CHAPTER 12 CRYPTOGRAPHY OF A GRAY LEVEL IMAGE USING A MODIFIED HILL CIPHER

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177 CHAPTER 12 CRYPTOGRAPHY OF A GRAY LEVEL IMAGE USING A MODIFIED HILL CIPHER

178 12.1 Introduction The study of cryptography of gray level images [110, 112, 118] by using block ciphers has gained considerable impetus in the recent years. The transformation of an image from its original form to some other form, such that it cannot be deciphered what it is, is really an interesting one. In a recent investigation [125, 128], we have developed two large block ciphers by modifying the Hill cipher [5]. In these ciphers, the key is of size 512 bits and the plaintext is of size 2048 bits. In one of the chapters [112], the plaintext matrix is multiplied by the key on one side and by its modular arithmetic inverse on the other side. From the cryptanalysis and the avalanche effect, we have noticed that the cipher is a strong one and it cannot be broken by any cryptanalytic attack. In the present chapter, our objective is to develop a block cipher, and to use it for the cryptography of a gray level image. Here, we have taken a key containing 64 decimal numbers (as it was in [125]), and generated a key matrix of size 32 x 32 by extending the key in a special manner (discussed later), and applied it in the cryptography of a gray level image. In Section 2, we have developed a procedure for the cryptography of a gray level image. In Section 3, we have used an example and illustrated the process. Finally, in Section 4, we have drawn conclusions from the analysis.

179 12.2 Development of a Procedure for the Cryptography of a Gray level image Consider a gray level image whose gray level values can be represented in the form of a matrix given by P = [Pij], i = 1 to n, j = 1 to n. (2.1) Here, each Pij lies between 0-255. Let us choose a key k let it be represented in the form of a matrix given by K = [Kij], i = 1 to n, j = 1 to n, (2.2) where each Kij is in the interval [0, 255]. Let C = [Cij], i = 1 to n, j = 1 to n (2.3) be a matrix, obtained on encryption.

180 The process of encryption and the process of decryption, which are quite suitable, for the problem on hand, are given in Fig. 1. Here, Mix ( ) is a function used for mixing thoroughly the decimal numbers (on converting them into binary bits) arising in the process of encryption at each stage of iteration. IMix ( ) is a function which represents the reverse process of Mix ( ). For a detailed discussion of these functions, and the algorithms involved in the processes of encryption and decryption, we refer to [125].

181 12.3 Illustration of the cryptography of the image Let us choose a key Q consisting of 64 numbers. This can be written in the form of a matrix given by 175 173 27 65 32 65 17 76 232 84 72 69 32 185 69 82 27 179 102 33 83 97 73 32 Q = 65 84 143 69 105 153 213 163 (3.1) 184 28 49 5 69 31 166 109 208 185 77 234 207 171 71 80 237 249 101 57 95 191 37 132 127 107 32 85 117 254 165 87 The length of the secret key (which is to be transmitted) is 512 bits. On using this key, we can generate a new key E in the form Q R E = (3.2) S U where U = Q T, in which T denotes the transpose of a matrix, and R and S are obtained from Q and U as follows. On interchanging the 1 st row and the 8 th row of Q, the 2 nd row and the 7 th row of Q, etc., we get R. Similarly, we obtain S from U. Thus, we have 175 173 27 65 32 65 17 76 127 107 32 85 117 254 165 87 232 84 72 69 32 185 69 82 237 249 101 57 95 191 37 132 27 179 102 33 83 97 73 32 208 185 77 234 207 171 71 80 65 84 143 69 105 153 213 163 184 28 49 5 69 31 166 109 184 28 49 5 69 31 166 109 65 84 143 69 105 153 213 163 208 185 77 234 207 171 71 80 27 179 102 33 83 97 73 32 237 249 101 57 95 191 37 132 232 84 72 69 32 185 69 82 E = 127 107 32 85 117 254 165 87 175 173 27 65 32 65 17 76 (3.3) 76 82 32 163 109 80 132 87 175 232 27 65 184 208 237 127 17 69 73 213 166 71 37 165 173 84 179 84 28 185 249 107 65 185 97 153 31 171 191 254 27 72 102 143 49 77 101 32 32 32 83 105 69 207 95 117 65 69 33 69 5 234 57 85 65 69 33 69 5 234 57 85 32 32 83 105 69 207 95 117 27 72 102 143 49 77 101 32 65 185 97 153 31 171 191 254 173 84 179 84 28 185 249 107 17 69 73 213 166 71 37 165 175 232 27 65 184 208 237 127 76 82 32 163 109 80 132 87

182 The size of this matrix is 16 x 16. This can be further extended to a matrix L of size 32 x 32 where E F L = (3.4) G H where H = E T, in which T denotes the transpose of a matrix, and F and G are obtained from E and H as follows. On interchanging the 1 st row and the 16 th row of E, the 2 nd row and the 15 th row of E, etc., we get F. Similarly, we obtain G from H. Thus, we have L. Here, 175 232 27 65 184 208 237 127 76 82 32 163 109 80 132 87 173 84 179 84 28 185 249 107 17 69 73 213 166 71 37 165 27 72 102 143 49 77 101 32 65 185 97 153 31 171 191 254 65 69 33 69 5 234 57 85 32 32 83 105 69 207 95 117 32 32 83 105 69 207 95 117 65 69 33 69 5 234 57 85 65 185 97 153 31 171 191 254 27 72 102 143 49 77 101 32 17 69 73 213 166 71 37 165 173 84 179 84 28 185 249 107 F = 76 82 32 163 109 80 132 87 175 232 27 65 184 208 237 127 (3.5) 127 107 32 85 117 254 165 87 175 173 27 65 32 65 17 76 237 249 101 57 95 191 37 132 232 84 72 69 32 185 69 82 208 185 77 234 207 171 71 80 27 179 102 33 83 97 73 32 184 28 49 5 69 31 166 109 65 84 143 69 105 153 213 163 65 84 143 69 105 153 213 163 184 28 49 5 69 31 166 109 27 179 102 33 83 97 73 32 208 185 77 234 207 171 71 80 232 84 72 69 32 185 69 82 237 249 101 57 95 191 37 132 175 173 27 65 32 65 17 76 127 107 32 85 117 254 165 87 87 132 80 109 163 32 82 76 127 107 32 85 117 254 165 87 165 37 71 166 213 73 69 17 237 249 101 57 95 191 37 132 254 191 171 31 153 97 185 65 208 185 77 234 207 171 71 80 117 95 207 69 105 83 32 32 184 28 49 5 69 31 166 109 85 57 234 5 69 33 69 65 65 84 143 69 105 153 213 163 32 101 77 49 143 102 72 27 27 179 102 33 83 97 73 32 107 249 185 28 84 179 84 173 232 84 72 69 32 185 69 82 G = 127 237 208 184 65 27 232 175 175 173 27 65 32 65 17 76 (3.6) 76 82 32 163 109 80 132 87 87 165 254 117 85 32 107 127 17 69 73 213 166 71 37 165 132 37 191 95 57 101 249 237 65 185 97 153 31 171 191 254 80 71 171 207 234 77 185 208 32 32 83 105 69 207 95 117 109 166 31 69 5 49 28 184 65 69 33 69 5 234 57 85 163 213 153 105 69 143 84 65 27 72 102 143 49 77 101 32 32 73 97 83 33 102 179 27 173 84 179 84 28 185 249 107 82 69 185 32 69 72 84 232 175 232 27 65 184 208 237 127 76 17 65 32 65 27 173 175

183 175 232 27 65 184 208 237 127 76 17 65 32 65 27 173 175 173 84 179 84 28 185 249 107 82 69 185 32 69 72 84 232 27 72 102 143 49 77 101 32 32 73 97 83 33 102 179 27 65 69 33 69 5 234 57 85 163 213 153 105 69 143 84 65 32 32 83 105 69 207 95 117 109 166 31 69 5 49 28 184 65 185 97 153 31 171 191 254 80 71 171 207 234 77 185 208 17 69 73 213 166 71 37 165 132 37 191 95 57 101 249 237 H = 76 82 32 163 109 80 132 87 87 165 254 117 85 32 107 127 (3.7) 127 237 208 184 65 27 232 175 175 173 27 65 32 65 17 76 107 249 185 28 84 179 84 173 232 84 72 69 32 185 69 82 32 101 77 49 143 102 72 27 27 179 102 33 83 97 73 32 85 57 234 5 69 33 69 65 65 84 143 69 105 153 213 163 117 95 207 69 105 83 32 32 184 28 49 5 69 31 166 109 254 191 171 31 153 97 185 65 208 185 77 234 207 171 71 80 165 37 71 166 213 73 69 17 237 249 101 57 95 191 37 132 87 132 80 109 163 32 82 76 127 107 32 85 117 254 165 87 Now we obtain the key matrix K given by E F K = (3.8) G J where J is obtained from H by rotating, circularly, two rows in the downward direction. 165 37 71 166 213 73 69 17 237 249 101 57 95 191 37 132 87 132 80 109 163 32 82 76 127 107 32 85 117 254 165 87 175 232 27 65 184 208 237 127 76 17 65 32 65 27 173 175 173 84 179 84 28 185 249 107 82 69 185 32 69 72 84 232 27 72 102 143 49 77 101 32 32 73 97 83 33 102 179 27 65 69 33 69 5 234 57 85 163 213 153 105 69 143 84 65 32 32 83 105 69 207 95 117 109 166 31 69 5 49 28 184 J = 65 185 97 153 31 171 191 254 80 71 171 207 234 77 185 208 (3.9) 17 69 73 213 166 71 37 165 132 37 191 95 57 101 249 237 76 82 32 163 109 80 132 87 87 165 254 117 85 32 107 127 127 237 208 184 65 27 232 175 175 173 27 65 32 65 17 76 107 249 185 28 84 179 84 173 232 84 72 69 32 185 69 82 32 101 77 49 143 102 72 27 27 179 102 33 83 97 73 32 85 57 234 5 69 33 69 65 65 84 143 69 105 153 213 163 117 95 207 69 105 83 32 32 184 28 49 5 69 31 166 109 254 191 171 31 153 97 185 65 208 185 77 234 207 171 71 80

184 The afore mentioned operations are performed for 1) enhancing the size of the key matrix to 32 x 32, and 2) obtaining the modular arithmetic inverse of K, in a trial and error manner. The modular arithmetic inverse of K is obtained as W X K 1 = (3.10) Y Z 97 88 109 254 81 212 23 61 171 225 56 38 244 234 62 169 163 152 204 236 138 46 91 44 239 129 54 97 146 219 207 35 191 33 253 162 131 177 149 144 119 106 196 208 240 35 109 99 116 73 140 191 39 144 174 0 38 143 149 250 83 43 214 62 102 190 142 112 91 139 39 172 220 109 172 198 153 20 227 87 56 69 34 173 252 79 231 38 127 155 34 138 18 208 162 43 245 27 141 216 146 40 156 144 213 241 76 84 1 20 219 202 W = 186 67 187 217 50 141 145 53 99 35 175 130 85 140 21 31 (3.11) 111 77 205 92 13 216 16 22 116 49 8 189 31 38 76 151 141 32 184 231 200 206 74 6 2 143 176 40 7 100 197 39 125 165 91 231 75 190 100 133 64 37 175 16 22 28 100 13 250 115 35 31 94 15 93 108 201 74 131 239 226 248 251 32 172 139 42 148 147 36 220 34 69 44 82 125 76 13 72 77 194 93 235 181 204 78 211 4 103 221 152 225 138 254 0 199 183 130 210 163 184 130 137 211 143 227 223 15 131 173 132 250 225 243 59 98 153 232 69 160 228 127 220 90 54 137 74 235 219 192 195 170 30 50 17 151 40 145 95 57 156 150 21 15 125 167 65 69 87 105 228 197 234 192 32 23 28 74 220 163 8 124 128 156 223 249 6 106 186 61 158 70 121 167 199 243 6 134 102 135 166 194 154 193 124 221 65 97 17 152 169 211 236 112 192 224 165 65 225 236 250 28 213 92 156 164 208 118 123 133 198 100 8 82 220 98 242 178 97 85 59 233 92 55 147 110 166 149 200 22 116 98 80 250 125 158 6 218 99 141 X = 20 36 15 49 218 152 42 106 4 185 203 136 153 148 192 47 (3.12) 180 61 144 158 140 12 23 193 250 167 252 144 162 120 46 171 197 118 54 100 241 21 59 216 42 100 38 242 196 67 85 41 190 193 105 40 54 201 222 45 211 169 139 217 115 65 158 59 43 88 200 67 154 142 142 29 119 226 219 176 73 13 238 223 56 56 61 39 0 71 208 116 140 19 166 240 156 119 170 217 128 232 66 69 105 79 170 52 219 179 129 177 206 73 236 214 223 25 30 23 37 99 208 135 81 107 111 98 13 143 152 85 76 147 194 28 95 76 199 134 46 27 183 120 25 162 218 190

185 169 62 234 244 38 56 225 171 61 23 212 81 254 109 88 97 35 207 219 146 97 54 129 239 44 91 46 138 236 204 152 163 99 109 35 240 208 196 106 119 144 149 177 131 162 253 33 191 62 214 43 83 250 149 143 38 0 174 144 39 191 140 73 116 87 227 20 153 198 172 109 220 172 39 139 91 112 142 190 102 43 162 208 18 138 34 155 127 38 231 79 252 173 34 69 56 202 219 20 1 84 76 241 213 144 156 40 146 216 141 27 245 Y = 31 21 140 85 130 175 35 99 53 145 141 50 217 187 67 186 (3.13) 151 76 38 31 189 8 49 116 22 16 216 13 92 205 77 111 39 197 100 7 40 176 143 2 6 74 206 200 231 184 32 141 13 100 28 22 16 175 37 64 133 100 190 75 231 91 165 125 32 251 248 226 239 131 74 201 108 93 15 94 31 35 115 250 77 72 13 76 125 82 44 69 34 220 36 147 148 42 139 172 199 0 254 138 225 152 221 103 4 211 78 204 181 235 93 194 250 132 173 131 15 223 227 143 211 137 130 184 163 210 130 183 235 74 137 54 90 220 127 228 160 69 232 153 98 59 243 225 192 219 15 21 150 156 57 95 145 40 151 17 50 30 170 195 167 125 163 220 74 28 23 32 192 234 197 228 105 87 69 65 124 8 243 199 167 121 70 158 61 186 106 6 249 223 156 128 134 6 211 169 152 17 97 65 221 124 193 154 194 166 135 102 112 236 118 208 164 156 92 213 28 250 236 225 65 165 224 192 133 123 55 92 233 59 85 97 178 242 98 220 82 8 100 198 110 147 141 99 218 6 158 125 250 80 98 116 22 200 149 166 Z = 36 20 47 192 148 153 136 203 185 4 106 42 152 218 49 15 (3.14) 61 180 171 46 120 162 144 252 167 250 193 23 12 140 158 144 118 197 41 85 67 196 242 38 100 42 216 59 21 241 100 54 193 190 59 158 65 115 217 139 169 211 45 222 201 54 40 105 88 43 223 238 13 73 176 219 226 119 29 142 142 154 67 200 56 56 217 170 119 156 240 166 19 140 116 208 71 0 39 61 232 128 214 236 73 206 177 129 179 219 52 170 79 105 69 66 25 223 85 152 143 13 98 111 107 81 135 208 99 37 23 30 147 76 190 218 162 25 120 183 27 46 134 199 76 95 28 194 From (3.8) and (3.10), we can readily find that K K 1 mod 256 = K 1 K mod 256 = I. (3.15) Let us consider the image of a hand, which is given below. Fig. 2. Image of a Hand

186 This image can be represented in the form of a binary matrix P given by 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 1 0 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 0 0 0 P = 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 (3.16) 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 where 1 denotes black and 0 denotes white.

187 On adopting the iterative procedure given in Fig. 1, we get the encrypted image C 1 0 0 1 0 0 1 1 1 1 1 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 1 0 1 0 1 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 1 1 0 1 1 0 0 0 0 0 1 1 1 1 0 0 0 1 0 0 0 1 1 1 0 1 1 1 1 1 1 0 0 1 0 0 1 1 0 0 1 0 0 1 0 0 0 0 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 1 0 1 0 0 1 1 0 1 1 0 1 1 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1 1 0 0 1 1 0 1 0 0 1 0 0 0 1 1 1 0 0 1 1 1 0 1 0 1 1 0 1 0 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 0 0 0 1 0 1 1 0 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 1 0 0 1 0 0 0 0 1 0 0 0 1 0 1 1 1 1 1 0 1 1 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 0 0 0 1 1 0 1 1 1 1 1 0 0 1 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 1 0 0 1 1 1 0 0 1 1 0 1 1 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1 1 0 0 1 1 1 1 0 0 1 1 0 1 1 1 0 0 1 0 1 0 0 0 1 1 1 0 1 0 0 0 C = 0 0 1 0 1 0 0 1 1 0 0 1 0 0 0 1 1 0 0 0 1 1 0 1 1 1 1 0 0 1 0 1 (3.17) 1 1 1 0 1 0 1 1 0 0 1 0 1 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 0 1 0 0 0 0 1 1 0 1 0 1 0 0 1 1 0 1 1 0 1 1 0 0 1 0 0 0 1 1 0 0 1 0 0 0 0 0 1 1 1 1 0 1 0 1 0 1 1 0 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 1 0 1 1 0 1 0 0 1 1 0 1 1 1 0 1 0 1 0 0 0 1 1 0 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 0 1 1 0 0 0 0 0 0 1 0 1 1 1 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 0 0 1 0 0 1 0 1 1 0 1 0 0 1 1 1 0 0 0 1 1 1 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 1 1 0 0 1 1 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 1 1 0 1 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 1 0 0 1 1 0 1 0 0 1 0 0 0 1 0 0 0 1 0 1 0 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 1 0 0 1 0 0 1 0 1 1 0 1 0 0 1 1 1 1 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 1 1 1 1 0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 1 1 1 1 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 1 1 1 1 0 1 0 1 0 0 0 0 1 1 1 0 1 1 0 0 1 1 1 1 0 0 1 0 1 1 1 1 1 0 0 1 1 1 1 0 0 On using (3.8), (3.10), (3.17), and the procedure for decryption (See Fig. 1.(b)), we get back the original binary image P, given by (3.16).

188 From the matrix C, on connecting each 1 with its neighbouring 1, we get an image which is in a zigzag manner (See Fig. 3). Fig. 3. Encrypted image of the hand It is interesting to note that, the original image and the encrypted image differ totally, and the former one, exhibits all the features very clearly, while the later one does not reveal anything. 6.4 Computations and Conclusions In this analysis, we have made use of a modified Hill cipher for encrypting a binary image. Here we have illustrated the procedure by considering a pair of examples: (1) the image of a hand, and (2) the image of upper half of a person. Here, we have noticed that, the encrypted image is totally different from the original image, and the security of the image is completely enhanced, as no feature of the original image can be traced out in any way from the encrypted image. This analysis can be extended for the images of signatures and thumb impressions.

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