Biological Inspired Methods
Copyright By PowCoder代写 加微信 powcoder
\noindent \textbf{New cluster centres:}\\
$\mathbf{m}_1 = \frac{\mathbf{x}_1 + \mathbf{x}_2 + \mathbf{x}_3}{3} = \left[ \begin{array}{c} 0 \\ 4 \end{array} \right]$, \\
$\mathbf{m}_2 = \frac{\mathbf{x}_4 + \mathbf{x}_5 + \mathbf{x}_6}{3} = \left[ \begin{array}{c} 4 \\ 0 \end{array} \right]$
X=[4 0 2 -2; 2 -2 4 0; 2 2 2 2];
m=mean(X,2);
Xm=X-repmat(m,[1, 4]);
C = (Xm(:,1)*Xm(:,1)’+Xm(:,2)*Xm(:,2)’+Xm(:,3)*Xm(:,3)’+Xm(:,4)*Xm(:,4)’)/4;
[V D] = svd(C);
XLT = V(:,1:2)’*(X-repmat(m,[1,4]))
\noindent $\mathbf{m} = \left[ \begin{array}{c} 1 \\ 1 \\ 2 \end{array} \right]$, $\mathbf{X}_m = \left[ \begin{array}{cccc} 3 & -1 & 1 & -3 \\ 1 & -3 & 3 & -1 \\ 0 & 0 & 0 & 0 \end{array} \right]$, $\mathbf{C} = \left[ \begin{array}{ccc} 5 & 3 & 0 \\ 3 & 5 & 0 \\ 0 & 0 & 0 \end{array} \right]$ \\
$\mathbf{V} = \left[ \begin{array}{ccc} -0.7071 & -0.7071 & 0 \\ -0.7071 & 0.7071 & 0 \\ 0 & 0 & 1 \end{array} \right]$, $\mathbf{D} = \left[ \begin{array}{ccc} 8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \end{array} \right]$ \\
$\hat{\mathbf{V}}^T = \left[ \begin{array}{ccc} -0.7071 & -0.7071 & 0 \\ -0.7071 & 0.7071 & 0 \end{array} \right]$ \\
$\mathbf{X}_{KL} = \left[ \begin{array}{cccc} -2.8284 & 2.8284 & -2.8284 & 2.8284 \\ -1.4142 & -1.4142 & 1.4142 & 1.4142 \end{array} \right]$ (using two components)
http://web.mit.edu/be.400/www/SVD/Singular_Value_Decomposition.htm
x1 x2 x3 x4
(–0.7071, –3.5355 )
m1: class 1
x1 and x3 belong to class 1
m2: class 2
x2 and x4 belong to class 2
\noindent From the 2D plot, we choose:\\ $\mathbf{m}_1 = \left[ \begin{array}{c} -2.8284 \\ 0 \end{array} \right]$ and $\mathbf{m}_2 = \left[ \begin{array}{c} 2.8284 \\ 0 \end{array} \right]$.
\noindent Transformed data:\\ $\mathbf{x}_{LT} = \hat{\mathbf{V}}^T \mathbf{x} = \left[ \begin{array}{ccc} -0.7071 & -0.7071 & 0 \\ -0.7071 & 0.7071 & 0 \end{array} \right] \left[ \begin{array}{c} 3 \\ -2 \\ 5 \end{array} \right] = \left[ \begin{array}{c} -0.7071 \\ -3.5355 \end{array} \right]$.
\noindent Computing the Euclidean distance:\\ $\Vert \mathbf{x}_{LT} – \mathbf{m}_1 \Vert = 4.1231$ and $\Vert \mathbf{x}_{LT} – \mathbf{m}_2 \Vert = 5$,\\ the new data is classified as class 1.
\noindent Form the 1D plot, we choose:\\ $\mathbf{m}_1 = \left[ \begin{array}{c} -2.8284 \end{array} \right]$ and $\mathbf{m}_2 = \left[ \begin{array}{c} 2.8284 \end{array} \right]$.\\
\noindent Transformed data:\\ $\mathbf{x}_{LT} = \hat{\mathbf{V}}^T \mathbf{x} = \left[ \begin{array}{ccc} -0.7071 & -0.7071 & 0 \end{array} \right] \left[ \begin{array}{c} 3 \\ -2 \\ 5 \end{array} \right] = \left[ \begin{array}{c} -0.7071 \end{array} \right]$.
\noindent As $-0.7071$ is closer to $m_1$ ($\Vert \mathbf{x}_{LT} – \mathbf{m}_1 \Vert = 2.1213$ and $\Vert \mathbf{x}_{LT} – \mathbf{m}_2 \Vert = 3.5355$), the new data is classified as class 1.
Distance between sample and (updated) cluster centre
Identify the cluster centre with the shortest distance from sample
Updated the identified cluster centre
Updated m2 is used
Updated m3 is used
= which class?
Distance between sample and (updated) cluster centre
Updated the identified cluster centre:
or create a new cluster centre
Created cluster centres
Identify the cluster centre with the shortest distance from sample
Distance less than threshold? If < threshold, update cluster centre, otherwise create a new cluster Note: In each iteration, the updated cluster centres are used. $\mathbf{m}_j \leftarrow \mathbf{m}_j + \eta (\mathbf{x} - \mathbf{m}_j )$ \noindent New sample: $\mathbf{x} = \left[ \begin{array}{c} 0 \\ -2 \end{array} \right]$\\ \noindent Euclidean distance:\\ $\Vert \mathbf{x} - \mathbf{m}_1 \Vert = 5.5509$,\\ $\Vert \mathbf{x} - \mathbf{m}_2 \Vert = 4.7170$.\\ \noindent As $\mathbf{x}$ is closer to $\mathbf{m}_2$, it is classified as class 2. \noindent \textbf{Normalise memberships:}\\ $\mathbf{\mu} = \left[ \begin{array}{cccccc} \textcolor{red}{\mu_{11}} & \textcolor{red}{\mu_{12}} & \textcolor{red}{\mu_{13}} & \textcolor{red}{\mu_{14}} & \textcolor{red}{\mu_{15}} & \textcolor{red}{\mu_{16}} \\ \textcolor{blue}{\mu_{21}} & \textcolor{blue}{\mu_{22}} & \textcolor{blue}{\mu_{23}} & \textcolor{blue}{\mu_{24}} & \textcolor{blue}{\mu_{25}} & \textcolor{blue}{\mu_{26}} \end{array} \right] = \left[ \begin{array}{cccccc} \textcolor{red}{1} & \textcolor{red}{0.5} & \textcolor{red}{0.5} & \textcolor{red}{0.5} & \textcolor{red}{0.5} & \textcolor{red}{0} \\ \textcolor{blue}{0} & \textcolor{blue}{0.5} & \textcolor{blue}{0.5} & \textcolor{blue}{0.5} & \textcolor{blue}{0.5} & \textcolor{blue}{1} \end{array} \right]$. \newcommand{\xone}{\left[ \begin{array}{c} -1 \\ 3 \end{array} \right]} \newcommand{\xtwo}{\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]} \newcommand{\xthree}{\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]} \newcommand{\xfour}{\left[ \begin{array}{c} 4 \\ -1 \end{array} \right]} \newcommand{\xfive}{\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]} \newcommand{\xsix}{\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]} Dataset is:$\left[ \begin{array}{cccccc} \mathbf{x}_1 & \mathbf{x}_2 & \mathbf{x}_3 & \mathbf{x}_4 & \mathbf{x}_5 & \mathbf{x}_6 \end{array} \right] = \left[ \begin{array}{cccccc} -1 & 1 & 0 & 4 & 3 & 5 \\ 3 & 4 & 5 & -1 & 0 & 1 \end{array} \right]$. \noindent \textbf{Normalise memberships:}\\ $\mathbf{\mu} = \left[ \begin{array}{cccccc} \mu_{11} & \mu_{12} & \mu_{13} & \mu_{14} & \mu_{15} & \mu_{16} \\ \mu_{21} & \mu_{22} & \mu_{23} & \mu_{24} & \mu_{25} & \mu_{26} \end{array} \right] = \left[ \begin{array}{cccccc} 1 & 0.5 & 0.5 & 0.5 & 0.5 & 0 \\ 0 & 0.5 & 0.5 & 0.5 & 0.5 & 1 \end{array} \right]$. \noindent $\mu_{11}' = \frac{\mu_{11}}{\mu_{11} + \mu_{21}} = \frac{1}{1+0} = 1$\\ $\mu_{21}' = \frac{\mu_{21}}{\mu_{11} + \mu_{21}} = \frac{0}{1+0} = 0$\\ $\mu_{11}' + \mu_{21}' = 1$ \newcommand{\xone}{\left[ \begin{array}{c} -1 \\ 3 \end{array} \right]} \newcommand{\xtwo}{\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]} \newcommand{\xthree}{\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]} \newcommand{\xfour}{\left[ \begin{array}{c} 4 \\ -1 \end{array} \right]} \newcommand{\xfive}{\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]} \newcommand{\xsix}{\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]} \begin{center} \textbf{Iteration 1} \end{center} \noindent \textbf{Update cluster centres:} $$\mathbf{m}_i = \frac{\displaystyle \sum_{j=1}^n \mu_{ij}^b\mathbf{x}_j}{\displaystyle \sum_{j=1}^n \mu_{ij}^b} = \frac{\displaystyle \sum_{j=1}^6 \mu_{ij}^2\mathbf{x}_j}{\displaystyle \sum_{j=1}^6 \mu_{ij}^2} = \frac{\mu_{i1}^2 \mathbf{x}_1 + \mu_{i2}^2 \mathbf{x}_2 + \mu_{i3}^2 \mathbf{x}_3 + \mu_{i4}^2 \mathbf{x}_4 + \mu_{i5}^2 \mathbf{x}_5 + \mu_{i6}^2 \mathbf{x}_6}{\mu_{i1}^2 + \mu_{i2}^2 + \mu_{i3}^2 + \mu_{i4}^2 + \mu_{i5}^2+ \mu_{i6}^2}$$ $$\mathbf{m}_1 = \frac{1^2\xone+0.5^2\xtwo+0.5^2\xthree+0.5^2\xfour+0.5^2\xfive+0^2\xsix}{1^2+0.5^2+0.5^2+ 0.5^2 +0.5^2 +0^2} =\left[ \begin{array}{c} 0.5 \\ 2.5 \end{array} \right]$$ $$\mathbf{m}_2 = \frac{0^2\xone+0.5^2\xtwo+0.5^2\xthree+0.5^2\xfour+0.5^2\xfive+1^2\xsix}{0^2+0.5^2+0.5^2+ 0.5^2 +0.5^2 +1^2}=\left[ \begin{array}{c} 3.5 \\ 1.5 \end{array} \right]$$ \noindent \textbf{Normalise memberships:}\\ $\mathbf{\mu} = \left[ \begin{array}{cccccc} \textcolor{red}{\mu_{11}} & \textcolor{red}{\mu_{12}} & \textcolor{red}{\mu_{13}} & \textcolor{red}{\mu_{14}} & \textcolor{red}{\mu_{15}} & \textcolor{red}{\mu_{16}} \\ \textcolor{blue}{\mu_{21}} & \textcolor{blue}{\mu_{22}} & \textcolor{blue}{\mu_{23}} & \textcolor{blue}{\mu_{24}} & \textcolor{blue}{\mu_{25}} & \textcolor{blue}{\mu_{26}} \end{array} \right] = \left[ \begin{array}{cccccc} \textcolor{red}{1} & \textcolor{red}{0.5} & \textcolor{red}{0.5} & \textcolor{red}{0.5} & \textcolor{red}{0.5} & \textcolor{red}{0} \\ \textcolor{blue}{0} & \textcolor{blue}{0.5} & \textcolor{blue}{0.5} & \textcolor{blue}{0.5} & \textcolor{blue}{0.5} & \textcolor{blue}{1} \end{array} \right]$. \newcommand{\xone}{\left[ \begin{array}{c} -1 \\ 3 \end{array} \right]} \newcommand{\xtwo}{\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]} \newcommand{\xthree}{\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]} \newcommand{\xfour}{\left[ \begin{array}{c} 4 \\ -1 \end{array} \right]} \newcommand{\xfive}{\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]} \newcommand{\xsix}{\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]} \begin{center} \textbf{Iteration 1} \end{center} \noindent \textbf{Update cluster centres:} $$\mathbf{m}_i = \frac{\displaystyle \sum_{j=1}^n \mu_{ij}^b\mathbf{x}_j}{\displaystyle \sum_{j=1}^n \mu_{ij}^b} = \frac{\displaystyle \sum_{j=1}^6 \mu_{ij}^2\mathbf{x}_j}{\displaystyle \sum_{j=1}^6 \mu_{ij}^2} = \frac{\mu_{i1}^2 \mathbf{x}_1 + \mu_{i2}^2 \mathbf{x}_2 + \mu_{i3}^2 \mathbf{x}_3 + \mu_{i4}^2 \mathbf{x}_4 + \mu_{i5}^2 \mathbf{x}_5 + \mu_{i6}^2 \mathbf{x}_6}{\mu_{i1}^2 + \mu_{i2}^2 + \mu_{i3}^2 + \mu_{i4}^2 + \mu_{i5}^2+ \mu_{i6}^2}$$ $$\mathbf{m}_1 = \frac{1^2\xone+0.5^2\xtwo+0.5^2\xthree+0.5^2\xfour+0.5^2\xfive+0^2\xsix}{1^2+0.5^2+0.5^2+ 0.5^2 +0.5^2 +0^2} =\left[ \begin{array}{c} 0.5 \\ 2.5 \end{array} \right]$$ $$\mathbf{m}_2 = \frac{0^2\xone+0.5^2\xtwo+0.5^2\xthree+0.5^2\xfour+0.5^2\xfive+1^2\xsix}{0^2+0.5^2+0.5^2+ 0.5^2 +0.5^2 +1^2}=\left[ \begin{array}{c} 3.5 \\ 1.5 \end{array} \right]$$ \textbf{Calculate membership values:} $$\mu_{ij} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^{\frac{2}{b-1}}}{ \displaystyle \sum_{r=1}^{c} \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_r \Vert \right)^{\frac{2}{b-1}}} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^2}{ \displaystyle \sum_{r=1}^{2} \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_r \Vert \right)^2} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^2}{( 1/ \Vert \mathbf{x}_j - \mathbf{m}_1 \Vert )^2+ ( 1/ \Vert \mathbf{x}_j - \mathbf{m}_2 \Vert )^2}$$ \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|} $j$ & $\mathbf{x}$ & $\Vert \mathbf{x} - \mathbf{m}_1 \Vert$ & $\Vert \mathbf{x} - \mathbf{m}_2 \Vert$ & $\left(\frac{1}{\Vert \mathbf{x} - \mathbf{m}_1 \Vert}\right)^2$ & $\left(\frac{1}{\Vert \mathbf{x} - \mathbf{m}_2 \Vert}\right)^2$ & $u_{1j}$ & $u_{2j}$ \\ 1&$\left[ \begin{array}{c} -1 \\3 \end{array} \right]$& 1.5811 & 4.7434 & 0.4 & 0.0444 & 0.9 & 0.1 \\ \hline 2&$\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]$& 1.5811 & 3.5355 & 0.4 & 0.08 & 0.8333 & 0.1667 \\ \hline 3&$\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]$& 2.5495 & 4.9497 & 0.1538 & 0.0408 & 0.7903 & 0.2097\\ \hline 4&$\left[ \begin{array}{c} 4 \\-1 \end{array} \right]$& 4.9497 & 2.5495 & 0.0408 & 0.1538 & 0.2097 & 0.7903 \\ \hline 5&$\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]$& 3.5355 & 1.5811 & 0.08 & 0.4 & 0.1667 & 0.8333\\ \hline 6&$\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]$& 4.7434 & 1.5811 & 0.0444 & 0.4 & 0.1 & 0.9 \\ \end{tabular} \end{center} \noindent \begin{align*} \mu_{11} &= \frac{ \left( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_1 \Vert \right)^2}{( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_1 \Vert )^2+ ( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_2 \Vert )^2} \\ \quad &= \frac{0.4}{0.4 + 0.0444} = 0.9 \end{align*} \begin{align*} \mu_{21} &= \frac{ \left( 1/ \Vert \mathbf{x}_2 - \mathbf{m}_1 \Vert \right)^2}{( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_2 \Vert )^2+ ( 1/ \Vert \mathbf{x}_1- \mathbf{m}_2 \Vert )^2} \\ \quad &= \frac{0.0444}{0.4 + 0.0444} = 0.1 \end{align*} \textcolor{red}{\left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^2} \textcolor{green}{( 1/ \Vert \mathbf{x}_j - \mathbf{m}_1 \Vert )^2 } \textbf{Calculate membership values:} $$\mu_{ij} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^{\frac{2}{b-1}}}{ \displaystyle \sum_{r=1}^{c} \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_r \Vert \right)^{\frac{2}{b-1}}} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^2}{ \displaystyle \sum_{r=1}^{2} \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_r \Vert \right)^2} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^2}{ ( 1/ \Vert \mathbf{x}_j - \mathbf{m}_1 \Vert )^2 + ( 1/ \Vert \mathbf{x}_j - \mathbf{m}_2 \Vert )^2}$$ \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|} $j$ & $\mathbf{x}$ & $\Vert \mathbf{x} - \mathbf{m}_1 \Vert$ & $\Vert \mathbf{x} - \mathbf{m}_2 \Vert$ & $\left(\frac{1}{\Vert \mathbf{x} - \mathbf{m}_1 \Vert}\right)^2$ & $\left(\frac{1}{\Vert \mathbf{x} - \mathbf{m}_2 \Vert}\right)^2$ & $u_{1j}$ & $u_{2j}$ \\ 1&$\left[ \begin{array}{c} -1 \\3 \end{array} \right]$& 1.5811 & 4.7434 & 0.4 & 0.0444 & 0.9 & 0.1 \\ \hline 2&$\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]$& 1.5811 & 3.5355 & 0.4 & 0.08 & 0.8333 & 0.1667 \\ \hline 3&$\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]$& 2.5495 & 4.9497 & 0.1538 & 0.0408 & 0.7903 & 0.2097\\ \hline 4&$\left[ \begin{array}{c} 4 \\-1 \end{array} \right]$& 4.9497 & 2.5495 & 0.0408 & 0.1538 & 0.2097 & 0.7903 \\ \hline 5&$\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]$& 3.5355 & 1.5811 & 0.08 & 0.4 & 0.1667 & 0.8333\\ \hline 6&$\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]$& 4.7434 & 1.5811 & 0.0444 & 0.4 & 0.1 & 0.9 \\ \end{tabular} \end{center} \noindent \begin{align*} \mu_{11} &= \frac{ \left( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_1 \Vert \right)^2}{( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_1 \Vert )^2+ ( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_2 \Vert )^2} \\ \quad &= \frac{0.4}{0.4 + 0.0444} = 0.9 \end{align*} \begin{align*} \mu_{21} &= \frac{ \left( 1/ \Vert \mathbf{x}_2 - \mathbf{m}_1 \Vert \right)^2}{( 1/ \Vert \mathbf{x}_1 - \mathbf{m}_2 \Vert )^2+ ( 1/ \Vert \mathbf{x}_1- \mathbf{m}_2 \Vert )^2} \\ \quad &= \frac{0.0444}{0.4 + 0.0444} = 0.1 \end{align*} \newcommand{\xone}{\left[ \begin{array}{c} -1 \\ 3 \end{array} \right]} \newcommand{\xtwo}{\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]} \newcommand{\xthree}{\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]} \newcommand{\xfour}{\left[ \begin{array}{c} 4 \\ -1 \end{array} \right]} \newcommand{\xfive}{\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]} \newcommand{\xsix}{\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]} \begin{center} \textbf{Iteration 2} \end{center} \noindent \textbf{Normalise memberships:}\\ $\mathbf{\mu} = \left[ \begin{array}{cccccc} \mu_{11} & \mu_{12} & \mu_{13} & \mu_{14} & \mu_{15} & \mu_{16} \\ \mu_{21} & \mu_{22} & \mu_{23} & \mu_{24} & \mu_{25} & \mu_{26} \end{array} \right] = \left[ \begin{array}{cccccc} 0.9 & 0.8333 & 0.7903 & 0.2097 & 0.1667 & 0.1 \\ 0.1 & 0.1667 & 0.2097 & 0.7903 & 0.8333 & 0.9 \end{array} \right]$. \noindent \textbf{Update cluster centres:} %\ifslides {\LARGE \fi $$\mathbf{m}_i = \frac{\displaystyle \sum_{j=1}^n \mu_{ij}^b\mathbf{x}_j}{\displaystyle \sum_{j=1}^n \mu_{ij}^b} = \frac{\displaystyle \sum_{j=1}^6 \mu_{ij}^2\mathbf{x}_j}{\displaystyle \sum_{j=1}^6 \mu_{ij}^b} = \frac{\mu_{i1}^b \mathbf{x}_1 + \mu_{i2}^b \mathbf{x}_2 + \mu_{i3}^b \mathbf{x}_3 + \mu_{i4}^b \mathbf{x}_4 + \mu_{i5}^b \mathbf{x}_5 + \mu_{i6}^b \mathbf{x}_6}{\mu_{i1}^b + \mu_{i2}^b + \mu_{i3}^b + \mu_{i4}^b + \mu_{i5} ^b+ \mu_{i6}^b}$$ %$$\mathbf{\mu}_i =\frac{\sum_{j=1}^n u_{ij}^2\mathbf{x}_j}{\sum_{j=1}^n u_{ij}^2}$$ \begin{align*} \mathbf{m}_1 &= \frac{0.9^2\xone+0.8333^2\xtwo+0.7903^2\xthree+0.2097^2\xfour+0.1667^2\xfive+0.1^2\xsix}{0.9^2+0.8333^2+0.7903^2+ 0.2097^2 +0.1667^2 +0.1^2}\\ &=\left[ \begin{array}{c} 0.0876 \\ 3.7529 \end{array} \right] \end{align*} \begin{align*} \mathbf{m}_2 &= \frac{0.1^2\xone+0.1667^2\xtwo+0.2097^2\xthree+0.7903^2\xfour+0.8333^2\xfive+0.9^2\xsix}{0.1^2+0.1667^2+0.2097^2+ 0.7903^2 +0.8333^2 +0.9^2}\\ &=\left[ \begin{array}{c} 3.9124 \\ 0.2471 \end{array} \right] \end{align*} %\ifslides } \fi \noindent \textbf{Calculate membership values:} $$\mu_{ij} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^{\frac{2}{b-1}}}{ \displaystyle \sum_{r=1}^{c} \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_r \Vert \right)^{\frac{2}{b-1}}} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^2}{ \displaystyle \sum_{r=1}^{2} \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_r \Vert \right)^2} = \frac{ \left( 1/ \Vert \mathbf{x}_j - \mathbf{m}_i \Vert \right)^2}{( 1/ \Vert \mathbf{x}_j - \mathbf{m}_1 \Vert )^2+ ( 1/ \Vert \mathbf{x}_j - \mathbf{m}_2 \Vert )^2}$$ \begin{center} \begin{tabular}{|c|c|c|c|c|c|c|c|} $j$ & $\mathbf{x}$ & $\Vert \mathbf{x} - \mathbf{m}_1 \Vert$ & $\Vert \mathbf{x} - \mathbf{m}_2 \Vert$ & $\left(\frac{1}{\Vert \mathbf{x} - \mathbf{m}_1 \Vert}\right)^2$ & $\left(\frac{1}{\Vert \mathbf{x} - \mathbf{m}_2 \Vert}\right)^2$ & $u_{1j}$ & $u_{2j}$ \\ 1&$\left[ \begin{array}{c} -1 \\3 \end{array} \right]$& 1.3228 & 5.6312 & 0.5715& 0.0315& 0.9477& 0.0523\\ \hline 2&$\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]$& 0.9453 & 4.7504 & 1.1191& 0.0443& 0.9619& 0.0381\\ \hline 3&$\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]$& 1.2502 & 6.1560 & 0.6398& 0.0264& 0.9604& 0.0396 \\ \hline 4&$\left[ \begin{array}{c} 4 \\-1 \end{array} \right]$& 6.1560 & 1.2502 & 0.0264& 0.6398& 0.0396& 0.9604\\ \hline 5&$\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]$& 4.7504 & 0.9453 & 0.0443& 1.1191& 0.0381& 0.9619 \\ \hline 6&$\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]$& 5.6312 & 1.3228 & 0.0315& 0.5715& 0.0523& 0.9477\\ \end{tabular} \end{center} \noindent \textbf{In iteration 1:}\\ $\mathbf{m}_1 = \left[ \begin{array}{c} 0.5 \\ 2.5 \end{array} \right]$, $\mathbf{m}_2 = \left[ \begin{array}{c} 3.5 \\ 1.5 \end{array} \right]$. \noindent \textbf{In iteration 2:}\\ $\mathbf{m}_1 = \left[ \begin{array}{c} 0.0876 \\ 3.7529 \end{array} \right]$, $\mathbf{m}_2 = \left[ \begin{array}{c} 3.9124 \\ 0.2471 \end{array} \right]$. \noindent Change is more than 0.5, algorithm does not converge yet. \newcommand{\xone}{\left[ \begin{array}{c} -1 \\ 3 \end{array} \right]} \newcommand{\xtwo}{\left[ \begin{array}{c} 1 \\ 4 \end{array} \right]} \newcommand{\xthree}{\left[ \begin{array}{c} 0 \\ 5 \end{array} \right]} \newcommand{\xfour}{\left[ \begin{array}{c} 4 \\ -1 \end{array} \right]} \newcommand{\xfive}{\left[ \begin{array}{c} 3 \\ 0 \end{array} \right]} \newcommand{\xsix}{\left[ \begin{array}{c} 5 \\ 1 \end{array} \right]} \begin{center} \textbf{Iteration 3} \end{center} \noindent \textbf{Normalise memberships:}\\ $\mathbf{\mu} = \left[ \begin{array}{cccccc} \mu_{11} & \mu_{12} & \mu_{13} & \mu_{14} & \mu_{15} & \mu_{16} \\ \mu_{21} & \mu_{22} & \mu_{23} & \mu_{24} & \mu_{25} & \mu_{26} \end{array} \right] = \left[ \begin{array}{cccccc} 0.9477 & 0.9619 & 0.9604 & 0.0396 & 0.0381 & 0.0523 \\ 0.0523 & 0.0381 & 0.0396 & 0.9604 & 0.9619 & 0.9477 \end{array} \right]$. \noindent \textbf{Update cluster centres:} %\ifslides {\LARGE \fi $$\mathbf{m}_i = \frac{\displaystyle \sum_{j=1}^n \mu_{ij}^b\mathbf{x}_j}{\displaystyle \sum_{j=1}^n \mu_{ij}^b} = \frac{\displaystyle \sum_{j=1}^6 \mu_{ij}^2\mathbf{x}_j}{\displaystyle \sum_{j=1}^6 \mu_{ij}^b} = \frac{\mu_{i1}^b \mathbf{x}_1 + \mu_{i2}^b \mathbf{x}_2 + \mu_{i3}^b \mathbf{x}_3 + \mu_{i4}^b \mathbf{x}_4 + \mu_{i5}^b \mathbf{x}_5 + \mu_{i6}^b \mathbf{x}_6}{\mu_{i1}^b + \mu_{i2}^b + \mu_{i3}^b + \mu_{i4}^b + \mu_{i5} ^b+ \mu_{i6}^b}$$ \begin{align*} \mathbf{m}_1 &= \frac{0.9477^2\xone+0.9619^2\xtwo+0.9604^2\xthree+0.0396^2\xfour+0.0381^2\xfive+0.0523^2\xsix} {0.9477^2+0.9619^2+0.9604^2+ 0.0396^2 +0.0381^2 +0.0522^2}\\ &=\left[ \begin{array}{c} 0.0187 \\ 4.0009 \end{array} \right] \end{align*} \begin{align*} \mathbf{m}_2 &= \frac{0.0523^2\xone+0.0381^2\xtwo+0.0396^2\xthree+0.9604 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com