Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 143 https://internationalpubls.com Encryption and Decryption using Hill Cipher and Integral Transforms Gobburi Rekha1, V. Srinivas2 1 Department of Humanities and Sciences, Malla Reddy College of Engineering and Technology 2 Department of Mathematics, Osmania university Article History: Received: 26-05-2024 Revised: 16-07-2024 Accepted: 27-07-2024 Abstract: We know that Hill Cipher is one of the regular methods that is used in cryptography for encryption and decryption of data using modular arithmetic under different modulus. Moreover, we have seen results using some integral transforms like Laplace transforms, Aboodh transforms and many more. This paper extracts the result of dual encryption and decryption with a combination of Hill Cipher and few integral transforms. Introduction: In many circumstances, the sender wants to keep the message private from the public or from unauthorized users. Here, we combine Hill Cipher with Laplace and Kamal transforms to encrypt the message and apply their inverse transforms to decrypt it. In contrast to cipher text, which is a coded version of plaintext, the original message written by the user is referred to as plaintext. Objectives: In this paper, we focus on two stages of encryption and two stages of decryption under modulo 255. Methods: Using Hill Cipher with different transformation techniques, we have encrypted and decrypted the data. One is Hill Cipher with Laplace transform technique while the other is Hill Cipher with Kamal transform technique. Results: In this paper, we see that the original text is transformed to cipher text which could be retrievable using some techniques. Conclusions: This is a novel method that uses dual techniques to safeguard data: first, it encrypts the encrypted cipher text, and then it decrypts it again. The data is protected all along the way because the key that is transmitted to the recipient is very big and difficult to crack. Keywords: Encryption, Decryption, Hill cipher, Laplace transform, Inverse Laplace transform, Kamal transform and Inverse Kamal transform. 1. Introduction As security of information plays a crucial role, we discuss a dual encryption and decryption of data using Hill cipher with combination of Laplace transform and Kamal transform. The study of encrypted messages is known as cryptography. In many circumstances, the sender wants to keep the message private from the public or from unauthorized users. Information is protected using encryption, and the original message is unlocked using decryption. Integral transformations have a wide range of uses, especially in the realm of cryptography. In cryptography, hyperbolic functions such as sine, cosine, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 144 https://internationalpubls.com exponential, polynomial, etc. are transformed using the Laplace, Elzaki, and Kamal methods. While Elzaki transform is used by [2] and Kamal Transform [9], [5, 6], [1] employ Laplace transform. Here, we combine Hill Cipher with Laplace and Kamal transforms to encrypt the message and apply their inverse transforms to decrypt it. In contrast to cipher text, which is a coded version of plaintext, the original message written by the user is referred to as plaintext. Encryption, also known as the process of rendering information unintelligible, decryption, or the process of creating cipher, is referred to as cryptography. There are two components to the encryption and decryption process: the algorithm and the key. The use of the key for encryption and decryption makes cryptography safer. We need the following well-known formulas because in this study we will cover the uses of Laplace transforms and Kamal transform. 2. Definitions Laplace Transform: If 𝑓(𝑑) is a function defined for all positive values of t, then the Laplace Transform of 𝑓(𝑑) is defined as: 𝐿 {𝑓(𝑑)} =𝐹(𝑠) =∫ e ∞ 0 βˆ’π‘ π‘‘ 𝑓(𝑑)𝑑𝑑, provided that the integral exists. Here the parameter S is a real or complex number. The corresponding inverse Laplace transform is πΏβˆ’1{𝐹(𝑠)} = 𝑓(𝑑). Some standard results of Laplace transform 𝐿 {1} = 1 s πΏβˆ’1 { 1 s } = 1 𝐿{𝑑 𝑛} = n! s n+1 πΏβˆ’1 { 1 s n+1 } = tn n! 𝐿{π‘’π‘Žπ‘‘} = 1 s βˆ’a πΏβˆ’1 { 1 sβˆ’ a } = π‘’π‘Žπ‘‘ 𝐿 {𝑑 𝑛𝑓(𝑑)} = (βˆ’1) 𝑛 d n ds n 𝐹(𝑠) πΏβˆ’1 { d n ds n 𝐹(𝑠)} = tn𝑓(𝑑) 𝑑𝑠 Kamal Transform: The Kamal transforms defined by the integral equation: 𝐾[𝑓(𝑑)] = 𝐺(𝑣) = ∫ f(t) ∞ 0 e βˆ’t v 𝑑𝑑, 𝑑 β‰₯ 0, π‘˜1 ≀ 𝑣 ≀ π‘˜2, where π‘˜1, π‘˜2 may be finite or infinite. Some standard results of Kamal transform 𝐾 [1] = 𝑣 πΎβˆ’1[𝑣] = 1 𝐾[𝑑] = 𝑣2 πΎβˆ’1[𝑣2] = 𝑑 𝐾[𝑑 𝑛] = 𝑛! 𝑣𝑛+1 πΎβˆ’1[𝑣 𝑛+1] = t n n! 𝐾[π‘‘π‘’π‘Žπ‘‘] = v 2 (1 βˆ’ av) 2 πΎβˆ’1 [ v 2 (1 βˆ’ av) 2 ] = π‘‘π‘’π‘Žπ‘‘ 3. Methods In this we discuss how to encrypt the original text into cipher text using the following steps. Encryption Algorithm: Step 1: Select the plain text and convert the text into numbers by assigning the values as A=1, B=2…. Z=26, Space=27, a=28, b=29, z=53.Arrange the plain text which transformed in numbers in matrix form of order nxn or mxn. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 145 https://internationalpubls.com Step 2: Let the matrix be denoted by P. Step 3: Define a key matrix whose inverse exists. Let A be the key matrix of order nxn which acts as encryption key and the inverse of the matrix A acts as decryption key. Step 4: Let us define 𝐸 ≑ 𝐴𝑃(π‘šπ‘œπ‘‘ 53) which is the first stage of encryption. Step 5: Let us define the function f(t) = βˆ‘ Gi n i=0 t2cosht , where Gi are constants.The values in E are taken as the coefficients in the function. Step 6: Apply Laplace transform for the function f(t). That is, 𝐿{𝑓(𝑑)} = 𝐺(𝑠) which is the second level of encryption and adjust the coefficients in G(s) under modulo 255 and transform them into symbols using the ASCII values. Hence, the sender sends the cipher text along with the key π‘Ÿi, where π‘Ÿi= (Fi – π‘ži) / 255. Now, we discuss how to decrypt the cipher text to original text using the following steps. Decryption Algorithm: Step 1: Transform the cipher text into numbers using the ASCII values. Now, using the cipher text which are transformed into numbers and the key we find the coefficients π‘ži using, qi = Fi + 255ri. Hence, G(s) = βˆ‘ qi s2i+3 n i=0 . Step 2: Apply inverse Laplace transform for G(s) on both sides. That is, Lβˆ’1{G(s)} = Lβˆ’1 {βˆ‘ qi si+3 n i=0 } which is the first level of decryption. Hence, we get Lβˆ’1{G(s)} = f(t). Step 3: Now, write the coefficients of f(t) in matrix form say E. Step 4: Let D ≑ Aβˆ’1E(mod 53) which is the second level of decryption. Step 5: Hence, we get D which is same as that of P. Therefore, on converting the values of the matrix D into symbols we get original plain text. 4. Results To see how the algorithm works let us consider the following illustrations. Example I. This illustration explains about Hill Cipher using Laplace integral transform. Step 1. Let the plain text to be sent is Environment Day. Now convert the plain text into numbers by assigning A=1, B=2…. Z=26, space = 27, a=28, b=29…. z=53. Therefore, the plaintext in the form of numbers is 5 41 49 36 45 42 41 40 32 41 47 27 4 28 52. Step 2: Arrange the numbers in matrix form as 𝑃 = ( 5 41 49 42 41 40 47 27 4 36 32 28 45 41 52 ). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 146 https://internationalpubls.com Step 3: Let A = ( 3 βˆ’3 4 2 βˆ’3 4 0 βˆ’1 1 ) be the encryption key and its inverse π΄βˆ’1 = ( 1 βˆ’1 0 βˆ’2 3 βˆ’4 βˆ’2 3 βˆ’3 ) be the decryption key. Step 4: I stage of encryption Let E ≑ AP(mod 53). ≑ ( 3 βˆ’3 4 2 βˆ’3 4 0 βˆ’1 1 ) ( 5 41 49 42 41 40 47 27 4 36 32 28 45 41 52 ) (mod 53). E ≑ ( 77 108 43 72 67 βˆ’6 5 βˆ’14 βˆ’36 124 88 βˆ’4 220 175 11 ) (mod 53). E = ( πŸπŸ’ 𝟐 πŸ’πŸ‘ πŸπŸ— πŸπŸ’ πŸ’πŸ• πŸ“ πŸ‘πŸ— πŸπŸ• πŸπŸ– πŸ‘πŸ“ πŸ’πŸ— πŸ– πŸπŸ” 𝟏𝟏 ). Step 5: II stage of encryption Now for the matrix E let us apply Laplace transform. Let 𝑓(𝑑) = βˆ‘ 𝑑2𝐺𝑖 14 𝑖=0 π‘π‘œπ‘ β„Žπ‘‘ , where π‘π‘œπ‘ β„Žπ‘‘ = 1 + 𝑑2 2! + 𝑑4 4! + 𝑑6 6! + 𝑑8 8! + β‹― … … Here, n=14. Therefore, π‘π‘œπ‘ β„Žπ‘‘ = 1 + 𝑑2 2! + 𝑑4 4! + 𝑑6 6! + 𝑑8 8! + 𝑑10 10! + 𝑑12 12! + 𝑑14 14! + 𝑑16 16! + 𝑑18 18! + 𝑑20 20! + 𝑑22 22! + 𝑑24 24! + 𝑑26 26! + 𝑑28 28! . 𝑓(𝑑) = 𝐺0 + 𝐺1 𝑑2 2! + 𝐺2 𝑑4 4! + 𝐺3 𝑑6 6! + 𝐺4 𝑑8 8! + 𝐺5 𝑑10 10! + 𝐺6 𝑑12 12! + 𝐺7 𝑑14 14! + 𝐺8 𝑑16 16! + 𝐺9 𝑑18 18! + 𝐺10 𝑑20 20! + 𝐺11 𝑑22 22! + 𝐺12 𝑑24 24! + 𝐺13 𝑑26 26! + 𝐺14 𝑑28 28! Where 𝐺0 = 24, 𝐺1 = 2, 𝐺2 = 43, 𝐺3 = 18, 𝐺4 = 8 , 𝐺5 = 19, 𝐺6 = 14, 𝐺7 = 47, 𝐺8 = 35, 𝐺9 = 16, 𝐺10 = 5, 𝐺11 = 39, 𝐺12 = 17, 𝐺13 = 49, 𝐺14 = 11. 𝑓(𝑑) = 𝑑2 [24 + 2 𝑑2 2! + 43 𝑑4 4! + 18 𝑑6 6! + 8 𝑑8 8! + 19 𝑑10 10! + 14 𝑑12 12! + 47 𝑑14 14! + 35 𝑑16 16! + 16 𝑑18 18! + 5 𝑑20 20! + 39 𝑑22 22! + 17 𝑑24 24! + 49 𝑑26 26! + 11 𝑑28 28! ]. 𝑓(𝑑) = [24𝑑2 + 2 𝑑4 2! + 43 𝑑6 4! + 18 𝑑8 6! + 8 𝑑10 8! + 19 𝑑12 10! + 14 𝑑14 12! + 47 𝑑16 14! + 35 𝑑18 16! + 16 𝑑20 18! + 5 𝑑22 20! + 39 𝑑24 22! + 17 𝑑26 24! + 49 𝑑28 26! + 11 𝑑30 28! ]. Apply Laplace transform for f(t). 𝐿{𝑓(𝑑)} = 𝐿 {24𝑑2 + 2 𝑑4 2! + 43 𝑑6 4! + 18 𝑑8 6! + 8 𝑑10 8! + 19 𝑑12 10! + 14 𝑑14 12! + 47 𝑑16 14! + 35 𝑑18 16! + 16 𝑑20 18! + 5 𝑑22 20! + 39 𝑑24 22! + 17 𝑑26 24! + 49 𝑑28 26! + 11 𝑑30 28! }. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 6s (2024) 147 https://internationalpubls.com 𝐿{𝑓(𝑑)} = 24 2! 𝑠3 + 2 4! 2!𝑠5 + 43 6! 4!𝑠7 + 18 8! 6!𝑠9 + 8 10! 8!𝑠11 + 19 12! 10!𝑠13 + 14 14! 12!𝑠15 +47 16! 14!𝑠17 + 35 18! 16!𝑠19 + 16 20! 18!𝑠21 + 5 22! 20!𝑠23 + 39 24! 22!𝑠25 + 17 26! 24!𝑠27 + 49 28! 26!𝑠29 + 11 30! 28!𝑠31 . 𝐿{𝑓(𝑑)} = 48 𝑠3 + 24 𝑠5 + 1290 𝑠7 + 1008 𝑠9 + 720 𝑠11 + 2508 𝑠13 + 2548 𝑠15 + 11280 𝑠17 + 10710 𝑠19 + 6080 𝑠21 + 2310 𝑠23 + 21528 𝑠25 + 11050 𝑠27 + 37044 𝑠29 + 9570 𝑠31 . Adjusting the values under modulo 255, we get 48 ≑ 48 mod 255 24 ≑ 24 mod 255 1290 ≑ 15 mod 255 1008 ≑ 243 mod 255 720 ≑ 210 mod 255 2508 ≑ 213 mod 255 2548 ≑ 253 mod 255 11280 ≑ 60 mod 255 10710 ≑ 0 mod 255 6080 ≑ 215 mod 255 2310 ≑ 15 mod 255 21528 ≑ 108 mod 255 11050 ≑ 85 mod 255 37044 ≑ 69 mod 255 9570 ≑ 135 mod 255 And the keys are 0,0,5,3,2,9,9,44,42,23,9,84,43,145,37. Hence, the receiver receives the cipher text 0CANSIóÒÕý