# The Cartesian Method and Fermat’s Last Theorem00footnotetext: The author is
grateful to Professor Danny Guerrero, PhD. for his valuable support.
Universidad Nacional Autónoma de Honduras, Department of Pure Mathematics.
Universidad Pedagógica Nacional Francisco Morazán, Department of Mathematics,
Tegucigalpa, Honduras<EMAIL_ADDRESS>This work
was carried out during the COVID-19 pandemic.

By Hector Ivan Nunez
<EMAIL_ADDRESS>
###### Abstract

We prove Fermat’s Last Theorem using the philosophical and mathematical
knowledge of 1637 when French mathematician Pierre de Fermat claimed to have a
proof of his conjecture. A partition of the set of positive real numbers and
the Cartesian method serve as the framework where mathematical artifices are
applied, leading us to discover the existence of at least an irrational number
whose structure is clearly revealed.

## 1\. Introduction

In 1637 the French Rene Descartes proposed a philosophical method of analysis,
today known as the Cartesian method[1]. Two important steps of this method
consist of dividing a complex problem into as many parts as possible until its
fundamental composition is found. Then, we are to solve those simple parts
that appear, group them and synthesize the problem. In that same year, Pierre
de Fermat conjectured on a mathematical structure that is remarkably similar
to the Pythagorean theorem. Naturally, the Fermat’s claim can be studied using
the Cartesian method and that is what our exposition is about. Fermat’s Last
Theorem is also known as the Fermat-Wiles Theorem. There is a lot of
literature[2] that tells its history that we will just state it:

> If $n$ is an integer greater than $2$, then there are no positive integers
> $a,b,c$, that satisfy the equation $a^{n}+b^{n}=c^{n}$.

We begin by developing a Pythagorean-geometrical representation for the
equation $a^{n}+b^{n}=c^{n}$. Additionally, we study the arithmetic structure
of that equation in the event that $a,b,c,n$ are positive integers and $n>2$.
Next, we divide the problem in two parts, one when $n$ is even and the other
when $n$ is odd, both with $n>2$ and $a,b,c$ positive integers. When $n$ is
even, we use the Fundamental Theorem of Arithmetic, and again we divide the
problem into simple parts. Using analogies to the Pythagorean theorem, we
investigate whether in these parts it is possible that a simplified form of
the equation $a^{n}+b^{n}=c^{n}$ is valid. The concept of Pythagorean triples
makes a final invitation to divide the case into several more parts once
again. The Pythagorean triples points out that such equation is not possible
since in the triple $(a,b,c)$ there are at least two positive integers and one
irrational number. From there, the remaining case for $n$ odd greater than $2$
is solved considering perfect square numbers, resulting in a proof of Fermat’s
Last Theorem.

Assumptions. The equation we are analyzing is $a^{n}+b^{n}=c^{n}$ where $a$,
$b$, $c$ and $n$ are arbitrary positive integers, $b\leq a<c$, $n>2$ and
$gcd(a,b,c)=1$. As it is simplified from any common factor, it is a primitive
Fermat’s equation. Zero, negative numbers and the trivial solution when
$n\in\\{1,2\\}$ are not considered since Fermat’s Last Theorem does not refer
to them. This paper aims to reach graduate and undergraduate audiences.

Set partitions and Notation. Let us define the set of rational numbers, called
fractional numbers. This set has all positive rational numbers excluding
positive integers. It is usually denoted by
$\mathbb{Q}^{+}\setminus\mathbb{Z}^{+}$, but it occurs so often that to
simplify the notation we will use just $\mathbb{Q}\setminus\mathbb{Z}$. All
set elements can be expressed as the fraction $p/q$ of two positive integers
such that $q\neq 0$, $q\neq 1$ and $gcd(p,q)=1$.

The set of irrational numbers are all positive real numbers which are neither
integers nor fractional numbers. Sometimes denoted by
$\mathbb{R}^{+}\setminus\mathbb{Q}^{+}$, $\neq\mathbb{Q}^{+}$ and others. We
will use $\mathbb{Q}^{c}$.

Therefore the universe of discourse has a partition of three disjoint sets,
$\mathbb{Z}^{+}$, $\mathbb{Q}\setminus\mathbb{Z}$ and $\mathbb{Q}^{c}$ such
that $\mathbb{Z}^{+}\cap\mathbb{Q}\setminus\mathbb{Z}=\emptyset$,
$\mathbb{Z}^{+}\cap\mathbb{Q}^{c}=\emptyset$,
$\mathbb{Q}\setminus\mathbb{Z}\cap\mathbb{Q}^{c}=\emptyset$ and
$\mathbb{Z}^{+}\cup\mathbb{Q}\setminus\mathbb{Z}\cup\mathbb{Q}^{c}=\mathbb{R}^{+}$.

The set $\mathbb{Z}^{+}$ has a partition of two subsets, well-known as
positive even integers and positive odd integers, and they will be called
simply even and odd. The subset of the bases of the equation
$a^{n}+b^{n}=c^{n}$ is called a triple, denoted by $(a,b,c)$.

## 2\. Mathematical Structure of Fermat’s Last Theorem

###  2.1 Geometric representation.

This Section 2.1 is optional. It is devoted to developing a geometric
representation of the equation $a^{n}+b^{n}=c^{n}$ in the space of real and
discrete numbers. That will help us to understand the dynamics of the four
variables $a,b,c,n$ from a geometrical point of view.

The equation $a^{n}+b^{n}=c^{n}$ can be written as a function of $n$ as
$c(n)=\sqrt[n]{a^{n}+b^{n}}$. We can graph this equation in the $(x,y)$
Cartesian plane, where $x,y\in\mathbb{R}$, $n$ is the independent variable and
$c$ is the dependent variable.

Let us graph the geometric representation of the equation
$c(n)=\sqrt[n]{a^{n}+b^{n}}$ in a Pythagorean way, as a triangle. In the first
quadrant of the $(x,y)$ Cartesian plane we draw an $ABC$ triangle, which has
one side of length $a$, and another of length $b$. The third side is set to a
length of $c=\sqrt[n]{a^{n}+b^{n}}$. The side $a$ is drawn from point $(0,0)$
to $(a,0)$, and side $b$ has one end fixed at $(a,0)$.

Let vertex $C$ be able to freely rotate according to the circle drawn by side
$b$ from point $(a,0)$. Notice that the location of vertex $C$ depends on the
value of $n$.

Since $b\leq a<c$, then vertex $C$ of $\Delta ABC$ cannot touch the circle
drawn by side $a$ from $(0,0)$. As $n$ rises in value, the vertex $C$
approaches this circular area. If $n$ goes down in value, vertex $C$ rotates
to the right approaching the $x$ axis when $n\rightarrow 1$. These and other
details are illustrated in Figure 1.

Let S be the space formed by the points $(x,y)$ such that
$a,b,c,n\in\mathbb{R}^{+}$ in the equation $c^{n}=a^{n}+b^{n}$ for $n>2$. That
space is shown in the shaded area of Figure 1, and is bounded between the arc
$\beta\gamma$ inclusive, and it is when $a=b$; arc $\gamma B$ not inclusive,
because $c>a$, and line $B\beta$ not inclusive, because $n>2$.

Certain relevant characteristics to Fermat’s Last Theorem appear in this area.
Suppose that $a$ and $b$ are positive integers. Then, there are $b$ arcs
$\beta\gamma$, and for each of them there could be$[\dagger]$ certain points
for the vertex $C$ such that $c$ is also a positive integer, but it is unknown
if $n$ is an integer in that point.

Figure 1: Fermat’s Last Theorem graphed as a more general representation than
Pythagoras’ theorem.

Since we have supposed that $a,b\in\mathbb{Z}^{+}$ and $c,n\in\mathbb{R}^{+}$,
from Figure 1 follows that the minimum $c$ is when $c\rightarrow a$ and
$n\rightarrow\infty$. The maximum $c$ in the arc $\beta\gamma$ occurs when
$a=b$ and $n\rightarrow 2$, hence $a<c<a\sqrt{2}$. Since the minimum integer
for $c$ is $a+1$, then for all $n>2$ we have $(a+1)\leq c<a\sqrt{2}$.
Furthermore, if $a=b$ then $c\in\mathbb{Q}^{c}$ because solving $c$ for
equation $a^{n}+b^{n}=c^{n}$ gives $c=a\sqrt[n]{2}$. Besides, observe that
$60$°$<\theta<90$°.

Remark. $\Delta ABC$ tends to be an equilateral triangle whenever $a=b$ and
$n\rightarrow\infty$. If $a>b$ and $n\rightarrow\infty$ then $\Delta ABC$
tends to be an isosceles triangle. If $1<n<2$, then $\Delta ABC$ is obtuse;
and for $n>2$ the $\Delta ABC$ is acute. At $n=2$, $\Delta ABC$ is a right
triangle.

###  2.2 Arithmetical Structure

Here, we develop some simple but useful concepts that are so often needed to
apply contradiction techniques in Section 3, that they deserve singling out.

######  Lemma 1.

Suppose equation $a^{n}+b^{n}=c^{n}$ is valid for $a,b,c,n$ positive integers
and $n>2$. Then a single element of the triple $(a,b,c)$ is even.

###### Proof.

Any odd raised to a positive integer power is odd, and any even raised to a
positive integer power is even. The sum of two odd numbers is even, and the
sum of an odd with an even is odd. Therefore, it is impossible that in the
equation $c^{n}=a^{n}+b^{n}$ we have a triple $(a,b,c)$ with all its elements
odd, because otherwise we would have that an odd number is equal to an even,
which is a contradiction.

Suppose that in the triple $(a,b,c)$ we have two even and one odd. If the odd
is $c$, then the sum $a^{n}+b^{n}$ is odd, which is a contradiction. If $c$ is
even, then the sum of an odd number and another even in $a^{n}+b^{n}$ is even,
which is also a contradiction. Therefore, in the triple $(a,b,c)$ there not
exist two even and one odd.

The case when the three elements of the triple $(a,b,c)$ are even is not
considered since in the Introduction (Assumptions) we have assumed that the
equation $c^{n}=a^{n}+b^{n}$ is simplified so that $a,b,c$ have no common
factors. ∎

######  Corollary 1.

Suppose equation $a^{n}+b^{n}=c^{n}$ is valid for $a,b,c,n$ positive integers
and $n>2$. Applying the Fundamental Theorem of Arithmetic, implies this
equation has one of the following three forms:

$\displaystyle(2^{k}d)^{n}$ $\displaystyle=a^{n}+b^{n}$ a, b, d are odd and
$k\in\mathbb{Z}^{+}$ (1) $\displaystyle c^{n}$
$\displaystyle=a^{n}+(2^{k}d)^{n}$ a, c, d are odd and $k\in\mathbb{Z}^{+}$
(2) $\displaystyle c^{n}$ $\displaystyle=(2^{k}d)^{n}+b^{n}$ b, c, d are odd
and $k\in\mathbb{Z}^{+}$ (3)

Notice that equations (2) and (3) look alike from the commutative property of
addition point of view. But we have stated that $b\leq a<c$, then to keep
generality and formality, the three equations will be evaluated individually,
by cases.

######  Lemma 2.

Suppose $c^{n}=a^{n}+b^{n}$ has $a,b,n$ as positive integers with $n>2$. Then
$c$ is not a fractional number. Furthermore, $c\in\mathbb{Z}^{+}$ or
$c\in\mathbb{Q}^{c}$.

###### Proof.

Assume that $c\in\mathbb{Q}\setminus\mathbb{Z}$ such that $c=p/q$. Then the
equation $c^{n}=a^{n}+b^{n}$ can be written as $(p/q)^{n}=a^{n}+b^{n}$. The
right-hand side of the equation is a positive integer since
$a,b,n\in\mathbb{Z}^{+}$; however, the left-hand side is a fractional number
since $(p/q)^{n}$ is irreducible, which is a contradiction. ∎

######  Corollary 2.

Suppose that two elements of the triple $(a,b,c)$ are positive integers. Then,
to satisfy equation $c^{n}=a^{n}+b^{n}$ with $n>2$, the third element is not a
fractional number.

## 3\. Fermat’s Last Theorem Proof

###  3.1 n is an even integer greater than $2$.

According to the Cartesian method, we should continue dividing the problem.
Let us divide the Corollary 1 equations in parts, one for $d=1$ and the other
for $d$ an odd integer greater than $1$. Once finding the fundamental
concepts, will work up to the most difficult. When synthesizing, it will be
proved that Fermat’s Last Theorem is valid when $n$ is even, $n>2$.

####  3.1.1 d=1

######  Lemma 3.

Let $n$ be an even integer greater than 2. Suppose $(a,b,c)$ is a triple with
two positive integers and the other element has the form $2^{k}$ with
$k\in\mathbb{R}$. If the equation $c^{n}=a^{n}+b^{n}$ holds, then
$2^{k}\in\mathbb{Q}^{c}$. In particular, Fermat’s Last Theorem is valid.

###### Proof.

Lemma 2 and Corollary 2 assert that
$2^{k}\notin\mathbb{Q}\setminus\mathbb{Z}$. By contradiction, let us assume
that $2^{k}\in\mathbb{Z}^{+}$, particularly $k\in\mathbb{Z}^{+}$. According to
Lemma 1, in $(a,b,c)$ there exist one even and two odd. Next, by cases, we
analyze the three equations of Corollary 1 with $d=1$.

Case 1. Notice that the left-hand side of Equation (1) with $d=1$ is
$c=2^{k}$, which is an even integer since $k\neq 0$ for $b\leq a<c$. Recall
that throughout this document $gcd(a,b,c)=1$. By Lemma 1 we know that $a$ and
$b$ are odd. Summarizing, the equation to be analyzed is with $n$ as an even
integer greater than $2$, $a$ and $b$ are odd, $c=2^{k}$ even,
$k\in\mathbb{Z}^{+}$, $b\leq a<2^{k}$, and it is as follows:

$\displaystyle 2^{kn}$ $\displaystyle=a^{n}+b^{n}.$ (4)

Writing Equation (4) similarly to Pythagoras’ theorem, gives:

$\displaystyle[2^{kn/2}]^{2}=[a^{n/2}]^{2}+[b^{n/2}]^{2}.$ (5)

Remark. Since $n$ is an even integer, and $a,b$ are odd, then $a^{n/2}$ and
$b^{n/2}$ are odd, and $2^{kn/2}$ even. It is evident that a necessary
condition for Equation (5) is that $(a^{n/2},b^{n/2},2^{kn/2})$ must be a
Pythagorean triple.

The equation for primitive Pythagorean triples[3] was ancestrally proved, and
requires that the hypotenuse $2^{kn/2}$ is odd, contradicting the fact that
$2^{kn/2}$ is even. Therefore $2^{k}\in\mathbb{Q}^{c}$.

Case 2. Suppose $b=2^{k}$ and $a,c\in\mathbb{Z}^{+}$. Then the equation to
analyze is:

$\displaystyle c^{n}=a^{n}+2^{kn}.$ (6)

Writing Equation (6) analogously to the Pythagorean theorem, gives:

$\displaystyle[c^{n/2}]^{2}=[a^{n/2}]^{2}+[2^{kn/2}]^{2}.$ (7)

Remark. Since $n$ is even, then Lemma 1 asserts that $c^{n/2},a^{n/2}$ are odd
and $2^{kn/2}$ even. It is evident that a necessary condition for Equation (7)
is that $(a^{n/2},2^{kn/2},c^{n/2})$ must be a Pythagorean triple. The
Pythagorean triples concept implies that the integers
$a^{n/2},2^{kn/2},c^{n/2}$ must satisfy the following conditions for two
positive integers $p$ and $q$ with $p>q$:

$\displaystyle c^{n/2}$ $\displaystyle=p^{2}+q^{2}$ (8) $\displaystyle
a^{n/2}$ $\displaystyle=p^{2}-q^{2}$ (9) $\displaystyle 2^{kn/2}$
$\displaystyle=2pq.$ (10)

Solving for $p$ in Equation (10), gives $p=2^{kn/2}/(2q)$. Substituting $p$ in
Equation (8) results that $c^{n/2}=[2^{kn/2}/(2q)]^{2}+q^{2}$, which reduces
to:

$\displaystyle c^{n/2}$ $\displaystyle=\frac{2^{kn-2}}{q^{2}}+q^{2}.$ (11)

First, consider Equation (11) when $q$ is even.

  1. 1.

Suppose $q^{2}=2^{kn-2}$. Squaring both sides of Equation (10) results that
$2^{kn-2}>q^{2}$, which is a contradiction.

  2. 2.

Assume $q$ is even such that $q^{2}\mid 2^{kn-2}$ and $2^{kn-2}\neq q^{2}$.
Then $c^{n/2}$ is even, contradicting the fact that $c^{n/2}$ is odd.

  3. 3.

If $q$ is even such that $q^{2}\nmid 2^{kn-2}$ then
$c^{n/2}\in\mathbb{Q}\setminus\mathbb{Z}$, which contradicts that $c^{n/2}$ is
odd.

Secondly, consider $q$ is odd.

  1. 1.

Suppose $q=1$. Solving Equation (9) gives $p^{2}=a^{n/2}+1$. Substituting in
Equation (8) and rewriting gives $c^{n/2}-a^{n/2}=2$. Such equality is
impossible since $a$ and $c$ are odd with $a<c$, and the minimum difference
between them is $2$, but when raising $a$ and $c$ to an integer power greater
than $1$, then the difference is greater than $2$. Therefore
$c^{n/2}-a^{n/2}>2$, which is a contradiction.

  2. 2.

Assume $q$ is odd greater than $1$. Then
$c^{n/2}\in\mathbb{Q}\setminus\mathbb{Z}$ since $q^{2}\nmid 2^{kn-2}$,
contradicting the fact that $c^{n/2}$ is odd.

Thus if $a$ and $c$ are odd, then Equation (7) is impossible with the other
element as $2^{k}$. Therefore, $2^{k}\in\mathbb{Q}^{c}$.

Case 3. Suppose $a=2^{k}$ and $b,c\in\mathbb{Z}^{+}$. Then the equation to
analyze is $c^{n}=2^{kn}+b^{n}$. Similarly to Case 2, and using the
commutative property of sum, it is concluded by contradiction that
$2^{k}\in\mathbb{Q}^{c}$. ∎

####  3.1.2 d is odd greater than 1

######  Lemma 4.

Let $n$ be even, greater than 2. Suppose $(a,b,c)$ is a triple with two
positive integers and the other element has the form $2^{k}d$ with
$k\in\mathbb{R}$, and $d$ is odd greater than $1$. If the equation
$c^{n}=a^{n}+b^{n}$ holds, then $2^{k}d\in\mathbb{Q}^{c}$. In particular,
Fermat’s Last Theorem is valid.

###### Proof.

Will proceed with a direct proof, by Cases.

Case 1. Suppose $c=2^{k}d$ such that $d$ is an odd integer greater than $1$
and $k\in\mathbb{R}$. Suppose $c^{n}=a^{n}+b^{n}$, with $a,b\in\mathbb{Z}^{+}$
and $n$ is an even integer greater than $2$. Notice that $2^{k}d$ can be
expressed as $2^{h}$, with $h\in\mathbb{R}$. Then $2^{k}d=2^{h}$. Using Lemma
3 (Case 1) with $c=2^{h}$ and $a,b\in\mathbb{Z}^{+}$ such that
$[2^{h}]^{n}=a^{n}+b^{n}$, it is concluded that $2^{h}\in\mathbb{Q}^{c}$.
Therefore, $2^{k}d\in\mathbb{Q}^{c}$. Furthermore, observe that if we apply
the transitive axiom of equality among $2^{h}$, $\sqrt[n]{a^{n}+b^{n}}$ and
$2^{k}d$, by Lemma 3 follows that $2^{k}d=2^{h}\in\mathbb{Q}^{c}$.

Case 2. Suppose $b=2^{k}d$, $d$ is odd greater than $1$ and $k\in\mathbb{R}$.
Suppose $c^{n}=a^{n}+b^{n}$ with $a,c\in\mathbb{Z}^{+}$, $n$ even greater than
$2$. Let’s set $2^{k}d=2^{h}$ with $h\in\mathbb{R}$. Applying Lemma 3 (Case 2)
follows that $2^{k}d\in\mathbb{Q}^{c}$.

Case 3. Suppose $a=2^{k}d$, $d$ is odd greater than $1$ and $k\in\mathbb{R}$.
Suppose $c^{n}=a^{n}+b^{n}$ with $b,c\in\mathbb{Z}^{+}$, $n$ even greater than
$2$. Let’s set $2^{k}d=2^{h}$ with $h\in\mathbb{R}$. Applying Lemma 3 (Case 3)
follows that $2^{k}d\in\mathbb{Q}^{c}$. ∎

####  3.1.3 Synthesis for $n$ as an even integer greater than $2$

######  Corollary 3.

Fermat’s Last Theorem is true for all $n$ even integers greater than $2$.

###  3.2 n is an odd integer greater than $2$.

Our approach to this case is via mathematical artifices, to make the Corollary
1 equations look like if $n$ were even and therefore enabling us to use what
has been proved in Section 3.1. Observe that each Lemma has a link to the next
one. It reflects the inductive reasoning in Cartesian lanes and waves, and
will continue like this until the end. Again, let’s address the solution by
cases. First we will consider perfect square numbers and afterwards will
generalize it for any positive integer.

####  3.2.1 Perfect square numbers within (a,b,c)

######  Lemma 5.

Let $n$ be an odd integer greater than 2. Suppose $(a,b,c)$ is a triple with
three perfect square numbers with one element of the form $2^{k}d$, where
$k\in\mathbb{R}$ and $d$ is odd. If the equation $c^{n}=a^{n}+b^{n}$ holds,
then $2^{k}d\in\mathbb{Q}^{c}$. In particular, Fermat’s Last Theorem is valid.

###### Proof.

According to Lemma 1, in $(a,b,c)$ there exist one even and two odd. Next, a
proof by contradiction is presented for the three equations of Corollary 1
when $n$ is odd, $n>2$.

Case 1. Suppose Equation (1) where $a,b$ are odd and $2^{k}d$ is even, and all
of them are perfect square numbers. Rewriting that equation similarly to
Pythagoras’ theorem, gives:

$\displaystyle[(2^{k}d)^{n/2}]^{2}=[a^{n/2}]^{2}+[b^{n/2}]^{2}.$ (12)

Remark. Since $a,b,2^{k}d$ are positive and perfect square numbers, if we
apply to each of them a square root, they are still integers. Therefore
$a^{n/2}$, $b^{n/2}$ are odd and $(2^{k}d)^{n/2}$ is even. It is evident that
a necessary condition for Equation (12) is that
$(a^{n/2},b^{n/2},(2^{k}d)^{n/2})$ must be a Pythagorean triple. Let us
consider a similar procedure as demonstrated in Section 3.1.

  1. 1.

Let $d=1$. This case is proved following a similar reasoning as shown in Lemma
3 (Case 1) under the here mentioned conditions. Therefore, by contradiction,
$2^{k}\in\mathbb{Q}^{c}$.

  2. 2.

Let $d>1$ and $d$ is odd. This case is proved following similar steps as
explained in Lemma 4 (Case 1). Therefore, by contradiction, we assert that
$2^{k}d\in\mathbb{Q}^{c}$.

  3. 3.

Synthesizing, $2^{k}d\in\mathbb{Q}^{c}$.

Case 2. Suppose Equation (2) where $a,c$ are odd and $2^{k}d$ is even, and all
of them are perfect square numbers. Suppose $n$ is an odd integer greater than
$2$. Let us follow an analogous procedure as shown in Case 1, invoking the
procedure of Lemma 3 (Case 2) and Lemma 4 (Case 2). For simplicity sake, we
conclude that $2^{k}d\in\mathbb{Q}^{c}$.

Case 3. Suppose Equation (3) where $b,c$ are odd and $2^{k}d$ is even, and all
of them are perfect square numbers. Let us follow an analogous procedure as
shown in Case 1, invoking the procedure of Lemma 3 (Case 3) and Lemma 4 (Case
3). Therefore, $2^{k}d\in\mathbb{Q}^{c}$. ∎

####  3.2.2 Arithmetical structure of equations with two perfect square
numbers

Before continuing with our proof, let us do some calculations related to the
equations contained in Lemma 5. We aim to understand the arithmetic structure
of equations of the form $c^{n}=a^{n}+b^{n}$ where $(a,b,c)$ has two odd
perfect square numbers and an irrational, particularly when $n$ is odd and
$n>2$. We are not interested in equations that have even-perfect square
numbers and an irrational, since they are not aligned to Lemma 5’s results.

Case 1. Consider the equation $(2^{\hat{k}}d)^{n}=\hat{a}^{n}+\hat{b}^{n}$
where $\hat{a},\hat{b}$ are perfect square numbers, $\hat{a},\hat{b},d,n$ are
odd, $n>2$. By Lemma 5 follows that $2^{\hat{k}}d\in\mathbb{Q}^{c}$.

Perfect square numbers $\hat{a},\hat{b}$ have the property that
$\hat{a}=a^{2}$ and $\hat{b}=b^{2}$ where $a,b$ are any positive odd integer.
Substituting, we have that $(2^{\hat{k}}d)^{n}=a^{2n}+b^{2n}$.

Solving for $a^{n}$ gives:

$\left[\frac{2^{\hat{k}}d}{a}\right]^{n}=a^{n}+\left[\frac{b^{2}}{a}\right]^{n}.$
(13)

Since $2^{\hat{k}}d\in\mathbb{Q}^{c}$ and $a$ is odd, then
$[(2^{\hat{k}}d)/a]$ can be expressed as $2^{h}d$ such that
$2^{h}d\in\mathbb{Q}^{c}$. Particularly, $2^{h}\in\mathbb{Q}^{c}$ and $d$ is
odd. Substituting in Equation (13) and summarizing, we have a valid equation
with the following structure:

$(2^{h}d)^{n}=a^{n}+\left[\frac{b^{2}}{a}\right]^{n}$ (14)

where $2^{h}d\in\mathbb{Q}^{c}$, $a.b,d,n$ are odd, $n>2$.

Remark. We claim that Equation (14) contains all the equations for $n$ odd,
$n>2$ as defined in Equation (1), but with $2^{k}d\in\mathbb{Q}^{c}$. This
will be proved in Case 1, next Lemma.

Case 2. Consider the equation $\hat{c}^{n}=\hat{a}^{n}+(2^{\hat{k}}d)^{n}$
where $\hat{c},\hat{a}$ are perfect square numbers, $\hat{c},\hat{a},d,n$ are
odd, $n>2$. Analogously to Case 1, we have a valid equation with the following
structure: $(c^{2}/a)^{n}=a^{n}+(2^{h}d)^{n}$, with $2^{h}d\in\mathbb{Q}^{c}$,
and $a,c$ are odd, and an analogous remark.

Case 3. Consider the equation $\hat{c}^{n}=(2^{\hat{k}}d)^{n}+\hat{b}^{n}$
where $\hat{c},\hat{b}$ are perfect square numbers, $\hat{c},\hat{b},d,n$ are
odd, $n>2$. Analogously to Case 1, we have a valid equation with the following
structure: $(c^{2}/b)^{n}=(2^{h}d)^{n}+b^{n}$, where
$2^{h}d\in\mathbb{Q}^{c}$, and $b,c$ are odd, and an analogous remark.

####  3.2.3 Any positive integer within (a,b,c)

######  Lemma 6.

Let $n$ be an odd integer greater than 2. Suppose $(a,b,c)$ is a triple with
two positive integers and the other element has the form $2^{k}d$ with
$k\in\mathbb{R}$ and $d$ is odd. If the equation $c^{n}=a^{n}+b^{n}$ holds,
then $2^{k}d\in\mathbb{Q}^{c}$. In particular, Fermat’s Last Theorem is valid.

###### Proof.

According to Lemma 1, in $(a,b,c)$ there exist one even and two odd. Next, a
proof by contradiction is presented for the three equations of Corollary 1
when $n$ is odd, $n>2$, by cases.

Case 1. Assume Equation (1) is valid, where $(a,b,2^{k}d)$ are integers. As
well, consider Equation (14), which has been proved and therefore we assert it
is valid. Let us compare the arithmetical structures of both equations.

  1. 1.

Suppose Equation (14) with $a=b$. Then $[b^{2}/a]=b$ and it is odd. We obtain
the equation $(2^{h}d)^{n}=a^{n}+b^{n}$, with $a=b$. By Lemma 5 we know that
$2^{h}d\in\mathbb{Q}^{c}$. We can confirm this, since with $a=b$, follows
$(2^{h}d)^{n}=2a^{n}$ and $2^{h}d=a\sqrt[n]{2}\in\mathbb{Q}^{c}$. Conversely,
If Equation (1) has $a=b$, then $2^{k}d=a\sqrt[n]{2}\in\mathbb{Q}^{c}$, which
contradicts that $2^{k}d$ is even. Therefore, $2^{k}d\in\mathbb{Q}^{c}$.

  2. 2.

Suppose Equation (14) with $a\mid b^{2}$ and $a\neq b$. Observe that
$[b^{2}/a]$ is odd. Let $f=[b^{2}/a]$. Then we have an equation of the form
$(2^{h}d)^{n}=a^{n}+f^{n}$.

Next, let us do a one-to-one comparison between Equation (14) and Equation (1)
under this conditions, as shown in Table 1.

Comparison | Valid Structure | Assumed as valid | Evaluation
---|---|---|---
| Equation (14) | Equation (1) | of
| $(2^{h}d)^{n}=a^{n}+f^{n}$ | $(2^{k}d)^{n}=a^{n}+b^{n}$ | Equation(1)
A |  $n$ is odd greater than $2$ |  $n$ is odd greater than $2$ | ✓
B | two addends raised to $n$ | two addends raised to $n$ | ✓
C | sum raised to $n$ | sum raised to $n$ | ✓
D |  $a$ is odd |  $a$ is odd | ✓
E |  $f$ is odd |  $b$ is odd | ✓
F |  $2^{h}d$ is irrational |  $2^{k}d$ is integer | ✗
Table 1: The valid structure for Equation (14) with $a\mid b^{2}$, versus the
structure for Equation (1).

Pay attention to line F, which shows the only difference between both
structures. Since we assert that $2^{h}d\in\mathbb{Q}^{c}$, follows that a
necessary condition to satisfy both equations is that
$2^{k}d\in\mathbb{Q}^{c}$, which is a contradiction. Therefore,
$2^{k}d\in\mathbb{Q}^{c}$.

Notice that this analysis could be applied to Item 1 above, and next Item 3 as
well.

  3. 3.

Suppose Equation (14) with $a\nmid b^{2}$. Then
$[b^{2}/a]\in\mathbb{Q}\setminus\mathbb{Z}$. Rewriting we have:

$\displaystyle(2^{h}d)^{n}$ $\displaystyle=a^{n}+(b^{2}/a)^{n}$ same as
Equation (14) (15) $\displaystyle(2^{h}d)^{n}-(b^{2}/a)^{n}$
$\displaystyle=a^{n}$ solving for $(b^{2}/a)$ (16)
$\displaystyle(2^{h}d)^{n}-(b^{2}/a)^{n}+b^{n}$ $\displaystyle=a^{n}+b^{n}$
$\displaystyle\text{adding $b^{n}$ to both sides}.$ (17)

Recall that a primitive Fermat’s equation has the form $c^{n}=a^{n}+b^{n}$
with $gcd(a,b,c)=1$ and within $(a.b.c)$ there are a single even number and
two odd. $n>2$.

Certainly, the right-hand side of Equation (17) is a member of a primitive
Fermat’s equation where $a,b,n$ are odd, $n>2$, and it is a positive even
integer.

Conversely, the left-hand side is a fractional number, since although
$2^{h}d\in\mathbb{Q}^{c}$ we have that the addend $(2^{h}d)^{n}$ is integer,
$b^{n}$ is integer and $(b^{2}/a)^{n}$ is fractional. Therefore, by
contradiction, Fermat’s Last Theorem is valid.

Supported by that fact and Corollary 2, in order to satisfy Equation (17), we
are allowed to replace the rascal $[(2^{h}d)^{n}-(b^{2}/a)^{n}+b^{n}]$ by
$(2^{k}d)^{n}$ where $2^{k}d\in\mathbb{Q}^{c}$.

Case 2. This case refers to Equation (2) when $n$ is odd and $n>2$.
Analogously to Case 1, applying the corresponding cases, lemmas and
statements, implies that $2^{k}d\in\mathbb{Q}^{c}$.

Case 3. This case refers to Equation (3) when $n$ is odd and $n>2$. Similarly
to Case 1, follows that $2^{k}d\in\mathbb{Q}^{c}$. ∎

####  3.2.4 Synthesis for $n$ as an odd integer greater than $2$

######  Corollary 4.

Fermat’s Last Theorem is true for all $n$ odd integers greater than $2$.

###  3.3 Synthesis for $n$ as an integer greater than $2$

######  Theorem 1.

Fermat’s Last Theorem: If $n$ is an integer greater than $2$, then there are
no positive integers $a,b,c$, that satisfy the equation $a^{n}+b^{n}=c^{n}$.

###### Proof.

In Section 3.1 we have proved that Fermat’s Last Theorem is valid when $n$ is
even greater than $2$. Whereas in Section 3.2 we have proved that it is also
valid when $n$ is odd greater than $2$. It happened that within $(a,b,c)$
there exists at least an irrational number of the form $2^{h}$. Finally,
synthesizing we assert that Fermat’s Last Theorem is valid. ∎

## 4\. Final Considerations

######  Conjecture 1.

Suppose $a,b,c$ are positive integers that satisfy equation
$a^{n}+b^{n}=c^{n}$. Then, $n$ is irrational.

######  Conjecture 2.

There are no fractional numbers $a,b,c,n$ such that $a^{n}+b^{n}=c^{n}$.

Let $a,b,n$ be integers and $n>2$. Beyond real numbers,
$c(n)=\sqrt[n]{a^{n}+b^{n}}$ is not a one-to-one function since has multiple
non real solutions. This follows from DeMoivre’s Theorem, named after French
mathematician Abraham de Moivre (1667-1754).

## Notes

$[\dagger]\ $If $b\in\\{1,2,3\\}$ then in arc $\beta\gamma$, Figure 1, there
not exist a vertex $C$ such that $c\in\mathbb{Z}^{+}$ since
$\sqrt[n]{a^{n}+b^{n}}\in\mathbb{Q}^{c}$ for all $a\in\mathbb{Z}^{+}$ and
$n>2$. The maximum amount of vertices C that exist in arc $\beta\gamma$ for
$a,b,c\in\mathbb{Z}^{+}$ and $n\in\mathbb{R}^{+}$ is $h$ such that
$h\leq\lfloor a(\sqrt[3]{2}-1)\rfloor\lessapprox\lfloor 0$.$2599\ a\rfloor$.

God alone is the author of all the motions in the world. Cogito, ergo sum.

— Rene Descartes

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# Investigating the Robustness and Properties of Detection Transformers (DETR)
Toward Difficult Images

Zhao Ning Zou∗
The Australian National University
Canberra, ACT
<EMAIL_ADDRESS>Yuhang Zhang∗
The Australian National University
Canberra, ACT
<EMAIL_ADDRESS>Robert Wijaya∗
The Australian National University
Canberra, ACT
<EMAIL_ADDRESS>
###### Abstract

Transformer-based object detectors (DETR) have shown significant performance
across machine vision tasks, ultimately in object detection. This detector is
based on a self-attention mechanism along with the transformer encoder-decoder
architecture to capture the global context in the image. The critical issue to
be addressed is how this model architecture can handle different image
nuisances, such as occlusion and adversarial perturbations. We studied this
issue by measuring the performance of DETR with different experiments and
benchmarking the network with convolutional neural network (CNN) based
detectors like YOLO and Faster-RCNN. We found that DETR performs well when it
comes to resistance to interference from information loss in occlusion images.
Despite that, we found that the adversarial stickers put on the image require
the network to produce a new unnecessary set of keys, queries, and values,
which in most cases, results in a misdirection of the network. DETR also
performed poorer than YOLOv5 in the image corruption benchmark. Furthermore,
we found that DETR depends heavily on the main query when making a prediction,
which leads to imbalanced contributions between queries since the main query
receives most of the gradient flow.

**footnotetext: equal contribution

##  1 Introduction

With the continuous development of deep learning, computer vision has reached
a new stage, and target detection, as one of the very vital core directions,
has also received attention and many applications based on object detection
algorithms.

Before the concept of deep learning was introduced, object detection was
mostly based on manual feature extraction. However, as manual feature
extraction methods often failed to meet various features in the targets,
traditional target detection algorithms could not meet people’s needs. After
the rise of deep learning, neural networks can automatically learn powerful
feature extraction and fitting capabilities from large amounts of data. Thus
many DL-based object detectors with excellent performance have emerged. These
detectors can be broadly classified into three categories: two-stage object
detection, one-stage object detection, and transformer-based object detection.

Faster R-CNN [1] is the most popular two-stage detector nowadays. It first
generates a proposal for the object bounding box in the image through a
network. It then extracts features from each candidate box, and uses them for
object classification and bounding box regression tasks to obtain the final
bounding box. On the other hand, the YOLO [2] series model is well-known as a
one-stage detector, which discards the anchor frame setting in two-stage and
extracts the prediction frames directly from the image. Furthermore, with the
increasing popularity of transformers applied in computer vision tasks, new
transformer-based object detectors, such as DETR [9], have also emerged.
Instead of using anchor frames and NMS, DETR uses an encoder-decoder structure
to classify each object in the image.

DETR is an end-to-end target detection network proposed by Facebook in 2020.
Compared to traditional RNNs, DETR uses multiple self-attentive structures,
and the parallel computing used therein allows DETR to extract relevance
efficiently in context. It is also one of the best-performing target detection
methods available.

This paper investigated the properties of DETR and compared the robustness of
the network using different interference to the image data, such as sticker
and occlusion. We find three main findings in our experiment (1) DETR can
handle a small amount of occlusion well compared to Faster R-CNN and YOLOv5,
but when there is too much information loss, the attention mechanism of DETR
is difficult to be useful. (2) the adversarial patch manages to produce a new
set of unnecessary keys, queries, and values in the network, which in most
cases, results in the misdirection of the network. (3) The image corruption
benchmark performance of DETR is lower than that for YOLO model. (4) We
observed a main query phenomenon in the DETR model and showed that is caused
the slow convergence problem of the model.

##  2 Related Works

Recent successes of Transformer-based models in computer vision tasks have
inspired several works [11, 12, 14, 15, 16, 5, 17, 18, 19] that study their
robustness against corrupted images and adversarial attacks. Some works [11,
12, 5, 18] claimed that transformers are more robust than CNNs in different
evaluation settings, including adversarial attacks, while others [13, 16, 17,
19] hypothesized that transformers are vulnerable as CNNs. In this study, we
aim to understand the robustness of the transformers-based mechanism for
object detector settings. We evaluated their robustness against patch masking,
adversarial attacks, and common natural corruption images.

Occlusion is one of the effects of target detector performance [6]. There are
two general types of occlusion: intra-class occlusion and inter-class
occlusion. Many problems such as pedestrian detection, stereo images, etc.
face problems caused by occlusion, whether for outdoor or indoor scene tasks.
Because when an object is occluded, some information is lost and the remaining
information may be difficult to recognise, studying the occlusion problem is
an inevitable step for target detection algorithms. [5] evaluated the
performance of Vision Transformer on occlusion. Therefore, we also analyzed
object detectors performance on occluded images.

Regarding a more realistic robustness benchmark, [7] proposed the ImageNet-C
dataset, which consists of 15 diverse corruption types, covering noise, blur,
weather, and digital categories. This dataset was first proposed for image
classification tasks. In [8], Claudio M. _et al._ evaluated object detectors’
performance in bad weather. They proposed several robustness benchmarks for
object detection named Pascal-C, Coco-C, and Cityscapes-C. In our work, we
utilized their image corruption generator and focused on the reasons behind
the performance drop of object detectors.

##  3 Methodology

###  3.1 Occlusion

Object occlusion is a major problem in target detection applications. Whether
it is pedestrian detection, object tracking, or autonomous driving, the object
to be detected may be occluded within or between classes. Such partial
occlusion reduces the features extracted from the objects and affects the
object detector’s precision.

Object occlusion has different implications for different tasks. In this
paper, we set the value of the image patches that need to be occluded to 0 to
reach the information loss effect in that part. Moreover, DETR is a network
based on a self-attentive model, so it is essential to study the impact of
regions containing more information on DETR. In this paper, we use two
different occlusion methods to test Faster R-CNN, YOLO, and DETR, (a) random
occluded, (b) salient occluded.

(a) Random occulted: We use dataset COCO128 to test the anti-jamming
capability of DETR. Since the image size is different in COCO128, we split the
image into 10 x 10 patches and randomly occlude these patches by setting the
pixel value of the occluded patches to 0 to simulate the loss of information.
The occlusion ratio is set to the number of occluded patches/total patches. In
our experiments, we will test images with masking ratios of 0.2, 0.4, 0.6, and
0.8, respectively.

(b) Salient occluded: The self-attentive mechanism in the transformer
contributes to the excellent performance of DETR. Therefore, it can
selectively extract information from the feature map for later classification
and box prediction. In a realistic study of how resistant DETR is to
interference, the focus should be on the effect of salient regions. In this
experiment, we treated the target with different occlusion rates to find the
reason why significant regions affect robustness. We considered the area
containing the top 20% of prospective information to be significant. We use
different occlusion rates to process the salient part of patches in the object
and feed the processed images into the three networks for comparison. Then we
use different occlusion rates to process the salient part of patches in the
object and feed the processed images into the three networks for comparison.

###  3.2 Adversarial Stickers

The adversarial attack is a popular method to disturb the output precision of
machine learning models. By adding a small portion of adversarial perturbation
to the image, the detectors can make a misleading judgment, resulting in a
significant decrease in performance. Some research utilized a similar approach
to evaluate the networks. In [3], Sharif _et al._ demonstrated that using
adversarial glasses is possible to fool facial recognition systems. These
glasses were designed to fit any face, allowing it to impersonate any person.
Another work [4] demonstrated different methods for constructing fake stop
signs to misclassify the models by creating a poster that looks like a stop
sign or modifying the stop signs to make it hard to recognizable. This
indicated that an adversarial patch is a reliable way to evaluate the
performance and robustness of a particular network.

We utilized the attack by completely replacing a portion of the image with a
sticker (patch). The stickers are masked with some constraints to fit in
different sizes of images by applying random values of translation and scaling
to place the stickers in different locations with different sizes. In
particular, given a three-dimensional image $x$, patch $p$, location $l$, and
transformation $t$ (in this case translation and scaling), we defined a patch
operator $S(p,x,l,t)$ which first apply the transformation to the patch, and
then following the transformation of patch $p$ to the location $l$ in the
image $x$.

This attack proposed to distract the way object detectors make their
predictions and draw the bounding boxes. In the case where an image contain
several objects, the network must be able to decide the most salient object in
the image. Thus, when a targeted object which some of its part overlapped by a
sticker, the detectors should be able to classified the object without
significantly reducing the accuracy of the targeted object and other objects
in the image.

DETR uses transformers to capture the global context of the image. Since the
adversarial stickers attack only change the pixel values of a certain region
in the image, it is essential to see whether the global scope attention of
DETR. Moreover, we want to investigate whether DETR is more robust to this
attack than CNN-based mechanism detectors such as YOLO and Faster-RCNN.

###  3.3 Benchmark

In a real-world scenario, the camera’s photo sometimes suffers from common
corruptions such as defocus blur, motion blur, or bad weather. Therefore, we
wanted to evaluate the performance of YOLOv5m, DETR R50, and DETR R101 on
corrupted images. The evaluations were done using 15 image corruptions with
five levels of severity in each corruption category, as shown in Figure 1.

Figure 1: The image corruptions used in Benchmark evaluation.

Apart from model comparison, we also investigated the reasons behind the
performance drop of DETR models. We analyzed the attention maps from the
transformer encoder and decoder to see how they changed as the image
corruption became more severe.

###  3.4 Query Properties

One of the critical innovations in DETR is the introduction of object queries
in the transformer decoder. It is implemented as learnable positional
encodings that are added to the decoder input at each layer. These queries
help the model decode a set of box coordinates and their class labels. The
number of the object query is usually smaller than 100 to reduce the number of
bounding boxes generated at the early stage and the computation cost. In [9],
queries have shown unique preferences on objects with a particular size and
position. Therefore it is believed that each query has several modes of
operation focusing on different areas and box sizes. In particular, the
transformer architectures allow the object query to utilize the context in the
whole image to detect large objects more precisely. In this paper, we further
analyzed the properties of queries and their relations with class labels. We
also presented the observation of main queries and their significant
contributions to the DETR’s model. Finally, we fine-tuned a DETR R50 model on
pascal-voc dataset to investigate the model’s transfer learning ability.

##  4 Evaluation

We record the experiment results with the primary COCO challenge metric, the
mean Average Precision (mAP). This metric averages precision-recall scores at
different Intersection-overUnion (IoU) threshold. Moreover, to mitigate bias
toward local patch, we ignore predicted boxes with less than 50% with the
target box. We did the same for original images for a fair comparison between
clean and attacked images.

###  4.1 Evaluation on Occlusion

We tested the robustness of Faster R-CNN, YOLOv5, and DETR on the COCO128
dataset with occlusion ratios (number of patches masked/total patches) of 0.2,
0.4, 0.6, and 0.8, and the results are shown in Figure 2. To prevent random
perturbations in random occlusion, four separate experiments were conducted,
and the evaluation values of the experimental results were taken. DETR
outperformed Faster R-CNN and YOLOv5 in several experiments, and the average
decrease in accuracy of DETR during the process of increasing the occlusion
ratio from 0.2 to 0.8 was $[52.92\%,73.91\%,69.04\%]$, while the figures for
Faster R-CNN and YOLOv5 were $[55.85\%,79.54\%,55.56\%]$ and
$[54.93\%,71.09\%,65.40\%]$. It can be seen that DETR performs well in terms
of resistance to interference from information loss.

Occlusion rate | 0.2 | 0.4 | 0.6 | 0.8
---|---|---|---|---
DETR | 0.342 | 0.161 | 0.042 | 0.013
YOLOv5 | 0.304 | 0.137 | 0.0396 | 0.0137
Faster-RCNN | 0.299 | 0.132 | 0.027 | 0.012
Table 1: The Mean Average Precision (mAP) result comparison of DETR, YOLOv5
and Faster-RCNN on images with different occlusion ratio.

Figure 2: The mAP of different occlusion rate on DETR, YOLO and Faster R-CNN

In the experiments with significant areas occluded, we also treated the
significant areas of the images with occluded rates of $0.2,0.4,0.6$, and
$0.8$, respectively.

Figure 3: The detection results of different occlusion rate on DETR, YOLO and
Faster R-CNN

At a significant region occluded rate of 0.2, DETR performed the best,
followed by YOLOv5 and Faster R-CNN. For example, DETR detected two classes,
potted plant and vase, and both had an accuracy of over 90%. The other two
networks detected only one class of targets with accuracies below 80%. When
the occlusion rate of the salient regions was raised to 0.4 and above, all
three networks showed false detections and failed to detect targets. From the
results, we can see that DETR can still perform the task of target detection
well when a small portion of important information is missing. We illustrate
why DETR has better robustness against occlusion by analyzing DETR’s attention
map.

As can be seen in Figure 4, when the detected object is partially occluded,
DETR’s attention will mainly focus on the not occluded part because the
transformer can combine global information to fill in the loss part by
surrounding pixels. However, it can be seen that when too much information is
missing due to too many occluded parts, the DETR’s attention gradually starts
to diverge, and it is unable to find the detected object accurately.

Figure 4: DETR detection results on different occlusion rate 0.2 (left top),
0.4 (right top), 0.6 (left bottom), 0.8 (right bottom).

###  4.2 Evaluation on Adversarial Stickers

We evaluate the three object detectors on images with adversarial stickers in
which the detectors utilize ResNet101 as the backbone. For the dataset, we
select 128 images from the MS COCO 2017 validation set [10]. The adversarial
patch was placed in the dataset images with random size and position. We
report the resulting mAPs in Table 2: mAP under attacked images measured by
the three object detectors. In terms of adversarial stickers evaluation, we
find that Faster-RCNN gives the lowest performance among all three detectors,
followed by DETR and YOLOv5.

| mAP | mAP50
---|---|---
DETR | 0.512 | 0.726
YOLOv5 | 0.548 | 0.743
Faster-RCNN | 0.495 | 0.674
Table 2: The Mean Average Precision (mAP) and mAP50 result comparison of DETR,
YOLOv5 and Faster-RCNN on images with adversarial stickers.

The resulting bounding box of DETR is also shown in Figure 5. As shown in the
upper row of Figure 5, the network misclassified the attacked image,
interpreting the stickers to be a person with 96% confidence instead of
entirely ignoring it. In the case the sticker not overlapping any item in the
image, the sticker will not affect the accuracy of other detected objects in
the image. However, in some cases where the stickers overlap the image, it can
affect the attention process resulting in lower accuracy of other detected
items in the image, as shown in the bottom row of Figure 5.

Figure 5: DETR detection result on original image (left) compare with the
image with the adversarial sticker (right). The upper row shows the case where
the sticker not overlapping other items in the image, while the lower row
illustrated the opposite case.

This experiment implies that the sticker is capable of producing new
unnecessary input features to be processed in the self-attention mechanism,
which misdirects the network’s attention. The impact of the sticker on the
self-attention mechanism of DETR is illustrated in Figure 6.

Figure 6: Self-attention mechanism for adversarial stickers patch settings. In
this case, $q$, $k$, and $v$ represent the projected queries, keys, and value
tokens of the input features. The mechanism involving dot-product attention
computing the dot-product between queries with all corresponding keys before
normalizing with softmax function to get the token attention weights. The
adversarial patch introduces a new input feature at $X_{2}$ which misdirects
the network’s attention to the adversarial patch.

As can be seen in Figure 6, the sticker patch manages to produce a new set of
keys, queries, and values in the network, which in most cases, results in the
misdirection of the network. In the case where the stickers not overlapping
other items in the image, the dot-product attention only computes the query
$q_{n}$ with the corresponding key $k_{n}$. However, in most cases when the
stickers overlap other items in the image, the queries of those particular
items (i.e., $q_{1},\dots,q_{n}$) are misdirected to the key token that
represents the adversarial sticker (in this case the $k_{2}$). This result in
increasing the attention weight to the corresponding adversarial patch key
token. In other words, the network misguided the attention from actual image
content to the adversarial stickers.

Figure 7: Visualization of self-attention weight on the adversarial patch.

To gain intuition on what the detector see on a particular attacked image, we
visualize the self-attention weight on the adversarial patch. As shown in the
Figure 7, the network consider the entire sticker object to make prediction,
counting it as the important object to predict.

###  4.3 Evaluation on Image Corruption Benchmark

We evaluated the average precision of YOLOv5m, DETR R50, and DETR R101 on 15
image corruption categories and five levels of severity each, as shown in
Figure 8. Although DETR models have better performance on original images,
their precision on corrupted images is generally lower than YOLOv5m. DETR
models performed much worse than YOLOv5m on images with impulse noise. Since
DETR only used ReLU activation, it weakened the model’s ability to counter
these extreme values. We visualized the attention maps at pixel (300, 450)
throughout the impulse noise test, shown in Figure 9. At the start, pixels
corresponding to the ”cat” had a strong correlation with our center point.
However, as the impulse noise increases, the attention map gradually shrinks
to the pixels near the center points. The encoder failed to correlate with
”cat” pixels. This finally leads to a decrease in precision.

Figure 8: The benchmark results for three object detectors.

Figure 9: The detection result from DETR on original image and and encoder
attention at (300, 450) as the impulse noise became more severe.

###  4.4 Query Properties

First, we evaluated the contribution of each query in object detection. Both
DETR R50 and DETR R101 models were tested on the MSCOCO validation dataset. We
collected all predictions with confidence larger than 0.8 and the
corresponding query ID to compute the query frequency, as shown in the middle
plots in Figure 10. We found an interesting phenomenon that both the DETR R50
and DETR R101 model have one main query (71 in DETR R50 and 68 in DETR R101)
that detect lots of objects. Each main query can account for 7.5% of the total
predictions. Intuitive thinking was that the main queries are responsible for
detecting ”person”, which is the most common object in the dataset. Therefore,
we investigated the relations between main queries and classes. We divided the
main queries’ frequency by the total frequency in each class to visualize
their contribution, as shown in the right plots. Both main queries have shown
very similar distribution over object categories. None of them has a high
contribution ($<$ 40%) in the ”person” category, and both of them has a high
contribution ($<$ 40%) in the ”airplane”, ”train”, ”cat” and ”bear” category.
This indicated that the main query has particular preferences on object
classes, but they are not detecting the most common object, ”person”.

Figure 10: Left columns: Class distribution from model’s prediction. Middle
columns: Frequency of each query in predictions. Right columns: Main queries’
contribution in detecting each object category.

Apart from class relations, we also analyzed bounding boxes’ location and size
predictions from main queries, shown in Figure 11. Both queries prefer to
detect medium to large objects in the center region. However, there are also a
large amount of small to medium-sized boxes circulating the center.

Figure 11: The predicted bounding boxes center location and size of main
queries. Yellow colored points refers to boxes with a large size.

To analyze the importance of the main query, we masked its outputs and
evaluated the precision change, as shown in Table 3. The main query indeed has
a massive influence on the model’s performance. Without predictions from the
main query, the average precision dropped by around 7 points. This shows that
DETR is heavy depending on the predictions from the main query. It will make
the model more vulnerable to attacks.

| mAP | mAP without main query
---|---|---
DETR R50 | 0.420 | 0.353
DETR R101 | 0.435 | 0.367
Table 3: The mAP of two models with and without predictions from the main
query.

####  4.4.1 Cause of The Main Query

In the following experiment, we investigated the cause of the main query and
the imbalanced query contribution problem in the DETR model. We fine-tuned a
DETR R50 model on the pascal-voc dataset. We first trained the model for 50
epochs to let it stabilize and then collected the gradient information that
was applied to each query in the following ten epochs. In Figure 12, we
computed the mean gradient magnitude applied on each query and the query
frequency. Query 71 remained as the main query in pascal-voc dataset. It
received the most gradient flow, and its magnitude is significantly larger
than any other query. We believe the cause for the imbalanced gradient flow is
due to the transformer structure. The DETR wanted to keep its transformer
module simple so every query is allowed to compute cross-attentions with the
whole encoder context. In the early stage of training, the main query
converged fast and made the best predictions among all queries though it might
not be precise. However, this is enough for the main query to receive most of
the gradient flow, while others had much fewer chances to update and compete
with the main query. It not only forced this single query to converge to most
of the detection tasks, but also affected other queries’ convergence progress.
Therefore, the imbalanced query contribution problem caused the slow
convergence and low performance of the DETR model.

Figure 12: The average gradient flow to each query during transfer learning
and the query frequency in pascal-voc-dataset.

####  4.4.2 Solutions to The Main Query

Our ideas against this problem are simple and direct, to suppress positive
predictions from the main query and encourage that from all remaining queries.
This can be achieved by two methods: random query drop and cross-attentions
through a smaller window.

To evaluate the performance of random query drop, we fine-tuned another DETR
R50 model on pascal-voc dataset for 30 epochs and compared the loss graph with
the previous fine-tuned model, as shown in Figure 13. We observed that the
model with random query drop has a lower testing loss and showed faster
convergence. Random query drop is only a prototype idea proposed by us, and we
hope it can be further evaluated on other transformers that use independent
queries like DETR. The second idea is similar to the one proposed in Swin
Transformer [20], where attention is computed inside a small window. This not
only reduces the computation cost but also prevents any single query from
being able to answer all detection tasks.

Figure 13: The loss graph comparison between two fine-tuned model with and
without random query drop.

##  5 Conclusion

In this work, we investigate the robustness and properties of transformer-
based detectors DETR by measuring its performance toward different image
nuisances and making a comparison with other detectors. Two widely adopted
CNN-based detectors like YOLOv5 and Faster-RCNN, are used to compare their
performance with DETR. These detectors are evaluated with two main test cases,
occlusion and adversarial stickers. We implement random and salient occlusion
to analyze how the network handles the specific regions impacted by occlusion,
specifically on the region where it contains essential information in the
image. In terms of adversarial stickers, this attack is utilized to change the
pixel values in some portion of the image to mislead the network. Moreover, we
benchmark the robustness of DETR and analyze the query properties to a set of
corruption images.

Our experiment implies that DETR performs well when it comes to resistance to
interference from information loss in occlusion images. Nevertheless, in the
case of the sticker patch, it manages to produce a new set of keys, queries,
and values in the network which in most cases results in the misdirection of
the network. This result in increasing the attention weight to the
corresponding adversarial patch key token, making the network misguided the
attention to the adversarial stickers. For the benchmark, the experiment
indicates that DETR precision on corrupted images is generally lower than
YOLOv5. We also found that DETR depends heavily on the predictions from the
main query. This impact the main query to receive most of the gradient flow,
which leads to imbalanced contributions among all queries.

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minimal $\rm cl$-reduction $(x_{1},\ldots,x_{k-1})$ of $\mathfrak{a}$, we have
that $(x_{1},\ldots,x_{k-1},x)^{\rm cl}=(\mathfrak{a}+(x))^{\rm cl}=I$.
However, this implies that $(x_{1},\ldots,x_{k-1},x)$ is a minimal $\rm
cl$-reduction of $I$ which is a contradiction by Proposition 9.8. ∎

The following two corollaries follow directly from Theorem 9.12.

######  Corollary 9.13.

Let $(R,\mathfrak{m})$ be a local excellent normal ring of characteristic
$p>0$ and $I$ be a tightly closed ideal. Suppose that the $*$-spread of $I$ is
$k$. Then the set of all $*$-prereductions of $I$ is

${\bf P}_{*}(I)=\\{(y_{1},\ldots,y_{k-1})+I^{*{\rm
sp}}\mid(y_{1},\ldots,y_{k})\text{ a minimal }*\text{-reduction of }I\\}.$

######  Corollary 9.14.

Let $(R,\mathfrak{m})$ be a local excellent normal ring of characteristic
$p>0$ and $I$ be a Frobenius closed ideal. Suppose that the $F$-spread of $I$
is $k$. Then the set of all $F$-prereductions of $I$ is

${\bf P}_{F}(I)=\\{(y_{1},\ldots,y_{k-1})+I^{F{\rm
sp}}\mid(y_{1},\ldots,y_{k})\text{ a minimal }F\text{-reduction of }I\\}.$

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