Paper Cryptography

Cryptography by using attributes of Magic Number 9 Mr. Amol G. Muley MTech (CSE) Email: [email protected] Mobile No.:09403593894 Abstract: Cryptography is an emerging technology in which two parties secure network communication by application of different encryption and decryption algorithms. The security of data during transmission by this method is based on important attributes of magic number nine. Sum of digits of any number which is divisible by nine is nine. So it is very difficult to find out key used for encryption and decryption. This method is simple to understand and implementation but robust to debug and crack. This method uses simple arithmetic and bitwise operations to provide network security. This method concentrates on robust encryption mechanism and irrational organization of cipher text. Key Words- Magic Number, Encryption, Decryption. 1.0 INTRODUCTION This encryption method is based on mathematical properties of magic number nine. Consider any number which is divisible by nine for example 99927 and if you perform sum of digits of given number it is nine (9+9+9+2+7=36= (3+6) =9). The development of this method is inspired by observing shortcomings of some standard cryptographic methods. This method uses key for encryption and decryption but by knowing mere key will not solve purpose of decryption. This method sends Secrete key with decrypted text and it also stores information about bits which is on for the corresponding bits from secrete key. By using text which gets after application of above method and information mentioned in last sentence is organized by changing bit positions to form final decrypted text. It uses simple arithmetic and bitwise operations. 2.0 RELATED WORK 64 bit plaintext is passed as input to this method; magic number is selected for performing certain bitwise and arithmetic operations. Magic Number is selected such that it is divisible by 9. Arithmetic addition operation is performed with the magic number and plain text to get 72 bit result. Decimal equivalent of 72 bit result is found out then by performing sum of all digits present in equivalent decimal number to get one digit number and suppose it is K21. Same way find out decimal equivalent of K1 and perform sum of digit operation to get one digit final number and suppose it is K11. Perform

bitwise OR operation to get 72-bit K3. Above mentioned things are explained with following example. K1(Hexadecimal)= 00000012 = 18(Decimal)=9=K11 ; K2(Hexadecimal)=00000019=25(Decimal)=7=K21 K3= K21 OR (Bitwise) K11. Now perform bitwise OR operation of K3 and K2 to get 72-bit K4.Find out K5. It is explained with following example. Plain Text (T) = 000000000000000---00111 Arithmetic Addition K1 = 000000000000000—10010 ----------------------------------------------------------------K2 = 000000000000000---11001 K21

= 000000000000000---00111 Arithmetic Addition followed by sum of digits (Result one digit) K11 = 000000000000000---01001 -----------------------------------------------------------------= 000000000000000---10000 (Result one digit which is equal to K21) K21 = 000000000000000---00111 Bitwise OR K11 = 000000000000000---01001 ---------------------------------------------------------------K3 = 000000000000000---01111(Result) Bitwise OR K2 = 000000000000000---11001 ----------------------------------------------------------------K4 = 000000000000000---11111(Result) K5

= 000000000000000---00110

First four 0th 0th 71st 71st bits are least bit bit bit bit significant of of of of bits of K21 K5 K4 K5 K4 0 1 1 1 0 0 --------------------1 0 1 First row shows details about bits and second row stores actual data

Figure1: Break Up of Final 80 bit Cipher Text. K5 only store information about bits which are off (from K2) for the corresponding bits which are on from K3. This information is useful at the time of decryption. Alternate bits from K4 and K5 are arranged to get 144 bit final cipher text. Information which is present in K5 is hidden and scattered and very much important in the process of decryption. Key we are using in this method is different, even code cracker gets Key;

he can not break code without getting value of K5. For encryption both Key (K1) and K5 are required. K1 will remain same for entire document and K5 keeps on changing for each block of size 64 bits. So it will be very tedious and difficult to identify K5 for each block of 64 bits. One more things I would like highlight upon is the formation of K4. First by performing arithmetic addition of plain text and K1 (Magic Number) K2 is calculated and then by performing arithmetic addition of sum of digits of K2(one digit sum) and sum of digits(one digit sum) of K1, K3 is calculated. If we observe meticulously nuances of sum of digits of K1 it is always nine. Any number can be considered which is Divisible by nine and perform sum of digits till we get single digit, this is explained by giving following examples. Suppose magic number is 99 99=9+9=27=2+7=9; Suppose magic number is 3636=3+6=9; Suppose magic number is 9999=9+9+9+9=36=3+6=9; suppose magic number is 27729=2+7+7+2+9=27=2+7=9. For any number sum of digits of K1 remain 9. So it is very difficult to back track Magic Number from K3. One more important fact related to magic number 9 , if we add ant number to 9 and perform sum of digits operation to get one digit number we get number which we added to magic number 9(sum of digits to get one digit number)

least significant four bits stored in final cipher text of size 148 bits stores value of K21 and which is sum of digits (one digit number) of actual data before encryption operation. Above explained (entire) method is graphically shown below

64 bit Plain Text

64 bit Magic Number (K1)

Arithmetic Addition

72 bit Result (K2)

Decimal Equivalent OR

K1

Decimal Equivalent

72 bit Result (K2)

K21

OR or OR

72 bit Result (K3)

72 bit Result (K4)

K5

Least Significant (four) bits from K21 and Alternate bits from K4 & K5

9 + 06 = 15 = 06(sum of digits of 15 to get one digit number) 9 + 17(sum of digits to get one digit number is 08) = 26 = 08

148 Bit Cipher Text

9+128(02) = 137(02) 9+1345(04) = 1354(04) 9+ 999(09) =1008(09) 9+1111(04) =1120(04) This important fact can be used in encryption and decryption process for (partial) verification of decrypted data with actual data. K21 stores sum of digits (one digit number) of actual data and (four) least significant bits are added in final 148 bit cipher text. If we observe arithmetic addition operation of K21 and K11, resultant value is equal to K21. By using above mentioned encryption method, encryption operation will be performed and by using decryption algorithm, decryption operation will be performed. At the time of encryption least significant four bits are stored in final cipher text of size 148 bits and it is shown in Figure1. Now after decryption operation, we are having actual data and it can be compared with least significant four bits stored in final cipher text of size 148 bits for partial verification. It is not possible to perform complete verification of actual data with decrypted data by above mentioned comparison operation but partial verification is possible by calculating sum of digits (one digit number) of decrypted data and comparing it with least significant four bits stored in final cipher text of size 148 bits which is input to decryption algorithm and it should match as

Figure2: Flowchart for Encryption Method For decryption 148 bit cipher text is collected and separated according the method adopted in decryption process, so that to get K4 and K5. Find out 1’s Complement of K5. Perform Bitwise AND Operation of K4 and 1’s compliment of K5 to get K2. Arithmetic Subtraction operation is performed with K2 and K1 to get original text. Above mentioned steps are explained by giving following example. K5

= 000000000000000---00110

K5 (1’S Complement) = 111111111111111---11001 Bitwise AND K4 = 000000000000000---11111 -------------------------------------------------------------------K2 = 000000000000000---11001 Arithmetic Subtraction K1 = 000000000000000—10010 -------------------------------------------------------------------Plain Text = 000000000000000---00111 Sum of digits (Result one digit) -------------------------------------------------------------------T1 = 000000000000000---00111 (T1= = K21) Comparison operation

000000000000000---00111 = 000000000000000---00111 Figure3: Flowchart for Decryption Method

148 Bit Cipher Text

Separate K4 & K5

K4

K5

3.0 CONCLUSION AND FUTURE WORK

AND

1’S Complement of K5

Method explained is simple to understand and easy to Debug and very difficult to think according to logic applied. This method is useful in implementation point of view. In this method only once we have calculated K5, twice it can be calculated and use in method for encryption. So that it will be more difficult to debug.

K1

K2

Arithmetic Subtraction

4.0 REFERENCES 64 bit Plain Text

T1

[1] Cryptography and Network Security by Behrouz A. Forouzan [2 [

(T1= = K21) Comparison operation