater than $\exp \left( 1 \right)$only if $f\left( {X_{1}^{{nomad}}} \right)>>f\left( {X_{2}^{{nomad}}} \right)$ as the first term produces exponential decay from $\exp \left( 1 \right)$, when ${d_1}>{d_2}$. But, if $f\left( {X_{1}^{{nomad}}} \right)>>f\left( {X_{2}^{{nomad}}} \right)$, then probably ${d_1}>>{d_2}$due to its linear relationship with $f\left( {X_{1}^{{nomad}}} \right)$, which leads $\exp \left( \bullet \right)$ towards zero, and hence $E_{2}^{{nomad}}$ becomes lesser than $\exp \left( 1 \right)$. Thus it is proved that $E_{1}^{{nomad}}>E_{2}^{{nomad}}$, when ${d_1}>{d_2}$ and$f\left( {X_{1}^{{nomad}}} \right)>f\left( {X_{2}^{{nomad}}} \right)$.

Lemma 4

$E_{1}^{{nomad}}$ is greater than $E_{2}^{{nomad}}$, if ${d_1}$ and $f\left( {X_{2}^{{nomad}}} \right)$ are greater than ${d_2}$ and $f\left( {X_{1}^{{nomad}}} \right)$, respectively.

Proof of Lemma 4

According to the lemma,

$$\text{max} \left( {{d_1},{d_2}} \right)={d_1}$$

(20)

$$\text{max} \left( {f\left( {X_{1}^{{nomad}}} \right),f\left( {X_{2}^{{nomad}}} \right)} \right)=f\left( {X_{2}^{{nomad}}} \right)$$

(21)

As $\frac{{f\left( {X_{2}^{{nomad}}} \right)}}{{f\left( {X_{1}^{{nomad}}} \right)}}>1$ from Eq. (21)

$$E_{1}^{{nomad}}=\exp (1)\frac{{f\left( {X_{2}^{{nomad}}} \right)}}{{f\left( {X_{1}^{{nomad}}} \right)}}>\exp (1)$$

(22)

$$E_{2}^{{nomad}}=\exp \left( {\frac{{{d_2}}}{{{d_1}}}} \right)<\exp (1)$$

(23)

Hence, it is proved that $E_{1}^{{nomad}}>E_{2}^{{nomad}}$, when ${d_1}>{d_2}$ and $f\left( {X_{2}^{{nomad}}} \right)>f\left( {X_{1}^{{nomad}}} \right)$.

Lemma 5

$E_{2}^{{nomad}}$ is greater than $E_{1}^{{nomad}}$, if ${d_2}$ and $f\left( {X_{1}^{{nomad}}} \right)$ are greater than ${d_1}$ and $f\left( {X_{2}^{{nomad}}} \right)$, respectively.

Lemma 6

$E_{2}^{{nomad}}$ is greater than $E_{1}^{{nomad}}$, if ${d_2}$ and $f\left( {X_{2}^{{nomad}}} \right)$ are greater than ${d_1}$ and $f\left( {X_{1}^{{nomad}}} \right)$, respectively.

Proof of Lemma 5 and 6

Lemma 5 is the vice versa of lemma 3 as $E_{2}^{{nomad}}>\exp \left( 1 \right)$ and $E_{1}^{{nomad}}<\exp \left( 1 \right)$. Lemma 6 is the vice versa of lemma 4 as $E_{2}^{{nomad}}=\exp \left( 1 \right)$ and probably $E_{1}^{{nomad}}<\exp \left( 1 \right)$ proved through Axiom 3.

Axiom 1

${C_r} \cdot L=L$ is an integer possibly between $1$ and $L - 1$, i.e.,${C_r} \cdot L \in \left( {1,L - 1} \right)$.

Axiom 2

$X_{{^{1}}}^{{nomad}}$ and $X_{2}^{{nomad}}$ are essentially different and hence ${d_1}$ is not always equal to ${d_2}$.

Axiom 3

$f\left( {X_{1}^{{nomad}}} \right)$ and $f\left( {X_{2}^{{nomad}}} \right)$ exhibit linear variation with respect to ${d_1}$ and ${d_2}$, respectively.

Theorem 1

${X^{cubs}}$ can be a subset of both ${X^{male}}$ and ${X^{female}}$ only if ${C_r}$ is selected in such a way that ${C_r} \cdot L=L$.

Proof of Theorem 1

It is known that $B$ has ${C_r} \cdot L$ number of ones in arbitrary vector positions and zeros in the remaining positions. Hence, ${X^{male}} \circ B$ have elements of ${X^{male}}$ and zeros from the positions where $B$ has ones and zeros, respectively. In contrast, ${X^{female}} \circ \bar {B}$ has elements of ${X^{female}}$ and zeros from the positions where $B$ has zeros and ones, respectively, as $\bar {B}$ is the one’s complement of $B$. Hence, it can be said that ${X^{male}} \circ B \subset {X^{male}}$ and ${X^{female}} \circ \bar {B} \subset {X^{female}}$. As ‘+’ operator in the mathematical representation of crossover operation given in [11] is equivalent to set union operation, the resultant ${X^{cubs}}$ are subset of both and only ${X^{male}}$ and ${X^{female}}$, i.e., ${X^{cubs}} \subset {X^{male}},{X^{female}}$.

Theorem 2

In a nomad coalition of only two lions, the evaluation score $E_{1}^{{nomad}}$ will be always greater than $E_{2}^{{nomad}}$ when $E_{1}^{{nomad}}$ is greater than or equal to exponential function of unity and vice versa.

Proof of Theorem 2

Let $E_{1}^{{nomad}}$ and $E_{2}^{{nomad}}$ be the evaluation scores of $X_{{^{1}}}^{{nomad}}$ and $X_{2}^{{nomad}}$, respectively. The evaluation scores can be calculated as

$$E_{1}^{{nomad}}=\exp \left( {\frac{{{d_1}}}{{\text{max} \left( {{d_1},{d_2}} \right)}}} \right)\frac{{\text{max} \left( {f\left( {X_{1}^{{nomad}}} \right),f\left( {X_{2}^{{nomad}}} \right)} \right)}}{{f\left( {X_{1}^{{nomad}}} \right)}}$$

(24)

$$E_{2}^{{nomad}}=\exp \left( {\frac{{{d_2}}}{{\text{max} \left( {{d_1},{d_2}} \right)}}} \right)\frac{{\text{max} \left( {f\left( {X_{1}^{{nomad}}} \right),f\left( {X_{2}^{{nomad}}} \right)} \right)}}{{f\left( {X_{2}^{{nomad}}} \right)}}$$

(25)

where ${d_1}$ is the Euclidean distance between $X_{{^{1}}}^{{nomad}}$ and ${X^{male}}$, ${d_2}$ is the Euclidean distance between $X_{2}^{{nomad}}$ and ${X^{male}}$.

From lemmas 3 and 4, it can be said that if $E_{1}^{{nomad}}=\exp \left( 1 \right)$ (according to lemma 3) and $E_{1}^{{nomad}}>\exp \left( 1 \right)$ (according to lemma 4), then $E_{2}^{{nomad}}<\exp \left( 1 \right)$. Similarly, lemmas 5 and 6 asserts $E_{2}^{{nomad}}>E_{1}^{{nomad}}$, if $E_{2}^{{nomad}} \geq \exp \left( 1 \right)$. Hence, the theorem states that it is not necessary to calculate both $E_{1}^{{nomad}}$ and $E_{2}^{{nomad}}$ to evaluate $X_{{^{1}}}^{{nomad}}$ and $X_{2}^{{nomad}}$, respectively. It is sufficient to calculate either of them and can be concluded by comparing it with $\exp \left( 1 \right)$.