IB Mathematics

Extended Essay

To what extent does an Arc Tan Encryption System able

to perform like Rivest-Shamir-Adleman?

Number of Words: 2343

2017 – 2018

Armin 12B (St. Francis)

Stella Maris International School

Gading Serpong, Indonesia

Table of Contents

Introduction. 3

Theory of Arc Tan Crypt. 4

Practical Encoding of a

Message using Arc TanCrypt. 10

Practical Decoding of a

Message using Arc TanCrypt. 13

Data Collection for

Encryption Test from Arc TanCrypt. 16

Introduction to RSA

(Rivest, Shamir, Adleman). 17

Practical Encoding of a

Message using RSA.. 18

Practical Decoding of a Message using RSA.. 19

Data Collection from RSA

for Encryption Test. 20

Conclusion. 20

Bibliography. 21

Introduction

With

the rapid development of network and multimedia technologies, privacy has

become a crucial part to the transmission of sensitive electronic data. Thus,

ensuring the privacy for the transmission of sensitive electronic data plays a

starring role in the eyes of many private users of several internet accounts

and further private activity online.

By

ensuring higher security in terms of cryptography for electronic data, we do

have to face algorithms and hours of mathematical studies and computer

engineering. By doing this Research, the author does have the intention to learn

more deeply about existing technologies and to propose a new improved method. By

proposing a new method later, the author intend to help people around the world

to feel safer whenever they must transfer private data through an untrusted

connection.

In

this essay, the author will first discuss about Tan Crypt which is the author’s

own original idea. The author intends to find out to what extend does an Arc

Tan Encryption System perform like Rivest-Shamir-Adleman. To find out the

answer, the author will compare both encryption systems at the end of this

essay in the conclusion section.

Theory of Arc Tan Crypt

all calculations are

in DEG-Mode

This system does not represent a public key cryptosystem.

It does make use of a single key that is used to encrypt and decrypt the

message. During the research, the author tried to find a way to encode data

very randomly using a simple key (or more than one key at the same time). So

the first step was to look around for something really unique in mathematics

which could be used for encoding data. An example that was found by the author

is…

Compared with elliptical graphs it does look more

compromising to be used for encryption when represented as graph as it can be

seen below.

Y – axis

X – axis

Figure 1:

When y = 789tan (2x+10)

When analyzing the graph, it can be seen that the line

is unregularly and seems very hard to find a general equation for this graph

directly. The only important thing at this point is just to find a type of

graph that does show a lot of random difference in y-values when the function

is altered by just a small number (e.g. 5,33,62).

Y – axis

X – axis

Figure 2:

When y = 12tan (-3x+8)

The

equation for decrypting the message can be easily found by the following steps

The first precaution in here is that when the Data is

encrypted and decrypted using the two formulas stated above, it does affect the

result. When the original data is compared with the encrypted and then

decrypted data, it is different. This problem is visualized below.

Compared to be different

In case the

calculation is done in the opposite way, which means that the encryption

formula and the decryption formula are exchanged with each other for the

following calculation process, the Original Data is the same with the Decrypted

Data. An graphical representation of this is shown below.

Compared to be the same

Practically, if the variable for ‘X’ has got a big

value of more than 6 digits (this value may vary due to different systems used,

e.g. GDC, Computer), it becomes hard to get the exact

same result after decryption as before doing the encryption.

A simple example can be shown below using simple and freely-assigned values for

the three keys…

Variable

Value assigned

A

2

B

3

X

010205000810

C

1

Z

12345678

= X

Decryption…

Z = 12345677.999999999999999999999998

The reason for this inaccuracy (.99999999999…)

is because the value for ? in ‘

‘ is bigger than

90 (if the angle is in degrees; e.g. radians -> 2?). The solution for this

problems is to make sure that the value for ‘?’ needs to be inversed. This step

makes sure that any value for ‘?’ is smaller than 1. Therefore, the formula for

encryption will change.

The objective for the following calculation is to meet

the objective or target stated above which is to decrease the size of the

number that needs to be used as the value for ‘?’.

Thus, two new formulas can be summarized from the

conclusion and calculation, as shown above, into the table below.

Encryption Formula

Decryption Formula

Thus, the validity for this formula is tested as shown

below using the same values as used in the example above.

Decryption…

X

= -0.32962962962962962962962962962959

Practical Encoding of a Message using Arc TanCrypt

For the first time, let a simple message to be encoded

using this cryptography system.

ABE HJ

1.

Now, it is needed

to convert this text into cipher which is using numbers. In this case, we will

use our own Encoding System but you can use any other Cipher – Codec (e.g.

ASCII) as long as the Decoder of the text (e.g. your friend) also use the same

Cipher – Codec

A

B

C

D

E

F

G

H

I

J

K

L

M

01

02

03

04

05

06

07

08

09

10

11

12

13

N

O

P

G

R

S

T

U

V

W

X

Y

Z

14

15

16

17

18

19

20

21

22

23

24

25

26

Note:

A space will be replaced by a zero in this Encoding System

Using this Table, we will convert the letters and all

the spaces into numbers; this is a single number, so it is named ‘Primary

Data’. All the numbers are combined into one single number

A

-> 01 | B

-> 02 | (space) -> 00 | …

ABE HJ

010205000810

Note:

This number is called Primary Data as

preferred by the author

2. Now

we will have to think of three different encryption keys, it has to be as

unique as possible by using different numbers (non-zero

real number,

) of

random difference. It doesn’t matter if it’s even or odd, it just has to be

unique and as long as possible

Key 1= -0.16329

Key 2= -1347.34323

Key

3= 3434313

3.

Now the author

will make use of the actual encryption Formula which is…

The

meaning or definition of those variables mentioned above like ‘A’ and ‘B’ are

written below in the table.

Variable

Definition

A

The value of the first key

B

The value of the second key

Z

The primary data

C

The value of the third key

X

The secondary data

The values are assigned to the corresponding variables

as shown in the table below

Variable

Value assigned

A

-0.16329

B

-1347.34323

Z

010205000810

C

3434313

X

Encrypted

Data (Secondary Data)

4.

So now all the

values are put into the formula and calculated

A B C X

5. The

result represents the final encrypted data or ‘Secondary Data’ which can now be

used to be transferred or shared to the receiver of the data

2548.9518443723587130981426922727

Note:

This number is called Secondary Data as

preferred by the author

Practical Decoding of a Message using Arc TanCrypt

Note: Same Message is used for decoding as

has been encoded in previous

section above

As

the receiver of the data, the author will use the following steps to decrypt

the Secondary Data back to the Primary Data.

1. To

decrypt the data, the receiver must make use of a decryption formula. The

decryption formula is as follows…

In this formula, there are some variables seen but

their meanings are still the same as before during the encryption process.

Below are the definitions of the variables.

Variable

Definition

Z

The secondary data

A

The value of the first key

C

The value of the third key

B

The value of the second key

X

The primary data

2.

Practically, the

receiver assigns all the known values, including the encryption key itself

which are three keys, to its corresponding values as defined above.

Variable

Value

X

2548.9518443723587130981426922727

A

-0.16329

C

3434313

B

-1347.34323

Z

Decrypted Data (Primary Data)

3.

Furthermore, all

the variables are put into the formula and the decrypted data can be retrieved.

4. This

step is only necessary if the length of the number is odd. For example,

Number

Length

Even

or odd?

Action

1234

4

Even

no

action

4567890

7

Odd

(Add zero before its number)

04567890

02340003243

11

Odd

002340003243

So in this case, the number ‘10205000810’ is an odd

number and therefore needs to have an additional ‘0’ (zero) on its left end.

010205000810

5.

Now the code can

be simply reconverted back into the original message.

01 -> A | 02

-> B | 00 -> (space) | …

010205000810

ABE HJ

Data Collection for Encryption Test from Arc

TanCrypt

Before the encryption test is being done later on,

there need to be more examples of different data to obtain more reliable test results.

Data ID

Content

of Data

Purpose

1

ABE

HJ

Simple

example

2

1234567890

Identic

Cumulative difference

3

AAAAAAAAAAAA

Repetitive

data

All steps as they have

been done in the sections above to encrypt the data using Arc TanCrypt (Step

1-5) are repeated and the corresponding results put in the table below; the

same corresponding keys are being used (Key 1= -0.16329; Key 2= -1347.34323;

Key 3= 3434313)

Data ID

Content

of Data

Encrypted

Data

1

ABE

HJ

2548.9518443723587130981426922727

2

ABCDEFGHIJK

2548.9520335559295211689030626686

3

AAAAAAAAAAAA

2548.9518443723587130141375417096

Introduction to RSA (Rivest, Shamir, Adleman)

(Page,

Don. https://www.pagedon.com/wp-content/uploads/2010/03/rsa-encryption.jpg.

28 March 2010. Document. 22 January 2018)

Above is a simple representation of

how RSA is applied as a public key cryptography. The first thing to do before

encrypting and decrypting a message is to generate the public key itself. There

are some rules and conditions that need to be considered before generating a

key for RSA which are listed down below in the box.

Choose two

primes p and q and let n = pq.

Let e ? Z be positive such

that gcd(e,?(n)) = 1.

Compute a

value for d ? Z such

that de?1 (mod ?(n)).

Our public

key is the pair (n,e) and our private key is the

triple (p,q,d).

For any

non-zero integer m