(IJACSA) International Journal of Advanced Computer Science and Applications, Vol. 3, No. 12, 2012 A Novel Feistel Cipher Involving a Bunch of Keys Supplemented with XOR Operation V.U.K Sastry K. Anup Kumar Dean R&D, Department of Computer Science and Associate Professor, Department of Computer Science Engineering, Sreenidhi Institute of Science & Tech. and Engineering, SNIST, Hyderabad, India Hyderabad, India Abstract—In this investigation, we have developed a novel block In order to satisfy (1.5), we chose ejk as an odd integer, cipher by modifying classical Feistel cipher. In this, we have used which lies in the interval [1-255], and thus we obtain djk also as a key bunched wherein each key has a multiplicative inverse. The an odd integer lying in the interval [1-255]. cryptanalysis carried out in this investigation clearly shows that this cipher cannot be broken by any attack. Here also we adopt an iterative procedure, and make use of the permutation process that consists of the interchange of the Keywords-encryption; decryption; cryptanalysis; avalanche effect; two halves of the plaintext , of course, represented in the form multiplicative inverse. of a pair of matrices. I. INTRODUCTION In the present analysis, our objective is to modify the Feistel cipher by including a bunch of keys. Here our interest is The study of the Feistel cipher [1-2] laid the foundation for to see how the different keys, occurring in the key bunch, the development of cryptography in the seventies of the last would influence the strength of the cipher. century. In the classical Feistel cipher the block size is 64 bits, and it is divided into two halves wherein each half is containing In what follows, we present the plan of the paper. In section 32 bits. The number of rounds in the iteration process is 16. 2, we introduce the development of the cipher and present the The basic equations governing the Feistel cipher can be written flowcharts and the algorithms corresponding to the cipher, in in the form section 3, we illustrate the cipher with an example and examine the avalanche effect. After that, we carry out the cryptanalysis Pi = Qi-1, (1.1) i i-1 i-1 i in section 4. Finally we present numerical computation and Q = P F (Q , K ), (1.2) draw conclusions. and II. DEVELOPMENT OF THE CIPHER Qi-1 = Pi, (1.3) i-1 i i i P = Q F (P , K ), (1.4) 2 i i th We consider a plaintext P containing 2m characters. On where P and Q are the blocks of the plaintext in the i using the EBCIDIC code this is written in terms of numbers round of the iteration process, F is a function chosen i th which are in the interval [0-255]. Now we write this in the form appropriately, and K is the key in the i round. In this analysis, of a pair of square matrices P0 and Q0, wherein each matrix is the XOR operation and the permutation that is performed by of size m. interchanging two halves of the data in the iteration process play a vital role in deciding the strength of the cipher. The basic equations governing the encryption of this block cipher are given by In the recent years, Sastry et al. [3-12] have offered several modifications to the Feistel cipher, and have studied various i i-1 [ Pjk ]= [ ejk Qjk ] mod 256, (2.1) aspects of this cipher, including different types of permutations and and substitutions. In all these investigations we have divided i i-1 i-1 the plaintext into a pair of matrices of equal size, and the key is [ Q ]= [e P ] mod 256 [Q ] , (2.2) taken in the form of a matrix. jk jk jk jk where j= 1 to m, k = 1 to m and i =1 to n, in which n is In the process of encryption, we take the key bunch as E, the number of rounds. and represent it in the form of a matrix given by E = [ejk]. The corresponding key bunch in the process of decryption is taken The corresponding equations of decryption are in the form, i-1 i as D = [djk]. Here for a given value of the key ejk, used in the [ Qjk ]= [ djk Pjk ] mod 256, (2.3) encryption, we determine the corresponding key djk, by using and the relation i-1 i i-1 [ Pjk ]=[djk( [Qjk ] [Qjk ] ) ] mod 256 (2.4) ( ejk x djk ) mod 256 =1, (1.5) where dkl is the multiplicative inverse of ekl, where j= 1 to m, k = 1 to m and i = n to 1, 92 | P a g e www.ijacsa.thesai.org (IJACSA) International Journal of Advanced Computer Science and Applications, Vol. 3, No. 12, 2012 i i th th here Pjk and Qjk are the j row k column elements of the left and right portions of the plaintext matrix, respectively, in Read Ciphertext C th the i round of the iteration process. and Key D On using the basic relations (2.1) - (2.4), governing the encryption and the decryption, the corresponding flowcharts for the encryption and the decryption can be written as shown Pn Qn below. Read Plaintext P and Key E for i = n to 1 for j =1 to m for k = 1 to m P0 Q0 i Pjk i for i = 1 to n Qjk i for j = 1 to m [djk Pjk ] mod 256 for k = 1 to m i-1 Pjk i-1 Q i-1 i-1 jk [e jk Pjk ] mod 256 [Qjk ] i-1 Q jk i i-1 [djk( [Qjk ] [Qjk ] ) ] mod 256 i-1 Pjk i-1 [ ejk Qjk ] mod 256 i i i P , Q Pjk i Q 0 0 jk P = P || Q Figure 2. The process of Decryption The algorithms for the encryption and the decryption are i i written as shown below. P , Q n n A. Algorithm for Encryption C = P || Q 1. Read P, E, and n 0 Figure 1. The Process of Encryption 2. P = Left half of P. Q0 = Right half of P. 3. for i = 1 to n 93 | P a g e www.ijacsa.thesai.org (IJACSA) International Journal of Advanced Computer Science and Applications, Vol. 3, No. 12, 2012 begin for j = 1 to m 066 114 111 116 begin 032 087 104 101 for k = 1 to m 0 begin P = (3.3) i i-1 032 119 101 114 [ Pjk ]= [ ejk Qjk ] mod 256, i i-1 i-1 [ Qjk ]= [ejk Pjk ] mod 256 [Qjk ] , 114 121 032 112 end end and end 6. C = Pn Qn /* represents concatenation */ 104 101 114 033 7. Write(C) 110 032 119 101 0 B. Algorithm for Decryption Q = (3.4) 1. Read C, D, and n. 101 032 118 101 n 2. P = Left half of C Qn = Right half of C 111 111 114 044 3. for i = n to 1 begin Let us take the encryption key bunch matrix E in the form for j = 1 to m 071 053 011 061 begin for k = 1 to m begin 117 069 057 051 i-1 i E = (3.5) [Qjk ] = [ djk Pjk ] mod 256, 121 139 101 043 i-1 i i-1 [Pjk ]=[djk ([Qjk ] [Qjk ]] mod 256 099 095 111 035 end and end end 6. P = P0 Q0 /* represents concatenation */ 119 029 163 021 7. Write (P) 221 141 009 251 In what follows we illustrate the cipher with a suitable example. D = (3.6) 201 035 109 131 III. ILLUSTRATION OF THE CIPHER Consider the plaintext given below. 075 159 143 139 Brother! When we were very poor, by looking at some On using the algorithm for encryption, given in section 2, corrupt politicians and employees who earned crores and we get the ciphertext C in the form crores, we use to think how to come up in life. Though you were a graduate having technical skills, you joined naxalites 088 166 064 218 222 060 064 088 thinking that only unethical rich are totally responsible for the ruination of our country. After the death of our father, I joined 140 078 014 104 028 204 176 036 as a police, our uncle started liquor business! He has earned C = (3.7) crores and crores. Though I have become a police inspector, I 088 094 002 182 244 188 202 108 am helpless. I am not able to control anything! I do not know when India will change! Write a letter. Do come back. (3.1) 240 120 038 118 208 224 146 196 Let us focus our attention on the first 32 characters of the plaintext (3.1). This is given by On using the keys in D, and applying the decryption algorithm on (3.6), we get back the original plaintext P given Brother! When we were very poor, (3.2) by (3.2) On using EBCIDIC code, we write (3.2) in the form of a Let us now study the avalanche effect. On changing the first pair of square matrices given by row , second column element of P0 from 114 to 115, we get a change of one binary bit in the plaintext. 94 | P a g e www.ijacsa.thesai.org (IJACSA) International Journal of Advanced Computer Science and Applications, Vol. 3, No. 12, 2012 On applying the encryption algorithm on this modified 2 (2.4m ) -7 plaintext, using the same key bunch matrix E, we get the 2 ciphertext C in the form 10 x 10 (2.4 m -15) ---------------------- years = 3.12 x 10 years 104 144 028 204 176 122 222 228 365 x 24 x 60 x 60 172 244 236 162 210 024 158 030 In this analysis, we have taken m=4.