paper-2 generalized hill cipher involving multiple keys, mixing and key dependent substitution
TRANSCRIPT
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Generalized Hill Cipher Involving Multiple Keys, Mixing and Key
Dependent SubstitutionV. U. K. Sastry
1and Ch. Samson
2
1Dept. of Computer Science & Engineering., SNIST,
Hyderabad, India,
[email protected]. of Information Technology, SNIST,
Hyderabad, India,
Abstract
In this paper we have developed a block cipher, called generalized Hill cipher, basing upon the classical Hill
cipher. In this analysis, we have used several key matrices and a set functions called Mix( ), Substitute( ). Thefunction substitute( ) depends upon the key matrices. Here each one of the key matrices is having a modular
arithmetic inverse. The cryptanalysis carried out in this investigation shows that this cipher is a strong one on
account of the multiple keys and the key dependent substitution function.
Keywords: Encryption, Decryption, Generalized Hill cipher, Mixing, Key Dependent Substitution,
Cryptanalysis, Avalanche effect
1. Introduction
In the literature of cryptography, it is well known that the classical Hill cipher [1], which had its origin
several decades back, can be broken by the known plaintext attack. In order to overcome this drawback many
other block ciphers such as DES [2] and AES [3] were developed in cryptography. However, in the last one
decade several researchers [4-18] have modified the Hill cipher in various ways and found mechanisms tostrengthen the cipher.
In a recent investigation [19], we have generalized the Hill cipher by introducing several powers of a single
key matrix and including a couple of functions named Mix( ) and Substitute( ) in order to strengthen the cipher
by modifying plain text at various stages of the iteration process involved in the cipher. The cryptanalysis of this
cipher has clearly indicated that this cipher cannot be broken by all possible conventional attacks.
In the present paper our objective is to develop generalized Hill cipher by including several keys and by
supporting the analysis with the help mixing and key dependent substitution. Here our interest is to see how the
different keys would influence the cipher in strengthening the cipher.
In what follows we present the plan of the paper. In section 2, we deal with the development of the cipher
and provide the necessary flowcharts and algorithms for encryption and decryption. In section 3, we discuss an
illustration of the cipher and mention about the avalanche effect. We analyze the cryptanalysis in section 4.
Finally, we deal with the computations carried out in this analysis and draw conclusions in section 5.
2. Development of the Cipher
Consider a plaintext P. Let us represent this in the form of a square matrix given by
P = [Pij], i=1 to n and j= 1 to n. (2.1)
Let us decompose this into a set of square matrices wherein each matrix is of size m. This is possible only when
n is divisible by m. Let n = mq. Then we can write
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P1 P2 Pq
Pq+1 Pq+2 P2q
P = . (2.2)
Pq2-q+1 Pq2-q+2 Pq2
wherein each Pi, i= 1 to q2, is a square matrix of size m. For convenience let us write q2=s. In view of the
relation (2.2), we can write
Pi = Decompose (P), where i takes the values 1 to s. The reverse process of the function Decompose ( ) can be
denoted as Compose ( ). This implies that all the Pis arranged in row wise order, as indicated in (2.2), will lead to
the matrix P.
Consider a set of s key matrices K1, K2Ks wherein each one is a square matrix of size m. It may be noted
here that we can get a square matrix of size s by arranging all the afore mentioned keys (K1 to Ks) in a row wise
manner. Let us call this as K. In this analysis, we choose all the key matrices K1 to Ks such that each one has
modular arithmetic inverse. The various steps involved in the block cipher under consideration can be
summarized as follows.
Pi = Decompose (P), i= 1 to s, (2.3)
On following the basic idea of the Hill cipher, we write
Pi = (Ki Pi) mod N, i= 1 to s. (2.4)
Thus we can get s matrices of the modified plaintext. On using the function Compose ( ), we get
P = Compose (Pi), (2.5)
Then on using the function Mix ( ), we have
P = Mix (P). (2.6)
Finally on making use of the function Substitute ( ), we get
P=Substitute (P). (2.7)
On carrying out the iteration process wherein the relations (2.3) (2.7) are included, we get the final form of P.
This will lead to the ciphertext C. In this analysis, we take N=256 as we confine our attention to EBCDIC code.
It may be noted here that the details of the functions Mix( ) and Substitute( ) will be explained in section 3.
The decryption process is governed by the relations
C=Isubstitute (C), (2.8)
C= Imix (C), (2.9)
Ci=Decompose(C) , (2.10)
Ci= (Ki-1* Ci) mod N, i= 1 to s, (2.11)
C = Compose (Ci). (2.12)
Here the functions Isubstitute( ) and Imix( ) denote the reverse processes of Substitute( ) and Mix( ) respectively.
In (2.11), Ki-1
is the modular arithmetic inverse of the Ki which satisfies the relation
Ki Ki-1
mod N = I,
where I is the unit matrix.
On carrying out the process of iteration by including the above set of relations (2.8) (2.12), we get the finalform of C which will result in the plaintext P. The flowcharts and the algorithms depicting the encryption and the
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decryption processes are presented below.
Read P, n, m, r, N
Construct NT and ST
for i = 1 to s
Pi= (Ki* Pi) mod N
end
P = Mix (P)
for i = 1 to s
Read Ki
end
P=Substitute
C= P
Write C
Read C, n, m, r, N
C=Isubstitute(C,NT,ST)
C= Imix (C)
Ci=Decompose(C)
for i = 1 to s
Ci= (Ki-1* Ci) mod N
end
s = n2/m
2
C = Compose(Ci)
P = C
Write P
for i= 1 to r
P = Compose(Pi)
for i = 1 to s
Read Ki
end
Construct NT and STPi =Decompose
for i = 1 to s
Compute Ki-1
end
s = n2/m
2
for i= 1 to r
(b) Process of Decryption(a) Process of Encryption
Figure 1: Schematic diagram of the cipher
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Here NT and ST are tables of size 16x16. These are required in the development of the function Substitute ( ),
explained in section 3.
Algorithm for Encryption
1. Read P, n, m, r, N2. s= n2/m23. for i = 1 to s
Read Ki
end
4. Construct NT and ST5. for i = 1 to r
Pi =Decompose (P)
for i = 1 to s
Pi= (Ki* Pi) mod N
end
P = Compose (Pi)
P= Mix(P)
P=Substitute (P,NT,ST)
end
6. C = P7. Write CAlgorithm for Decryption
1. Read P,n, m, r, N
2. s= n2/m2
3. for i = 1 to s
Read Ki
end
4. for i = 1 to s
Compute Ki-1
end
5. Construct NT and ST
6. for i = 1 to r
C=Isubstitute (C, NT, ST)
C= Imix (C)
Ci=Decompose(C)
for i = 1 to s
Ci= (Ki-1* Ci) mod N
endC = Compose (Ci)
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end
7. P = C
8. Write P
3. Illustration of the CipherLet us consider the plaintext given below.
Dear friend! We both joined in the central intelligence department at the same time. When you wrote about Iron
mining, I wondered about the approach of that fellow and I wrote to you I know only about data mining and
knowledge mining, and nowhere have I heard about such sort of mining. But in the recent past in our area I have
come across the sand mining which is going on. Some people are taking away Lorries of sand and selling it in
the nearest city. Do you know who are the people supporting all this activity? All the local politicians and the
police officers are encouraging this and suggesting ways and means by which more money can be achieved by
the persons who initiated this activity. They are trying very seriously and contacting me even offering bribe. I do
not know I can keep up my ethical values and live like a sincere man. I strongly feel that I must resign for this
post. Had I joined in a college as a lecturer, I would have been very glad in my life. Please suggest me what to
do. Yours Ashok. (3.1)
Let us now consider the plaintext comprising the first 256 characters of (3.1). It is given by
Dear friend! We both joined in the central intelligence department at the same time. When you wrote about Iron
mining, I wondered about the approach of that fellow and I wrote to you I know only about data mining and
knowledge mining, and nowhere have I he
On making use of the EBCDIC code, we have the above plaintext P in the form
We now have the sixteen key matrices K i, i=1 to 16. Here we use a single matrix K for writing all the 16
matrices. Hence
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Here we have written the first four matrices K1 to K4 (each one of size 4x4) in the first four rows, one after
another. We have placed K5 to K8 in the next four rows. Similarly we have written K9 toK12 and K13 toK16 in the
subsequent set of rows (Considering 4 rows at a time). We have represented the 16 matrices in the above form
for convenience.
On adopting the afore mentioned approach, the corresponding modular arithmetic inverses of the 16 key
matrices are given by the following single matrix.
In this matrix also the first four rows contain K1-1
to K4-1
, the next four rows contain K5-1
to K8-1
and so on. Here
it is to be noted that all these modular arithmetic inverses [21] are obtained separately.
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Let us now discuss how the functions Mix ( ) and Substitute ( ) are developed. In the process of encryption, on
carrying out the operation of the function Compose ( ), we get the plaintext P in the form
P = [Pij], i=1 to 16 and j= 1 to 16. (3.3)
The matrix in (3.3) can be readily brought to the form of another matrix of size 8x32, by placing the bottom 8
rows of P on the right side of the top 8 rows. Then P can be written in the form
P = [Pij], i=1 to 8 and j= 1 to 32. (3.4)
On representing each element of this matrix in terms of binary bits, we get a matrix of size 8x256. This is given
by
P111 P112 ... P118 . P1321 P1322 P1328
P211 P212 ... P218 . P2321 P1322 P2328
P = . (3.5)
.
P811 P812 P818 . P8321 P8322 P8328
The binary bits P111 P212 ... P811, lying in the first columncan be written in the form of a decimal number.
Similarly the binary bits in the subsequent columns can also be written in terms of decimal numbers. Thus we
get 256 decimal numbers which can be written in the form of a 16x16 matrix given by
P = [Pij], i=1 to 16 and j= 1 to 16. (3.6)
The Pij in (3.6) are obtained due to mixing process.
Let us now develop the process involved in the function Substitute ( ). In this analysis, as we are dealing with
the plaintext matrix of size 16x16, wherein each number can be represented in terms of 8 binary bits, let us
construct a table, denoted as NT, which includes the numbers 0 to 255 in a sequential manner. This can be
written in the form
NT (u, v) =16(u-1) + (v-1), u=1 to 16 and v=1 to 16. (3.7)
Let us form a square matrix of size 16, called ST, by including all the elements which are occurring in the keys
K1 to K16, one after another, in a row wise manner excluding elements which are getting repeated if any. After
placing all these elements in ST, we fill up the remaining places in ST with the missing numbers in the set 0 to
255. Thus we form ST which is given in Table 1, called substitution table.
Let us now see the process of substitution. This can be achieved by using the relation given below.
If P (i,j) = NT(u,v), then S (i, j) =ST (u, v), where S (i, j) is the result of substitution. This is valid for all i and j.
In other words, the above relation can be mentioned as follows. If the ith
row jth
column element of P is equal to
the uth row vth column element of the matrix NT, then the ith row jth column element of S(i,j),obtained as a result
of substitution, is equal to the uth row vth column element of the ST.
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196 224 77 140 38 25 105 152 204 5 47 87 45 69 184 153
63 167 219 146 50 9 163 80 15 134 250 232 120 102 61 44
65 205 101 121 155 216 169 113 143 79 145 235 17 126 162 218
24 135 183 171 36 75 21 125 97 243 124 248 128 168 215 89
170 181 95 142 249 154 201 12 213 85 31 189 237 13 114 81
66 123 19 156 41 172 94 141 26 117 76 78 67 138 111 93
148 203 207 53 28 119 39 23 43 175 56 49 71 161 98 246
255 222 91 106 220 34 133 139 147 233 164 127 177 151 254 192
115 252 256 225 58 16 73 55 236 239 185 64 159 14 231 131
217 221 165 193 59 68 33 52 137 51 62 241 35 173 107 29
182 174 186 103 1 2 3 4 6 7 8 10 11 18 20 22
27 30 32 37 40 42 46 48 54 57 60 70 72 74 82 83
84 86 88 90 92 96 99 100 104 108 109 110 112 116 118 122
129 130 132 136 144 149 150 157 158 160 166 176 178 179 180 187
188 190 191 194 195 197 198 199 200 202 206 208 209 210 211 212
214 223 226 227 228 229 230 234 238 240 242 244 245 247 251 253
Table 1: Key Dependent Substitution
Now on using the encryption algorithm given in section 2, we get the ciphertext C in the form
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By taking the ciphertext C given by (3.8) as input and executing the decryption algorithm, we get back the
original plaintext.
In order to have an insight into the strength of the cipher; let us examine the avalanche effect. On changing the
first row tenth column element of (3.2) from 149 to 148, we have a one binary bit change in the plaintext. On
using the modified plaintext, the keys K1 to K16, and applying the encryption algorithm, given in section 2, we
get the corresponding ciphertext given by
On comparing (3.8) and (3.9), after converting them into their binary form, we notice that they differ by 1064
binary bits out of 2048 bits. From this we notice that the avalanche effect is quite good.
Let us now examine the effect of one bit change in the key. Here we have a set of keys (K 1 to K16). On
changing the first row third column element of K1 from 77 to 76, we have a one bit change, as a whole, in the
bunch of the keys. On using the modified bunch of keys and the original plaintext given by (3.2), and applying
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the encryption algorithm given in section 2, we get the ciphertext C in the form
On comparing the ciphertexts (3.8) and (3.10) in their binary form, we find that they differ by 1059 bits out
of 2048 bits. This also indicates that the avalanche effect is quite significant. In view of the above results, we
conclude that this cipher is expected to have very good strength.
4. Cryptanalysis
The different types of cryptanalytic attacks which are available in the literature of cryptography are
1. Cipertext only attack (Brute force attack),
2. Known plaintext attack,
3. Chosen plaintext attack and
4. Chosen ciphertext attack.
Generally every cipher is to be developed so that it withstands the first two attacks at least [1]. However, one has
to visualize intuitively that the cipher cannot be broken by the last two attacks (3 and 4).
Let us examine the brute force attack. In this analysis, the size of the key bunch (16 keys) is 256 decimal
numbers (2048 binary bits). In view of this fact, the size of the key space is
22048 = (210)20.48 (103)20.48 =1061.44.
If we assume that the time required for the computation of the cipher with one value of the key is 10-7
seconds,
then the time required for the execution of the cipher with all the possible keys in the key space is approximately
equal to
years=46.445161.44
7.4461
103.1210103.12606024365
1010=
As this is a formidable quantity, we find that it is simply impossible to break the cipher by the brute force attack.
In the case of the known plaintext attack, we know as many pairs of the plaintext and the ciphertext as werequire for attacking the cipher. Thus we have several pairs of P1, P2P16 and the corresponding C1, C2C16. If
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we confine our attention to only to one round of iteration process, i.e., if we take r =1, we have the following
relations occurring in the encryption process.
Pi = KiPi mod N, i=1, 2 16, (3.11)
P = Compose (Pi) (3.12)
P= Mix (P) (3.13)
P=Substitute(P,NT,ST) (3.14)
C= P. (3.15)
As we know C, we get P occurring on the right hand side of (3.15). Thus the P on the left hand side of (3.14) is
known to us. As the substitution table ST is a key dependent one, we cannot form this table and determine P
occurring on the right hand side of (3.14). Thus we cannot move in the upward direction and determine the set of
keys Ki occurring in the equation (3.11). Hence the cipher cannot be broken by the known plaintext attack. Here
the key dependent substitution table is playing a prominent role in strengthening the cipher.
In view of the several functions such as Mix( ) and Substitute( ) used in this analysis, it is not intuitively
possible to choose either the plaintext or the ciphertext and break this cipher by chosen plaintext attack or chosen
ciphertext attack. In the light of the above analysis, we conclude that this cipher cannot be broken by any attack.
5. Computations and Conclusions
In this investigation, we have developed a block cipher, basing upon the classical Hill cipher by introducing a
set of keys (16 keys) wherein each key is a square matrix of size 4. In this analysis, we have utilized a set of
functions named Mix( ) and Substitute( ) in each round of the iteration process. The function Mix( ) mixes
thoroughly the binary bits of the plaintext in each round of the iteration process. The function Substitute( ) is a
key dependent function, and it depends upon all the 16 keys involved in the analysis. Here the decryption
process includes the modular arithmetic inverses of all the 16 keys. The cryptanalysis clearly shows that this
cipher cannot be broken as the substitution carried out in this analysis is key dependent.
Computer programs for encryption and decryption are written by using MATLAB [20]. The plaintext (3.1),
mentioned in section 3, is divided into four of blocks wherein each block is consisting of 256 characters.
However the last block containing 249 characters is appended with 7 blank characters in order to make it a
complete block having 256 characters.
On using the encryption algorithm, we have obtained the ciphertext corresponding to all these blocks. Here
we have presented the ciphertext of the entire plaintext, excluding the first block which is already presented in
(3.8). The ciphertext which we have here is given below.
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253 241 25 134 223 167 157 19 112 207 74 241 212 190 254 231
222 239 55 107 190 20 232 98 172 227 242 232 129 51 174 210
44 27 25 109 190 131 81 189 79 194 110 41 208 55 241 195
70 15 169 185 68 95 85 90 225 29 226 250 34 73 108 160
239 183 95 209 198 240 191 157 53 131 251 192 103 1 29 150
30 210 39 250 13 239 20 61 106 182 189 90 65 99 14 167
222 203 135 3 93 45 59 129 99 240 56 41 40 91 147 157
254 141 212 127 216 45 72 218 15 92 60 88 22 47 247 107
100 152 55 191 76 133 137 252 18 86 198 197 89 118 111 46
244 188 208 190 207 151 106 237 17 235 194 195 74 229 44 163
151 30 105 201 110 79 85 21 18 205 123 197 144 236 2 68
9 242 127 208 186 153 168 14 221 25 132 96 95 163 100 246
188 20 30 145 1 211 3 193 28 103 219 127 221 137 192 146
132 205 222 234 157 95 78 12 67 54 172 15 95 228 196 50
192 209 85 151 126 121 83 32 222 72 248 1 65 201 23 167
255 120 232 247 118 160 51 21 223 188 256 86 80 111 121 15
125 140 2 51 24 134 211 85 72 96 172 133 1 9 179 213
57 233 25 191 199 124 249 197 227 75 43 46 213 189 117 109
193 56 182 13 188 59 225 248 77 235 48 149 223 169 83 66
241 214 194 35 80 240 225 153 226 254 30 128 25 193 76 23
30 33 234 16 70 0 142 16 64 130 27 180 12 20 50 19
52 111 183 166 46 66 224 144 49 63 204 30 165 203 61 59
135 57 41 186 52 139 242 39 71 9 29 164 119 57 69 206
126 199 202 200 138 245 200 175 167 235 218 14 14 228 196 1
208 225 210 121 73 175 74 101 238 184 19 129 151 59 148 29
211 149 204 28 38 130 126 56 8 148 199 12 108 68 55 113
15 121 220 89 145 138 202 23 17 136 21 222 182 193 107 54
199 115 86 30 197 127 53 201 47 60 199 92 25 160 3 127
92 180 148 165 172 200 124 24 116 147 31 90 154 107 162 247
186 75 62 33 175 107 159 192 210 73 78 138 249 202 4 121
82 9 108 163 182 51 171 231 86 41 138 50 101 143 139 41
130 31 106 210 22 90 215 176 193 223 252 69 84 127 192 106
206 81 163 73 26 73 15 253 23 12 2 252 130 154 26 81
81 197 17 87 35 61 185 91 225 155 235 118 146 217 173 67
27 11 12 86 45 122 150 186 204 20 126 218 187 65 120 82
199 195 176 94 110 111 22 134 223 250 235 57 177 63 8 3027 157 99 119 56 248 204 75 204 254 215 250 158 19 71 165
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97 42 10 169 99 137 199 86 23 112 172 216 157 215 45 97
70 241 59 11 86 211 34 34 22 199 8 228 155 12 119 173
19 109 62 10 4 70 83 138 30 53 202 119 199 193 33 232
29 130 50 37 147 211 162 223 53 124 194 193 244 161 181 202
204 50 140 42 205 69 134 65 25 119 117 139 223 8 37 188
161 221 26 3 20 96 159 203 243 203 235 79 161 249 142 123
111 17 57 67 224 44 111 36 141 142 167 11 45 93 250 83
168 223 253 200 115 134 149 66 21 228 88 213 101 103 85 206
185 190 219 196 32 34 177 79 256 19 238 255 169 105 153 115
12 141 149 31 162 117 216 231 214 39 241 110 213 9 25 119
125 151 243 224 162 55 11 254 210 56 236 83 3 161 2 197
The strength of the cipher is quite significant as we have several keys involved in the cipher, and the substitution
table that we have utilized in this analysis is key dependent. The cryptanalysis in this investigation clearly
indicates that this cipher is a strong one and it cannot be broken by any attack.
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[16] Aruna Varanasi, V.U.K.Sastry, S.Udaya Kumar, A Modern Hill Cipher Involving a Pair of Keys,Modular Arithmetic Addition and Substitution,International Journal of Advanced Research in Computer
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[17] V. U. K. Sastry, D. S. R. Murthy, S. Durga Bhavani,A Block Cipher Involving a Key Applied on Both theSides of the Plain Text, International Journal of Computer and Network Security (IJCNS), Vol. 1, No.1,
pp. 27 30, Oct 2009.
[18] Aruna Varanasi, V.U.K.Sastry, S.Udaya Kumar, A Modern Hill Cipher Involving a Pair of Keys, XORoperation and Substitution, International Journal of Advanced Research in Computer Science, Vol.2
No.3, pp. 496-500, May-June 2011.
[19] V. U. K. Sastry, Ch.Samson, A Generalized Hill Cipher Involving Different Powers of a Key, Mixingand Substitution, International Journal of Advanced Research in Computer Science, May-June 2012(Sent
for publication).
[20] http://www.mathworks.com/products/matlab[21] William H. Press, Brian P. Flannery, Saul A. Teukolsky, William T. Vetterling, Numerical Recipes in C:
The Art of Scientific Computing, Second Edition, 1992, Cambridge University Press, pp. 36-39.
http://www.mathworks.com/products/matlabhttp://www.mathworks.com/products/matlabhttp://www.mathworks.com/products/matlab -
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International Journal of Computational Intelligence and Information Security, July 2012 Vol. 3, No. 6
ISSN: 1837-7823
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Dr. V. U. K. Sastry is presently working as Professor in the Dept. of Computer Science
and Engineering (CSE), Director (SCSI), Dean (R & D), SreeNidhi Institute of Science and Technology
(SNIST), Hyderabad, India. He was Formerly Professor in IIT, Kharagpur, India and worked in IIT, Kharagpur
during 1963 1998. He guided 12 PhDs, and published more than 70 research papers in various international
journals. Hereceived the best Engineering College Faculty Award in Computer Science and Engineering for the
year 2008 from the Indian Society for Technical Education (AP Chapter), and Cognizant- Sreenidhi Best faculty
award for the year 2012. His research interests are Network Security & Cryptography, Image Processing, Data
Mining and Genetic Algorithms.
Mr.Ch. Samson obtained his Diploma from Govt. Polytechnic, Hyderabad in 1994, B. E.
from Osmania University in 1998 and M. E from SRTM University in 2000. Presently he is pursuing Ph.D. from
JNTUH, Hyderabad since 2009. He published 9 research papers in international journals and two papers in
conferences. He is currently working as Associate Professor and Associate Head in the Dept. of Information
Technology (IT), SNIST since June 2005. His research interests are Image Processing, Image Cryptography and
Network Security.