# Contraction Principles in $M_{s}$-metric Spaces

N. Mlaiki1, N. Souayah${}^{2},$ K. Abodayeh${}^{3},$ T. Abdeljawad4
Department of Mathematical Sciences, Prince Sultan University1,3,4
Department of Natural Sciences, Community College, King Saud University2
Riyadh, Saudi Arabia 11586
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###### Abstract

In this paper, we give an interesting extension of the partial $S$-metric
space which was introduced [4] to the $M_{s}$-metric space. Also, we prove the
existence and uniqueness of a fixed point for a self mapping on an
$M_{s}$-metric space under different contraction principles.

##  1 Introduction

Many researchers over the years proved many interesting results on the
existence of a fixed point for a self mapping on different types of metric
spaces for example, see ([2], [3], [5], [6], [7], [8], [9], [10], [12], [17],
[18], [19].) The idea behind this paper was inspired by the work of Asadi in
[1]. He gave a more general extension of almost any metric space with two
dimensions, and that is not just by defining the self ”distance” in a metric
as in partial metric spaces [13, 14, 15, 16, 11], but he assumed that is not
necessary that the self ”distance” is less than the value of the metric
between two different elements.

In [4], an extension of $S$-metric spaces to a partial $S$-metric spaces was
introduced. Also, they showed that every $S$-metric space is a partial
$S$-metric space, but not every partial $S$-metric space is an $S$-metric
space. In our paper, we introduce the concept of $M_{s}$\- metric spaces which
an extension of the partial $S$-metric spaces in which we will prove some
fixed point results.

First, we remind the reader of the definition of a partial $S$-metric space

######  Definition 1.

[4] Let $X$ be a nonempty set. A partial S-metric on $X$ is a function
$S_{p}:X^{3}\rightarrow[0,\infty)$ that satisfies the following conditions,
for all $x,y,z,t\in X:$

  * (i)

$x=y$ if and only if $S_{p}(x,x,x)=S_{p}(y,y,y)=S_{p}(x,x,y)$,

  * (ii)

$S_{p}(x,y,z)\leq S_{p}(x,x,t)+S_{p}(y,y,t)+S_{p}(z,z,t)-S_{p}(t,t,t)$,

  * (iii)

$S_{p}(x,x,x)\leq S_{p}(x,y,z)$,

  * (iv)

$S_{p}(x,x,y)=S_{p}(y,y,x).$

The pair $(X,S_{p})$ is called a partial S-metric space.

Next, we give the definition of an $M_{s}$-metric space, but first we
introduce the following notation.

Notation.

  1. 1.

$m_{{s}_{x,y,z}}:=\min\\{m_{s}(x,x,x),m_{s}(y,y,y),m_{s}(z,z,z)\\}$

  2. 2.

$M_{{s}_{x,y,z}}:=\max\\{m_{s}(x,x,x),m_{s}(y,y,y),m_{s}(z,z,z)\\}$

######  Definition 2.

An $M_{s}$-metric on a nonempty set X is a function
$m_{s}:X^{3}\rightarrow\mathbb{R}^{+}$ such that for all $x,y,z,t\in X$, the
following conditions

are satisfied:

  1. 1.

$m_{s}(x,x,x)=m_{s}(y,y,y)=m_{s}(x,x,y)$ if and only if $x=y,$

  2. 2.

$m_{{s}_{x,y,z}}\leq m_{s}(x,y,z),$

  3. 3.

$m_{s}(x,x,y)=m_{s}(y,y,x),$

  4. 4.

$(m_{s}(x,y,z)-m_{{s}_{x,y,z}})\leq(m_{s}(x,x,t)-m_{{s}_{x,x,t}})+(m_{s}(y,y,t)-m_{{s}_{y,y,t}})+(m_{s}(z,z,t)-m_{{s}_{z,z,t}}).$

The pair $(X,m_{s})$ is called an $M_{s}$-metric space. Notice that the
condition $m_{s}(x,x,x)=m_{s}(y,y,y)=m_{s}(z,z,z)=m_{s}(x,y,z)\Leftrightarrow
x=y=z$ implies (1) above.

It is a straightforward to verify that every partial $S$-metric space is an
$M_{s}$-metric space but the converse is not true. The following is an example
of an $M_{s}$-metric which is not a partial $S$-metric space.

######  Example 1.1.

Let $X=\\{1,2,3\\}$ and define the $M_{s}$-metric space $m_{s}$ on $X$ by
$m_{s}(1,2,3)=6$,
$m_{s}(1,1,2)=m_{s}(2,2,1)=m_{s}(1,1,1)=8$,
$m_{s}(1,1,3)=m_{s}(3,3,1)=m_{s}(3,3,2)=m_{s}(2,2,3)=7$,
$m_{s}(2,2,2)=9,\text{ and }m_{s}(3,3,3)=5$.
It is not difficult to see that $(X,m_{s})$ is an $M_{s}$-metric space, but
since $m_{s}(1,1,1)\not\leq m_{s}(1,2,3)$ we deduce that $m_{s}$ is not a
partial $S$-metric space.

######  Definition 3.

Let $(X,m_{s})$ be an $M_{s}$-metric space. Then:

  1. 1.

A sequence $\\{x_{n}\\}$ in $X$ converges to a point $x$ if and only if

$\lim_{n\rightarrow\infty}(m_{s}(x_{n},x_{n},x)-m_{{s}{x_{n},x_{n},x}})=0.$

  2. 2.

A sequence $\\{x_{n}\\}$ in $X$ is said to be $M_{s}$-Cauchy sequence if and
only if

$\lim_{n,m\rightarrow\infty}(m_{s}(x_{n},x_{n},x_{m})-m_{{s}{x_{n},x_{n},x_{m}}}),\text{
and
}\lim_{n\rightarrow\infty}(M_{{s}{x_{n},x_{n},x_{m}}}-m_{{s}{x_{n},x_{n},x_{m}}})$

exist and finite.

  3. 3.

An $M_{s}$-metric space is said to be complete if every $M_{s}$-Cauchy
sequence $\\{x_{n}\\}$ converges to a point $x$ such that

$\lim_{n\rightarrow\infty}(m_{s}(x_{n},x_{n},x)-m_{{s}{x_{n},x_{n},x}})=0\text{
and
}\lim_{n\rightarrow\infty}(M_{{s}{x_{n},x_{n},x}}-m_{{s}{x_{n},x_{n},x}})=0.$

A ball in the $M_{s}-$metric $(X,m_{s})$ space with center $x\in X$ and radius
$\eta>0$ is defined by

$B_{s}[x,\eta]=\\{y\in X\mid m_{s}(x,x,y)-m_{{s}{x,x,y}}\leq\eta\\}.$

The topology of $(X,M_{s})$ is generated by means of the basis
$\beta=\\{B_{s}[x,\eta]:\eta>0\\}$.

######  Lemma 1.2.

Assume $x_{n}\rightarrow x$ and $y_{n}\rightarrow y$ as $n\rightarrow\infty$
in an $M_{s}-$metric space $(X,m_{s})$. Then,

$\lim_{n\rightarrow\infty}(m_{s}(x_{n},x_{n},y_{n})-m_{sx_{n},x_{n},y_{n}})=m_{s}(x,x,y)-m_{sx,x,y}$

###### Proof.

The proof follows by the inequality $(4)$ in Definition 2. Indeed, we have

$\displaystyle|(m_{s}(x_{n},x_{n},y_{n})-m_{sx_{n},x_{n},y_{n}})-(m_{s}(x,x,y)-m_{sx,x,y})|$
$\displaystyle\leq$ $\displaystyle
2[(m_{s}(x_{n},x_{n},x)-m_{sx_{n},x_{n},x})$ (1) $\displaystyle+$
$\displaystyle(m_{s}(y_{n},y_{n},y)-m_{sy_{n},y_{n},y})]$

∎

##  2 Fixed Point Theorems

In this section, we consider some results about the existence and the
uniqueness of fixed point for self mappings on an $M_{s}$-metric space, under
different contractions principles.

######  Theorem 1.

Let $(X,m_{s})$ be a complete $M_{s}$-metric space and $T$ be a self mapping
on $X$ satisfying the following condition:

$m_{s}(Tx,Tx,Ty)\leq km_{s}(x,x,y),$ (2)

for all $x,y\in X,$ where $k\in[0,1).$ Then T has a unique fixed point $u$.
Moreover, $m_{s}(u,u,u)=0.$

###### Proof.

Since $k\in[0,1)$, we can choose a natural number $n_{0}$ such that for a
given $0<\epsilon<1,$ we have $k^{n_{0}}<\dfrac{\epsilon}{8}.$ Let
$T^{n_{0}}\equiv F$ and $F^{i}x_{0}=x_{i}$ for all natural numbers $i$, where
$x_{0}$ is arbitrary. Hence, for all $x,y\in X$, we have

$m_{s}(Fx,Fx,Fy)=m_{s}(T^{n_{0}}x,T^{n_{0}}x,T^{n_{0}}y)\leq
k^{n_{0}}m_{s}(x,x,y).$

For any $i$, we have

$\displaystyle m_{s}(x_{i+1},x_{i+1},x_{i})$
$\displaystyle=m_{s}(Fx_{i},Fx_{i},Fx_{i-1})$ $\displaystyle\leq
k^{n_{0}}m_{s}(x_{i},x_{i},x_{i-1})$ $\displaystyle\leq
k^{n_{0}+i}m_{s}(x_{1},x_{1},x_{0})\rightarrow 0\ as\ i\rightarrow\infty.$

Similarly, by (2) we have $m_{s}(x_{i},x_{i},x_{i})\rightarrow 0$ as
$i\rightarrow\infty$. Thus, we choose $l$ such that

$m_{s}(x_{l+1},x_{l+1},x_{l})<\frac{\epsilon}{8}\text{ and
}m_{s}(x_{l},x_{l},x_{l})<\frac{\epsilon}{4}.$

Now, let $\eta=\frac{\epsilon}{2}+m_{s}(x_{l},x_{l},x_{l})$. Define the set

$B_{s}[x_{l},\eta]=\\{y\in X\mid
m_{s}(x_{l},x_{l},y)-m_{{s}{x_{l},x_{l},y}}\leq\eta\\}.$

Note that, $x_{l}\in B_{s}[x_{l},\eta]$. Therefore
$B_{s}[x_{l},\eta]\neq\emptyset.$ Let $z\in B_{s}[x_{l},\eta]$ be arbitrary.
Hence,

$\displaystyle m_{s}(Fz,Fz,Fx_{l})$ $\displaystyle\leq
k^{n_{0}}m_{s}(z,z,x_{l})$
$\displaystyle\leq\frac{\epsilon}{8}[\frac{\epsilon}{2}+m_{{s}{z,z,x_{l}}}+m_{s}(x_{l},x_{l},x_{l})]$
$\displaystyle<\frac{\epsilon}{8}[1+2m_{s}(x_{l},x_{l},x_{l})].$

Also, we know that
$m_{s}(Fx_{l},Fx_{l},x_{l})=m_{s}(x_{l+1},x_{l+1},x_{l})<\dfrac{\epsilon}{8}.$
Therefore,

$\displaystyle m_{s}(Fz,Fz,x_{l})-m_{{s}{Fz,Fz,x_{l}}}$ $\displaystyle\leq
2[(m_{s}(Fz,Fz,Fx_{l})-m_{{s}{Fz,Fz,Fx_{l}}})]+(m_{s}(Fx_{l},Fx_{l},x_{l})-m_{{s}{Fx_{l},Fx_{l},x_{l}}})$
$\displaystyle\leq 2m_{s}(Fz,Fz,Fx_{l})+m_{s}(Fx_{l},Fx_{l},x_{l})]$
$\displaystyle\leq
2\frac{\epsilon}{8}(1+2m_{s}(x_{l},x_{l},x_{l}))+\frac{\epsilon}{8}$
$\displaystyle=\frac{\epsilon}{4}+\frac{\epsilon}{8}+\frac{\epsilon}{2}m_{s}(x_{l},x_{l},x_{l})$
$\displaystyle<\frac{\epsilon}{2}+m_{s}(x_{l},x_{l},x_{l}).$

Thus, $Fz\in B_{b}[x_{l},\eta]$ which implies that $F$ maps
$B_{b}[x_{l},\eta]$ into itself. Thus, by repeating this process we deduce
that for all $n\geq 1$ we have $F^{n}x_{l}\in B_{b}[x_{l},\eta]$ and that is
$x_{m}\in B_{b}[x_{l},\eta]$ for all $m\geq l.$ Therefore, for all $m>n\geq l$
where $n=l+i$ for some $i$

$\displaystyle m_{s}(x_{n},x_{n},x_{m})$
$\displaystyle=m_{s}(Fx_{n-1},Fx_{n-1},Fx_{m-1})$ $\displaystyle\leq
k^{n_{0}}m_{s}(x_{n-1},x_{n-1},x_{m-1})$ $\displaystyle\leq
k^{2n_{0}}m_{s}(x_{n-2},x_{n-2},x_{m-2})$ $\displaystyle\vdots$
$\displaystyle\leq k^{in_{0}}m_{s}(x_{l},x_{l},x_{m-i})$ $\displaystyle\leq
m_{s}(x_{l},x_{l},x_{m-i})$
$\displaystyle\leq\frac{\epsilon}{2}+m_{{s}{x_{l},x_{l},x_{m-i}}}+m_{s}(x_{l},x_{l},x_{l})$
$\displaystyle\leq\frac{\epsilon}{2}+2m_{s}(x_{l},x_{l},x_{l}).$

Also, we have $m_{s}(x_{l},x_{l},x_{l})<\frac{\epsilon}{4},$ which implies
that $m_{s}(x_{n},x_{n},x_{m})<\epsilon$ for all $m>n>l,$ and thus
$m_{s}(x_{n},x_{n},x_{m})-m_{{s}{x_{n},x_{n},x_{m}}}<\epsilon$ for all
$m>n>l.$ By the contraction condition (2) we see that the sequence
$\\{m_{s}(x_{n},x_{n},x_{n})\\}$ is decreasing and hence, for all $m>n>l$, we
have

$\displaystyle M_{{s}{x_{n},x_{n},x_{m}}}-m_{{s}{x_{n},x_{n},x_{m}}}$
$\displaystyle\leq M_{{s}{x_{n},x_{n},x_{m}}}$
$\displaystyle=m_{s}(x_{n},x_{n},x_{n})$ $\displaystyle\leq
km_{s}(x_{n-1},x_{n-1},x_{n-1})$ $\displaystyle\vdots$ $\displaystyle\leq
k^{n}m_{s}(x_{0},x_{0},x_{0})\rightarrow 0\text{ as }n\rightarrow\infty.$

Thus, we deduce that

$\lim_{n,m\rightarrow\infty}(m_{s}(x_{n},x_{n},x_{m})-m_{{s}{x_{n},x_{n},x_{m}}})=0,\text{
and
}\lim_{n\rightarrow\infty}(M_{{s}{x_{n},x_{n},x_{m}}}-m_{{s}{x_{n},x_{n},x_{m}}})=0$

Hence, the sequence $\\{x_{n}\\}$ is an $M_{s}$-Cauchy. Since $X$ is complete,
there exists $u\in X$ such that

$\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n},u)-m_{{s}{x_{n},x_{n},u}}=0,~{}~{}\lim_{n\rightarrow\infty}M_{sx_{n},x_{n},u}-m_{{s}{x_{n},x_{n},u}}=0$

The contraction condition (2) implies that
$m_{s}(x_{n},x_{n},x_{n})\rightarrow 0$ as $n\rightarrow\infty$. Moreover,
notice that

$\lim_{n\rightarrow\infty}M_{sx_{n},x_{n},u}-m_{{s}{x_{n},x_{n},u}}=\lim_{n\rightarrow\infty}|m_{s}(x_{n},x_{n},x_{n})-m_{s}(u,u,u)|=0,$

and hence $m_{s}(u,u,u)=0$. Since $x_{n}\rightarrow u$, $m_{s}(u,u,u)=0$ and
$m_{s}(x_{n},x_{n},x_{n})\rightarrow 0$ as $n\rightarrow\infty$ then
$\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n},u)=\lim_{n\rightarrow\infty}m_{sx_{n},x_{n},u}=0$.
Since $m_{s}(Tx_{n},Tx_{n},Tu)\leq km_{s}(x_{n},x_{n},u)\rightarrow 0$ as
$n\rightarrow\infty$, then $Tx_{n}\rightarrow Tu$.

Now, we show that $Tu=u.$ By Lemma 1.2 and that $Tx_{n}\rightarrow Tu$ and
$x_{n+1}=Tx_{n}\rightarrow u$, we have

$\displaystyle\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n},u)-m_{{s}{x_{n},x_{n},u}}$
$\displaystyle=0$
$\displaystyle=\lim_{n\rightarrow\infty}m_{s}(x_{n+1},x_{n+1},u)-m_{{s}{x_{n+1},x_{n+1},u}}$
$\displaystyle=\lim_{n\rightarrow\infty}m_{s}(Tx_{n},Tx_{n},u)-m_{{s}{Tx_{n},Tx_{n},u}}$
$\displaystyle=m_{s}(u,u,u)-m_{{s}{Tu,Tu,u}}$
$\displaystyle=m_{s}(Tu,Tu,u)-m_{{s}{Tu,Tu,u}}.$

Hence, $m_{s}(Tu,Tu,u)=m_{{s}{Tu,Tu,u}}=m_{s}(u,u,u)$ , but also by the
contraction condition (2) we see that $m_{{s}{Tu,Tu,u}}=m_{s}(Tu,Tu,Tu)$ .
Therefore, (2) in Definition 2 implies that $Tu=u.$

To prove the uniqueness of the fixed point $u,$ assume that $T$ has two fixed
points $u,v\in X;$ that is, $Tu=u$ and $Tv=v.$ Thus,

$m_{s}(u,u,v)=m_{s}(Tu,Tu,Tv)\leq km_{s}(u,u,v)<m_{s}(u,u,v),$

$m_{s}(u,u,u)=m_{s}(Tu,Tu,Tu)\leq km_{s}(u,u,u)<m_{s}(u,u,u),$

and

$m_{s}(v,v,v)=m_{s}(Tv,Tv,Tv)\leq km_{s}(v,v,v)<m_{s}(v,v,v),$

which implies that $m_{s}(u,u,v)=0=m_{s}(u,u,u)=m_{s}(v,v,v),$ and hence $u=v$
as desired. Finally,assume that $u$ is a fixed point of $T.$ Then applying the
contraction condition (2) with $k\in[0,1)$, implies that

$\displaystyle m_{s}(u,u,u)$ $\displaystyle=m_{s}(Tu,Tu,Tu)$
$\displaystyle\leq km_{s}(u,u,u)$ $\displaystyle\vdots$ $\displaystyle\leq
k^{n}m_{s}(u,u,u)$

Taking the limit as $n$ tends to infinity, implies that is $m_{s}(u,u,u)=0.$ ∎

In the following result, we prove the existence and uniqueness of a fixed
point for a self mapping in $M_{s}$-metric space, but under a more general
contraction.

######  Theorem 2.

Let $(X,m_{s})$ be a complete $M_{s}$-metric space and $T$ be a self mapping
on $X$ satisfying the following condition:

$m_{s}(Tx,Tx,Ty)\leq\lambda[m_{s}(x,x,Tx)+m_{s}(y,y,Ty)],$ (3)

for all $x,y\in X,$ where $\lambda\in[0,\frac{1}{2}).$ Then T has a unique
fixed point $u,$ where $m_{s}(u,u,u)=0.$

###### Proof.

Let $x_{0}\in X$ be arbitrary. Consider the sequence $\\{x_{n}\\}$ defined by
$x_{n}=T^{n}x_{0}$ and $m_{s_{n}}=m_{s}(x_{n},x_{n},x_{n+1}).$ Note that if
there exists a natural number $n$ such that $m_{s_{n}}=0,$ then $x_{n}$ is a
fixed point of $T$ and we are done. So, we may assume that $m_{s_{n}}>0,$ for
$n\geq 0.$ By (3), we obtain for any $n\geq 0,$

$\displaystyle m_{s_{n}}$
$\displaystyle=m_{s}(x_{n},x_{n},x_{n+1})=m_{s}(Tx_{n-1},Tx_{n-1},Tx_{n})$
$\displaystyle\leq\lambda[m_{s}(x_{n-1},x_{n-1},Tx_{n-1})+m_{s}(x_{n},x_{n},Tx_{n})]$
$\displaystyle=\lambda[m_{s}(x_{n-1},x_{n-1},x_{n})+m_{s}(x_{n},x_{n},x_{n+1})]$
$\displaystyle=\lambda[m_{s_{n-1}}+m_{s_{n}}].$

Hence, $m_{s_{n}}\leq\lambda m_{s_{n-1}}+\lambda m_{s_{n}},$ which implies
$m_{s_{n}}\leq\mu m_{s_{n-1}},$ where $\mu=\frac{\lambda}{1-\lambda}<1$ as
$\lambda\in[0,\frac{1}{2}).$ By repeating this process we get

$m_{s_{n}}\leq\mu^{n}m_{s_{0}}.$

Thus, $\lim_{n\rightarrow\infty}m_{s_{n}}=0.$ By (3), for all natural numbers
$n,m$ we have

$\displaystyle m_{s}(x_{n},x_{n},x_{m})$
$\displaystyle=m_{s}(T^{n}x_{0},T^{n}x_{0},T^{m}x_{0})=m_{s}(Tx_{n-1},Tx_{n-1},Tx_{m-1})$
$\displaystyle\leq\lambda[m_{s}(x_{n-1},x_{n-1},Tx_{n-1})+m_{s}(x_{m-1},x_{m-1},Tx_{m-1})]$
$\displaystyle=\lambda[m_{s}(x_{n-1},x_{n-1},x_{n})+m_{s}(x_{m-1},x_{m-1},x_{m})]$
$\displaystyle=\lambda[m_{s_{n-1}}+m_{s_{m-1}}].$

Since $\lim_{n\rightarrow\infty}m_{s_{n}}=0,$ for every $\epsilon>0$, we can
find a natural number $n_{0}$ such that $m_{s_{n}}<\frac{\epsilon}{2}$ and
$m_{s_{m}}<\frac{\epsilon}{2}$ for all $m,n>n_{0}.$ Therefore, it follows that

$m_{s}(x_{n},x_{n},x_{m})\leq\lambda[m_{s_{n-1}}+m_{s_{m-1}}]<\lambda[\frac{\epsilon}{2}+\frac{\epsilon}{2}]<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\
\ \text{for all }n,m>n_{0}.$

This implies that

$m_{s}(x_{n},x_{n},x_{m})-m_{{s}{x_{n},x_{n},x_{m}}}<\epsilon\ \ \text{for all
}n,m>n_{0}.$

Now, for all natural numbers $n,m$ we have

$\displaystyle M_{{s}{x_{n},x_{n},x_{m}}}$
$\displaystyle=m_{s}(Tx_{n-1},Tx_{n-1},Tx_{n-1})$
$\displaystyle\leq\lambda[m_{s}(x_{n-1},x_{n-1},Tx_{n-1})+m_{s}(x_{n-1},x_{n-1},Tx_{n-1})]$
$\displaystyle=\lambda[m_{s}(x_{n-1},x_{n-1},x_{n})+m_{s}(x_{n-1},x_{n-1},x_{n})]$
$\displaystyle=\lambda[m_{s_{n-1}}+m_{s_{n-1}}]$ $\displaystyle=2\lambda
m_{s_{n-1}}.$

As $\lim_{n\rightarrow\infty}m_{s_{n-1}}=0,$ for every $\epsilon>0$ we can
find a natural number $n_{0}$ such that $m_{s_{n}}<\frac{\epsilon}{2}$ and for
all $m,n>n_{0}.$ Therefore, it follows that

$M_{{s}{x_{n},x_{n},x_{m}}}\leq\lambda[m_{s_{n-1}}+m_{s_{n-1}}]<\lambda[\frac{\epsilon}{2}+\frac{\epsilon}{2}]<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\
\ \text{for all }n,m>n_{0},$

which implies that

$M_{{s}{x_{n},x_{n},x_{m}}}-m_{{s}{x_{n},x_{n},x_{m}}}<\epsilon\ \ \text{for
all }n,m>n_{0}.$

Thus, $\\{x_{n}\\}$ is an $M_{s}$-Cauchy sequence in $X.$ Since $X$ is
complete, there exists $u\in X$ such that

$\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n},u)-m_{{s}{x_{n},x_{n},u}}=0.$

Now, we show that $u$ is a fixed point of $T$ in $X.$ For any natural number
$n$ we have,

$\displaystyle\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n},u)-m_{{s}{x_{n},x_{n},u}}$
$\displaystyle=0$
$\displaystyle=\lim_{n\rightarrow\infty}m_{s}(x_{n+1},x_{n+1},u)-m_{{s}{x_{n+1},x_{n+1},u}}$
$\displaystyle=\lim_{n\rightarrow\infty}m_{s}(Tx_{n},Tx_{n},u)-m_{{s}{Tx_{n},Tx_{n},u}}$
$\displaystyle=m_{s}(Tu,Tu,u)-m_{{s}{Tu,Tu,u}}.$

This implies that $m_{s}(Tu,Tu,u)-m_{{s}{u,u,Tu}}=0,$ and that is
$m_{s}(Tu,Tu,u)=m_{{s}{u,u,Tu}}.$ Now, assume that

$m_{s}(Tu,Tu,u)=m_{s}(Tu,Tu,Tu)\leq 2\lambda m_{s}(u,u,Tu)=2\lambda
m_{s}(Tu,Tu,u)<m_{s}(u,u,Tu).$

Thus,

$m_{s}(Tu,Tu,u)=m_{s}(u,u,u)\leq m_{s}(Tu,Tu,Tu)\leq 2\lambda
m_{s}(u,u,Tu)<m_{s}(u,u,Tu)$

Therefore, $Tu=u$ and thus $u$ is a fixed point of $T.$

Next, we show that if $u$ is a fixed point, then $m_{s}(u,u,u)=0.$ Assume that
$u$ is a fixed point of $T,$ then using the contraction (3), we have

$\displaystyle m_{s}(u,u,u)$ $\displaystyle=m_{s}(Tu,Tu,Tu)$
$\displaystyle\leq\lambda[m_{s}(u,u,Tu)+m_{s}(u,u,Tu)]$
$\displaystyle=2\lambda m_{s}(u,u,Tu)$ $\displaystyle=2\lambda m_{s}(u,u,u)$
$\displaystyle<m_{s}(u,u,u)\ \ \text{since }\lambda\in[0,\frac{1}{2});$

that is, $m_{s}(u,u,u)=0.$

Finally, To prove uniqueness, assume that $T$ has two fixed points say $u,v\in
X.$ Hence,

$m_{s}(u,u,v)=m_{s}(Tu,Tu,Tv)\leq\lambda[m_{s}(u,u,Tu)+m_{s}(v,v,Tv)]=\lambda[m_{s}(u,u,u)+m_{s}(v,v,v)]=0,$

which implies that $m_{s}(u,u,v)=0=m_{s}(u,u,v)=m_{s}(u,u,v),$ and hence $u=v$
as required.

∎

In closing, the authors would like to bring to the reader’s attention that in
this interesting $M_{s}$-metric space it is possible to add some control
functions in both contractions of Theorems 1, and 2.

######  Theorem 3.

Let $(X,m_{s})$ be a complete $M_{s}$-metric space and $T$ be a self mapping
on $X$ satisfying the following condition: for all $x,y,z\in X$

$m_{s}(Tx,Ty,Tz)\leq m_{s}(x,y,z)-\phi(m_{s}(x,y,z)),$ (4)

where $\phi:[0,\infty)\rightarrow[0,\infty)$ is a continuous and non-
decreasing function and $\phi^{-1}(0)=0$ and $\phi(t)>0$ for all $t>0$. Then
$T$ has a unique fixed point in $X$.

###### Proof.

Let $x_{0}\in X$. Define the sequence $\\{x_{n}\\}$ in $X$ such that
$x_{n}=T^{n-1}x_{0}=Tx_{n-1}$, for all $n\in{\rm I}{\rm N}$. Note that if
there exists an $n\in{\rm I}{\rm N}$ such that $x_{n+1}=x_{n}$, then $x_{n}$
is a fixed point for $T$. Without lost of generality, assume that $x_{n+1}\neq
x_{n}$, for all $n\in{\rm I}{\rm N}$. Now

$\displaystyle m_{s}(x_{n},x_{n+1},x_{n+1})$ $\displaystyle=$ $\displaystyle
m_{s}(Tx_{n-1},Tx_{n},Tx_{n})$ (5) $\displaystyle\leq$ $\displaystyle
m_{s}(x_{n-1},x_{n},x_{n})-\phi(m_{s}(x_{n-1},x_{n},x_{n}))$
$\displaystyle\leq$ $\displaystyle m_{s}(x_{n-1},x_{n},x_{n}).$

Similarly, we can prove that $m_{s}(x_{n-1},x_{n},x_{n})\leq
m_{s}(x_{n-2},x_{n-1},x_{n-1}).$ Hence, $m_{s}(x_{n},x_{n+1},x_{n+1})$ is a
monotone decreasing sequence. Hence there exists $r\geq 0$ such that

$\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n+1},x_{n+1})=r.$

Now, by taking the limit as $n\rightarrow\infty$ in the inequality (5), we get
$r\leq r-\phi(r)$ which leads to a contradiction unless $r=0$. Therefore,

$\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n+1},x_{n+1})=0.\\\ $

Suppose that $\\{x_{n}\\}$ is not an $M_{s}$-Cauchy sequence. Then there
exists an $\epsilon>0$ such that we can find subsequences $x_{m_{k}}$ and
$x_{n_{k}}$ of $\\{x_{n}\\}$ such that

$m_{s}(x_{n_{k}},x_{m_{k}},x_{m_{k}})-m_{sx_{n_{k}},x_{m_{k}},x_{m_{k}}}\geq\epsilon.$
(6)

Choose $n_{k}$ to be the smallest integer with $n_{k}>m_{k}$ and satisfies the
inequality (6). Hence,
$m_{s}(x_{n_{k}},x_{m_{k-1}},x_{m_{k-1}})-m_{sx_{n_{k}},x_{m_{k-1}},x_{m_{k-1}}}<\epsilon.$
Now,

$\displaystyle\epsilon$ $\displaystyle\leq$ $\displaystyle
m_{s}(x_{m_{k}},x_{n_{k}},x_{n_{k}})-m_{sx_{m_{k}},x_{n_{k}},x_{n_{k}}}$
$\displaystyle\leq$ $\displaystyle
m_{s}(x_{m_{k}},x_{n_{k-1}},x_{n_{k-1}})+2m_{s}(x_{n_{k-1}},x_{n_{k-1}},x_{n_{k-1}})-m_{sx_{m_{k}},x_{n_{k-1}},x_{n_{k-1}}}$
$\displaystyle\leq$
$\displaystyle\epsilon+2m_{s}(x_{n_{k-1}},x_{n_{k-1}},x_{n_{k-1}})$
$\displaystyle<$ $\displaystyle\epsilon,$

as $n\rightarrow\infty$. Hence, we have a contradiction.
Without lost of generality, assume that
$m_{sx_{n},x_{n},x_{m}}=m_{s}(x_{n},x_{n},x_{n})$. Then we have

$\displaystyle 0\leq m_{sx_{n},x_{n},x_{m}}-m_{sx_{n},x_{n},x_{m}}$
$\displaystyle\leq$ $\displaystyle M_{sx_{n},x_{n},x_{m}}$ $\displaystyle=$
$\displaystyle m_{s}(x_{n},x_{n},x_{n})$ $\displaystyle=$ $\displaystyle
m_{s}(Tx_{n-1},Tx_{n-1},Tx_{n-1})$ $\displaystyle\leq$ $\displaystyle
m_{s}(x_{n-1},x_{n-1},x_{n-1})-\phi(m_{s}(x_{n-1},x_{n-1},x_{n-1}))$
$\displaystyle\leq$ $\displaystyle m_{s}(x_{n-1},x_{n-1},x_{n-1})$
$\displaystyle\vdots$ $\displaystyle\leq$ $\displaystyle
m_{s}(x_{0},x_{0},x_{0})$

Hence,
$\displaystyle\lim_{n\rightarrow\infty}m_{sx_{n},x_{n},x_{m}}-m_{sx_{n},x_{n},x_{m}}$
exists and finite. Therefore, $\\{x_{n}\\}$ is an $M_{s}$-Cauchy sequence.
Since $X$ is complete, the sequence $\\{x_{n}\\}$ converges to an element
$x\in X$; that is,

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}m_{s}(x_{n},x_{n},x)-m_{sx_{n},x_{n},x}$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}m_{s}(x_{n+1},x_{n+1},x)-m_{sx_{n+1},x_{n+1},x}$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}m_{s}(Tx_{n},Tx_{n},x)-m_{sTx_{n},Tx_{n},x}$
$\displaystyle=$ $\displaystyle m_{s}(Tx,Tx,x)-m_{sTx,Tx,x}.$

Similarly to the proof of Theorem 2, it is not difficult to show that this
implies that, $Tx=x$ and so $x$ is a fixed point.
Finally, we show that $T$ has a unique fixed point. Assume that there are two
fixed points $u,v\in X$ of $T$. If we have $m_{s}(u,u,v)>0$, then Condition
(4) implies that

$m_{s}(u,u,v)=m_{s}(Tu,Tu,Tv)\leq
m_{s}(u,u,v)-\phi(m_{s}(u,u,v))<m_{s}(u,u,v),$

and that is a contradiction. Therefore, $m_{s}(u,u,v)=0$ and similarly
$m_{s}(u,u,u)=M_{s}(v,v,v)=0$ and thus $u=v$ as desired. ∎

In closing, is it possible to define the same space but without the symmetry
condition, (i.e. $m_{s}(x,x,y)\neq m_{s}(y,y,x)?$) If possible, what kind of
results can be obtained in such space?

## References

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# Mast Kalandar at SemEval-2024 Task 8: On the Trail of Textual Origins:
RoBERTa-BiLSTM Approach to Detect AI-Generated Text

Jainit Sushil Bafna
IIIT Hyderabad
<EMAIL_ADDRESS>
&Hardik Mittal
IIIT Hyderabad
<EMAIL_ADDRESS>
Suyash Sethia11footnotemark: 1
IIIT Hyderabad
<EMAIL_ADDRESS>
&Manish Shrivastava
IIIT Hyderabad
<EMAIL_ADDRESS>
&Radhika Mamidi
IIIT Hyderabad
<EMAIL_ADDRESS>Equal contribution.

###### Abstract

Large Language Models (LLMs) have showcased impressive abilities in generating
fluent responses to diverse user queries. However, concerns regarding the
potential misuse of such texts in journalism, educational, and academic
contexts have surfaced. SemEval 2024 introduces the task of Multigenerator,
Multidomain, and Multilingual Black-Box Machine-Generated Text Detection,
aiming to develop automated systems for identifying machine-generated text and
detecting potential misuse. In this paper, we i) propose a RoBERTa-BiLSTM
based classifier designed to classify text into two categories: AI-generated
or human ii) conduct a comparative study of our model with baseline approaches
to evaluate its effectiveness. This paper contributes to the advancement of
automatic text detection systems in addressing the challenges posed by
machine-generated text misuse. Our architecture ranked 46th on the official
leaderboard with an accuracy of 80.83 among 125.

Mast Kalandar at SemEval-2024 Task 8: On the Trail of Textual Origins:
RoBERTa-BiLSTM Approach to Detect AI-Generated Text

Jainit Sushil Bafna IIIT Hyderabad<EMAIL_ADDRESS>Hardik
Mittal††thanks: Equal contribution. IIIT Hyderabad
<EMAIL_ADDRESS>

Suyash Sethia11footnotemark: 1 IIIT Hyderabad
<EMAIL_ADDRESS>Manish Shrivastava IIIT Hyderabad
<EMAIL_ADDRESS>Radhika Mamidi IIIT Hyderabad
<EMAIL_ADDRESS>

##  1 Introduction

The task of classifying text as either AI-generated or human-generated holds
significant importance in the field of natural language processing (NLP). It
addresses the growing need to distinguish between content created by
artificial intelligence models and that generated by human authors, a
distinction crucial for various applications such as content moderation,
misinformation detection, and safeguarding against AI-generated malicious
content. This task is outlined in the task overview paper by Wang et al.
(2024), emphasizing its relevance and scope in the NLP community.

Our system employs a hybrid approach combining deep learning techniques with
feature engineering to tackle the classification task effectively.
Specifically, we leverage a BiLSTM (Bidirectional Long Short-Term Memory)
Schuster and Paliwal (1997) neural network in conjunction with RoBERTa Liu et
al. (2019), a pre-trained language representation model, to capture both
sequential and contextual information from the input sentences. This hybrid
architecture enables our system to effectively capture nuanced linguistic
patterns and semantic cues for accurate classification.

Participating in this task provided valuable insights into the capabilities
and limitations of our system. Quantitatively, our system achieved competitive
results, ranking 46 relative to other teams in terms of accuracy and F1 score.
Qualitatively, we observed that our system struggled with distinguishing
between sentences generated by AI models trained on specific domains or
datasets with highly similar linguistic patterns.

We have released the code for our system on GitHub111https://github.com/Mast-
Kalandar/SemEval2024-task8, facilitating transparency and reproducibility in
our approach.

##  2 Related Works

a)  Model/Source chatGPT cohere davinci dolly human wikihow 3000 3000 3000
3000 15499 wikipedia 2995 2336 3000 2702 14497 reddit 3000 3000 3000 3000
15500 arxiv 3000 3000 2999 3000 15498 peerread 2344 2342 2344 2344 2357
b)  Model/Source bloomz human wikihow 500 500 wikipedia 500 500 reddit 500 500
arxiv 500 500 peerread 500 500

Table 1: Table a) contains statistics about the train split. Table b) contains
statistics about the validation split from the M4 dataset

In the field of detecting machine-generated text, numerous methodologies and
models have been examined. A distinguished methodology is the application of
the RoBERTa Classifier, which enhances the RoBERTa language model through
fine-tuning for the specific purpose of identifying machine-generated text.
The proficiency of pre-trained classifiers like RoBERTa in this domain has
been affirmed through various studies, including those conducted by Solaiman
et al. (2019) and additional research by Zellers et al. (2019); Ippolito et
al. (2019); Bakhtin et al. (2020); Jang et al. (2020); Uchendu et al. (2021).
Concurrently, the XLM-R Classifier exploits the multilingual training of the
XLM-RoBERTa model to effectively recognize machine-generated text in various
languages, as demonstrated by Conneau et al. (2019).

Alternatively, the exploration of logistic regression models that incorporate
GLTR (Giant Language model Test Room) features has been undertaken. These
models strive to discern subtleties in text generation methodologies by
analyzing token probabilities and distribution entropy, as investigated by
Gehrmann et al. (2019). Furthermore, detection efforts have utilized
stylometric and NELA (News Landscape) features, which account for a broad
spectrum of linguistic and structural characteristics, including syntactic,
stylistic, affective, and moral dimensions, as reported by Li et al. (2014)
and Mitchell et al. (2023). Additionally, proprietary frameworks like GPTZero,
devised by Princeton University, focus on indicators such as perplexity and
burstiness to analyze texts for machine-generated content identification.
Although the specific technical details are sparingly disclosed, the reported
effectiveness of GPTZero in identifying outputs from various AI language
models highlights its significance in the ongoing development of machine-
generated text detection strategies Ouyang et al. (2022); Brown et al. (2020);
Radford et al. (2019); Touvron et al. (2023a).

##  3 Background

Model | Accuracy | F1 | Precision | Recall | Params*
---|---|---|---|---|---
Full RoBERTa fine tune | 80.68 | 80.54 | 81.55 | 80.68 | 124M
LoRA with RoBERTa (Freezed) | 81.59 | 81.06 | 85.64 | 81.59 | 0.7M
LoRA with LongFormer | 75.34 | 75.14 | 76.16 | 75.34 | 6M
BiLSTM with RoBERTa (Un-Freezed) | 70.77 | 61.15 | 91.19 | 46.00 | 18M
GRU with RoBERTa (Freezed) | 74.65 | 80.54 | 81.55 | 80.68 | 3M
BiLSTM with RoBERTa (Freezed) | 82.52 | 82.14 | 83.96 | 80.40 | 4M
Table 2: The performance of the models tried on the dev set of the dataset.
*The params only accounts for trainable unfreezed parameters.

###  3.1 Dataset

For the machine-generated text, the researchers used various multilingual
language models like ChatGPTOpenAI (2024), textdavinci-003OpenAI (2022),
LLaMaTouvron et al. (2023b), FlanT5Chung et al. (2022), CohereCohere (2024),
Dolly-v2databricks (2022), and BLOOMzMuennighoff et al. (2023). These models
were given different tasks like writing Wikipedia articles, summarizing
abstracts from arXiv, providing peer reviews, answering questions from Reddit
and Baike/Web QA, and creating news briefs. As evident from Table 1, the
training set lacks any sentences generated by the Bloomz model, which stands
as the sole model represented in the validation set. This deliberate choice
ensures a robust assessment of our model’s generalization capabilities across
all machine-generated outputs, regardless of the specific model generating
them. By exposing our model to diverse machine-generated sentences during
training, including those from unseen models like Bloomz in the validation
set, we aim to evaluate its ability to effectively generalize to novel inputs
and make reliable predictions across the spectrum of machine-generated text.

###  3.2 Task

We focused on Subtask-A of the SemEval Task 8 which involves developing a
classifier to differentiate between monolingual sentences generated by
artificial intelligence (AI) systems and those generated by humans. This
classification task is essential for distinguishing the origin of text and
understanding whether it was produced by AI models or by human authors.

####  3.2.1 Objective

The primary objective is to build a robust classifier capable of accurately
distinguishing between AI-generated and human-generated sentences. The
classifier should generalize well across various AI models and domains,
ensuring consistent performance regardless of the specific model or domain
from which the text originates.

The goal was to design a model that not only performs this task with high
accuracy but also adapts to various AI models and domains. It’s crucial for
the classifier to accurately identify the origin of sentences, regardless of
the technology used to generate them or their subject matter, ensuring broad
applicability and effectiveness

##  4 System Overview

Figure 1: Our proposed architecture of BiLSTM with freezed RoBERTa

Based on our observation (See Appendix A), we discovered that language
modeling task encodes the various features required for detection of AI
written text. So we used pretrained RoBERTa in most of our architectures so
exploit this power of language models.

###  4.1 Full RoBERTa Finetune

The Full RoBERTaLiu et al. (2019) Finetune model, chosen as our baseline,
boasted an extensive architecture and possessed the highest parameter count
among the models under evaluation. Serving as a comprehensive starting point,
this model allowed us to assess the effectiveness of subsequent enhancements
in comparison.

###  4.2 LoRA with RoBERTa (Frozen)

Incorporating Low Rank Adapters Hu et al. (2021), we applied fine-tuning
techniques to the RoBERTa model while strategically freezing all layers. This
approach enabled us to adapt the model to our specific task domain, leveraging
pre-trained representations effectively.

###  4.3 LoRA with LongFormer

The limitation of RoBERTa’s context length (max 512 tokens) posed challenges
for handling lengthy sentences in our dataset. To address this, we
investigated LongFormer Beltagy et al. (2020), a model designed to efficiently
manage longer contexts. Despite employing LoRA for fine-tuning, the model’s
performance on the validation set fell short of expectations, indicating
potential difficulties in generalization.

###  4.4 RoBERTa (2 Layers unfreezed) + BiLSTM

Expanding upon RoBERTa’s capabilities, we introduced a hybrid architecture by
unfreezing two layers and integrating a BiLSTM network Schuster and Paliwal
(1997). RoBERTa served as the primary encoder for sentence representations,
with the subsequent BiLSTM layer trained to classify based on the last hidden
state.

###  4.5 RoBERTa (Frozen) + GRU

In our endeavor to augment RoBERTa’s capabilities, we devised a hybrid
architecture by integrating a Gated Recurrent Unit (GRU) Chung et al. (2014)
network with the frozen RoBERTa model. Within this framework, RoBERTa served
as the encoder for generating sentence representations, while a subsequent GRU
layer was incorporated for sequential processing and classification tasks.
This amalgamation aimed to leverage the strengths of both RoBERTa’s contextual
understanding and GRU’s recurrent dynamics, contributing to enhanced
performance on our target task.

###  4.6 RoBERTa (Frozen) + BiLSTM

In our pursuit of enhancing RoBERTa’s capabilities, we devised a hybrid
architecture by coupling a Bidirectional Long Short-Term Memory (BiLSTM)
network with the RoBERTa model Liu et al. (2019). In this setup, RoBERTa
functioned as the encoder for sentence representations, while a subsequent
BiLSTM layer was employed for classification, utilizing the last hidden state
for decision-making. For a detailed visual representation of the model’s
architecture, please refer to the accompanying Figure 1.

We explored various methodologies (refer to Table 2 for detailed performance
metrics) before selecting the optimal approach as our final model.
Subsequently, we assessed the performance of the chosen model, RoBERTA
(Freezed) + BiLSTM, on the test dataset.

##  5 Experiments

###  5.1 Preprocessing

All textual data underwent standard preprocessing steps, including
tokenization, lowercasing, and punctuation marks. Additionally, specific
domain-related preprocessing, such as handling special characters or domain-
specific terms, was performed as necessary.

###  5.2 Hyperparameter Tuning

Hyperparameters were tuned using a combination of grid search and random
search techniques. We explored various hyperparameter combinations to identify
the optimal configuration for each model variant.
The configuration for LSTM and GRU used in Table 2 is hidden_size=256,
layers=2, dropout=0.2, with LoRA rank being 20 has been found as the best
configuration for the models. For RoBERTa+LSTM model’s feedforward had a
single weight matrix of dimension 512*2.

##  6 Results

Model | Accuracy | F1 | Precision | Recall | Params*
---|---|---|---|---|---
Full RoBERTa fine tune+ | 88.47 | 88.44 | 93.36 | 84.02 | 124M
LoRA with RoBERTa (Freezed) | 80.91 | 80.18 | 83.88 | 80.14 | 0.7M
LoRA with LongFormer | 63.39 | 57.51 | 72.45 | 61.67 | 6M
BiLSTM with RoBERTa (Un-Freezed) | 80.80 | 80.19 | 83.08 | 80.12 | 18M
GRU with RoBERTa (Freezed) | 84.71 | 84.33 | 86.53 | 84.13 | 3M
BiLSTM with RoBERTa (Freezed) | 80.83 | 80.83 | 74.65 | 96.16 | 4M
Table 3: The performance of the models tried on the test set of the dataset.
* The params only accounts for trainable unfreezed parameters.
\+ Baseline mentioned in task overview paper

We tested our models on various models on the test set. The results can be
viewed in (Table: 3).
Ranking: Our BiLSTM+RoBERTa model achieved a ranking of 46 out of 125
participants in the competition, demonstrating its competitive performance (as
shown in Table 3). These results highlight the effectiveness of various
models, including BiLSTM+RoBERTa and GRU+RoBERTa, in addressing the task
objectives. We submitted BiLSTM+RoBERTa based on its strong performance on the
validation set. However, after testing all models listed in Table 3, we found
that GRU+RoBERTa achieved a significantly better result, with an accuracy
increase of approximately 4%.

##  7 Conclusion

In conclusion, our BiLSTM+RoBERTa model effectively tackled the task,
achieving competitive results, thanks to its deep learning and pre-trained
language model. While a similar model with unfrozen RoBERTa boasted higher
precision, its complexity came at the cost of increased parameters.

Impressively, our model ranked 46th out of 125 competition entries (Table 3),
showcasing its potential alongside approaches like GRU+RoBERTa. Interestingly,
post-competition analysis revealed GRU+RoBERTa’s superior accuracy (by about
4%). This highlights the value of exploring diverse architectures and
hyperparameter tuning for peak performance.

Moving forward, there are several avenues for future work to explore. Firstly,
further experimentation with different model architectures, including
alternative combinations of encoders and classifiers, could potentially yield
improvements in performance. Additionally, fine-tuning hyperparameters and
exploring advanced techniques for model optimization may enhance the
robustness and generalization capabilities of our system. Furthermore,
incorporating additional contextual information or domain-specific knowledge
could potentially augment the model’s understanding and performance on
specific tasks. Overall, our findings contribute to the ongoing research
efforts in natural language processing and provide valuable insights for
future developments in this domain.

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## Appendix A

### A. Setup

In this study, we implemented a methodology aimed at distinguishing human-
generated sentences from machine-generated ones within a training dataset. To
achieve this, we initially segregated the dataset into two distinct subsets:
one containing human-generated sentences and the other comprising machine-
generated ones. Subsequently, we trained separate models utilizing these
segregated datasets. Specifically, we employed two distinct models for this
task : i) Bidirectional Long Short-Term Memory (BiLSTM) model, ii) RoBERTa
model.

Following the training phase, we proceeded to evaluate the performance of both
models on a validation dataset. During this evaluation, we measured the loss
incurred by each model when tasked with discerning between human-generated and
machine-generated sentences. This evaluation process was crucial for assessing
the efficacy and generalization capabilities of the trained models in
accurately distinguishing between the two types of sentences.

### B. Results

The results are in form of graphs in Figure 2

((a)) Model Trained on : Human Sentences, Losses Computed on : Human Sentences

((b)) Model Trained on : Human Sentences, Losses Computed on : Machine
Sentences

((c)) Model Trained on : Machine Sentences, Losses Computed on : Human
Sentences

((d)) Model Trained on : Machine Sentences, Losses Computed on : Machine
Sentences

Figure 2: Overall Results on Models trained on Human and Machine Generated
Sentences and Losses Calculated on Human and Machine Generated Sentences

We noted a consistent pattern across both sets of models – those trained on
human-generated sentences and those trained on machine-generated sentences.
Specifically, we observed that the losses incurred by human-generated
sentences on the validation set exhibited a wider distribution with higher
variance, while the losses associated with machine-generated sentences
displayed a narrower distribution with lesser variance.

This observation leads to a compelling inference regarding the predictive
nature of the model losses for each type of data. The wider distribution and
higher variance in losses for human-generated sentences suggest a greater
level of unpredictability associated with these sentences. In contrast, the
narrower distribution and lesser variance in losses for machine-generated
sentences indicate a higher level of predictiveness in the model’s performance
on these sentences.

This finding sheds light on the inherent characteristics of human-generated
versus machine-generated sentences, particularly regarding their
predictability when processed by the trained models. Such insights are crucial
for understanding the intricacies of model behavior and the challenges posed
by different types of data in natural language processing tasks.