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\markboth{Nahrul Hayati et al.} %Jika lebih dari dua penulis, tuliskan sebagai Nama Penulis Pertama dkk.
{The Application of Discrete Hidden Markov Model on Crosses of Tetraploid Plant}
	
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\title{THE APPLICATION OF DISCRETE HIDDEN MARKOV MODEL ON CROSSES OF TETRAPLOID PLANT}
	
\author{NAHRUL HAYATI$^{a}$\footnote{corresponding author}, EKO SULISTYONO$^{b}$, VITRI APRILLA HANDAYANI$^{c}$\\}
	
\address{$^{a,b,c}$ Mathematics Department, Batam Institute of Technology\\
The Vitka City Complex, Batam, 29425, Indonesia\\
email : \email{nahrul@iteba.ac.id, eko@iteba.ac.id, vitri@iteba.ac.id}}

\maketitle
\setcounter{page}{1} %bagian ini diedit oleh editor Jurnal Matematika UNAND
	
\begin{abstract}
\begin{center}
Accepted ..... \quad Revised ..... \quad Published ..... %tanggal-tanggal tersebut \textbf{dikosongkan} saja
\end{center}

\bigskip
		
\textbf{Abstract}. % Dalam bahasa Inggris
\textit{In plant heredity, the phenotype is the result of observation that can be directly observed, while the genotype is the underlying hidden factor that underlies the expression of the phenotype. The genotype is an important aspect that needs to be understood to explain the pattern of trait inheritance and predict trait inheritance in subsequent generations. The discrete hidden Markov model is a model generated by pair of an unobserved Markov chain and an observation process. This model can be applied to tetraploid plant crosses by modeling genotypes as hidden state and phenotypes as the obeservation process. The probability of dominant phenotype in monohybrid, dihybrid and trihybrid crosses occurring over ten generations during that period is as follows 61,305\%, 37,583\%, and 23,041\%. Furthermore, as more traits are crossed, the probability of dominant phenotype appearing within ten generations decreases. When the dominant phenotype occurs over ten generations, the same genotype can be obtained in monohybrid, dihybrid, and trihybrid crosses, which is heterozygous in the first and second generations, while from the third to the tenth generation it is homozygous dominant.}

\end{abstract}
	
\keywords{Hidden Markov model, Tetraploid}

\section{Introduction}
The heredity of plants or the inheritance of traits from parents to offspring is a fundamental aspect in the field of genetics and plant breeding. Since the discovery of the laws of inheritance formulated by Gregor Mendel in 1865, the science of genetics has developed rapidly and become an important foundation in efforts to increase productivity and quality of cultivated plants \cite{A}. 

In the study of plant heredity, the phenotype (physical appearance) is the result of observation that can be directly observed, while the genotype (genetic composition) is the underlying hidden factor that underlies the expression of the phenotype. Nevertheless, the genotype is an important aspect that needs to be understood to explain the pattern of trait inheritance and predict trait inheritance in subsequent generations \cite{B}.

The discrete hidden Markov model is a statistical approach that can be used to model stochastic processes with hidden states that cannot be directly observed, but can be inferred from observations on observable variables. In plant heredity, this model can be applied by modeling genotypes as hidden state and phenotypes as the observation process. In addition, this model is also widely used in other field of biomathematics, including \cite{C}, \cite{D}, \cite{E}, \cite{F}, \cite{G}, \cite{H}.  

Hayati \cite{B} in her research was interested in applying this discrete hidden Markov model to diploid plant crossings. In her research, it can be known that the model can be used to explain the characteristics of true breeding in diploid plant crossings. In this study, the discrete hidden Markov model is applied to tetraploid plant crossings. This is because tetraploid plants often have advantages such as larger organ size, higher plant vigor, and greater genetic diversity compared to diploid plants \cite{I}.

\section{Discrete Hidden Markov Model}
The hidden Markov model (HMM) is a model generated by a pair of an unobserved Markov chain $X=\{X_k\}_{k\in N}$ and an observation process $Y=\{X_k\}_{k\in N}$. The Markov chain $X_k$ is influenced by $X_{k+1}$, while the observation $Y_k$ is influenced by $X_k$. The characteristics of HMM as follows

\begin{enumerate}
	\item The transition probability matrix  $ \textbf{\textit{A}} =(a_{ij})_{N \times N} $  for  $i,j=1,2,\cdots,N;$ with
	\begin{center}
		$a_{ij}=P(X_{k+1}=j|X_k=i), \; a_{ij} \geq 0 $ and $ \sum_{j=1}^{N}a_{ij}=1. $
	\end{center}
	The probability matrix of the initial state \boldmath $ \pi $ \unboldmath $ = (\pi_i)_{N \times 1} $ for $ i=1,2,\cdots,N; $ with
	\begin{center}
		$\pi_i=P(X_1=i),$  and  $ \sum_{i=1}^{N}\pi_i=1 .$
	\end{center}
	
	\item The emission probability matrix $ \textbf{\textit{B}} =(b_i(j))_{N \times M} $ for $ i=1,2,\cdots,N; \: j=1,2,\cdots,M;$ with
	\begin{center}
		$b_i(j)=P(Y_k=j|X_k=i),\; b_i(j) \geq 0 $ and $\sum_{j=1}^{M}b_i(j)=1$.
	\end{center}
	
	\item It is assumed that $\{Y_k|X_k\}$ are independent.
\end{enumerate}

Based on those characteristics, parameters are obtained as the characteristics of HMM, namely
\begin{center}
	$ \lambda $ = (\textbf{\textit{A}},\textbf{\textit{B}},\boldmath $ \pi $).
\end{center}

There are three main problems in HMM, namely calculating the probability of the emergence of an observation sequence using the forward and backward algorithms, determining the optimal hidden state sequence using Veterbi algorithm, and re-estimating the HMM parameters using the Baum-Welch algorithm so that the probability $P(Y_1=y_1,\cdots,Y_K=y_K|\lambda)$ is maximum \cite{J}.

\subsection{Forward and Backward Algoritms}
There are three stages in the forward algorithm, namely:
\begin{enumerate}
	\item Initialization stage: 
	%\begin{center}
	$\alpha_1(j)=\pi_j b_j (y_1)$, for $j=1,2,\cdots,N.$
	%\end{center}
	\item Induction stage:
	%\begin{center}
	$\alpha_{k+1}(j)=(\sum_{i=1}^{N}\alpha_k(i)a_{ij})b_j(y_{k+1}),$ \\for $j=1,2,\cdots,N;$ $\:k=1,2,\cdots,K-1.$
	%\end{center} 
	\item Termination stage:
	%\begin{center}
	$P(Y_1=y_1,\cdots,Y_K=y_K|\lambda)=\sum_{j=1}^{N}\alpha_K(j).$
	%\end{center}
\end{enumerate}
There are three stages in the backward algorithm, namely:
\begin{enumerate}
	\item Initialization stage:
	%\begin{center}
	$\beta_K(j)=1$, for  $j=1,2,\cdots,N.$
	%\end{center}
	\item Induction stage:
	%\begin{center}
	$\beta_k(j)=\sum_{i=1}^{N}b_i(y_{k+1})\beta_{k+1}(j)a_{ji},$ \\for $ \:j=1,2,\cdots,N;$ $ \:k=K-1,K-2,\cdots,1.$
	%\end{center}
	\item Termination stage:
	%\begin{center}
	$P(Y_1=y_1,\cdots,Y_K=y_K|\lambda)=\sum_{j=1}^{N}\pi_jb_j(y_1)\beta_1(j).$
	%\end{center}
\end{enumerate}

\subsection{Veterbi Algorithm}
There are four stages in the Veterbi algorithm, namely:
\begin{enumerate}
	\item Initialization stage:
	%\begin{center}
	$\delta_1(j)=\pi_j b_j (y_1),$ and $\psi_1(j)=\emptyset$
	%\end{center}
	\item Recursion stage:
	%\begin{center}
	$\delta_k(j)\:=b_j(y_k)\max_{1\leq i \leq N}\{a_{ij}\delta_{k-1}(i)\},$ and \\
	$\psi_k(j)=\arg\max_{1\leq i \leq N}\{a_{ij}\delta_{k-1}(i)\},$
	for $k=2,3,\cdots,K-1$.
	%\end{center}
	\item Termination stage:
	%\begin{center}
	$P^\ast=\max_{1\leq i \leq N}\{\delta_K(i)\},$ and
	${x_K}^\ast=\arg\max_{1\leq i \leq N}\{\delta_K(i)\}.$
	%\end{center}
	\item Backtracking stage:
	%\begin{center}
	${x_k}^\ast=\psi_{k+1}({x_{k+1}}^\ast),\:\mbox{for}\; k=K-1,K-2,\cdots,1.$
	%\end{center}
\end{enumerate}
\subsection{Baum-Welch Algorithm}
Defined variables 
%\begin{center}
\[\xi_k(i,j)=\frac{\alpha_k(i)a_{ij}b_j(y_{k+1})\beta_{k+1}(j)}{\sum_{j=1}^{N}\alpha_k(j)\beta_k(j)}\mbox{ and } \gamma_k(i)=\sum_{j=1}^{N}\xi_k(i,j),\]
%\end{center}
for $i=1,2,\cdots,N$ and $k=1,2,\cdots,K$. The re-estimated parameters is as follows:
\begin{enumerate}
	\item $\hat{\pi}_i=\gamma_1(i), \quad \mbox{for} \; i=1,2,\cdots,N.$
	\item $\hat{a}_{ij}=\frac{\sum_{k=1}^{K-1}\xi_k(i,j)}{\sum_{k=1}^{K-1}\gamma_k(i)},$
	for $i=1,2,\cdots,N$ and $j=1,2,\cdots K.$ 
	\item $\hat{b}_i(j)=\frac{\sum_{k=1,\;s.t\; y_k=j}^{K}\gamma_k(i)}{\sum_{k=1}^{K}\gamma_k(i)}.$
\end{enumerate}

\section{Crosses of Tetraploid Plant}
In 1865, Gregor Mendel formulated the theory of inheritance based on experiments with garden peas. He stated that parents pass on to their offspring discrete genes that maintain their identity from generation to generation. Genes have alternative forms or alleles. An organism that has a pair of identical alleles for a character is called homozygous for the gene controlling that character. An organism that has two different alleles for a gene is called heterozygous for that gene, and the expression of one allele (the dominant one) masks the effect of the other (the recessive allele). Because of the different effect of dominant and recessive alleles, the traits of an organism do not always reveal its genetic composition. Thus, a distinction is made between the appearance or obeservable trait of an organism, called the phenotype, and its genetic makeup, called genotype \cite{A}.

To obtain new varieties with superior traits, plant breeding or crossing of plants is performed. A cross involving a single pair of contrasting traits is called a monohybrid cross. A cross involving two pairs of contrasting traits is called a dihybrid cross, and one involving three pairs of contransting traits is called a trihybrid cross. These crosses begin with a cross between a pair of true-breeding (homozygous) parental line $(P_1)$ to produce the first filial generation $(F_1)$. Then, a cross is made among the $F_1$ individuals $(P_2)$ to produce the second filial generation $(F_2)$. To determine all possible genotype of phenotype combination in such crosses, a Punnett square can be used \cite{A}
.

In tetraploid plant, the crosses that occur result in offspring that are also tetraploid with four complete sets of chromosomes in their somatic cells. Tetraploid crosses can cause genetic variation and can produce new adventageous traits in plant breeding \cite{I}.tentang apa yang dikerjakan dalam penelitian, serta apa hasil yang diperoleh. Perhatikan cara penulisan teorema, lema, akibat, proposisi, definisi, contoh berikut.

\subsection{Monohybrid Crosses}

$P_1 \qquad \;: \; \; FFFF \; \times \; \; ffff $\\
$F_1 \qquad \;: \qquad \quad FFff $\\
$P_2 \qquad \;: \; \; FFff \; \times \; \; FFff$\\
$\mbox{Gamete} :\qquad FF, Ff, ff $\\
$F_2 \qquad \;:$ 

\begin{table}[h]
\renewcommand{\arraystretch}{1.2}
\label{F2_monohybrid}
\centering
\begin{tabular}{|c|c|c|c|}
	\hline
	{} & $FF$ & $Ff$ & $ff$\\
	\hline
	$\quad\enspace FF \enspace\quad$ & $\quad\enspace FFFF \enspace\quad$ & $\quad\enspace FFFf \enspace\quad$ &
	$\quad\enspace FFff \enspace\quad$ \\
	\hline
	$ Ff $ & $FFFf$ & $FFff$ & $Ffff$ \\
	\hline
	$ ff $ & $FFff$ & $Ffff$ & $ffff$ \\
	\hline
\end{tabular}
\end{table}

The $k^{th}$ descendant genotype denoted by $X_k$ random variable and state space of a Markov chain $\{X_k\}$ is
\begin{center}
	$S_X=\{FFFF, FFFf, FFff, Ffff, ffff\}.$
\end{center}
The $k^{th}$ descendant phenotype denoted by $Y_k$ random variable and state space of an observation process $\{Y_k\}$ is
\begin{center}
	$S_Y=\{F, f\}.$
\end{center}
The Markov chain $X_k$ is influenced by $X_{k-1}$ and the observation process $Y_k$ is influenced by $X_k$. Figure 1 shows the relationship between $X_k,Y_k,$ and $X_{k-1}$ for a case when the phenotype sequence is always dominant $F$.

\begin{figure}[h]
	\centering
	\includegraphics[width=4.3in]{Figure2}
	\caption{Monohybrid Crosses of Tetraploid Plant}
\end{figure}

Based on Figure 1 it is known that the first generation has a heterozygote genotype $FFff$, so the first generation has a dominant phenotype $F$. The dominant phenotype $F$ on the second generation is influenced by unknown genotype whether it is heterozygote $FFFf$, $FFff$, $Ffff$ or homozygote $FFFF$.  

The characteristics of HMM for monohybrid crosses are transition probability matrix \boldmath$A_1$, emission probability matrix $B_1$, and probability matrix of the initial state \boldmath$\pi_1$. The first is determining transition probability matrix $A_1$. The dimension of matrix $A_1$ is 5 $\times$ 5, that is determined by number of state space \unboldmath$S_X$. The entries of matrix \boldmath$A_1$ are obtained by punnett square of state space \unboldmath$S_X$.

\begin{table}[h]
	\renewcommand{\arraystretch}{1.2}
	\label{A_monohybrid}
	\centering
	
	\begin{tabular}{|ccccccccc|}
		\hline
		\multicolumn{4}{|c|}{$\qquad \qquad FFFF \qquad \qquad$} & {$\times$} & \multicolumn{4}{|c|}{$\qquad \qquad FFFF \qquad \qquad $} \\
		%\hline
		\cline{1-4}\cline{6-9} \multicolumn{4}{|c|}{$(FF)$} &  & \multicolumn{4}{|c|}{$(FF)$}\\
		\hline
		\multicolumn{9}{|c|}{$FFFF$ 100\%}\\
		\hline
		\multicolumn{9}{c}{$a_{FFFF,j}=[1\quad 0\quad 0\quad 0\quad 0]$}\\
		
		\multicolumn{9}{c}{ } \\
		\hline
		\multicolumn{4}{|c|}{$FFFf$} & {$\times$} & \multicolumn{4}{|c|}{$FFFf$} \\
		\cline{1-4}\cline{6-9} \multicolumn{4}{|c|}{$\qquad(3FF,3Ff)$} &  & \multicolumn{4}{|c|}{$(3FF,3Ff)$}\\
		\hline
		\hline
		\multicolumn{3}{|c|}{} & \multicolumn{3}{|c|}{$3FF$} & \multicolumn{3}{|c|}{$3Ff$}\\
		\hline
		\multicolumn{3}{|c|}{$3FF $} & \multicolumn{3}{|c|}{$ 9FFFF $} & \multicolumn{3}{|c|}{$ 9FFFf $}\\
		\hline
		\multicolumn{3}{|c|}{$\qquad \enspace 3Ff \enspace \qquad $} & \multicolumn{3}{|c|}{$\qquad \enspace 9FFFf \qquad \enspace $} & \multicolumn{3}{|c|}{$9FFff$}\\
		\hline
		\multicolumn{9}{|c|}{$\quad FFFF (\frac{1}{4}), FFFf (\frac{1}{2}), FFff (\frac{1}{4}), Ffff (0), ffff (0) \quad$ }\\
		\hline
		\multicolumn{9}{c}{$a_{FFFf,j}=[\frac{1}{4}\quad \frac{1}{2}\quad \frac{1}{4} \quad 0 \quad 0]$}\\
		\multicolumn{9}{c}{ } \\
		\hline
		\multicolumn{4}{|c|}{$FFff$} & {$\times$} & \multicolumn{4}{|c|}{$FFff$} \\
		\cline{1-4}\cline{6-9} \multicolumn{4}{|c|}{$(1FF,4Ff,1ff)$} &  & \multicolumn{4}{|c|}{$(1FF,4Ff,1ff)$}\\
		\hline
	\end{tabular}
\end{table}

\begin{table}[h]
	\renewcommand{\arraystretch}{1.2}
	\label{A_monohybrid}
	\centering
	
	\begin{tabular}{|ccccccccc|}
		\hline
		\multicolumn{3}{|c|}{} & \multicolumn{2}{|c|}{$1FF$} & \multicolumn{2}{|c|}{$4Ff$} &
		\multicolumn{2}{|c|}{$1ff$} \\
		\hline
		\multicolumn{3}{|c|}{$1FF$} & \multicolumn{2}{|c|}{$\quad 1FFFF \quad$} & \multicolumn{2}{|c|}{$\quad 4FFFf \quad$} &
		\multicolumn{2}{|c|}{$\quad 1FFff \quad$} \\
		\hline
		\multicolumn{3}{|c|}{$4Ff$} & \multicolumn{2}{|c|}{$4FFFf$} & \multicolumn{2}{|c|}{$16FFff$} &
		\multicolumn{2}{|c|}{$4Ffff$} \\
		\hline
		\multicolumn{3}{|c|}{$1ff$} & \multicolumn{2}{|c|}{$1FFff$} & \multicolumn{2}{|c|}{$4Ffff$} &
		\multicolumn{2}{|c|}{$1ffff$} \\
		\hline 
		\multicolumn{9}{|c|}{$\quad FFFF (\frac{1}{36}), FFFf (\frac{2}{9}), FFff (\frac{1}{2}), Ffff (\frac{2}{9}), ffff (\frac{1}{36}) \quad$ }\\
		\hline
		\multicolumn{9}{c}{$a_{FFff,j}=[\frac{1}{36}\quad \frac{2}{9}\quad \frac{1}{2} \quad \frac{2}{9} \quad \frac{1}{36}]$}\\
		
		\multicolumn{9}{c}{ } \\
		\hline
		\multicolumn{4}{|c|}{$Ffff$} & {$\times$} & \multicolumn{4}{|c|}{$Ffff$} \\
		\cline{1-4}\cline{6-9} \multicolumn{4}{|c|}{$(3Ff,3ff)$} &  & \multicolumn{4}{|c|}{$(3Ff,3ff)$}\\
		\hline
		\hline
		\multicolumn{3}{|c|}{} & \multicolumn{3}{|c|}{$3Ff$} & \multicolumn{3}{|c|}{$3ff$}\\
		\hline
		\multicolumn{3}{|c|}{$3Ff $} & \multicolumn{3}{|c|}{$ 9FFff $} & \multicolumn{3}{|c|}{$ 9Ffff $}\\
		\hline
		\multicolumn{3}{|c|}{$\qquad \enspace 3ff \enspace \qquad $} & \multicolumn{3}{|c|}{$\qquad \enspace 9Ffff \qquad \enspace $} & \multicolumn{3}{|c|}{$9ffff$}\\
		\hline
		\multicolumn{9}{|c|}{$\quad FFFF (0), FFFf (0), FFff (\frac{1}{4}), Ffff (\frac{1}{2}), ffff (\frac{1}{4}) \quad$ }\\
		\hline
		\multicolumn{9}{c}{$a_{Ffff,j}=[0\quad 0\quad \frac{1}{4} \quad \frac{1}{2} \quad \frac{1}{4}]$}\\
		
		\multicolumn{9}{c}{ } \\
		\hline
		\multicolumn{4}{|c|}{$ffff$} & {$\times$} & \multicolumn{4}{|c|}{$ffff$} \\
		%\hline
		\cline{1-4}\cline{6-9} \multicolumn{4}{|c|}{$(ff)$} &  & \multicolumn{4}{|c|}{$(ff)$}\\
		\hline
		\multicolumn{9}{|c|}{$ffff$ 100\%}\\
		\hline
		\multicolumn{9}{c}{$a_{ffff,j}=[0\quad 0\quad 0\quad 0\quad 1]$}		
	\end{tabular}
\end{table}
\newpage
So, the transition probability matrix \boldmath$A_1$  is
\begin{center}
	$A_1$ \unboldmath  $=\left[{\begin{array}{ccccc}
			a_{FFFF,FFFF} & a_{FFFF,FFFf} & a_{FFFF,FFff} & a_{FFFF,Ffff} & a_{FFFF,ffff}\\
			a_{FFFf,FFFF} & a_{FFFf,FFFf} & a_{FFFf,FFff} & a_{FFFf,Ffff} & a_{FFFf,ffff}\\
			a_{FFff,FFFF} & a_{FFff,FFFf} & a_{FFff,FFff} & a_{FFff,Ffff} & a_{FFff,ffff}\\
			a_{Ffff,FFFF} & a_{Ffff,FFFf} & a_{Ffff,FFff} & a_{Ffff,Ffff} & a_{Ffff,ffff}\\
			a_{ffff,FFFF} & a_{ffff,FFFf} & a_{ffff,FFff} & a_{ffff,Ffff} & a_{ffff,ffff}
	\end{array}}\right]$
\end{center}

\renewcommand{\arraystretch}{1.2} 
\unboldmath
\begin{center}
	$=\left[\begin{array}{ccccc}
		1 & 0 & 0 & 0 & 0\\
		\frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0\\
		\frac{1}{36} & \frac{2}{9} & \frac{1}{2} & \frac{2}{9} & \frac{1}{36}\\
		0 & 0 & \frac{1}{4} & \frac{1}{2} & \frac{1}{4}\\
		0 & 0 & 0 & 0 & 1
	\end{array}\right].\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \; \; $
\end{center}

The second is determining emission probability matrix \boldmath$B_1$. The dimension of matrix $B_1$ is 5 $\times$ 2, that is determined by number of state space \unboldmath $S_X$ and $S_Y$. The first column of matrix  \boldmath $B_1$ \unboldmath tells that type $FFFF,FFFf,FFff$ and $Ffff$ of genotype will be observed as dominant phenotype $F$. The second column of matrix \boldmath$B_1$ tells that type \unboldmath$ffff$ of genotype will be observed as recessive phenotype $f$.

\boldmath
\begin{center}
	$B_1$ \unboldmath $=\left[{\begin{array}{cc}
			b_{FFFF}(F) & b_{FFFF}(f) \\
			b_{FFFf}(F) & b_{FFFf}(f) \\
			b_{FFff}(F) & b_{FFff}(f) \\
			b_{Ffff}(F) & b_{Ffff}(f) \\
			b_{ffff}(F) & b_{ffff}(f) 
	\end{array}}\right]=\left[{\begin{array}{cc}
			1 & 0 \\
			1 & 0 \\
			1 & 0 \\
			1 & 0 \\
			0 & 1 \end{array}}\right].$
\end{center}

The third is determining probability matrix of the initial state \boldmath$\pi_1$. The dimension of matrix $\pi_1$ is 5 $\times$ 1, that is determined by nuber of state space \unboldmath$S_X$. Since Mendel start the parental generation with heterozygote plants and $P_2$ has only one heterozygote i.e. $FFff$, therefore the initial value of $\pi_i$ is one for type $FFff$ of genotype and zero to the other.

\boldmath
\begin{center}
	$\pi_1$ \unboldmath $=\left[{\begin{array}{c}
			\pi_{FFFF} \\
			\pi_{FFFf} \\
			\pi_{FFff} \\
			\pi_{Ffff} \\
			\pi_{ffff} 
	\end{array}}\right]=\left[{\begin{array}{c}
			0  \\
			0  \\
			1  \\
			0  \\
			0  \end{array}}\right].$
\end{center}

\subsection{Dihybrid Crosses}

\unboldmath$P_1 \qquad \;: \; \; FFFFGGGG \; \times \; \; ffffgggg $\\
$F_1 \qquad \;: \qquad \qquad \quad FFffGGgg $\\
$P_2 \qquad \;: \; \; FFffGGgg \; \times \; \; FFffGGgg$\\
$\mbox{Gamete} :\quad FFGG, FFGg, FFgg, FfGG, FfGg, Ffgg, ffGG, ffGg, ffgg$\\
$F_2 \qquad \;:$

\begin{table}[h]
	\renewcommand{\arraystretch}{1.2}
	\label{F2_monohybrid}
	\centering
	\footnotesize
	\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
		\hline 
		& $FFGG$ & $FFGg$ & $FFgg$ & $FfGG$ & $FfGg$ & $Ffgg$ & $ffGG$ & $ffGg$ & $ffgg$ \\ 
		\hline 
		$FFGG$ & $FFFF$ & $FFFF$ & $FFFF$ & $FFFf$ & $FFFf$ & $FFFf$ & $FFff$ & $FFff$ & $FFff$ \\ 
		%\hline 
		& $GGGG$ & $GGGg$ & $GGgg$ & $GGGG$ & $GGGg$ & $GGgg$ & $GGGG$ & $GGGg$ & $GGgg$ \\ 
		\hline 
		$FFGg$ & $FFFF$ & $FFFF$ & $FFFF$ & $FFFf$ & $FFFf$ & $FFFf$ & $FFff$ & $FFff$ & $FFff$ \\ 
		%\hline 
		& $GGGg$ & $GGgg$ & $Gggg$ & $GGGg$ & $GGgg$ & $Gggg$ & $GGGg$ & $GGgg$ & $Gggg$ \\ 
		\hline 
		$FFgg$ & $FFFF$ & $FFFF$ & $FFFF$ & $FFFf$ & $FFFf$ & $FFFf$ & $FFff$ & $FFff$ & $FFff$ \\ 
		%\hline 
		& $GGgg$ & $Gggg$ & $gggg$ & $GGgg$ & $Gggg$ & $gggg$ & $GGgg$ & $Gggg$ & $gggg$ \\ 
		\hline 
		$FfGG$ & $FFFf$ & $FFFf$ & $FFFf$ & $FFff$ & $FFff$ & $FFff$ & $Ffff$ & $Ffff$ & $Ffff$ \\ 
		%\hline 
		& $GGGG$ & $GGGg$ & $GGgg$ & $GGGG$ & $GGGg$ & $GGgg$ & $GGGG$ & $GGGg$ & $GGgg$ \\ 
		\hline 
		$FfGg$ & $FFFf$ & $FFFf$ & $FFFf$ & $FFff$ & $FFff$ & $FFff$ & $Ffff$ & $Ffff$ & $Ffff$ \\ 
		%\hline 
		& $GGGg$ & $GGgg$ & $Gggg$ & $GGGg$ & $GGgg$ & $Gggg$ & $GGGg$ & $GGgg$ & $Gggg$ \\ 
		\hline 
		$Ffgg$ & $FFFf$ & $FFFf$ & $FFFf$ & $FFff$ & $FFff$ & $FFff$ & $Ffff$ & $Ffff$ & $Ffff$ \\ 
		%\hline 
		& $GGgg$ & $Gggg$ & $gggg$ & $GGgg$ & $Gggg$ & $gggg$ & $GGgg$ & $Gggg$ & $gggg$ \\ 
		\hline 
		$ffGG$ & $FFff$ & $FFff$ & $FFff$ & $Ffff$ & $Ffff$ & $Ffff$ & $ffff$ & $ffff$ & $ffff$ \\ 
		%\hline 
		& $GGGG$ & $GGGg$ & $GGgg$ & $GGGG$ & $GGGg$ & $GGgg$ & $GGGG$ & $GGGg$ & $GGgg$ \\ 
		\hline 
		$ffGg$ & $FFff$ & $FFff$ & $FFff$ & $Ffff$ & $Ffff$ & $Ffff$ & $ffff$ & $ffff$ & $ffff$ \\ 
		%\hline 
		& $GGGg$ & $GGgg$ & $Gggg$ & $GGGg$ & $GGgg$ & $Gggg$ & $GGGg$ & $GGgg$ & $Gggg$ \\ 
		\hline 
		$ffgg$ & $FFff$ & $FFff$ & $FFff$ & $Ffff$ & $Ffff$ & $Ffff$ & $ffff$ & $ffff$ & $ffff$ \\ 
		%\hline 
		& $GGgg$ & $Gggg$ & $gggg$ & $GGgg$ & $Gggg$ & $gggg$ & $GGgg$ & $Gggg$ & $gggg$ \\ 
		\hline 
	\end{tabular}  
\end{table}

The $k^{th}$ descendant genotype denoted by $X_k$ random variable and state space of a Markov chain $\{X_k\}$ is

\begin{center}
	$S_X=\{FFFFGGGG$, $FFFFGGGg$, $FFFFGGgg$, $FFFFGggg$, $FFFFgggg$,\\ 
	$ \qquad FFFfGGGG$, $FFFfGGGg$, $FFFfGGgg$, $FFFfGggg$, $FFFfgggg$,\\
	$ \quad  FFffGGGG$, $FFffGGGg$, $FFffGGgg$, $FFffGggg$, $FFffgggg$,\\
	$FfffGGGG$, $FfffGGGg$, $FfffGGgg$, $FfffGggg$, $Ffffgggg$, \\
	$ffffGGGG$, $ffffGGGg$, $ffffGGgg$, $ffffGggg$, $ffffgggg\}. \;$
\end{center}
The $k^{th}$ descendant phenotype denoted by $Y_k$ random variable and state space of an observation process $\{Y_k\}$ is
\begin{center}
	$S_Y=\{FG, Fg, fG, fg\}.$
\end{center}
The Markov chain $X_k$ is influenced by $X_{k-1}$ and the observation process $Y_k$ is influenced by $X_k$. The characteristics of HMM for dihybrid crosses can be obtained in the same way with the previous crosses.

\renewcommand{\arraystretch}{1.2} 
\boldmath
\begin{center}
	$A_2$
	$=\left[\begin{array}{ccccc}
		1A_1 & 0 & 0 & 0 & 0\\
		\frac{1}{4}A_1 & \frac{1}{2}A_1 & \frac{1}{4}A_1 & 0 & 0\\
		\frac{1}{36}A_1 & \frac{2}{9}A_1 & \frac{1}{2}A_1 & \frac{2}{9}A_1 & \frac{1}{36}A_1\\
		0 & 0 & \frac{1}{4}A_1 & \frac{1}{2}A_1 & \frac{1}{4}A_1\\
		0 & 0 & 0 & 0 & 1A_1
	\end{array}\right],
	\; B_2=\left[{\begin{array}{cc}
			B_1 & 0 \\
			B_1 & 0 \\
			B_1 & 0 \\
			B_1 & 0 \\
			0 & B_1 \end{array}}\right], 			
	\; \pi_2=\left[{\begin{array}{c}
			0  \\
			0  \\
			\pi_1  \\
			0  \\
			0  \end{array}}\right].$
\end{center}

\subsection{Trihybrid Crosses}

\unboldmath$P_1 \qquad \;: \; \; FFFFGGGGHHHH \; \times \; \; ffffgggghhhh $\\
$F_1 \qquad \;: \qquad \qquad \quad FFffGGggHHhh $\\
$P_2 \qquad \;: \; \; FFffGGggHHhh \; \times \; \; FFffGGggHHhh$\\
$\mbox{Gamete} :\; \; FFGGHH, FFGGHh, FFGGhh, FFGgHH, FFGgHh, FFGghh,$\\
$\mbox{\qquad \qquad \; \;} FFggHH, FFggHh, FFgghh, FfGGHH, FfGGHh, FfGGhh,$\\
$\mbox{\qquad \qquad \; \;} FfGgHH, FfGgHh, FfGghh, FfggHH, FfggHh, Ffgghh,$\\
$\mbox{\qquad \qquad \; \;} ffGGHH, ffGGHh, ffGGhh, ffGgHH, ffGgHh, ffGghh,$\\
$\mbox{\qquad \qquad \; \;} ffggHH, ffggHh, ffgghh$\\

The $k^{th}$ descendant genotype denoted by $X_k$ random variable and state space of a Markov chain $\{X_k\}$ is

%\begin{center}
$S_X=\{FFFFGGGGHHHH$, $FFFFGGGGHHHh$, $FFFFGGGGHHhh$,\\
$ \mbox{\qquad \qquad \; \;} FFFFGGGGHhhh$, $FFFFGGGGhhhh$, $FFFFGGGgHHHH$,\\
$ \mbox{\qquad \qquad \; \;} FFFFGGGgHHHh$, $FFFFGGGgHHhh$, $FFFFGGGgHhhh$,\\
$ \mbox{\qquad \qquad \; \;} FFFFGGGghhhh$, $FFFFGGggHHHH$, $FFFFGGggHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFFGGggHHhh$, $FFFFGGggHhhh$, $FFFFGGgghhhh$,\\
$ \mbox{\qquad \qquad \; \;} FFFFGgggHHHH$, $FFFFGgggHHHh$, $FFFFGgggHHhh$,\\
$ \mbox{\qquad \qquad \; \;} FFFFGgggHhhh$, $FFFFGggghhhh$, $FFFFggggHHHH$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFFggggHHHh$, $FFFFggggHHhh$, $FFFFggggHhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFFgggghhhh$, $FFFfGGGGHHHH$, $FFFfGGGGHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfGGGGHHhh$, $FFFfGGGGHhhh$, $FFFfGGGGhhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfGGGgHHHH$,	$FFFfGGGgHHHh$, $FFFfGGGgHHhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfGGGgHhhh$,	$FFFfGGGghhhh$, $FFFfGGggHHHH$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfGGggHHHh$, $FFFfGGggHHhh$, $FFFfGGggHhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfGGgghhhh$,	$FFFfGgggHHHH$, $FFFfGgggHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfGgggHHhh$, $FFFfGgggHhhh$, $FFFfGggghhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfggggHHHH$, $FFFfggggHHHh$, $FFFfggggHHhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFFfggggHhhh$, $FFFfgggghhhh$, $FFffGGGGHHHH$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffGGGGHHHh$, $FFffGGGGHHhh$, $FFffGGGGHhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffGGGGhhhh$, $FFffGGGgHHHH$,	$FFffGGGgHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffGGGgHHhh$, $FFffGGGgHhhh$,	$FFffGGGghhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffGGggHHHH$, $FFffGGggHHHh$, $FFffGGggHHhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffGGggHhhh$, $FFffGGgghhhh$,	$FFffGgggHHHH$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffGgggHHHh$, $FFffGgggHHhh$, $FFffGgggHhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffGggghhhh$, $FFffggggHHHH$, $FFffggggHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} FFffggggHHhh$, $FFffggggHhhh$, $FFffgggghhhh$,\\
$ \mbox{\qquad \qquad \; \;} FfffGGGGHHHH$, $FfffGGGGHHHh$, $FfffGGGGHHhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FfffGGGGHhhh$, $FfffGGGGhhhh$, $FfffGGGgHHHH$,\\	
$ \mbox{\qquad \qquad \; \;} FfffGGGgHHHh$, $FfffGGGgHHhh$, $FfffGGGgHhhh$,\\	
$ \mbox{\qquad \qquad \; \;} FfffGGGghhhh$, $FfffGGggHHHH$, $FfffGGggHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} FfffGGggHHhh$, $FfffGGggHhhh$, $FfffGGgghhhh$,\\	
$ \mbox{\qquad \qquad \; \;} FfffGgggHHHH$, $FfffGgggHHHh$, $FfffGgggHHhh$,\\ 
$ \mbox{\qquad \qquad \; \;} FfffGgggHhhh$, $FfffGggghhhh$, $FfffggggHHHH$,\\ 
$ \mbox{\qquad \qquad \; \;} FfffggggHHHh$, $FfffggggHHhh$, $FfffggggHhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} Ffffgggghhhh$, $ffffGGGGHHHH$, $ffffGGGGHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffGGGGHHhh$, $ffffGGGGHhhh$, $ffffGGGGhhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffGGGgHHHH$,	$ffffGGGgHHHh$, $ffffGGGgHHhh$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffGGGgHhhh$,	$ffffGGGghhhh$, $ffffGGggHHHH$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffGGggHHHh$, $ffffGGggHHhh$, $ffffGGggHhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffGGgghhhh$,	$ffffGgggHHHH$, $ffffGgggHHHh$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffGgggHHhh$, $ffffGgggHhhh$, $ffffGggghhhh$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffggggHHHH$, $ffffggggHHHh$, $ffffggggHHhh$,\\ 
$ \mbox{\qquad \qquad \; \;} ffffggggHhhh$, $ffffgggghhhh	 \}. \;$\\
%\end{center}

The $k^{th}$ descendant phenotype denoted by $Y_k$ random variable and state space of an observation process $\{Y_k\}$ is
\begin{center}
	$S_Y=\{FGH, FGh, FgH, Fgh, fGH, fGh, fgH, fgh\}.$
\end{center}
The Markov chain $X_k$ is influenced by $X_{k-1}$ and the observation process $Y_k$ is influenced by $X_k$. The characteristics of HMM for trihybrid crosses can be obtained in the same way with the previous crosses.
\newpage
\renewcommand{\arraystretch}{1.2} 
\boldmath
\begin{center}
	$A_3$
	$=\left[\begin{array}{ccccc}
		1A_2 & 0 & 0 & 0 & 0\\
		\frac{1}{4}A_2 & \frac{1}{2}A_2 & \frac{1}{4}A_2 & 0 & 0\\
		\frac{1}{36}A_2 & \frac{2}{9}A_2 & \frac{1}{2}A_2 & \frac{2}{9}A_2 & \frac{1}{36}A_2\\
		0 & 0 & \frac{1}{4}A_2 & \frac{1}{2}A_2 & \frac{1}{4}A_2\\
		0 & 0 & 0 & 0 & 1A_2
	\end{array}\right],
	\; B_3=\left[{\begin{array}{cc}
			B_2 & 0 \\
			B_2 & 0 \\
			B_2 & 0 \\
			B_2 & 0 \\
			0 & B_2 \end{array}}\right], 			
	\; \pi_3=\left[{\begin{array}{c}
			0  \\
			0  \\
			\pi_2  \\
			0  \\
			0  \end{array}}\right].$
\end{center}

Based on Subsection 3.1, 3.2, and 3.3, it can be known the general form of the characteristics of HMM on crosses of tetraploid plant as follows

\renewcommand{\arraystretch}{1.2} 
\boldmath
\begin{center}
	$A_{k+1}$
	$=\left[\begin{array}{ccccc}
		1A_{k} & 0 & 0 & 0 & 0\\
		\frac{1}{4}A_{k} & \frac{1}{2}A_{k} & \frac{1}{4}A_{k} & 0 & 0\\
		\frac{1}{36}A_{k} & \frac{2}{9}A_{k} & \frac{1}{2}A_{k} & \frac{2}{9}A_{k} & \frac{1}{36}A_{k}\\
		0 & 0 & \frac{1}{4}A_{k} & \frac{1}{2}A_{k} & \frac{1}{4}A_{k}\\
		0 & 0 & 0 & 0 & 1A_{k}
	\end{array}\right],
	B_{k+1}=\left[{\begin{array}{cc}
			B_k & 0 \\
			B_k & 0 \\
			B_k & 0 \\
			B_k & 0 \\
			0 & B_k \end{array}}\right], 			
	\pi_{k+1}=\left[{\begin{array}{c}
			0  \\
			0  \\
			\pi_k  \\
			0  \\
			0  \end{array}}\right].$
\end{center}

\section{The Application of Discrete Hidden Markov Model}
The first problem in HMM is calculating the probability of the emergence of an observation sequence using the forward and backward algorithms. The following is the probability that model $\lambda=(\boldmath A,B,\pi)$ produces the observation sequence \unboldmath$O=\{F,F,F,F,F,F,F,F,F,F\}$ in monohybrid crosses of tetraploid plants.
\begin{table}[h]
	\renewcommand{\arraystretch}{1.2}
	\label{forward}
	\centering
	\caption{The Forward and Backward Algorithm of Monohybrid Crosses}
	\begin{tabular}{|c|c|c|c|c|c|c|c|c|}	
		\hline 
		t & 1 & 2 & 3 & 4 & $\cdots$ & 8 & 9 & 10 \\ 
		\hline 
		$\alpha_t(1)$ & 0 & 0,02778 & 0,09722 & 0,16281 & $\cdots$ & 0,33720 & 0,36434 & 0,38695 \\ 
		\hline 
		$\alpha_t(2)$ & 0 & 0,22222 & 0,22222 & 0,19136 & $\cdots$ & 0,09303 & 0,07752 & 0,06460 \\ 
		\hline 
		$\alpha_t(3)$ & 1 & 0,5 & 0,36111 & 0,29167 & $\cdots$ & 0,13954 & 0,11628 & 0,09690 \\ 
		\hline 
		$\alpha_t(4)$ & 0 & 0,22222 & 0,22222 & 0,19136 & $\cdots$ & 0,09303 & 0,07752 & 0,06460 \\ 
		\hline 
		$\alpha_t(5)$ & 0 & 0 & 0 & 0 & $\cdots$ & 0 & 0 & 0 \\ 
		\hline 
		\multicolumn{8}{|c|}{$P(O|\lambda)$} & 0,61305 \\ 
		\hline 
		\multicolumn{9}{c}{ } \\
		\multicolumn{9}{c}{ } \\
	\end{tabular} 
	
	\begin{tabular}{|c|c|c|c|c|c|c|c|c|}	
		\hline 
		t & 10 & 9 & 8 & 7 & $\cdots$ & 3 & 2 & 1 \\ 
		\hline 
		$\beta_t(1)$ & 1 & 1 & 1 & 1 & $\cdots$ & 1 & 1 & 1 \\ 
		\hline 
		$\beta_t(2)$ & 1 & 1 & 0,99306 & 0,97222 & $\cdots$ & 0,87015 & 0,85077 & 0,83430 \\ 
		\hline 
		$\beta_t(3)$ & 1 & 0,97222 & 0,90278 & 0,83719 & $\cdots$ & 0,66280 & 0,63566 & 0,61305 \\ 
		\hline 
		$\beta_t(4)$ & 1 & 0,75 & 0,61806 & 0,53472 & $\cdots$ & 0,37405 & 0,35273 & 0,33528 \\ 
		\hline 
		$\beta_t(5)$ & 1 & 0 & 0 & 0 & $\cdots$ & 0 & 0 & 0 \\ 
		\hline 
		\multicolumn{8}{|c|}{$P(O|\lambda)$} & 0,61305 \\ 
		\hline 
	\end{tabular} 
\end{table}

The calculation in the forward and backward algorithms yields the same probability value. Therefore, the probability of the dominant phenotype $F$ occurring over ten generations during that period is 0,61305 or 61,305\%. 

After knowing that the probability of observation is 0,61305, the next problem is determining the optimal hidden state sequence in this case, which is genotype $\{FFFF, FFFf, FFff, Ffff, ffff\}$. This second problem can be solved using the Veterbi algorithm.
\begin{table}[h]
	\renewcommand{\arraystretch}{1.2}
	\label{forward}
	\centering
	\caption{The Veterbi Algorithm of Monohybrid Crosses}
	\begin{tabular}{|c|c|c|c|c|c|c|c|c|}	
		\hline 
		t & 1 & 2 & 3 & 4 & $\cdots$ & 8 & 9 & 10 \\ 
		\hline 
		$\delta_t(1)$ & 0 & 0,02778 & 0,05556 & 0,05556 & $\cdots$ & 0,05556 & 0,05556 & 0,05556 \\ 
		\hline 
		$\delta_t(2)$ & 0 & 0,22222 & 0,11111 & 0,05556 & $\cdots$ & 0,00347 & 0,00174 & 0,00087 \\ 
		\hline 
		$\delta_t(3)$ & 1 & 0,5 & 0,25 & 0,125 & $\cdots$ & 0,00781 & 0,00391 & 0,00195 \\ 
		\hline 
		$\delta_t(4)$ & 0 & 0,22222 & 0,11111 & 0,05556 & $\cdots$ & 0,00347 & 0,00174 & 0,00087 \\ 
		\hline 
		$\delta_t(5)$ & 0 & 0 & 0 & 0 & $\cdots$ & 0 & 0 & 0 \\ 
		\hline 
		\multicolumn{8}{|c|}{$P^\ast$} & 0,05556 \\ 
		\hline 
		\multicolumn{8}{|c|}{${x_5}^\ast$} & 1 ($FFFF$) \\ 
		\hline
		
		\multicolumn{9}{c}{ } \\
		\hline
		t & 1 & 2 & 3 & 4 & $\cdots$ & 8 & 9 & 10 \\ 
		\hline 
		$\psi_t(1)$ & 0 & 3 & 2 & 1 & $\cdots$ & 1 & 1 & 1 \\ 
		\hline 
		$\psi_t(2)$ & 0 & 3 & 2 & 2 & $\cdots$ & 2 & 2 & 2 \\ 
		\hline 
		$\psi_t(3)$ & 1 & 3 & 3 & 3 & $\cdots$ & 3 & 3 & 3 \\ 
		\hline 
		$\psi_t(4)$ & 0 & 3 & 3 & 3 & $\cdots$ & 3 & 3 & 3 \\ 
		\hline 
		$\psi_t(5)$ & 0 & 3 & 4 & 4 & $\cdots$ & 4 & 4 & 4 \\  
		\hline
	\end{tabular} 
\end{table}

Based on the calculation from the Veterbi algorithm, it can be known that when the dominant phenotype $F$ over ten generations occurs, the most optimal sequence of hidden state is
\[x^\ast=\{FFff,FFFf,FFFF,FFFF,FFFF,\cdots,FFFF,FFFF\}.\]

The last problem is re-estimating the HMM parameters using the Baum-Welch algorithm. The process of re-estimating the HMM parameters is carried out to obtain a new parameter \boldmath$\hat{\lambda}=(\hat{A},\hat{B},\hat{\pi})$. These new parameters are what result in \unboldmath $P(O|\hat{\lambda})\ge P(O|\lambda)$.

\begin{center}
	\boldmath $\hat{A_1}$ \unboldmath $ =\left[{\begin{array}{ccccc}
			1 & 0 & 0 & 0 & 0\\
			0,27415 & 0,51136 & 0,21449 & 0 & 0\\
			0,04018 & 0,28986 & 0,52535 & 0,14461 & 0\\
			0 & 0 & 0,43363 & 0,56637 & 0\\
			0 & 0 & 0 & 0 & 0
	\end{array}}\right],$
	\boldmath
	\; $\hat{B_1}$ \unboldmath $=\left[{\begin{array}{cc}
			1 & 0 \\
			1 & 0 \\
			1 & 0 \\
			1 & 0 \\
			0 & 1 \end{array}}\right], $
	\boldmath			
	\; $\hat{\pi_1}$ \unboldmath $=\left[{\begin{array}{c}
			0  \\
			0  \\
			1  \\
			0  \\
			0  \end{array}}\right].$
\end{center}

The problems of HMM for dihybrid and trihybrid crosses can be solved in the same way with monohybrid crosses. In the dihybrid crosses, it is known that the probability of dominant phenotype $FG$ occurring over ten generations during that period is 0,37583 or 37,583\%. Next, when the dominant phenotype $FG$ over ten generations occurs, the most optimal sequence of hidden state is
\[x^\ast=\{FFffGGgg,FFFfGGGg,FFFFGGGG,\cdots,FFFFGGGG\}.\]

In the trihybrid crosses, it is known that the probability of dominant phenotype $FGH$ occurring over ten generations during that period is 0,23041 or 23,041\%. Next, when the dominant phenotype $FGH$ over ten generations occurs, the most optimal sequence of hidden state is
\[x^\ast=\{FFffGGggHHhh,FFFfGGGgHHHh,FFFFGGGGHHHH,\cdots\}.\]

\section{Conclusions}
The discrete HMM can be applied to tetraploid plant crosses. The probability of dominant phenotype in monohybrid, dihybrid and trihybrid crosses occurring over ten generations during that period is as follows 61,305\%, 37,583\%, and 23,041\%. Furthermore, as more traits are crossed, the probability of dominant phenotype appearing within ten generations decreases. When the dominant phenotype occurs over ten generations, the same genotype can be obtained in monohybrid, dihybrid, and trihybrid crosses, which is heterozygous in the first and second generations, while from the third to the tenth generation it is homozygous dominant.

%\section{Ucapan Terima kasih}
%Tuliskan ucapan terima kasih kepada pihak-pihak yang telah membantu dalam penulisan makalah ini (\textbf{tanpa gelar}). 

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\end{document}
