This networking assignment sample solution learn you about encryption and description technique via Cipher Algorithm help understanding security mechanism

# Introduction

This assignment has been made to understand the concepts of security and apply them on the given assignment. It will enhance the learner skills. The RSA, DSA and several other encryption algorithms has been applied to it.

## Task 1 Symmetric and Asymmetric Cipher

**Symmetric Encryption**

Symmetric Encryption is an oldest method that is a method of encryption in which a text of a message is converted into some another form to prevent it from message decoding by unintended user. A secret key is used to encode or decode the message at both the ends. [Also Read: – Big Mike Web Programming Assignment]

**Asymmetric Encryption**

Asymmetric method is similar to symmetric encryption but it has key pair that is a public key and a private key. A public key can be freely provided to the person who wants to send a message over the network and only the intended recipient can decipher the message through his private key. [Also Read: – IT Programming Database Assignment Help]

**Task 1.1**

image 1

(3+3+0+0+4+6+2) % 10 = 8

The Caesar Cipser text of the last name will be

Plain text = karbhari

Caesar Cipser text will be = sizjpizq

(To calculate the encrypted text there is a need to shift the text by 8)

**Task 1.2**

image 2

Created hash text using online tool the number was s3300462. After removing‘s’ the number left 3300462. The rest text after converting into hash code looks as follows: [Also Read: – App Stores Marketing Assignment Help]

**SHA 256**

b689e970f18eb4983dfcb19ea813251d4277019fc2a7776acaf742baa2480dac

P=1266631

Q=426971

P is prime? Yes

Q is prime? Yes

N: 540814704701

K: 540813011100

GCD(E,K): GCD(7,540813011100) is 1

Number of bits per group: 39

**Message in encoded form:**

Cipher = (Msg)^{ e} mod N

1001100010011010010010011100010100011100000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100011110100101011100100010010100011101011110000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100011110100101011100100010010100011101011110000100111001011101101111010010111100010000100111001011101101111010010111100011110100101011100100010010100011101011110000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100011110100101011100100010010100011101011111110100101011100100010010100011101011110000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100011110100101011100100010010100011101011110000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100011110100101011100100010010100011101011110000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100011110100101011100100010010100011101011110000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100010000100111001011101101111010010111100011110100101 01110010001001010001110101111000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001000010011100101110110111101001011110001111010010101110010001001010001110101111000010011100101110110111101001011110001000010011100101110110111101001011110001

**Task 1.3**

**Introduction to RSA**

RSA named after three persons that are Ron Rivest, Adi Shamir and Leonard Adleman and it is an asymmetric cryptography mechanism that uses two different keys that are one public key and one private key. Algorithm will be explained below through the example.

image 3

Read more about Online Assignment Help

Created hash text using online tool the number was s3300462. After removing‘s’ the number left 3300462. The rest text after converting into hash code looks as follows:

**SHA 256**

b689e970f18eb4983dfcb19ea813251d4277019fc2a7776acaf742baa2480dac

**Numeric form of student registration number**

098 054 056 057 101 057 055 048 102 049 056 101 098 052 057 056 051 100 102 099 098 049 057 101 097 056 049 051 050 053 049 100 052 050 055 055 048 049 057 102 099 050 097 055 055 055 054 097 099 097 102 055 052 050 098 097 097 050 052 056 048 100 097 099

The task has assumed two prime numbers

Let p = 11 and q = 7

N = 77 (11*7)

PHI = 60 (10*6)

E = 7

D = 7

Public key (h, e) = (77, 7)

Private Key (h, d) = (77, 7)

**The encryption of the given number**

P1 = (98)7 MOD (101) = 86812553324672 MOD (101) = 09

P2 = (54)7 MOD (101) = 1338925209984 MOD (101) = 24

P3 = (56)7 MOD (101) = 1727094849536 MOD (101) = 00

P4 = (57)7 MOD (101) = 1954897493193 MOD (101) = 73

P5 = (101)7 MOD (101) = 107213535210701 MOD (101) = 51

P6 = (57)7 MOD (101) = 1954897493193 MOD (101) = 73

P7 = (55)7 MOD (101) = 1522435234375 MOD (101) = 18

P8 = (48)7 MOD (101) = 587068342272 MOD (101) = 44

P9 = (102)7 MOD (101) = 114868566764928 MOD (101) = 42

P10 = (49)7 MOD (101) = 678223072849 MOD (101) = 29

P11 = (56)7 MOD (101) = 1727094849536 MOD (101) = 00

P12 = (101)7 MOD (101) = 107213535210701 MOD (101) = 51

P13 = (98)7 MOD (101) = 86812553324672 MOD (101) = 09

P14 = (52)7 MOD (101) = 1028071702528 MOD (101) = 75

P15 = (57)7 MOD (101) = 1954897493193 MOD (101) = 73

P16 = (56)7 MOD (101) = 1727094849536 MOD (101) = 00

P17 = (51)7 MOD (101) = 897410677851 MOD (101) = 36

P18 = (100)7 MOD (101) = 100000000000000 MOD (101) = 10

P19 = (102)7 MOD (101) = 114868566764928 MOD (101) = 42

P20 = (99)7 MOD (101) = 93206534790699 MOD (101) = 82

P21 = (98)7 MOD (101) = 86812553324672 MOD (101) = 09

P22 = (49)7 MOD (101) = 678223072849 MOD (101) = 29

P23 = (57)7 MOD (101) = 1954897493193 MOD (101) = 73

P24 = (101)7 MOD (101) = 107213535210701 MOD (101) = 51

P25 = (97)7 MOD (101) = 80798284478113 MOD (101) = 46

P26 = (56)7 MOD (101) = 1727094849536 MOD (101) = 00

P27 = (49)7 MOD (101) = 678223072849 MOD (101) = 29

P28 = (51)7 MOD (101) = 897410677851 MOD (101) = 36

P29 = (50)7 MOD (101) = 781250000000 MOD (101) = 15

P30 = (53)7 MOD (101) = 1174711139837 MOD (101) = 36

P31 = (49)7 MOD (101) = 678223072849 MOD (101) = 29

P32 = (100)7 MOD (101) = 100000000000000 MOD (101) = 10

P33 = (52)7 MOD (101) = 1028071702528 MOD (101) = 75

P34 = (50)7 MOD (101) = 781250000000 MOD (101) = 15

P35 = (55)7 MOD (101) = 1522435234375 MOD (101) = 18

P36 = (55)7 MOD (101) = 1522435234375 MOD (101) = 18

P37 = (48)7 MOD (101) = 587068342272 MOD (101) = 44

P38 = (49)7 MOD (101) = 678223072849 MOD (101) = 29

P39 = (57)7 MOD (101) = 1954897493193 MOD (101) = 73

P40 = (102)7 MOD (101) = 114868566764928 MOD (101) = 42

P41 = (99)7 MOD (101) = 93206534790699 MOD (101) = 82

P42 = (50)7 MOD (101) = 781250000000 MOD (101) = 15

P43 = (97)7 MOD (101) = 80798284478113 MOD (101) = 46

P44 = (55)7 MOD (101) = 1522435234375 MOD (101) = 18

P45 = (55)7 MOD (101) = 1522435234375 MOD (101) = 18

P46 = (55)7 MOD (101) = 1522435234375 MOD (101) = 18

P47 = (54)7 MOD (101) = 1338925209984 MOD (101) = 24

P48 = (97)7 MOD (101) = 80798284478113 MOD (101) = 46

P49 = (99)7 MOD (101) = 93206534790699 MOD (101) = 82

P50 = (97)7 MOD (101) = 80798284478113 MOD (101) = 46

P51 = (102)7 MOD (101) = 114868566764928 MOD (101) = 42

P52 = (55)7 MOD (101) = 1522435234375 MOD (101) = 18

P53 = (52)7 MOD (101) = 1028071702528 MOD (101) = 75

P54 = (50)7 MOD (101) = 781250000000 MOD (101) = 15

P55 = (98)7 MOD (101) = 86812553324672 MOD (101) = 09

P56 = (97)7 MOD (101) = 80798284478113 MOD (101) = 46

P57 = (97)7 MOD (101) = 80798284478113 MOD (101) = 46

P58 = (50)7 MOD (101) = 781250000000 MOD (101) = 15

P59 = (52)7 MOD (101) = 1028071702528 MOD (101) = 75

P60 = (56)7 MOD (101) = 1727094849536 MOD (101) = 00

P61 = (48)7 MOD (101) = 587068342272 MOD (101) = 44

P62 = (100)7 MOD (101) = 100000000000000 MOD (101) = 10

P63 = (97)7 MOD (101) = 80798284478113 MOD (101) = 46

P64 = (99)7 MOD (101) = 93206534790699 MOD (101) = 82

**Final Encrypted Text =**

0924007351731844422900510975730036104282092973351460029361536291075151818442973428221546181818244682464218751509464615750044104682

## Task 2 Transposition Cipher

**Task 2.1**

Transposition Cipher is a method of cryptography in which the data is encrypted in such a way that no one can read this data or data can be safe. It is very simple and effective encryption technique which follows a simple rule of mixing characters to convert the plain text into the cipher text. Unauthorized user will not be able to access the data and the data will not be available in a readable format. [Also Read: – Network Security Research Report Assignment Help]

**The given cipher data is:**

mt pliloayvyty e h l tesishriviasmelp ml

In row column format the data will be shown as below:

mt p

lilo

ayvy

ty e

h l

tesi

shri

vias

melp

ml

After swapping row with each other the data will be as follows:

tmp

ilol

yayv

yte

h l

etis

hsir

ivsa

empl

m l

The cipher text after encryption in column format:

tiyy ehiemmlathtsvm poye iisp lv lsrall

Thus the final output is the below cipher text:

**tiyy ehiemmlathtsvm poye iisp lv lsrall**

Read more about Assignment Help Australia

##### Conclusion

The learner has understood the concepts of encryption and decryption. It helped in understanding of security mechanisms such as hash algorithms, RSA and so on which will help them in future purposes.

##### References

Asecuritysite.com, (2016). RSA. [online] Available at: http://asecuritysite.com/encryption/rsa_2.

Csgnetwork.com, (2016). *Powers And Power Roots Calculator*. [online] Available at: http://www.csgnetwork.com/powerrootpowercalc.html.

*Top Grade Assignment Help provide technical assignment writing service based on case study requirements in affordable prices and we are providing most flexible online assignment writing help, so book your Assignment with us, **order now** *