The document discusses the concept of encryption, which secures data by converting plain text into encoded text through a specific algorithm, ensuring privacy during transmission. It outlines the process of both encryption and decryption, detailing the variables and mathematical operations involved. Additionally, it mentions the use of a skip variable, sender information attachment, and user registration options for password protection in encrypted files.

2. Team members
• Mohit Kundu
(3313 -61 -0028)
• Sabyaschi Ghosh
(3313 -61 -0031)
• Nilanjit Roy
(3313 -61 -0029)

3. Encryption is a method of converting an original message of
regular text into encoded text. The text is encrypted by
means of an algorithm (type of formula). If information is
encrypted, there would be a low probability that anyone
other than the receiving party who has the key to the code or
access to another confidential process would be able to
decrypt (translate) the text and convert it into plain,
comprehensible text.

4. 1. Data security
Encryption ensures data security by encrypting the data that can’t be
accessed by others. This way the data can also be sent over unsecured
channels as security is present in the encrypted data itself.
2. Data Integrity
Encryption ensures that if the data has been modified during transit or
intentionally by a hacker, the same cannot be decoded at the recipient's end,
providing complete security of the data.

5. Encryption Decryption

7. 1. Start.
2. Input the plain text.
3. Assume a variable K and initialized with a constant value.
4. Assume a variable Kmax and initialized with a constant value greater than K.
5. Take a character from the plain text and repeat 5 to 13 steps while the all characters
of the plain text have not been taken.
6. Check the value of the variable K whether its value less than Kmax or not.
7. If the value of K is equal or greater than Kmax then reset the value of K with initial
value.
8. Assume one more variable N.
9. Do the following 3 steps (i.e. 10, 11, 12) for N =1 to K.
10. Convert the character to their corresponding ASCII (decimal) equivalent.
11. Add {5*(-1)N+3N} with the ASCII (decimal) value of the character.
12. Convert the Modified ASCII value to its equivalent character and store it as
encrypted text.
13. Increment the value of K by 1.
14. Stop

8. 1. Please Start..
2. Input the Encrypted text.
3. Initialized a variable SKP, where the value of SKP is equal to (K-1) (K variable is
initialized at the time of encryption algorithm initialization.).
4. Take the a character from the encrypted text and repeat 5 to 9 steps while the all
characters of the encrypted text have not been taken.
5. Convert the character to their corresponding ASCII (decimal) equivalent.
6. Add 2 to the ASCII value of the character.
Modified ASCII value = Original ASCII value of character + {5*(-1)N+3N}
Original ASCII value of character = Modified ASCII value - {5*(-1)N+3N}
For each character of plain text, the value of first bit of each bit Stream is = {5*(-
1)1+3*1} (where N=1) = -5+3 = - 2
Original ASCII value of character = Modified ASCII value - {5*(-1)N+3N}
Original ASCII value of character = Modified ASCII value - (- 2)
Original ASCII value of character = Modified ASCII value +2
Convert the Modified ASCII value to its equivalent character and store it in a New
plain text.
Check the value of SKIP less than (Kmax-1) or not.

9. 7. If the value of SKIP is equal or greater than (Kmax-1) then reset the value of SKIP
with initial value..
8. Skip next SKP (Where SKP is a variable) character.
9. Increment the value of SKP by 1.
10. Stop

11. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
67
• No of character for each character of plain text (k-1)
: 4
• Value of N : 1
• ASCII Value of plain character : 67
+ 5 * (-1)1 + 3*1
• ASCII Value of cipher character :
= 67
= 67 + (-5) + 3
= 67 + (-2)
= 65
65
A
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )

12. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
67
• No of character for each character of plain text (k-1)
: 4
• Value of N : 2
• ASCII Value of plain character : 67
+ 5 * (-1)2 + 3*2
• ASCII Value of cipher character :
= 67
= 67 + (+5) + 6
= 67 + (+11)
= 78
65 78
A N
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )

13. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
67
• No of character for each character of plain text (k-1)
: 4
• Value of N : 3
• ASCII Value of plain character : 67
+ 5 * (-1)3 + 3*3
• ASCII Value of cipher character :
= 67
= 67 + (-5) + 9
= 67 + (+4)
= 71
65 78 71
A N G
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )

14. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
67
• No of character for each character of plain text (k-1)
: 4
• Value of N : 4
• ASCII Value of plain character : 67
+ 5 * (-1)4 + 3*4
• ASCII Value of cipher character :
= 67
= 67 + (+5) + 12
= 67 + (+17)
= 84
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )

15. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567
• No of character for each character of plain text (k-1)
: 5
• Value of N : 1
• ASCII Value of plain character : 65
+ 5 * (-1)1 + 3*1
• ASCII Value of cipher character :
= 65
= 65 + (-5) + 3
= 65 + (-2)
= 63
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63
?

16. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567
• No of character for each character of plain text (k-1)
: 5
• Value of N : 2
• ASCII Value of plain character : 65
+ 5 * (-1)2 + 3*2
• ASCII Value of cipher character :
= 65
= 65 + (+5) + 6
= 65 + (+11)
= 76
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76
? L

17. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567
• No of character for each character of plain text (k-1)
: 5
• Value of N : 3
• ASCII Value of plain character : 65
+ 5 * (-1)3 + 3*3
• ASCII Value of cipher character :
= 65
= 65 + (-5) + 9
= 65 + (+4)
= 69
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69
? L E

18. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567
• No of character for each character of plain text (k-1)
: 5
• Value of N : 4
• ASCII Value of plain character : 65
+ 5 * (-1)4 + 3*4
• ASCII Value of cipher character :
= 65
= 65 + (+5) + 12
= 65 + (+17)
= 82
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R

19. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567
• No of character for each character of plain text (k-1)
: 5
• Value of N : 5
• ASCII Value of plain character : 65
+ 5 * (-1)5 + 3*5
• ASCII Value of cipher character :
= 65
= 65 + (-5) + 15
= 65 + (+10)
= 75
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R
75
K

20. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567 66
• No of character for each character of plain text (k-1)
: 6
• Value of N : 1
• ASCII Value of plain character : 66
+ 5 * (-1)1 + 3*1
• ASCII Value of cipher character :
= 66
= 66 + (-5) + 3
= 66 + (-2)
= 64
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R
75
K
64
@

21. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567 66
• No of character for each character of plain text (k-1)
: 6
• Value of N : 2
• ASCII Value of plain character : 66
+ 5 * (-1)2 + 3*2
• ASCII Value of cipher character :
= 66
= 66 + (+5) + 6
= 66 + (+11)
= 77
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R
75
K
64 77
@ M

22. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567 66
• No of character for each character of plain text (k-1)
: 6
• Value of N : 3
• ASCII Value of plain character : 66
+ 5 * (-1)3 + 3*3
• ASCII Value of cipher character :
= 66
= 66 + (-5) + 9
= 66 + (+4)
= 70
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R
75
K
64 77 70
@ M F

23. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567 66
• No of character for each character of plain text (k-1)
: 6
• Value of N : 4
• ASCII Value of plain character : 66
+ 5 * (-1)4 + 3*4
• ASCII Value of cipher character :
= 66
= 66 + (+5) + 12
= 66 + (+17)
= 83
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R
75
K
64 77 70 83
@ M F S

24. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567 66
• No of character for each character of plain text (k-1)
: 6
• Value of N : 5
• ASCII Value of plain character : 66
+ 5 * (-1)5 + 3*5
• ASCII Value of cipher character :
= 66
= 66 + (-5) + 15
= 66 + (+10)
= 76
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R
75
K
64 77 70 83
@ M F S
76
L

25. C A B
• The variable K, which value is lies between 5 to 9 and it also determine the number of
character to be generated for a single
character of the plain text.
6567 66
• No of character for each character of plain text (k-1)
: 6
• Value of N : 6
• ASCII Value of plain character : 66
+ 5 * (-1)6 + 3*6
• ASCII Value of cipher character :
= 66
= 66 + (+5) + 18
= 66 + (+23)
= 89
65 78 71 84
A N G T
• The variable N, which value lies between
1 to (K-1)
( Let CAB is a plain text )
63 76 69 82
? L E R
75
K
64 77 70 83
@ M F S
76
L
89
Y

29. 65
• The SKIP variable is used to skip
no encrypted character.
• For first character of a bit stream
the value of N always be 1
• For first time the value of SKIP
variable is 0.
No of Character to SKIP :
Value of N :
Original ASCII value of character :
- {5*(-1)1+3*1}
0
- {(-5)+3}= 65
= 65 - (-2)
= 65 + 2
= 67
= 65
1
A N G T ? L E R K @ M F S L Y
C

30. A N G T ? L E R K @ M F S L Y
65 63
• The SKIP variable is used to skip
no encrypted character.
• For first character of a bit stream
the value of N always be 1
• For first time the value of SKIP
variable is 0.
No of Character to SKIP :
Value of N :
Original ASCII value of character :
- {5*(-1)1+3*1}= 63
- {(-5)+3}= 63
= 63 - (-2)
= 63 + 2
= 65
3
1
C A

31. A N G T ? L E R K @ M F S L Y
65 63 64
• The SKIP variable is used to skip
no encrypted character.
• For first character of a bit stream
the value of N always be 1
• For first time the value of SKIP
variable is 0.
No of Character to SKIP :
Value of N :
Original ASCII value of character :
- {5*(-1)1+3*1}= 64
- {(-5)+3}= 64
= 64 - (-2)
= 64 + 2
= 66
4
1
C A B

36. Sender X Y Z
= Can decode
= Can’t decode
Encrypted text
Encrypted text
Sender X Y Z

37. Encoder
Password
Plain Text
Spacial
Character
Encrypted Text

38. • It does Support trail version & demo version.
• Register user can set password to the encrypted file.
• It automatically attched the sender information with this encrypted file.
• It requires low memory space (Primary memory as well as secondary memory.

39. • Our application is devoloped in C enviroment only for windows xp platform.
• This application only works on text type file(.txt) .
• We can not use ‘tiled’ (~) character in password, because tiled character is
used to separate text file password and original text.
•There is no provision to know if the encrypted file is modified or some text
missing.

40. • At present our application works on only text type of
document(.txt), but in future we are thinking that we also works on
video(.mp4), audio(.mp3) and image type of file(.jpg) format.
• Now our application generates 4-8 bit character for each single
character in plain text, but we are thinking to change it in 32-64
character for each single character in plain text.
• If someone changes our cipher text file, hen there may be creating
some problem to get the original file by decryption. But in future we
are trying to develop that; if someone modify the cipher text file
then it will automatically find out the error and correct it.

41. [1] Bement A. L. et. al. (2004), Standards for Security Categorization of Federal
Information and Information Systems, Computer Security Division, Information
Technology Laboratory, National Institute of Standards and Technology
Gaithersburg, MD 20899-8900.
[2] Ayushi, (2010), A Symmetric Key Cryptographic Algorithm, International Journal
of Computer Applications (0975 - 8887) Volume 1. No. 15.
[3] Atul Kahate, (2008) Cryptography and Network Security, Tata McGraw-Hill
Education, pg. 47.
[4] Ijaz Ali Shoukat , Kamalrulnizam Abu Bakar and Mohsin Iftikhar, “A Survey about
the Latest Trends and Research Issues of Cryptographic Elements”, p 141,
International Journal of Computer Science Issues, Vol. 8, Issue 3, No. 2, May 2011,
ISSN 1694 0814.
[5] Sukalyan Som, Saikat Ghosh, “A Survey of Traditional or Character Oriented
Symmetric Key Cryptography”, International Journal of Advanced Research in
Computer Science, Vol. 2, No. 4, July-August 2011

42. International Journal of Computer Applications (0975 – 8887)
Volume 39– No.8, February 2012
ISBN: 978-93-80866-31-7

43. SRIJONI MAITRA
ARUN KUMAR CHAKRABARTI
SAMIT MAZUNDER
SONALI GUPTA