text:
\begin{document}

\title{Why the Firefly Algorithm Works?}

\author{Xin-She Yang$^1$, Xing-shi He$^2$ \\[10pt]
1)  School of Science and Technology, Middlesex University, London NW4 4BT, UK.\\
2) College of Science, Xi'an Polytechnic University, Xi'an, China.
}

\date{}

\maketitle

\begin{center}
\parbox{5.2in}{
\noindent {\small {\bf Citation Detail}: X.-S. Yang, X.S. He, Why the firefly algorithm works?
in: {\it Nature-Inspired Algorithms and Applied Optimization} (Edited by X.-S. Yang), Springer,  pp. 245-259 (2018).
\href{https://doi.org/10.1007/978-3-319-67669-2_11}{https://doi.org/10.1007/978-3-319-67669-2\_11}
}  }
\end{center}

\abstract{Firefly algorithm is a nature-inspired optimization algorithm and there have been significant developments since its appearance about ten years ago. This chapter summarizes the latest developments about the firefly algorithm and its variants as well as their diverse applications. Future research directions are also highlighted.

}

\section{Introduction}

Nature-inspired computation has become a new paradigm in optimization, machine learning, data mining and computational intelligence with a diverse range of applications.
The essence of nature-inspired computing is the nature-inspired algorithms such as
genetic algorithm (GA) \cite{Holland}, particle swarm optimization (PSO) \cite{Kennedy} and firefy algorithm (FA) \cite{Yang}. Most nature-inspired algorithms use some characteristics of swarm intelligence \cite{Fisher}, and an overview of swarm intelligence to nature-inspired computation was recently carried out by Yang \cite{YangIEEE}.

Among nature-inspired algorithms, firefly algorithm (FA) was developed by Xin-She Yang in late 2007 and early 2008 \cite{Yang}, and it is almost ten years since its development. Significant developments have been made in the last few years, and thus this chapter intends to provide a state-of-the-art review of FA and its variants with an emphasis on the most recent studies.

Therefore, this chapter is organized as follows. Section 2 introduces the fundamentals of the firefly algorithm, and Section 3 explains why FA works well in practice. Section 4 highlights the main differences between FA and PSO, and Section 5 summarizes some of the recent variants of FA.
Section 6 reviews some of the diverse applications of FA and its variants. Finally, Section 7 concludes with discussion of future research directions.

\section{Firefly Algorithm}

The bioluminescence flashes of fireflies are an amazing sight in the summer sky in tropical and temperate regions. It is estimated that there are about 2000 species of fireflies and most species produce short, rhythmic flashes. Each species can have different flashing patterns and rhythms, and one of the main functions of such flashing light acts as a signaling system to communicate with other fireflies. The rate of flashing, intensity of the flashes and the amount of time between flashes form part of the signaling system \cite{Lewis}, and female fireflies respond to a male's unique flashing pattern. Some tropical fireflies can even synchronize their flashes, leading to self-organized behaviour.

As light intensity in the night sky decreases as the distance from the flashing source increases, the range of visibility can be typically a few hundred metres, depending on weather conditions. The attractiveness of a firefly is usually linked to the brightness of its flashes and the timing accuracy of its flashing patterns.

\subsection{The Standard Firefly Algorithm}

 Based on the above characteristics, Xin-She Yang developed the firefly algorithm (FA) \cite{Yang,YangFA}. Inside FA, the attractiveness of a firefly is determined by its brightness.
 Due to exponential decay of light absorption and inverse-square law of light variation with distance, a highly nonlinear term is used to simulate the variation of light intensity or attractiveness.

 In the FA, the main algorithmic equation for the position $\x_i$ (as a solution vector to a problem) is
\be  \x_i^{t+1} =\x_i^t + \beta_0 e^{-\gamma r^2_{ij} }
(\x_j^t-\x_i^t) + \alpha \; \ff{\epsilon}_i^t, \label{FA-equ-100} \ee
where $\alpha$ is a scaling factor controlling the step sizes of the random walks,
while $\gamma$ is a scale-dependent parameter controlling the visibility of the fireflies (and thus
search modes). In addition, $\beta_0$ is the attractiveness constant when the
distance between two fireflies is zero (i.e., $r_{ij}=0$). In the above equation,
the second term on the right-hand side (RHS) is the nonlinear attractiveness
 which varies with distance, while the third term is a randomization term and
 $\ff{\epsilon}_i^t$ means that the random number vectors should be drawn from
 a Gaussian distribution at each iteration.

This system is a nonlinear system, which may lead to rich characteristics
in terms of algorithmic behaviour.
Loosely speaking, FA belongs to the category of swarm intelligence (SI) based algorithms,
and all SI-based algorithms use some aspects of swarming intelligence \cite{Fisher}.

It is worth pointing out that the distance $r_{ij}$ between firefly $i$ and
firefly $j$ can be defined as their Cartesian distance. However, for some problems
such as the internet routing problems, this `distance' can be defined as time delay.
For certain combinatorial problems, it can be defined even as Hamming distance \cite{Osaba}.
In addition, since the brightness of a firefly is associated with the objective landscape
with its position as the indicator, the attractiveness of a firefly seen by others, depending on their relative positions and relative brightness. Thus, the beauty is in the eye of the beholder. Consequently, a pair comparison is needed for comparing all fireflies. The main steps of FA can be summarized as the pseudocode in Algorithm~\ref{Alg-1}.

\begin{algorithm}
\hrule
Initialize all the parameters ($\alpha, \beta, \gamma, n$)\;
Initialize randomly a population of $n$ firefies\;
Evaluate the fitness of the initial population at $\x_i$ by $f(\x_i)$ for $i=1,...,n$\;
\While{($t<$ MaxGeneration)}{
\For{All fireflies ($i=1:n$)}
{\For{All other fireflies ($j=1:n$) (inner loop)}{
\If{Firefly $j$ is better/brighter than $i$}
{Move firefly $i$ towards $j$ according to Eq.~(\ref{FA-equ-100})\;}}
Evaluate the new solution and accept the new solution if better\; }
Rank and update the best solution found so far\;
Update iteration counter $t \gets t+1$\;
Reduce $\alpha$ (randomness strength) by a factor\;
} \hrule
\caption{Firefly algorithm. \label{Alg-1}}
\end{algorithm}

Furthermore, $\alpha$ is a parameter controlling the strength of the randomness or perturbations in FA. The randomness should be gradually reduced to speed up the overall convergence. Therefore, we can use
\be \alpha=\alpha_0 \theta^t, \ee
where $\alpha_0$ is the initial value and $0<\theta<1$ is a reduction factor. In most cases, we can use $\theta=0.9$ to $0.99$, depending on the type of problems and the desired quality of solutions.

In fact, since FA is a nonlinear system, it has the ability to automatically
subdivide the whole swarm into multiple subswarms. This is because short-distance
attraction is stronger than long-distance attraction, and the division of
swarm is related to the mean range of attractiveness variations.
After division into multi-swarms, each subswarm can potentially swarm
around a local mode. Consequently, FA is naturally suitable for multimodal
optimization problems. Furthermore, there is no explicit use of the best solution $\ff{g}^*$,
thus selection is through the comparison of relative brightness
according to the rule of `beauty is in the eye of the beholder'.

\subsection{Special Cases of FA}

To gain more insight, let us analyze the FA system more carefully.
By looking at Eq.~(\ref{FA-equ-100}) closely, we can see that $\gamma$ is an important scaling parameter \cite{Yang,YangFA}.

\subsubsection{Case A: $\gamma=0$}

At one extreme, we can set $\gamma=0$, which means that there is no exponential decay and thus the visibility is very high. In this case, all fireflies can see each other in the whole domain and we
have \be \x_i^{t+1}=\x_i^t +\beta_0 (\x_j^t-\x_i^t) +\alpha \epsilon_i^t. \ee

\begin{itemize}

\item If $\gamma=0$, $\alpha=0$ and $\beta_0$ is fixed, then FA becomes a variant of differential evolution (DE) without crossover \cite{Storn,Yang2014}.
In this special case, if we replace $\x_j$ by the best solution in the group $\ff{g}^*$,
this reduced FA is equivalent to
a special case of accelerated particle swarm optimization (APSO) \cite{Yang,Yang2014}.

\item If $\beta_0=0$, FA is equivalent to the basic simulated annealing (SA) with $\alpha$ as the cooling schedule \cite{Yang2014}. In addition, if $\epsilon_i$ is further replaced by $\epsilon \x_i$, this special case is equivalent to the pitch adjustment of the harmony search (HS) algorithm.

\end{itemize}

Thus, it is clear that DE, APSO, SA and HS are special cases of the standard FA. In other words,
FA can be considered as a good combination of APSO, HS, SA and DE enhanced in a nonlinear system.
It is no surprise that FA can outperform these algorithms for many applications.

\subsubsection{Case B: $\gamma \gg 1$}

At the other extreme when $\gamma \gg 1$,  the visibility range is very short.
Fireflies are essentially flying in a dense fog and they cannot see each other clearly. Thus, each firefly flies independently and randomly. In fact, the exponential term $\exp[-\gamma r_{ij}^2]$
will decrease significantly if $\gamma r_{ij}^2=1$, which means that the radius $R$ or range of influence can be defined by
\be R=\frac{1}{\sqrt{\gamma}}. \ee
Therefore, a good value of $\gamma$ should be linked to the scale or limits of the design variables so that the fireflies within a range are visible to each other.

For a given objective landscape, if the average scale of the domain is $L$, then $\gamma$ can be estimated by
\be \gamma=\frac{1}{L^2}. \ee
If there is no prior knowledge about its possible scale, we can start with $\gamma=1$ for most problems, and then increase or decrease it when necessary. In theory, $\gamma \in [0, \infty)$,
but in practice, we can use $\gamma=O(1)$, which means that we can use $\gamma=0.001$ to $1000$ for most problems we may meet.

\subsection{Discrete FA}
The standard FA was designed to solve continuous optimization problems. In order to solve  discrete optimization problems, some discretization techniques should be used. For example, one way of converting a continuous variable $x$ to a binary one is to use the sigmoidal function
\be S(x)=\frac{1}{1+e^{-x}}, \ee
where $S \rightarrow 1$ for $x \rightarrow \infty$, while $S \rightarrow 0$
for $x \rightarrow -\infty$. However, this S-shaped function requires a large range to get a proper conversion. In practice, many researchers use an additional rule with a random threshold. A common technique is to use a random number $r \in [0,1]$. If $S>r$, then $S=1$, otherwise, $S=0$. Obviously, once we have $S \in \{0, 1\}$, we can use $u=2S-1$ to get $u \in \{+1, -1\}$ if needed.

Another way of conversion is to use random permutation. For example, a set of a uniformly distributed random number such as  $r=[0.3,\; 0.9, ..., \;0.7]$ can be converted to integers.
On the other hand, an interesting conversion technique is to use a modulus function by
\be u = \lfloor x +k \rfloor \;\; \textrm{mod} \;\; m, \ee
to convert $x$ to an integer $u$. Here, $k$ and $m>0$ are integers.

There are other methods for discretization, including random keys, random permutation, Hamming-distance based method, tanh(x), and others \cite{Rodrig}.

Many studies using FA have demonstrated how the algorithm works and the effectiveness of the algorithm. Interested readers can refer to the book by Yang \cite{Yang2014} and reviews \cite{Fister,Tila2}.  Now let us explain in more detail why the algorithm works.

\section{Why the Firefly Algorithm Works?}

In the above descriptions, we have explained the main steps of the FA and how it works.
We now try to summarize why it works so well in practice.

The exact reasons why FA works may require further mathematical analysis, specially for the variants to be introduced later. As it still needs a theoretical framework to explain the working mechanisms  of FA and all other algorithms, we do not intend to figure out all the reasons why an algorithm works. However, from both empirical observations and the analysis of the algorithm structure, we can summarize the following four reasons why the FA works \cite{Yang2014}:

\begin{itemize}
\item From the special cases discussed in the previous section, we know that APSO, SA, HS and DE are special cases of FA, and thus FA can be considered as a good combination of all these algorithms. Therefore, it is no surprise that FA can work more efficiently than these algorithms.

\item Due to the nonlinear attraction mechanism in FA, the short-distance attraction is stronger than long-distance attraction; therefore, the whole swarm can automatically subdivide into multiple subswarms. Each swarm can potentially swarm around a local mode, and among all the local modes, there is always a global optimal solution. Consequently, the multiswarm nature of FA enables FA to find multiple optimal solutions simultaneously and FA is naturally suitable for solving nonlinear, multimodal optimization problems.
    Therefore, for a given problem with $m$ modes, if the number of fireflies $n$ is much higher than $m$ (i.e., $n \gg m$), then all the optima (including the global best) can be found simultaneously.

\item The influence radius or range is controlled by $\gamma$. As a small value of $\gamma$ means higher influence and higher visibility, while a higher value of $\gamma$ reduces its influence and visibility. Therefore, we can tune $\gamma$ to control the subdivision of the swarm. If $\gamma=0$, there is no subdivision and all fireflies belong to a single swarm.
    A moderate value of $\gamma$ leads to multiswarms, while a much higher value of $\gamma$ may lead to individual random walks without a swarm. As a result, the diversity and properties of the population are linked to $\gamma$. This nonlinearity provides much richer dynamic characteristics.

\item In comparison with PSO and other algorithms, FA does not use velocities explicitly, which means that FA does not have any drawbacks associated with velocities. In addition, FA does not use $\ff{g}^*$ in its equation. The use of $\ff{g}^*$ can potentially lead to premature convergence if the initial $\ff{g}^*$ lies in the wrong region, which will attract all other agents towards it. Therefore, FA can avoid any disadvantage associated with $\ff{g}^*$.

\end{itemize}

It is worth pointing out that all these parameters have to be tuned properly. For example, $\alpha$ as the strength of the random walks must be reduced gradually; otherwise, the convergence may be slowed down by too much randomness. Similarly, a proper value of $\gamma$ has to be tuned to allow a good set of subswarms to emerge automatically \cite{Yang, YangCSFA}.

\section{FA is not PSO}

Though FA and PSO are both swarm intelligence based algorithms, they thus share some similarity;  however, FA is not PSO because they have some significant differences.
Apart from the different inspiration from nature, we briefly summarize here the main differences between FA and PSO:

\begin{itemize}
\item FA is a nonlinear system due to the nonlinear attraction term $\beta_0 \exp(-\gamma r_{ij}^2)$, while PSO is a linear system because its updating equations are linear in terms of $\x_i$ and $\v_i$. The nonlinear dynamic nature of FA can lead to much richer characteristics in terms of algorithmic behaviour and population properties.

\item The strong nonlinearity of FA means that FA has an ability of multi-swarming, while PSO cannot. Thus,  FA can find multiple optimal solutions simultaneously and consequently deal with multimodal problems more effectively.

\item PSO uses velocities, but FA does not. Thus, FA does not have the drawbacks associated with velocity initialization and instability for high velocities of particles.

\item FA has some scaling control (via $\gamma$), while PSO has no scaling control. Such scaling control can give FA more flexibility.

\end{itemize}

All these differences enable FA to search the design spaces more effectively for multimodal objective landscapes.

\section{Variants of FA}

Since the development of FA in 2008, it has been applied to many applications \cite{YangCSFA}.
A comprehensive review was done by Fister et al. in 2013 \cite{Fister}, covering the literature up to 2013. Yang and He provided another review from a different perspective in 2013 \cite{YangHe}. More recently, Tilahun et al. provided an updated review on the continuous versions of the firefly algorithm and its variants \cite{Tila2}, and the discrete versions of the firefly algorithms were also reviewed by Tilahun and Ngnotchouye in 2017 \cite{Tila}.

Despite the success of the standard FA, many variants have been developed to enhance its performance in the last few years. Again many of these variants have been reviewed by Fister et al. \cite{Fister} and Tilahun et al. \cite{Tila2}, and we will not repeat their coverage. Instead, here we will focus only on the most recent variants that have just appeared  in the last few years.

Though there are a diverse range of variants of FA, they can be loosely put into the following
six major variants/categories:
\begin{itemize}

\item {\bf Discrete FA}: The standard FA was designed to solve problems in the continuous domains. To solve discrete or combinatorial optimization problems, some modifications are needed.
For example, Marichelvam et al. developed a discrete FA for solving hybrid flow shop scheduling problems \cite{Marich,Marich2}, while Osaba developed a discrete FA for solving vehicle routing problems with recycling policy \cite{Osaba}. In addition, Poursalehi et al. used an effective discrete FA for optimizing fuel reload design of nuclear reactors \cite{Poursa}.
Zhang et al. used a discrete double-population for assembly sequence planning \cite{ZhangZ}.
These variants can be used to solve scheduling and planning problems as well as routing problems.

\item {\bf Adaptive FA}: In the standard FA, parameters are fixed, and it may be advantageous to use adaptively varying parameter values. Baykasoglu and Ozsoydan developed an adaptive firefly algorithm with chaos to solve mechanical design problems \cite{Bayk}, and G\'alvex and Iglesias \cite{Galvez} developed a memetic self-adaptive FA for shape fitting \cite{Galvez}.

\item {\bf Modified/Enhanced FA}: Researchers have designed various ways to modify and enhance the performance of FA. For example, Cheung et al. developed a non-homogeneous FA \cite{Cheung}, while Chou and Ngo developed a modified FA for multidimensional structural optimization \cite{Chou}. Darwish combined FA with a Bayesian classifer for solving classification problems \cite{Darwish}, while Fister et al. used quaternion to represent the solutions of FA in higher dimensions \cite{Fister2}.

    In addition, He and Huang used a modified FA for multilevel thresholding of color image segmentation \cite{HeLF}. Gupta used a modifed FA for controller design \cite{Gupta}. Tesch and Kaczorowska used a rotational FA for arterial cannula shape optimization \cite{Tesch}. Verma et al. developed an opposition and dimensional based modified firefly algorithm \cite{Verma}. Furthermore, Wang et al. developed a modified FA based on light intensity difference \cite{WangB}, while Wang et al. modified FA with neighborhood attraction \cite{WangH} and Yu et al. developed a variable step size FA \cite{Yu}.

    Additionally, Zhou et al. used an information-fusing FA for wireless sensor placement for structural monitoring \cite{Zhou}, and Zhou et al. combined FA with Newton's method to identify boundary conditions for transient heat conduction problems \cite{ZhouHL}.

\item {\bf Chaotic FA}: Some of the parameters in the FA can be replaced by the outputs of some chaotic maps, which may be able to enhance the exploration ability of the FA. For example,
Gandomi et al. developed a chaotic FA in 2013 \cite{Gandomi}, while Gokhale and Kale used a tent map for their chaotic FA \cite{Gokhale}. Also, Zouache et al. developed a quantum-inspired FA for discrete optimization problems \cite{Zou}, and Dhal et al. developed a chaotic FA for enhancing image contrast \cite{Dhal}.  Chaos-based FA variants were reviewed by Fister et al. \cite{Fister3}.

\item {\bf Hybrid FA}: Hybridization can be a good way to create new algorithm tools by combing
the advantages of each algorithm involved in the hybrid. For example, Aleshab and Abdullah developed a hybrid FA with a probabilistic neural network for solving classification problems \cite{Alwesh}, and Zhang et al. developed a hybrid by combing FA with DE and achieved
improved performance and accuracy \cite{ZhangLina}.

\item {\bf Multiobjective FA}: The standard FA was for single objective optimization
and Yang extended the standard FA to multiobjective firefly algorithm (MOFA) for design optimization \cite{YangMOFA}. In addition, Eswari and Nickolas developed a modified multiobjective FA for task scheduling \cite{Eswari}, while Wang et al. developed a hybrid multiobjective FA for big data optimization \cite{WangH2}, and Zhao et al. developed a decomposition-based multiobjective FA for RFID network planning with uncertainty \cite{Zhao}.

\end{itemize}

\section{Applications of FA and its Variants}

The applications of FA and its variants are diverse, a quick Google scholar search gives
more than 7000 outputs, and it is not possible to cover all these applications here.
It is not our intention to review even a good fraction of the applications in the current literature. For comprehensive reviews, interested readers can refer to \cite{Fister,Tila,Tila2}.  Here, our emphasis will be on the recent, new applications that can be representative in areas from engineering design to energy systems and from scheduling to image processing. For example, FA has been applied in the design of radial expanders in organic Rankine cycles \cite{Bahad},  design optimization of steel frames \cite{Carbas}, distributed generation system \cite{Chaur}, beam design \cite{Erdal}, wavelet neural network optimization \cite{Zainu}, hysteresis model identification \cite{Zaman}, detection of TEC seismo-ionospheric anomalies \cite{Akhoon} and structural search in chemistry \cite{Avenda}.

In the area of clustering and classification, Senthinath et al. compared and evaluated the performance of clustering using FA \cite{Senthin}. Gope et al. used FA for rescheduling of real power for congestion management concerning pumped storage hydro-units \cite{Gope}.
Long et al. used FA for heart disease predictions \cite{Long}.

For applications in design and optimization, Mohanty applied FA for designing shell and tube heat exchangers \cite{Mohanty}, while Shukla and Singh used FA to select parameters for advanced machining processes \cite{Shukla}. Hung applied FA in OFDM systems \cite{Hung} and
Kamarian et al. used FA for thermal buckling optimization of composite plates \cite{Kamar}, while
Jafari and Akbari used FA to optimize micrometre-scale resonator modulators \cite{Jafa}, and
Othman et al. used a supervised FA to achieve optimal placement of distributed generators \cite{Othman}. Also, Singh et al. combined FA with least-squares method
to estimate power system harmonics \cite{Singh}, and Kaur and Ghosh used a fuzzy FA for network reconfiguration of unbalanced distribution networks \cite{Kaur}.

In the area of energy engineering and energy systems, Ghorbani et al. used FA for prediction of gas flow rates from gas condensate reservoirs \cite{GhorbH}, and Massan et al. used FA to solve wind turbine applications \cite{Massan}. Wang et al. used an FA-BP neural network to forecast electricity price \cite{WangD} and Rastgou and Moshtagh used FA for multi-stage transmission expansion planning \cite{Rastgou}. In addition,  Satapathy et al. used a hybrid HS-FA based approach to improve the stability of PV-BESS diesel generator-based microgrid \cite{Sata}.

For image processing,  Kanimozhi and Latha used FA for region-based image retrieval \cite{Kani},
and Rajinikanth and Couceiro used an FA-based approach for color image segmentation \cite{Raji}. S\'achez et al. used FA to optimize modular granular
neural networks for human recognition \cite{Sanchez}.
Rahebi and Hardalac used FA for optic disc detection in retinal images \cite{Rahebi}, while
Gao et al. used FA for visual tracking \cite{Gao}, and Zhang et al. used a discrete FA for endmember extraction from hyperspectral images \cite{Zhang}.

In the area of time series and forecasting,
Xiao et al. used a combined model for electrical load forecasting \cite{Xiao} and
Ghorbani et al. used FA for capacity prediction in combination with support vector machine \cite{Ghorb}. In addition, Ibrahim and Khatib used a hybrid model for solar radiation prediction \cite{Ibrah}.

In the area of planning and navigation, Ma et al. used FA for planning navigation paths \cite{Ma}, and Patle et al. used FA to optimize mobile robot navigation \cite{Patle}.

In deep learning and software engineering, Rosa et al. used FA for learning parameters in deep belief networks \cite{Rosa}. Srivatsava used an FA-based approach for generating optimal software test sequences \cite{Sriva}, and  Kaushik et al. integrated FA in artificial neural network for predicting software costs accurately \cite{Kaush}.

Other applications include nanoscale structural optimization by Kougianos and Mohanty \cite{Kougi}, protein complex identification  by Lei et al. \cite{Lei},  protein structure prediction by Maher et al. \cite{Maher}. Also,
Nekouie and Yaghoobi used FA to carry out multimodal optimization \cite{Nekou}, and
Sundari et al. used an improved FA for programmed PWM in multilevel inverters with adjustable DC sources \cite{Sundari}.

\section{Conclusions and Future Directions}

As we have seen from the above reviews and discussions, FA and its variants have been successfully
applied in a wide spectrum of real-world applications. Despite its success, there are still some interesting future research directions concerning FA, and we will summarize them as follows.

\begin{enumerate}

\item {\it Theory}: Though we know FA and its variants work well, we do not have solid theoretical proof why they work and under exactly what conditions. A recent study by He et al. proved the global convergence of the flower pollination algorithm \cite{He2017}. It can be expected that the same methodology can be applied to analyze the firefly algorithm and other algorithms. Therefore, more theoretical analysis is needed.

\item {\it Adaptivity}: All bio-inspired algorithms including FA have parameters, and tuning of these parameters can be tedious. Ideally, algorithms should be able to tune their parameters using a self-tuning framework \cite{YangSTA} and also adapt their values to suit for a given type of problems. Future work can focus on the parameter adaptivity of FA and its variants.

\item {\it Hybrid}: Though there are many different variants of FA, it is no doubt that more hybrid variants will appear in the future. At the moment, hybridization is by trial and error, and it is not clear yet how to achieve a better hybrid by combining different algorithms, which needs more research and further insight.

\item {\it Co-evolution}: Simple hybridization can often work well; however, co-evolution can be more advantageous by co-evolving two or more algorithms together so as to allow the successful characteristics of an algorithm to enhance the co-evolutionary algorithm structure. It is not clear how to carry out co-evolution of algorithms.

\item {\it Applications}: In addition to the diverse range of applications reviewed in this chapter, there are more research opportunities of applying FA and its variants. Future applications can focus on the area in big data, deep learning and large-scale problems.
    Big data in combination with machine learning techniques such as deep nets can be an active research area for many years to come, and the nature-inspired algorithms can expect to play an important role in this area.

\end{enumerate}

As we can see that FA and its variants have been very successful in many applications, there are more opportunities for future research and applications. The authors hope that this work can inspire future research in the above mentioned directions with more real-world applications.


TITLE: How many different group homomorphisms from $\mathbb{Z}/\mathbb{3Z}$ into $SL_{n}(7)$ are there?
QUESTION [3 upvotes]: I have the following question:

How many different group homomorphisms from $\mathbb{Z}/\mathbb{3Z}$ into $SL_{n}(7)$ are there?

I think that the answer is the number of elements of order $3$ in $SL_{n}(7)$.
For $n=1$, we have just trivial homomorphism. And for $n=2$, $|SL_{2}(7)|=336$ There must be a method to answer this question. Thanks!

REPLY [2 votes]: The number of homomorphisms is actually the number of elements of order $3$ plus one, because of the trivial homomorphism.
It might be easier to do the counting in ${\rm GL}_n(7)$, and then decide how many of your conjugacy classes of elements of order $3$ have determinant $1$.
Every element of order $3$ can be diagonalized in ${\rm GL}_n(7)$, and its conjugacy class is determined by its multiset of eigenvalues, which are all equal to $1,2$ or $4$.
So the classes can be parametrized by pairs $a,b$, the multiplicities of eigenvalues $2$ and $4$, where $1 \le a+b \le n$, and the centralizer of the such an element (in ${\rm GL}_n(7)$) is isomorphic to ${\rm GL}_a(7) \times {\rm GL}_b(7) \times {\rm GL}_{n-a-b}(7)$. So you can work out the order of this class.
To get determinant $1$, you need $3|(a-b)$. It would not be too hard to calculate all this for a fixed $n$, but a general formula would be  a bit of a mess.
Why do you want to do this anyway?


TITLE: Find a limit without l'Hospital: $\lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$
QUESTION [9 upvotes]: Find $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$.
$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ =
 $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \ln(1+\frac{1}{2^x})}{\ln3^x + \ln(1+\frac{1}{3^x})}\right)$ =
 $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \frac{1}{2^x}\ln(1+\frac{1}{2^x})^{2^{x}}}{\ln3^x + \frac{1}{3^x}\ln(1+\frac{1}{3^x})^{3^{x}}}\right)$ = $\cdots$
This method is not working.

REPLY [1 votes]: We have
$$
\ln(1+t)=t+o(t^2).
$$
Thus, since $2^x\to0$ and $3^x\to0$ when $x\to-\infty$,
$$
\frac{\ln(1+2^x)}{\ln(1+3^x)}=\frac{2^x+o(2^{2x})}{3^x+o(3^{2x})}
=\left(\frac23\right)^x\,\frac{1+o(2^x)}{1+o(3^x)}.
$$
The second fraction goes to $1$ when $x\to-\infty$. So the limit will be equal to
$$
\lim_{x\to-\infty}\left(\frac23\right)^x=\lim_{x\to-\infty}\exp\left(x\ln\left(\frac23\right)\,\right)=+\infty.
$$


TITLE: Prove that the equation $x+\ln x=5$ has at least one root, and find it with precision of 2 digits
QUESTION [0 upvotes]: I need to prove that the equation $x+\ln x=5$ has at least one root, and to find that root in precision of two digits after the decimal point.
I know how to prove the root's existence using the intermediate value theorem:
I defined $f(x)=x+\ln x-5$, and $f(1)=-4<0$ and $f(e^5)=e^5>0$ so the root exists.
I don't know how to find it, and how to prove that the precision is good enough.

REPLY [0 votes]: Consider that you look for the zero's of function $$f(x)=x+\log(x)-5$$ We have
$$f'(x)=1+\frac 1x$$ which is aloways positive in the real domain because of the logarithm. So, $f(x)$ is an increasing function with only one zero.
Now, to make life easier, use inspection for $x=e^k$ $(k=0,1,2,\cdots)$ and you will see that the root is between $e$ $(f(e)=e-4\approx -1.28 <0)$ and $e^2$ $(f(e^2)=e^2-3\approx 4.39>0)$. So, the root is more than likely closer to $e$ than to $e^2$.
If you do not want (or cannot) use Lambert function, you need a numerical method such as Newton for example. So, using $x_0=e$ the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 2.71828 \\
 1 & 3.65529 \\
 2 & 3.69340 \\
 3 & 3.69344
\end{array}
\right)$$
You could also expand $f(x)$ as a Taylor series built around $x=e$. This would give
$$f(x)=(e-4)+\left(1+\frac{1}{e}\right) (x-e)-\frac{(x-e)^2}{2 e^2}+\frac{(x-e)^3}{3
   e^3}+O\left((x-e)^4\right)$$ Using the expansion to $O\left((x-e)^2\right)$, this would give (as an approximation) $x \sim \frac{5 e}{1+e} \approx 3.65529$  which is, by definition, the first iterate of Newton method. Using the expansion to $O\left((x-e)^3\right)$ gives a quadratic equation is $(x-e)$ and the solution would be $x \sim e \left(2+e-\sqrt{e (4+e)-7}\right)\approx 3.70329$ which is better.
Starting from the above expansion, you could also use series reversion which would give
$$x=e+t+\frac{t^2}{2 e (1+e)}+\frac{(1-2 e) t^3}{6 e^2 (1+e)^2}+O\left(t^4\right) \qquad \text{where} \qquad t=\frac{e (f(x)-e+4)}{1+e}$$ Making $f(x)=0$ that is to say $t=\frac{(4-e) e}{1+e}$ would give
$$x=e+\frac{(4-e) e}{1+e}+\frac{(e-4)^2 e}{2 (1+e)^3}+\frac{(e-4)^3 e (2 e-1)}{6 (1+e)^5}+\cdots \approx 3.69277$$ which is almost the solution.


TITLE: The Helium mass fraction from the Big Bang Nucleosynthesis
QUESTION [3 upvotes]: In Perkin's book Particle Astrophysics (page 144): I do not understand how one comes to the following expression (the second equality with $r$) for the Helium mass fraction due to the Big Bang Nucleosynthesis:
$$Y= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}= \frac{2r}{1+r} $$
where $r=N_\text{n}/N_\text{p}$.
The first equality follows from the fact that He is (approximately) 4 times heavier than H:
$$Y = \frac{m_\text{He}}{m_\text{He}+m_\text{H}}= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}.$$
However I can't derive the second equality relating $Y$ to $r$:
$N_\text{He}= 2N_\text{p} + 2N_\text{n}$ and  $N_\text{H}= N_\text{p} + N_\text{n}$
$$Y= \frac{4N_\text{He}}{4N_\text{He}+N_\text{H}}= \frac{8(N_\text{n}+N_\text{p})}{9(N_\text{n}+N_\text{p})}\quad???$$

REPLY [5 votes]: The problem is you have the wrong relations between $\{N_\mathrm{H}, N_\mathrm{He}\}$ and $\{N_\mathrm{p}, N_\mathrm{n}\}$. Every hydrogen contains 1 proton, and every helium contains 2, so $N_\mathrm{p} = N_\mathrm{H} + 2 N_\mathrm{He}$. The neutrons are only contributed to by helium in the accounting: $N_\mathrm{n} = 2 N_\mathrm{He}$. Inverting these relations yields
\begin{align}
N_\mathrm{H} & = N_\mathrm{p} - N_\mathrm{n} \\
N_\mathrm{He} & = \frac{1}{2} N_\mathrm{n}.
\end{align}
It looks like you got the direction wrong, in the sense that there should be fewer helium nuclei than protons or neutrons (the 2's are on the wrong side). Also, "hydrogen" means ${}^1\mathrm{H}$ not ${}^2\mathrm{D}$ unless otherwise stated.


TITLE: Calculate the area of a particular pentagon-circumference intersection.
QUESTION [0 upvotes]: I'm trying to answer the following question:
"Given a regular pentagon with side length equal to $r$ and a circle of radius $r$ that intersects the pentagon in two of its consecutive vertices and has its centre outside the pentagon, calculate the area of the intersection of the two figures as a function of $r$."
I don't even know where to start, any tip is really really much appreciated!
Thanks in advice!

REPLY [0 votes]: Okay, I think I got the formula:
Basically I added the area of the pentagon, the area of the circle and subtracted the circular sector relative to the angle of 60 degrees.
$r^2\phi+r^2\pi-(\frac{\pi r^2}{6}-\frac{\sqrt3r^2}{4})$
then
$r^2(\phi+(\pi-\frac{\pi}{6}\ ) + \frac{\sqrt3}{4})$
and finally
$r^2(\frac{5}{6}\pi+\phi+\frac{\sqrt3}{4}) \approx r^2(4,7709)$
Hope this is right, and that maybe one day it can be useful to someone!


TITLE: What is the most general group possible?
QUESTION [3 upvotes]: I have read that free groups are the "most general" groups given some generators (from Wikipedia):

The construction of a free product is similar in spirit to the construction of a free group (the most general group that can be made from a given set of generators).

I'd like to know: is a free group the most general possible type of group? That is, are there any groups that cannot be described in group presentation (i.e. with a set of generators and a complete set of relations)? If so, what would such a group look like, and how would it relate to a free group? (Can you get that group by somehow restricting a free group? Is it more general than a free group? Are they not comparable?)

REPLY [3 votes]: Every group has a presentation.  Let $G$ be any group, and let $F$ be the free group on the underlying set of $G$.  Then by the universal property of free groups, there is a homomorphism $p:F\to G$ which sends each generator to the corresponding element of $G$.  Since we have a generator for each element of $G$, this homomorphism is surjective.  Letting $R$ be the kernel of $p$, we then have a presentation of $G$ as the quotient of the free group $F$ by the set of relations $R$.


TITLE: How to show that the event that a prisoner does not go free is not measurable
QUESTION [9 upvotes]: I was reading this webpage a few months ago about the following problem-

A countable infinite number of prisoners are placed on the natural numbers, facing in the positive direction (ie, everyone can see an infinite number of prisoners).  Hats will be placed and each prisoner will be asked what his hat color is.  However, to complicate things, prisoners cannot hear previous guesses or whether they were correct.  In this new situation, what is the best strategy?

(I won't link the best strategy in case someone wants to give it a go but note that my question is about the solution)
and I recall my friend and I were trying to come up with a formal argument for why the probability is ill-defined. We kept going in circles so we left it in the end.
Recently though I stumbled upon the page again, and saw Terence Tao's comment; I copy the relevant paragraph,

If we have an infinite number of prisoners, with the hats assigned randomly (thus, we are working on the Bernoulli space ${\Bbb Z}_2^{\Bbb N}$), and one uses the strategy coming from the axiom of choice, then the event $E_j$ that the $j^{th}$ prisoner does not go free is not measurable, but formally has probability 1/2 in the sense that $E_j$ and its translate $E_j + e_j$ partition ${\Bbb Z}_2^{\Bbb N}$ where $e_j$ is the $j^{th}$ basis element, or in more prosaic language, if the $j^{th}$ prisoner’s hat gets switched, this flips whether the prisoner gets to go free or not. The “paradox” is the fact that while the $E_j$ all seem to have probability 1/2, each element of the event space lies in only finitely many of the $E_j$. This can be seen to violate Fubini’s theorem – if the $E_j$ are all measurable. Of course, the $E_j$ are not measurable, and so one’s intuition on probability should not be trusted here.

It feels like he concludes the non-measurability of $E_j$ from a violation of Fubini, but I don't see it. Can someone flesh this argument out for me? It has been nagging me for a long time now and I would be very grateful :)

REPLY [3 votes]: The problem, as far as I can see, is with this sentence

The “paradox” is the fact that while the $E_j$ all seem to have probability 1/2, each element of the event space lies in only finitely many of the $E_j$. This can be seen to violate Fubini’s theorem – if the $E_j$ are all measurable.

I don't understand it either. Say, we denote by $n(\omega)$ the number of sets where $\omega$ belongs to. Then, by double counting (= Fubini),  $$\mathsf E[n(\omega)] = \sum_{j\ge 1} \mathsf P(E_j) = +\infty.$$ However, this in no way contradicts the finiteness of $n(\omega)$.
So I dare to doubt (this part of) Terence Tao's argument, and will give my own, adding more details about the objects involved.
We have $\Omega = \{0,1\}^{\mathbb N} = \{\omega = (\omega_1,\omega_2,\dots)\mid \forall\, i\ge 1\ \omega_{i}\in\{0,1\}\}$. The $\sigma$-algebra $\mathcal F$ of events is generated by cylindrical sets $C_{a_1,a_2,\dots,a_N} = \{\omega\mid \omega_i = a_i,i=1,\dots,N\}$, and the probability $\mathsf P$ is obtained as the extension of $\mathsf P(C_{a_1,\dots,a_N}) = 2^{-N}$.
Define the flipping operators: $$
T_i\omega = (\omega_1,\dots,\omega_{i-1},1-\omega_i,\omega_{i+1}, \omega_{i+2},\dots).$$
Obviously, they preserve the probability $\mathsf P$.
Now assume that all the events $E_j = \{j\text{th prisoner escapes}\}$ are in $\mathcal F$. It is obvious that $T_j E_j = \Omega\setminus E_j$, so $\mathsf P(E_j) = 1/2$, as $T_j$ is $P$-preserving. Thus, by Fatou's lemma,
$$
\mathsf P(\text{all but finite number of prisoners escape}) = \mathsf P(\liminf_{j\to\infty} E_j)\le \liminf_{j\to\infty} \mathsf P(E_j) = 1/2.\tag{1}
$$
However, the event $E:= \{\text{all but finite number of prisoners escape}\} = \liminf_{j\to\infty} E_j$ is a tail event, that is, it does not depend on any finite number of $\omega_j$'s1. Since $\omega_j$'s are independent, Kolmogorov's 0-1 law implies that $\mathsf P(E) \in\{0,1\}$. Consequently, $\mathsf P(E)=0$, contradicting the sure escape.

1 Here I use that the $j$th prisoner sees only hats in front of him and does not hear what is going on behind him. However, (1) stays valid (thanks to PhoemueX for pointing this out) even in the case where the $j$th prisoner knows everything except his own hat.


TITLE: Reflexive Generalized Inverse
QUESTION [0 upvotes]: Definition:
G is a generalized inverse of A if and only if AGA=A.G is said to be reflexive if and only if GAG=G.
I was trying to solve the problem:
If A is a matrix and G be it's generalized inverse then G is reflexive if and only if rank(A)=rank(G).
One direction of the proof is easy but I am finding it difficult to prove the second part.I have proved that GAG is infact a generalized inverse but why it should be equal to G but not any other generalized inverse is not very apparent to me

REPLY [1 votes]: Suppose that $G$ is a generalized inverse (i.e. $AGA = A$) such that $A$ and $G$ have the same rank. Let $A = PQ$ be a rank decomposition, so that $P$ has $r$ columns and $Q$ has $r$ rows with $r$ equal to the rank of $A$. Similarly, let $G = BC$ denote a rank decomposition of $G$. We have
$$
AGA = A \implies P(QB)(CP)Q = PQ.
$$
Because $P$ has independent columns and $Q$ has independent rows, we can conclude from the above that
$$
P[(QB)(CP) - I_r]Q = 0 \implies (QB)(CP) - I_r = 0 \implies (QB)(CP) = I_r,
$$
which is to say that $(QB)^{-1} = (CP)$. With that, it follows that
$$
GAG = B(CP)(QB)C = BC = G,
$$
which is to say that $G$ is a reflexive inverse, which was what we wanted.


TITLE: Find the probability of an event given the probability of the union and the probability of the complement
QUESTION [0 upvotes]: "There are two independent events, A and B, and we know that the probability that both occurs simultaneously is $\frac{1}{6}$ and the probability that none of them occurs is $\frac{1}{3}$. So, what is the probability that just one of them occurs?"
I need to solve this assignment's question and I'm struggling to figure out how to do that. I think I need to know P(A$\bigcap$B) to solve it but I cannot find that!
Does anyone know how to solve it?

REPLY [3 votes]: Case I. You only need to know the probability that one of the two events occurs, not the two actual probabilities.
In that case, if both occur with probability $\frac{1}{6}$, and neither occurs with probability $\frac{1}{3}$, then the only remaining possibility is that exactly one occurs, and that must be what?
Case II. You actually need to determine the two probabilities $P(A)$ and $P(B)$ (though you can't know which is which).
The fact that the two events are independent means that the probability that both occur simultaneously can be written as $P(A)P(B)$.  You know that equals $\frac{1}{6}$.  The probability that neither occurs is $[1-P(A)][1-P(B)]$, and you know that equals $\frac{1}{3}$.
You should now be able to solve for the two probabilities, although you can't tell which is which.


text:
\begin{document}

\maketitle

\begin{abstract}
In this paper, we define a lassoing on a link, a local addition of a trivial knot to a link.
Let $K$ be an $s$-component link with the Conway polynomial non-zero.
Let $L$ be a link which is obtained from $K$ by $r$-iterated lassoings.
The complete splitting number $\mathrm{split}(L)$ is greater than or equal to $r+s-1$,
and less than or equal to $r+\mathrm{split}(K)$.
In particular, we obtain from a knot by $r$-iterated component-lassoings an algebraically completely splittable link $L$ with $\mathrm{split}(L)=r$.
Moreover, we construct a link $L$ whose unlinking number is greater than $\mathrm{split}(L)$.
\end{abstract}

\section{Introduction}

\noindent The splittability of a link is one of the basic concepts in knot theory.
For example, the splittability interacts with polynomial invariants:
the Alexander polynomial and the Conway polynomial take zero for a splittable link.
Jones polynomial and skein polynomial have specific formulae with respect to the split sum.
Moreover, the splittabilities of links or spatial graphs are studied and applied to other subjects: chemistry, biology, psychology, etc.
For example, Kawauchi proposed a model of prion proteins as a spatial graph \cite{kawauchip2}, and
Yoshida studied its splittability which concerns with the study of prion diseases: mad cow disease, scrapie, Creutzfeldt-Jakob disease, etc. \cite{yoshida}.
Another example is about a model of human mind which is also proposed by Kawauchi \cite{kawauchip}, \cite{kawauchip2};
by considering one's mind as a knot and by considering a mind relation of $n$ persons as an $n$-component link,
the models ``mind knots'' and ``mind links'' are studied.
The splittability of a link corresponds to the ``self-releasability'' of a mind link.

For a two-component link, Adams defined the splitting number which represents how distant the link is from a splittable link \cite{adams}.
In this paper, we define for an $n$-component link $L$ $(n=2,3,4,\dots )$ the complete splitting number $\mathrm{split}(L)$ which represents how distant the link is from a completely splittable link.
The \textit{unlinking number} $u(L)$ of a link $L$ is the minimal number of crossing changes in any diagram of $L$ which are needed to obtain the trivial link $L$.
Since a trivial link is completely splittable, we have $\mathrm{split}(L)\le u(L)$.
Lassoing is a crossing-changing and loop-adding local move as shown in Figure \ref{quoitting}
(we give the precise definitions of completely splittable, complete splitting number, and a lassoing in Section \ref{csn}).

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=50mm]{quoitting.eps}
\caption{}
\label{quoitting}
\end{center}
\end{figure}

\noindent For any $r$-component link $L=L_1\cup L_2\cup...\cup L_r$ $(r=1,2,3,\dots )$ with the Conway polynomial
$\nabla(L)\ne 0$, there are $\ell$-iterated lassoings from $L$ to an
algebraically completely splittable link $L^*$ with $\nabla (L^*)\ne 0$ where
$\ell =\sum_{i<j}|Link(L_i,L_j)|$ (we define an algebraically completely splittable link in Section \ref{csn}).
For any $s$-component link $K=K_1\cup K_2\cup...\cup K_s$ $(s\ge 1)$ with
$\nabla(K)\ne 0$, there are $(\ell +u)$-iterated lassoings from $K$ to an
algebraically completely splittable link L with trivial components such that
$\nabla(L)\ne 0$ where
$\ell =\sum_{i<j}|Link(K_i,K_j)|$ and $u=\sum_{i=1}^s u(K_i)$.
In this paper, we show the following theorem:

\phantom{x}
\begin{thm}
Any link $L$ obtained from any $s$-component link $K=K_1\cup K_2\cup...\cup K_s$ $(s=1,2,3,\dots )$ with
$\nabla(K)\ne 0$ by $r$-iterated lassoings $(r=0,1,2,\dots )$ satisfies
$$r+split(K)\ge split(L)\ge r+s-1.$$
\label{mainthm}
\end{thm}
\phantom{x}

\noindent We have the following corollaries:

\phantom{x}
\begin{cor}
For any $s$-component link $K=K_1\cup K_2\cup...\cup K_s$ $(s=1,2,3,\dots )$ with $\mathrm{split}(K)=s-1$,
and any integer $r\ge \ell +u$ where $\ell =\sum_{i<j}|Link(K_i,K_j)|$ and $u=\sum_{i=1}^s u(K_i)$,
there are $r$-iterated lassoings from $K$ to an algebraically completely splittable link $L$ with trivial components such that
$\mathrm{split}(L)=r+s-1$.
\end{cor}
\phantom{x}

\phantom{x}
\begin{cor}
Let $K$ be a knot. Let $L$ be a link which is obtained from $K$ by $r$-iterated lassoings ($r=1,2,3,\dots $).
Then $L$ has $\mathrm{split}(L)=r$.
\label{knot-lasso}
\end{cor}
\phantom{x}

\noindent We define a \textit{component-lassoing} to be the lassoing at a self-crossing point of a diagram.
We have the following corollary:

\phantom{x}
\begin{cor}
Every link $L$ obtained from a knot $K$ by $r$-iterated component-lassoings $(r=1,2,3,\dots )$
is an $(r+1)-$component algebraically completely splittable link with $\mathrm{split}(L)=r$.
\label{comp-lasso-cor}
\end{cor}
\phantom{x}

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=30mm]{7-2-6.eps}
\caption{}
\label{7-2-6}
\end{center}
\end{figure}

\noindent For example, the link $7^2_6$ depicted in Figure \ref{7-2-6} which is a link obtained from a trefoil knot by a single component-lassoing,
has the linking number zero and $\mathrm{split}(7^2_6)=1$.
We also remark that $u(7^2_6)=2$ (\cite{kohn}).
Adams also showed in \cite{adams} that there is a two-component link, each component of which is trivial, but such that its splitting number is less than its unlinking number, like the link $7^2_6$.
We show in Section \ref{n-t} that for any integer $r>0$ and any knot $K$ with Nakanishi's index $e(K)>2r$,
any link $L$ obtained from $K$ by $r$-iterated lassoings is a link such that $\mathrm{split}(L)<u(L)$, i.e., $L$ is non-trivial by any $r$ crossing changes.

\section{Complete splitting number}
\label{csn}

\noindent Let $L=L_1\cup L_2\cup \dots \cup L_r$ be a link consisting of sublinks $L_i$ $(i=1,2,\dots ,r)$.
A link $L$ is \textit{splittable into} $L_1,L_2,\dots ,L_r$ if there exist mutually disjoint 3-balls $B_i$ ($i=1,2,\dots ,r$)
in $S^3$ such that $L_i \subset B_i$.
For example, the link $M$ in Figure \ref{splitinto} is splittable into $M_1$ and $M_2$
whereas the link $N$ is not splittable into $N_1$ and $N_2$.
A link $L$ is \textit{splittable} if $L$ is splittable into subdiagrams $L_1$ and $L_2$, where $L=L_1\cup L_2$, $L_1,L_2\neq \phi$.
For example, the link $M$ in Figure \ref{splitinto} is a splittable link.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=60mm]{splitinto.eps}
\caption{}
\label{splitinto}
\end{center}
\end{figure}

\noindent A link $L$ is \textit{completely splittable} if $L$ is splittable into all the knot components of $L$.
In particular, a knot is assumed as a non-splittable link but a completely splittable link.
A link $L$ is \textit{algebraically completely splittable} if every two knot components $K_i$ and $K_j$ of $L$ have the linking number
$\mathrm{Link}(K_i,K_j)=0$.
For example, the link $E$ in the left hand of Figure \ref{compsplit} is not completely splittable
but algebraically it is completely splittable.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=60mm]{compsplit.eps}
\caption{}
\label{compsplit}
\end{center}
\end{figure}

\noindent The \textit{complete splitting number} $\mathrm{split}(D)$ of a link diagram $D$ is the minimal number of crossing changes
which are needed to obtain a diagram of a completely splittable link from $D$.
For example, the link diagram $F$ in the right hand in Figure \ref{compsplit} has $\mathrm{split}(F)=1$.
As a relation to the warp-linking degree $ld(D)$ of $D$, we have $\mathrm{split}(D)\leq ld(D)$,
where the warp-linking degree is a restricted warping degree which can be calculated directly or by using matrices \cite{shimizu-d}, \cite{shimizu}.
The \textit{complete splitting number} $\mathrm{split}(L)$ of a link $L$ is the minimal number of crossing changes in any diagram of the link which are needed to obtain a completely splittable link.

\noindent Let $p$ be a crossing point of a link diagram $D$.
We put a lasso around $p$, i.e., we apply a crossing change at $p$, and add a loop alternately around the crossing
as shown in Figure \ref{quoitting}.
Then, we obtain another link diagram $D'$.
The diagram $D'$ is said to be obtained from $D$ by \textit{lassoing} at $p$.
Let $L'$ be the link which has the diagram $D'$.
The link $L'$ is said to be obtained from $L$ by a \textit{lassoing}.
For example, we obtain the Borromean ring from the Hopf link by a lassoing (see Figure \ref{lasso-ex}).

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=80mm]{lasso-ex.eps}
\caption{}
\label{lasso-ex}
\end{center}
\end{figure}

\noindent A link $L'$ is said to be obtained from $L$ by \textit{$r$-iterated lassoings}
if $L'$ is obtained from $L$ by lassoings $r$ times iteratively.
For example, the link $L$ in Figure \ref{2-lasso} is a link obtained from a trivial knot by two-iterated lassoings.
Since a lassoing depends on the choice of a crossing point and the choice of a diagram of the link,
we may have many types of link by a lassoing.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=145mm]{2-lasso.eps}
\caption{}
\label{2-lasso}
\end{center}
\end{figure}

\section{Conway polynomial}

\noindent Let $\nabla (L;z)$ be the Conway polynomial of a link $L$ with an orientation.
We have the following lemma:

\phantom{x}
\begin{lem}
We have
\begin{align*}
\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=20pt]{11.eps}
         \end{minipage}
;z\right) =-z^3\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=18pt]{12.eps}
         \end{minipage}
;z\right) ,
\end{align*}

\begin{align*}
\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=20pt]{21.eps}
         \end{minipage}
;z\right) =-z^3\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=18pt]{12.eps}
         \end{minipage}
;z\right) ,
\end{align*}

\begin{align*}
\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=20pt]{31.eps}
         \end{minipage}
;z\right) =z^3\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=18pt]{32.eps}
         \end{minipage}
;z\right) ,
\end{align*}

\begin{align*}
\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=20pt]{41.eps}
         \end{minipage}
;z\right) =z^3\nabla \left( \begin{minipage}{20pt}
              \includegraphics[width=18pt]{32.eps}
         \end{minipage}
;z\right) .
\end{align*}
\phantom{x}
\begin{proof}
We obtain the first equality by the skein relations in Figure \ref{conway}.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=115mm]{conway.eps}
\caption{}
\label{conway}
\end{center}
\end{figure}

\noindent The other equalities are similarly obtained.
\end{proof}
\label{lem-conway}
\end{lem}
\phantom{x}

\begin{eg}
The link diagram $D$ in Figure \ref{conway-ex} is obtained from a diagram of a trefoil knot by 2-iterated lassoings.
Then we have $\nabla (L)=z^3\times z^3\times \nabla (3_1)=z^6(1+z^2)$, where $L$ is a link represented by $D$, and $3_1$ is a trefoil knot.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=25mm]{conway-ex.eps}
\caption{}
\label{conway-ex}
\end{center}
\end{figure}
\end{eg}
\phantom{x}

\noindent We remark that for a link $L'$ with $\nabla (L')=0$, there are no lassoings from $L'$ to $L$ with $\nabla (L)\neq 0$.
We have the following corollary:

\phantom{x}
\begin{cor}
Let $L$ be a link obtained from a link $L'$ with $\nabla (L')\neq 0$, in particular from any knot $K$,
by $r$-iterated lassoings ($r=1,2,3,\dots $).
Then we have $\nabla (L)\neq 0$.
\label{knot-quoit}
\end{cor}
\phantom{x}

\noindent Let $\Lambda$ be the integral Laurent polynomial ring, i.e., $\Lambda =\mathbb{Z}[t,t^{-1}]$.
With respect to the one-variable Alexander polynomial, we have the following corollary by Lemma \ref{lem-conway}
by substituting $t^{\frac{1}{2}}-t^{-\frac{1}{2}}$ for $z$:

\phantom{x}
\begin{cor}
Let $L'$ be a link which is obtained from a link $L$ by a lassoing. Then we have
$$\Delta _{L'}(t){\dot =}(t-1)^3\Delta _L(t), $$
where $\Delta _L(t)$ is the one-variable Alexander polynomial of $L$,
and ${\dot =}$ means equal up to multiplications of the units of $\Lambda$.
\end{cor}
\phantom{x}

\noindent We show an example.

\phantom{x}
\begin{eg}
We have
$$\Delta (5^2_1)\doteq \Delta (7^2_8)\doteq \Delta (8^2_{15})\doteq (t-1)^3\Delta (0),$$
$$\Delta (7^2_6)\doteq \Delta (9^2_{55})\doteq \Delta (9^2_{56})\doteq (t-1)^3\Delta (3_1),$$
$$\Delta (8^2_{13})\doteq (t-1)^3\Delta (4_1),$$
$$\Delta (9^2_{31})\doteq (t-1)^3\Delta (5_1),$$
$$\Delta (9^2_{32})\doteq \Delta (9^2_{33})\doteq (t-1)^3\Delta (5_2),$$
where $\Delta (L)=\Delta _L(t)$.
All the two-component links with the crossing number nine or less which are obtained from knots by lassoings
have been listed above.
\end{eg}
\phantom{x}

\noindent Up to multiplications of $t-1$, the one-variable Alexander polynomial of any link is the Alexander polynomial of an algebraically completely
splittable link consisting of trivial components:

\phantom{x}

\begin{cor}
Let $(t-1)^mf(t)$ be the Alexander polynomial of a link, where $m$ is a non-negative integer, $f(t)\in \Lambda$, and $f(1)\neq 0$.
Then, there exists a non-negative integer $n$ such that the Laurent polynomial $(t-1)^{m+3n}f(t)$ is the Alexander polynomial of
an algebraically completely splittable link consisting of trivial components.

\begin{proof}
We can change a crossing by a lassoing.
\end{proof}
\end{cor}
\phantom{x}

\section{Proof of Theorem \ref{mainthm}}

\noindent In this section, we prove Theorem \ref{mainthm}.
Before the proof, we define some notions which are due to \cite{kawauchi} to prove Theorem \ref{mainthm}.
For the integral Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$,
a \textit{multiplicative set} of $\Lambda$ is a subset $S\subset \Lambda -\{ 0\} $ which satisfies the following three conditions:
the units $\pm t^i$ ($i\in \mathbb{Z}$) are in $S$,
the product $gg'$ of any elements $g$ and $g'$ of $S$ is in $S$, and
every prime factor of any element $g\in S$ is in $S$.
For the quotient field $Q(\Lambda)$ of $\Lambda$ and a multiplicative set $S$ of $\Lambda$,
$\Lambda _S=\{ f/g \in Q(\Lambda ) | f \in \Lambda , g \in S \}$ is a subring of $Q(\Lambda)$.
For a finitely generated $\Lambda$-module $H$, let $H_S$ be the $\Lambda _S$-module $H\otimes _{\Lambda}\Lambda _S$, and
$e_S(H)$ the minimal number of $\Lambda _S$-generators of $H_S$.
We take $e_S(H)=0$ when $H=0$.
We call $e_{S}(H)$ the $\Lambda _{S}$-\textit{rank} of $H$.
Let $L$ be an oriented link in $S^3$, and $E(L)=cl(S^3-L)$ the compact exterior of $L$.
Let $\tilde{E}(L)\rightarrow E(L)$ be the infinite cyclic covering
which is induced from the epimorphism $\gamma _L:\pi _1(E(L))\rightarrow \mathbb{Z}$
sending each oriented meridian of $L$ to $1\in \mathbb{Z}$.
Then we can regard $H_1(\tilde{E}(L))$ as a finitely $\Lambda$-module.
We denote $e_S(H_1(\tilde{E}(L)))$ by $e_S(L)$.
Let $L$, $L'$ be links which have the same number of components.
By Theorem 2.3 in \cite{kawauchi}, we immediately have
\begin{align}
d^X(L,L')\geq \vert e_s(L)-e_s(L')\vert ,
\label{ka23}
\end{align}
where $d^X(L,L')$ denotes the $X$-distance between $L$ and $L'$.
We prove Theorem \ref{mainthm}.

\phantom{x}
\noindent {\it Proof of Theorem \ref{mainthm}.}
Let $L$ be a link which is obtained from a link $K=K_1\cup K_2\cup \dots \cup K_s$ with $\nabla (K)\ne 0$
by $r$-iterated lassoings $(r=1,2,3,\dots )$.
Let $L'$ be a completely splittable link which is obtained from $L$ by $m$ crossing changes,
where $m=\mathrm{split}(L)=d^X(L,L')$.
We set $S=\Lambda -\{0\} $.
Since $L'$ is completely splittable and the number of components of $L'$ is $r+s$, we have
\begin{align}
e_s(L')=r+s-1.
\label{esl0}
\end{align}
The Alexander polynomial of $L$ is non-zero because the Conway polynomial of $L$ is non-zero by Corollary \ref{knot-quoit}.
Hence we have
\begin{align}
e_s(L)=0.
\label{esl1}
\end{align}
By substituting the equalities (\ref{esl0}), (\ref{esl1}) and $d^X(L,L')=\mathrm{split}(L)$ into the inequality (\ref{ka23}), we have
$$\mathrm{split}(L)\geq r+s-1.$$
From the $r$-iterated lassoings, we have
$$r+\mathrm{split}(K)\ge \mathrm{split}(L).$$
Hence we have the inequality
$$r+\mathrm{split}(K)\ge \mathrm{split}(L)\ge r+s-1.$$
\hfill$\square$\\
\phantom{x}

\noindent As the contraposition of Theorem \ref{mainthm}, we have the following corollary:

\phantom{x}
\begin{cor}
Let $K=K_1\cup K_2\cup \dots \cup K_s$ be an $s$-component link.
If $K$ has $\mathrm{split}(K)<s-1$, then $\nabla (K)=0$.
\end{cor}

\section{Non-triviality}
\label{n-t}

\noindent In this section, we discuss the non-trivialities of completely splittable links which are obtained from $L$
in Corollary \ref{knot-lasso} by $r$ crossing changes $(r=1,2,\dots )$.
For a link $L$ obtained from a knot $K$ by $r$-iterated lassoings, we have the following theorem:

\phantom{x}
\begin{thm}
If a link $L$ is obtained from a knot $K$ with $e(K)>2r$ by $r$-iterated lassoings $(r=1,2,3,\dots )$,
then we have $\mathrm{split}(L)=r$ and $u(L)>r$.
\label{split-and-u}
\end{thm}
\phantom{x}

\noindent We remark that in Theorem \ref{split-and-u} the link L is an algebraically completely splittable link if the $r$-iterated lassoings are all component-lassoings.
Before proving Theorem \ref{split-and-u}, we have the following Lemma:

\phantom{x}
\begin{lem}
Let $L_0=K_1+K_2+\dots +K_{r}$ be a completely splittable link with $r$ components.
Then we have
$$u(L_0)=\sum _{i=1}^ru(K_i).$$

\phantom{x}
\begin{proof}
We have $u(L_0)=u+u_1+u_2+\dots +u_r$,
where $u$ is the number of non-self crossing changes and
$u_i$ is the number of crossing changes on $K_i$ which are needed to obtain the trivial link from $L_0$.
Then we have
$$u(L_0)=u+u_1+\dots +u_r\ge u_1+\dots +u_r\ge \sum _{i=1}^ru(K_i).$$
Since $L_0$ is completely splittable, we have
$$u(L_0)\le \sum _{i=1}^ru(K_i).$$
Therefore the equality holds.
\end{proof}
\label{lem-u}
\end{lem}
\phantom{x}

\noindent We show Theorem \ref{split-and-u}

\phantom{x}
\noindent {\it Proof of Theorem \ref{split-and-u}.}
Let $L_0=K_1+K_2+\dots +K_{r+1}$ be a completely splittable link
which is obtained from $L$ by $r$ crossing changes.
For the integral Laurent polynomial ring $\Lambda =\mathbb{Z}[t,t^{-1}]$,
we take the multiplicative set $S$ of $\Lambda$ so that $S$ is the set of units of $\Lambda$.
Then $e_S(L)$ is equivalent to Nakanishi's index $e(L)$ \cite{kawauchi}.
Since we can consider $L_0=K_1+K_2+\dots +K_{r+1}$ to be a connected sum $O^{r+1}\# K_1\# K_2\# \dots \# K_{r+1}$, we have
\begin{align*}
H_1(\tilde{E}(L_0))&\cong H_1(\tilde{E}(O^{r+1}))\oplus H_1(\tilde{E}(K_1\# K_2\# \dots \# K_{r+1})) \\
                   &\cong \Lambda ^r \oplus H_1(\tilde{E}(K_1\# K_2\# \dots \# K_{r+1})).
\end{align*}
And by \cite{kawauchi2}, we have $e(L_0)=r+e(K_1\# K_2\# \dots \# K_{r+1})$.
By substituting this into the inequality (\ref{ka23}), we have
\begin{align*}
d^X(L,L_0)\geq |e(L)-e(L_0)|\geq e(L)-r-e(K_1\# K_2\# \dots \# K_{r+1}).
\end{align*}
Recall that $d^X(L,L_0)=\mathrm{split}(L)=r$. Then we have
\begin{align}
r\geq e(L)-r-e(K_1\# K_2\# \dots \# K_{r+1}).
\label{byconnectedsum}
\end{align}
Next, we consider another completely splittable link $K+O^r$ which is obtained from $L$ by the $r$ anti-lassoings (see Figure \ref{anti-lasso}).

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=50mm]{anti-lasso.eps}
\caption{}
\label{anti-lasso}
\end{center}
\end{figure}

\noindent Since $K+O^r=O^{r+1}\# K$, we have
\begin{align*}
H_1(\tilde{E}(L_0))\cong \Lambda ^r \oplus H_1(\tilde{E}(K)).
\end{align*}
And by \cite{kawauchi2}, we have $e(K+O^r)=r+e(K)$. Hence we have
\begin{align}
r\geq r+e(K)-e(L)
\label{byantilassoings}
\end{align}
by \cite{kawauchi}. By summing the inequalities (\ref{byconnectedsum}) and (\ref{byantilassoings}), we have
$$2r\geq e(K)-e(K_1\# K_2\# \dots \# K_{r+1}).$$
From Lemma \ref{lem-u}, we have
$$u(L_0)=\sum _{i=1}^{r+1}u(K_i)\geq u(K_1\# K_2\# \dots \# K_{r+1})\geq e(K_1\# K_2\# \dots \# K_{r+1})\geq e(K)-2r.$$
Hence $L_0$ is non-trivial if $e(K)>2r$.
\hfill$\square$\\
\phantom{x}

\noindent For a knot which has Nakanishi's index large enough, we can construct a link such that the unlinking number is
greater than the complete splitting number. Here is an example.

\phantom{x}
\begin{eg}
Since the knot $K$ in Figure \ref{connected-tref} which is the connected sum of $2r+1$ trefoil knots has Nakanishi's index $e(K)=2r+1$,
any link $L$ obtained from $K$ by $r$-iterated lassoings has the unlinking number more than $r$ whereas $\mathrm{split}(L)=r$.

\begin{figure}[!ht]
\begin{center}
\includegraphics[width=70mm]{connected-tref.eps}
\caption{}
\label{connected-tref}
\end{center}
\end{figure}
\end{eg}

\begin{ack}
The author is grateful to Reiko Shinjo for pointing out Adams' paper.
She is also grateful to the members of Friday Seminar on Knot Theory in Osaka City University
who assisted her in valuable comments and the tender encouragement.
She especially thanks her advisor, Professor Akio Kawauchi for his help, support, and the valuable advices.
She is partly supported by JSPS Research Fellowships for Young Scientists.
\end{ack}

config: arxiv


TITLE: Relationship between $G$-conjugacy and $N_G(D)$-conjugacy
QUESTION [1 upvotes]: Let $G$ be a group, $D$ a $p$-subgroup of $G$. Suppose $x,y$ both lie in $C_G(D)$ the centraliser of $D$, that is, $xd=dx$ and $yd=dy$ for all $d\in D$.
I am solving a problem about the centre of the group algebra $kG$, which I have reduced to showing that: $x,y$ are conjugate in $G$, that is $\exists g\in G$ such that $gxg^{-1}=y$, if and only if $x,y$ are conjugate in $N_G(D)$, where $N_G(D)$ is the normaliser in $G$ of $D$.
It is clear that if $x,y$ are conjugate in $N_G(D)$ then they are conjugate in $G$. If $D$ is the trivial subgroup these two notions agree, but I'm not sure how to do more than this.
Plausibly useful things are that $N_G(D)$ acts on $C_G(D)$ by conjugation, but I can't get an argument about the size of the two conjugacy classes to work.
Any hints much appreciated, I'm sure I'm missing something obvious.

REPLY [2 votes]: I am afraid that this is not true in general.
For a counterexample, let $G = A_6$, and let $P = \langle x,y \rangle \in {\rm Syl}_2(G)$, where we can take $x = (1,2)(3,4)$, $y = (3,5)(4,6)$.
Then $P$ is dihedral of order $8$, and we let $D := Z(P) = \langle (3,4)(5,6) \rangle$.
Then $C_G(D) = N_G(D) = P$, and $x$ and $y$ are conjugate in $G$ but not in $P$.


TITLE: What is wrong with this cardinality-based proof of the Cantor–Schröder–Bernstein theorem?
QUESTION [2 upvotes]: Let $A$ and $B$ be sets, not necessarily finite. Let $f: A\rightarrow B$ be injective, and let $g:B\rightarrow A$ be injective.
The Cantor–Schröder–Bernstein theorem then says there must be a bijection between $A$ and $B$.
I once got the proof of this assigned on a problem set, and when I presented the following proof, the TA said that it was insufficient. Take $\vert X\vert$ to be the cardinality of a set $X$.
$f$ injective $\implies \vert A\vert\le\vert B\vert$
$g$ injective $\implies \vert A\vert\ge\vert B\vert$
Therefore, $\vert A\vert=\vert B\vert$, and there is a bijection between $A$ and $B$. $\square$
I have never understood why this proof is insufficient but assume it has something to do with $\le$, $\ge$, and $=$ when it comes to comparing infinite cardinalities.
What is inadequate about that proof?

REPLY [1 votes]: The main issue is that you are assuming that the comparison of cardinalities is an order (in particular, that it is antisymmetric). However, that is precisely what the Cantor-Bernstein-Schroeder Theorem establishes. So you are trying to prove the theorem by invoking the theorem sub silentio.
To be precise, the following are fairly common ways to define comparison of cardinalities:
Definition. We say two sets $A$ and $B$ are equipollent (roughly, "equally powerful") if and only if there exists a bijection $f\colon A\to B$. We denote this by writing $A\sim B$.
Equipollence is an equivalence relation. That is, it is (i) Symmetric: $A\sim A$ for all sets $A$. (ii) Reflexive: for all sets $A$ and $B$, if $A\sim B$ then $B\sim A$. And (iii) Transitive: for all sets $A$, $B$, and $C$, if $A\sim B$ and $B\sim C$, then $A\sim C$.
Then we define some weaker conditions:
Definition. For sets $A$ and $B$, we write $A\preceq B$ if and only if there exists an injection $i\colon A\to B$. We write $A\succeq B$ if and only if there exists a surjection $s\colon A\to B$. We write $A\prec B$ if and only if $A\preceq B$ and $A$ is not equipollent to $B$; and we write $A\succ B$ if and only if $A\succeq B$ and $A$ is not equipollent to $B$.
Now, we would like several things to be true:

We would like $\preceq$ to define a (partial) order; that is, it should be reflexive, transitive, and antisymmetric, which in this case means: if $A\preceq B$ and $B\preceq A$, then $A\sim B$.

We would like $\succeq$ to define a (partial) order.

We would like $A\succeq B$ to be equivalent to $B\preceq A$.

For 1, the only difficult thing is antisymmetry. That is precisely the Cantor-Bernstein-Schroeder Theorem. This is what you invoked in your argument, so it cannot be used to prove the Cantor-Bernstein-Schroeder Theorem. Cantor-Bernstein-Schroeder is a theorem of $\mathsf{ZF}$ set theory: you do not need the Axiom of Choice to prove it.
For 2, likewise, the only tricky part is antisymmetry. It turns out the Dual Cantor-Bernstein-Schroeder Theorem cannot be proven in $\mathsf{ZF}$: it implies that every infinite set has a countable subset, and this is known to be independent of $\mathsf{ZF}$. See for example this math.overflow post.
Point 3 is not even true in general for a tricky reason: it is easy to verify that $\varnothing\preceq X$ for every set $X$; but if $X$ is not empty, then it is not true that $X\succeq\varnothing$!
But even if we exclude the empty set, so that we say "for nonempty sets $A$ and $B$, $A\preceq B$ is equivalent to $B\succeq A$," then this is not provable in $\mathsf{ZF}$ set theory. (You gloss over this problem by using $\geq$ as meaning the same as $\leq$ but in the opposite order; one should not do that with cardinalities!). If $A\preceq B$ and $A\neq\varnothing$, then we can prove that $B\succeq A$: given $i\colon A\to B$ injective, let $a_0\in A$ be fixed, and define $s\colon B\to A$ by letting $s(b)=a$ if $b\in i(A)$ and $i(a)=b$, and defining $s(b)=a_0$ otherwise.
However, in general we cannot show that if there is a surjection $s\colon B\to A$, then there must be an injection $i\colon A\to B$ (that is, in general we cannot show that $B\succeq A$ implies $A\preceq B$). We need the Axiom of Choice for this to hold.
There is also the problem of whether any two sets can be compared: given two sets $A$ and $B$, must there always be an injection $i\colon A\to B$ or an injection $j\colon B\to A$? This is called "Bernstein's Theorem", and it is in fact equivalent to the Axiom of Choice.
Now, in $\mathsf{ZFC}$ set theory, with the Axiom of Choice, you can prove that every set can be bijected with a unique cardinal (which are defined as ordinals that cannot be bijected with any strictly smaller ordinal). Then $|A|$ represents the unique cardinal that $A$ can be bijected with. We define cardinal order by saying $\kappa\leq \lambda$ if and only if $\kappa=\lambda$ or $\kappa\in\lambda$ (the order of the ordinals), and then we can prove that $A\preceq B$ if and only if $|A|\leq|B|$.
Assuming all of that is overkill for Cantor-Bernstein-Schroeder. Which is why your argument is either secretly circular, or it actually assumes a lot of things that you were perhaps not aware of in trying to do a "simpler" proof.