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journal of xi an shiyou university natural science edition issn 1673 064x atechniqueforencodinganddecodingusing matrixtheory 1 2 3 4 t ranjani a manshath v maheshwari and v balaji 1department of mathematics ...

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              Journal of Xi' an Shiyou University, Natural Science Edition                                         ISSN : 1673-064X
                   ATECHNIQUEFORENCODINGANDDECODINGUSING
                                                          MATRIXTHEORY
                                                       1                2                   3                4
                                           T.Ranjani , A.Manshath , V.Maheshwari and V.Balaji
                               1Department of Mathematics, D.K.M College for Women, Vellore - 632001.
                                              2Department of Mathematics and Actuarial science,
                          B.S.Abdur Rahman Crescent Institute of Science and Technology, Chennai - 600 048.
                                                           3Department of Mathematics,
                             Vels Institute of Science, Technology and Advanced studies, Chennai - 600117.
                                 4Department of Mathematics, Sacred Heart College, Tirupattur - 635601.
                                                          E-mail: pulibala70@gmail.com
                                        Abstract
                       In this paper, we have disccussed an algorithm for
                  encryption and decryption using matrix theory and we
                  have worked two examples for this algorithm.
                  KeyWords: Encryption, Decryption, Key Matrix.
                  2010AMSSubClass: 94A60,94B27,94B40
                  1. Introduction
                       Theartandtechnologyofconcealingthemessages
                  to introduce secrecy in information safety is diagnosed               Figure 1. Procedure for encoding
                  as cryptography. The technique of disguising a message
                  in such a way on hide its substance is coding. An en-
                  crypted message is cipher text. The approach of turning
                  cipher text into plain text is decryption. The encryp-
                  tion process is comprised of a Algorithm, with a key
                  [3]. The key is the value independent of the plaintext.
                  In [5], they have developed a technique for encryption
                  and decryption using matrix theory.
                  1.1. Prerequisite
                                                                                        Figure 2. Procedure for decoding
                  Theorem1.1 A text message of strings of some length
                  size L can be converted into a matrix called a message
                  matrix R of size n > m and n is the least such that m×        2. Results and Discussions
                  n ≥ L depending upon the length of the message with
                  the help of suitably chosen numerical and zeros. [2]          Illustration 2.1 The message which we are going to
                                                                                send to the receiver is
                  1.2. Algorithms                                               ”LETUSMAKEITSIMPLEINDEPENDENTLY.”
                                                                                Nowletusconvert the message into numbers ,
                    VOLUME 16 ISSUE 9                                     48-51                            http://xisdxjxsu.asia/
               Journal of Xi' an Shiyou University, Natural Science Edition                                            ISSN : 1673-064X
                  12 5 20 27 21 19 27 13 1 11 5 27 9 20                             Theencodedmessagetobesentis
                  27 19 9 13 16 12 5 27 9 14 4 5 16 5 14 4 5                    82 85 90 126 97 261 56 17 213 102 113 85 130
                  14 20 12 25 28                                                128 165 76 61 149 55 32 152 87 65 189 62 69
                                                                                50 45 30 109 93 94 109 146 137 210
                                                                                To get the original message receiver should multiply by
                  Setting these numbers into a matrix form R as,                T−1                            
                                                                                               82    85    90
                         12   5   20
                                                                                                             
                         27   21  19                                                            126    97    261
                                                                                                             
                                                                                             56     17    213
                         27   13   1                                                                           
                                                                                                             
                                                                                              102 113      85
                         11   5   27                                                                           
                                                                                                                                
                                                                                              130 128 165
                         9    20  27                                                                           
                                                                                                              −24       20    −5
                                                                                               76    61    149
                         19   9   13                                                     −1                                       
                  R=                                                          R=RTT =                         18 −15 4
                                                                                               55    32    152
                         16   12   5                                                                           
                                                                                                                  5     −4     1
                                                                                               87    65    189
                         27   9   14                                                                           
                                                                                                             
                                                                                               62    69    50
                         4    5   16                                                                           
                                                                                                             
                       5 14 4                                                                45     30    109
                                                                                             93     94    109
                                    
                         5    14  20
                         12   25  28                                                            146 137 210
                                                                                Therefore, the decoded message is,
                  Now let us assume a non singular matrix T as,                 12 5 20 27 21 19 27 13 1 11 5 27 9 20 27 19 9
                                                                              13 16 12 5 27 9 14 4 5 16 5 14 4 5 14 20 12
                       1   0  5                                                 25 28
                                                                 −1
                  T= 2 1 6 as an encryption key then T                 is       Hence, we received the original plaintext by chang-
                   3 4 0                                                      ing the numbers into alphabets. We get the original
                    −24     20    −5                                            message as ”LET US MAKEITSIMPLEINDEPEN-
                   18     −15     4 
                                       . We now multiplied matrix R with        DENTLY.”
                      5     −4     1
                  a non singular matrix T to get the encoded matrix Q .
                                          
                               12    5   20
                                          
                               27   21   19                                     NOTE:WehaveusedMatlabformatrixmultiplication.
                                          
                                          
                               27   13    1
                                          
                                          
                               11    5   27
                                          
                             9 20 27                                        3.  CongruenceModuloMethod
                                           1 0 5
                                          
                               19    9   13
                                                     
                  Q=RT=                    2 1 6
                               16   12    5
                                           3 4 0                              Definition 3.1 Let g be a positive integer, we say that
                                          
                               27    9   14
                                                                              miscongruentton(modg)ifg(m-n)wheremandnare
                             4      5   16
                                                                              integers i.e., m = n+sg and s ∈ z, we write
                             5 14 4
                                                                              m≡n(modg)iscalledcongruencerelation,thenumber
                             5 14 20                                          g is the modulus of congruence.[1], [4]
                               12   25   28
                                      82      85    90                        Definition 3.2 Inverseofanintegerl tomodulogisl−1
                                                                                             −1                     −1
                                                                              such that [l.l] ≡1(modg),wherel        is called inverse
                                        126    97   261
                                                                              of l.
                                      56      17   213
                                                       
                                                       
                                        102 113      85
                                                       
                                                       
                                        130 128 165
                                                                              4. Results and discussions
                                      76      61   149
                                  Q=                   
                                      55      32   152
                                                       
                                      87      65   189                        Illustration 4.1 First we are going to assign numbers
                                                                              from 1 to 26 to the 26 alphabets starting from A to
                                      62      69    50
                                                                              Z.Since we are going to use congruence method so let
                                      45      30   109
                                                                              ustakematrixmodulo28. Considerthemessagethatas
                                      93      94   109                        plain text is
                                        146 137 210                             ”LETUSMAKEITSIMPLEINDEPENDENTLY.”
                         VOLUME 16 ISSUE 9                              48-51                                       http://xisdxjxsu.asia/
              Journal of Xi' an Shiyou University, Natural Science Edition                                              ISSN : 1673-064X
                                                                                                                        
                                                                                      1   0  5   11            146             6
                                                                                                                        
                    Alphabet   A     B    C     D     E       F       G               2   1  6   5  mod(28)= 189 mod(28)= 21
                    Number     1     2    3     4     5       6       7               3   4  0   27             53            25
                              -27   -26   -25  -24   -23     -22      -21                                     F
                    Alphabet   H     I    J     K     L       M       N                                        
                                                                                                            = U =FUY
                    Number     8     9    10    11   12      13       14                                       Y
                              -20   -19   -18  -17   -16     -15      -14                                               
                    Alphabet   O     P    Q     R     S       T       U               1   0  5   9             144             4
                    Number     15   16    17    18   19      20       21                                                
                                                                                      2   1  6   20 mod(28)= 200 mod(28)= 4
                              -13   -12   -11  -10    -9      -8      -7              3   4  0   27            107            23
                    Alphabet   V    W     X     Y     Z    spacebar  DOT                                      D
                    Number     22   23    24    25   26      27       0                                     =D=DDW
                               -6   -5    -4    -3    -2      -1      0                                        W
                                                                                                                        
                  Now let us assign the numbers to the above words by                 1   0  5   19             84             0
                                                                                                                        
                  using above table , and we are going to arrange it in               2   1  6   9  mod(28)= 125 mod(28)= 13
                  3×1matrix.                                                          3   4  0   13             93             9
                                                                                                      .
                                                                                                               
                          12            27            27            11                                      = M =.MI
                                                                                                        I
                  LET= 5 ;US= 21 ; MA= 13 ; KE = 5 ;
                          20            19             1            27                                                  
                                                                               1  0   5  16             41            13
                          9             19             16           27                                                  
                                                                               2  1   6  12 mod(28)= 74 mod(28)= 18
                  IT = 20 ;SIM= 9 ;PLE= 12 ;IN= 9 ;                                    3  4   0   5             96            12
                          27            13              5           14                                         
                                                                                                          M
                              4                   5                  5                                      =R=MRL
                                                                                                          L
                  DEP =       5 ; END = 14 ; ENT = 14 ;
                              16                  4                 20                                                  
                                                                                  1   0  5   27             97            13
                          12                          1   0   5                                                         
                                                                                      2   1  6   9  mod(28)= 147 mod(28)= 7
                                                                                  3   4  0   14            117             5
                  LY.= 25 LetthekeymatrixT= 2 1 6                                                              
                           0                        3   4   0                                                M
                                −24    20    −5                                                             =G=MGE
                  andT−1= 18         −15     4                                                               E
                                 5     −4     1                                                                         
                                                                                  1   0  5   4              84             0
                                                                                                                        
                           −24     20    −5                 4   20   23               2   1  6   5  mod(28)= 109 mod(28)= 25
                    −1                                                            3   4  0   16             32             4
                  T    = 18 −15 4 mod(28)= 18 13 4                                                            .
                             5     −4     1                 5   24    1                                        
                  Nowwemultiplied the column vector corresponding to                                        = Y =.YD
                  key matrix,                                                                                  D
                                                        
                    1  0   5  12             112            0     .                                                     
                                                                             1  0   5   5             25            25
                    2  1   6   5  mod(28)= 149 mod(28)= 9 = I =.I.                                                      
                    3  4   0  20             56             0     .                    2  1   6  14 mod(28)= 48 mod(28)= 20
                                                                                       3  4   0   4             71            15
                                                                                                               
                                                                                                                Y
                                                                                                    =T=YTO
                         1  0   5  27             122            10                                             O
                                                           
                         2  1   6  21 mod(28)= 189 mod(28)= 21
                         3  4   0  19             165            25                                                     
                                                                                    1   0  5   5             105            21
                                                                                                                        
                                                  J                                   2   1  6   14 mod(28)= 144 mod(28)= 4
                                                                                    3   4  0   20             71            15
                                              = U =.UY
                                                  Y                                                            
                                                                                                               U
                                                                                                            =D=UDO
                                                                                                               O
                                                                                                                
                         1  0   5  27             32             4                    1   0  5   12             12            12
                                                                                                                
                         2  1   6  13 mod(28)= 73 mod(28)= 17                         2   1  6   25 mod(28)= 49 mod(28)= 21
                         3  4   0   1             133            21                   3   4  0   0             136            24
                                                D                                                           L
                                                                                                             
                                              = Q =DQU                                                      = U =LUX
                                                  U                                                            X
                      VOLUME 16 ISSUE 9                                     48-51                                 http://xisdxjxsu.asia/
               Journal of Xi' an Shiyou University, Natural Science Edition                                         ISSN : 1673-064X
                                                                                                                    
                  Hencethemessagetobesentis,                                       4  20   23   .             4   20  23    0
                                                                                                                    
                  ”.I.JUYDQUFUYDDW.MIMRLMGE.YDYTO                                 18  13   4    Y mod(28)= 18     13   4   25 mod(28)
                  UDOLUX”                                                          5  24   1    D             5   24   1    4
                  Bymultiplying the inverse of key matrix T, receiver can                                   4
                  decrypt the message easily.                                                             =5=DEP
                                                                                                     16
                      4   20  23   .             4   20  23   0                                                     
                                                       
                     18   13   4   I mod(28)= 18     13   4   9 mod(28)            4  20   23   Y             4   20  23   25
                      5   24   1   .             5   24   1   0                                                     
                                                                                  18  13   4    T mod(28)= 18     13   4   21 mod(28)
                                                                                 5  24   1    O             5   24   1   15
                                                12                                                           
                                             =5=LET                                                       5
                                                20                                                        = 14 =END
                                                                                                      4
                     4   20  23    J             4   20  23   10                                                    
                                                       
                    18   13   4   U mod(28)= 18      13  4    21 mod(28)          4   20   23   U             4   20  23   21
                     5   24   1   Y              5   24  1    25                                                    
                                                                                18  13   4    D mod(28)= 18     13   4    4  mod(28)
                                                27                                5   24   1    O             5   24   1   15
                                                                                                           
                                             = 21 =US                                                         5
                                                19                                                           
                                                                                                          = 14 =ENT
                                                                                                      20
                     4   20  23   D              4   20  23   4                                                     
                                                       
                    18   13   4   Q mod(28)= 18      13  4    17 mod(28)          4   20   23   L             4   20  23   12
                     5   24   1   U              5   24  1    21                                                    
                                                                                18  13   4    U mod(28)= 18     13   4   21 mod(28)
                                                27                                5   24   1    X             5   24   1   24
                                                                                                           
                                             = 13 =MA                                                         12
                                                 1                                                           
                                                                                                          = 25 =LY.
                                                                                                      0
                     4   20  23   F              4   20  23   6                 Finally, we decrypyted the original message ”LET US
                                                       
                    18   13   4   U mod(28)= 18      13  4    21 mod(28)        MAKEITSIMPLEINDEPENDENTLY.”
                     5   24   1   Y              5   24  1    25
                                                
                                                11
                                             =5=KE
                                                27                              5. conclusion
                                                       
                     4   20  23   D              4   20  23   4                     This paper introduces the method for sending the
                                                       
                    18   13   4   D mod(28)= 18      13   4   4  mod(28)        secret messages. The key matrix and congruence mod-
                     5   24   1   W              5   24   1   23
                                               9                              ulo should be understood to decrypt the message more
                                                                              securely between the receiver and the sender.
                                             = 20 =IT
                                                27
                                                                        Acknowledgment
                     4   20  23    .             4   20  23   0
                                                       
                    18   13   4   M mod(28)= 18      13   4   13 mod(28)            The corresponding author (Dr.V.Balaji) for finan-
                     5   24   1    I             5   24   1   9
                                                                              cial assistance No.FMRP5766/15(SERO/UGC).
                                                19
                                             =9=SIM
                                                13                              References
                                                       
                     4   20  23   M              4   20  23   13                 [1] W.  Edwin Clark,     Elementary  Number Theory,
                                                       
                    18   13   4   R mod(28)= 18      13   4   18 mod(28)             University of South Florida (2002).
                     5   24   1   L              5   24   1   12                 [2] Koblitz, Algebraic aspects of Cryptography, Springer-
                                                
                                                16                                   Velag, Berlin Heidelberg, Newyork.
                                                
                                             = 12 =PLE                           [3] A. Menzes, P. Van oorschot and S. Vanstoe, Hand book
                                                 5                                   of applied Cryptography, CRC Press, (1997).
                                                                         [4] P. Shanmugam and C. Loganathan, Involuntory Matrix
                     4   20  23   M              4   20  23   13                     in Cryptography, IJRRAS, 6(4)(2011).
                                                       
                    18   13   4   G mod(28)= 18      13   4   7  mod(28)         [5] L.Vinothkumar   and   V.Balaji,   Encryption  and
                     5   24   1   E              5   24   1   5
                                                                                   Decryption Technique Using Matrix Theory, Journal of
                                                27                                   computational Mathematics, vol.3, Issue-2.2019;1-7.
                                             =9=IN
                                                14
                       VOLUME 16 ISSUE 9                                   48-51                                       http://xisdxjxsu.asia/
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...Journal of xi an shiyou university natural science edition issn x atechniqueforencodinganddecodingusing matrixtheory t ranjani a manshath v maheshwari and balaji department mathematics d k m college for women vellore actuarial b s abdur rahman crescent institute technology chennai vels advanced studies sacred heart tirupattur e mail pulibala gmail com abstract in this paper we have disccussed algorithm encryption decryption using matrix theory worked two examples keywords key amssubclass introduction theartandtechnologyofconcealingthemessages to introduce secrecy information safety is diagnosed figure procedure encoding as cryptography the technique disguising message such way on hide its substance coding en crypted cipher text approach turning into plain encryp tion process comprised with value independent plaintext they developed prerequisite decoding theorem strings some length size l can be converted called r n least that results discussions depending upon help suitably chosen nume...

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