Triple Encryption Scheme Using two Independent Keys

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Triple Encryption Scheme Using two

Independent Keys

A. CHANDRA SEKHAR Department of Mathematics,GIT GITAM University, Visakhapatnam, India

[email protected] Ch.PRGATHI

Department of Mathematics,GIT GITAM University, Visakhapatnam, India

[email protected] S.ASHOK KUMAR Department of Mathematics,GIT GITAM University, Visakhapatnam, India

[email protected]

Abstract :

To strengthen the Data Encryption Standards (DES) in recent research works multiple encryptions were introduced. In multiple encryptions generation of two independent keys plays a vital role. In this paper we proposed a triple encryption scheme using two independent keys is introduced.

Keywords: Fibonacci numbers, Pell numbers, Affine transformation, Vigenere transformation

1. INTRODUCTION

Multiple encryption is a process of encrypting the information that is already encrypted [1], [6]. Super-encryption is simply the use of multiple ciphers, usually in multiple steps, as a singular Super-encryption scheme and is a very important technique and many modern strong encryption algorithms can be regarded as resulting from super-encryption using a number of a comparatively weak algorithm.

2. FIBONACCI NUMBERS

The Fibonacci sequence is 1, 1, 2, 3, 5, 8. . . [1][6] Where each entry is formed by adding the two previous ones, starting with 1 and 1 as the first two terms. Fibonacci numbers can be generated from following recurrence relation

F

_n₊₁

=

F

_n

+

F

_n₋₁with

F

₁

=

F

₂

=

1

.

3. Pell Numbers

The Pell numbers [6] are defined by the recurrence relation

1 2

0

1

2

n

n n

if n

P

if n

P

₋

P

₋

other wise



=



=



=



₊



In words, the sequence of Pell numbers starts with 0 and 1, and then each Pell number is the sum of twice the previous Pell number and the Pell number before that. The first few terms of the sequence are 0,1,2,5,12,29,70,169, 408,985, 2378, 5741, 13890,...

4. AffineCipher

An affine enciphering transformation [1] is

C

≡

aP b

+

(

mod

N

)

where the pair (a, b) is the encrypting key and gcd(a,N)=1. If y = E(x) = (ax+b) mod26,[5]then we can “solve for x in terms of y” and so _E−1_{( )}_y _{that is, if}

(

)

y≡ ax+b

mod 26

theny− ≡b ax

(

mod 26

)

or equivalentlyax≡

(

y b−

)

_{mod 26 .}

5. Vigenerecipher

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6. PROPOSED WORK

Encryption algorithm:

Step-1: Alice creates plaintext P = p1 p2, p3 … pm

Step-2: Alice uses the offset rule with Fibonacci numbers F=f1,f2,f3…fn to each value in sequential order to get

the1st ciphertext C1.

Step-3: Alice encrypts the ciphertext C1 by using inverse affine transformation

( )

1 1

E

−

y

=

a

−

(

y b

−

)

mod 26

to get ciphertext C2 where for a and b are secret.

Step-4: Alice applies the offset rule to C2 with Fibonacci numbers to get ciphertext C3

Step-5: Alice sends ciphertext C3 to Bob.

Decryptionalgorithm:

Step-1: Bob receives the encrypted message C3.

Step-2: Bob use reverse offset rule with Fibonacci number to C3 get plaintext P2.

Step-3: Bob compute first plain text P2 with affine transformation E(x) = (ax+b) mod 26, Gcd(a,N)=1 where for

a and b are kept secret, from the first level decrypted message P1

Step-4: Bob use reverse offset rule to P1 with Fibonacci number to get the original plaintext message P.

VIGENERE CIPHER Encryption algorithm:

Step-1: Alice creates plaintexts P = p1 p2, p3…… pm

Step-2: Alice uses the offset rule with Pell numbers F=f1,f2,f3…fn to each value in sequential order to get the1st

ciphertext is C1.

Step-3: Alice use reverse offset rule with vigenere key to C1 get ciphertext C2.

Step-4: Alice applies the offset rule with to C2 with to Pell number get ciphertext C3

Step-5: Alice sends ciphertext C3 to Bob.

Decryptionalgorithm:

Step-1: Bob receives the encryption message C3.

Step-2: Bob use reverse offset rule with Pell number with C3 to get plaintext P2.

Step-3: Bob use offset rule with vigenere key to P2 get plaintext P1.

Step-4: Bob use reverse offset rule to P1 with Pell number to get the original plaintext message P.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

EXAMPLE

Step-1: Let the plain text be P is GITAMUNIVERSITY

Step-2: Alice uses offset rule with Fibonacci numbers to plaintext get ciphertext C1

Offset rule with Fibonacci

numbers

I A M U N I V E R S I T Y

6 +

1 1 19

+ 2

0 + 3

12 + 5

20 + 8

13 + 13

8 + 21

21 + 34

4 + 55

17 + 89

18 + 144

8 + 233

19 + 377

24 + 610

7 21 3 17 28 26 29 55 59 106 162 241 396 634

Mod 26 7 9 21 3 17 2 0 3 3 7 2 6 7 6 10

First encryption

message

H V D R C A D D H C G H G K

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Step-4: Alice again use offset rule with Fibonacci numbers to C2 get ciphertext C3

Offset rule with Fibonacci

numbers

P F X J R O Y J J P O U P U A

15 5 23 9 17 14 24 9 9 15 14 20 15 20 0 15

+ 1

5 +

23 2

9 + 3

17 + 5

14 + 8

24 + 13

9 + 21

9 + 34

15 + 55

14 + 89

20 + 144

15 + 233

20 + 377

0 + 610 16 25 12 22 22 37 30 43 70 103 164 248 397 610 Mod 26 16 6 25 12 22 22 11 4 17 18 25 8 14 7 12 Third encryption

message

Q G Z M W W L R S Z I O H M

Step-5: Alice send third encrypted message C3 is QGZMWWLERSZIOHM

Decryption algorithm:

Step-1: Bob receive message C3is QGZMWWLERSZIOHM

Step-2: Reverse Offset rule with the first decrypted plaintext P2.

Reverse offset rule with Fibonacci

numbers

Q G Z M W W L E R S Z I O H M

16 6 25 12 22 22 11 4 17 18 25 8 14 7 12

16 - 1

6 - 1

25 - 2

12 - 3

22 - 5

22 -

11 - 13

4 - 21

17 - 34

18 - 55

25 - 89

8 - 144

14 - 233

7 - 377

12 - 610 15 5 23 9 17 14 -2 -17 -17 -37 -64 -136 -219 -360 -598 Mod 26 15 5 23 9 17 14 24 9 9 15 14 20 15 20 0 First decryption

message

F X J R O Y J J P O U P U A

Step-3 Bob compute Inverse of Affine transformation E(x)=(ax+b) mod 26 for a = 5 & b= 10 to P2 get

plaintext P1

Step-4: Bob use Reverse Offset rule with Fibonacci number to P1 get plaintext P

Message H J V R C A D D H C G H G K

y 7 9 21 3 2 0 3 3 7 2 6 7 6 10

y-10 -3 -1 11 -7 7 -8 -7 -7 -3 -8 -4 -3 -4 0

21(y-10) -21 231 -147 -168 -210 -147 -147 -63 -168 -84 -63 -84 0 21 (y-10)

mod26

15 5 23 9 17 14 24 9 9 15 14 20 15 20 0

Second encrypted message

P F X J R O Y J J P O U P U A

x 15 5 23 9 17 14 24 9 9 15 14 20 15 20 0

5x+8 85 35 125 55 95 80 130 55 55 85 80 110 85 110 10

(5x+8) mod 26 7 9 21 3 17 2 0 3 3 7 2 6 7 6 10

Second decrypted

message is

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VIGENERECIPHER

Step-1: Let the plaintext P be PRIMENUMBERS

Step-2: Alice use offset rule with Pell number to plaintext get ciphertext C1

Offset rule with Pell number

P R I M E N U M B E R S 15 + 0 17 + 1 8 + 2 12 + 5 4 + 12 13 + 29 20 + 70 12 + 169 1 + 408 4 + 985 17 + 2378 18 + 5741 15 18 10 17 16 42 90 181 409 989 2395 5759

Mod 26 15 18 10 17 16 16 12 25 19 1 3 13

First encrypted message P S K R Q Q M Z T B D N Using Vigenere cipher for key

Step-3:Alice use reverse offset rule with vignerekey to first encrypted ciphertext get ciphertext C2

Reverse offset rule with key

P S K R Q Q M Z T B D N - 11 18 - 20 10 - 2 17 - 0 16 - 18 16 - 11 12 - 20 25 - 2 19 - 0 1 - 18 3 - 11 13 - 20

4 -2 8 -2 5 -8 23 19 -17 -8 -7

Mod 26 4 24 8 17 24 5 18 23 19 9 18 19

second encrypted message E Y I R Y F S X T J S T

Step-4: Alice use offset rule with Pell number to C2 get ciphertext C3.

Offset rule with Pell number

E Y I R Y F S X T J S T 4 + 0 24 + 1 8 + 2 17 + 5 24 + 12 5 + 29 18 + 70 23 + 169 19 + 408 9 + 985 18 + 2378 19 + 5741 4 25 10 22 36 34 88 192 427 994 2396 5760

Mod 26 4 25 10 22 10 8 10 10 11 6 4 14

Third encrypted message Z K W K I K K L G E O

Step-5: Alice send encrypted message C3 is EZKWKIKKLGEO

Decryption algorithm:

Message

H J V D R C A D D H C G H G K

7 9 21 3 17 2 0 3 3 7 2 6 7 6 10

Reverse Offset rule with Fibonacci number 7 - 1 9 - 1 21 - 2 3 - 3 17 - 5 2 - 8 0 - 13 3 - 21 3 - 34 7 - 5 5 2 - 89 6 - 144 7 - 233 6 - 377 10 - 610

6 8 0 12

-6 -13 -18 -31 -4 8 -87 -138 -226 -371 -600

Mod 26 6 8 19 0 12 2 0

13 8 21 4 17 18 8 19 24 Third decrypted

message is

G I T A M U N I V E R S I T Y

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Step-2: Bob use reverse offset rule with Pell number to C3 get plaintext P2

Reverse offset rule with Pell

number

E Z K W K I K K L G E O

4 - 0

25 - 1

10 - 2

22 - 5

10 - 12

8 - 29

10 - 70

10 - 169

11 - 408

6 -

4 - 2378

14 - 5741 4 24 8 17 2

-21

-60 -159

-397

-979

-2374 -5727 Mod 26 4 24 8 17 24 5 18 23 19 9 18 19 First decrypted

message

E Y I R Y F S X T J S T

Step-3: Bob use offset rule with Vigenere transformation to P2 get plaintext P1

offset rule with key

E Y I R Y F S X T J S T 4

+ 11

24 + 20

8 + 2

17 + 0

24 + 18

5 + 11

18 + 20

23 + 2

19 + 0

9 + 18

18 + 11

19 + 20 15 44 10 17 42 16 25 19 27 29 39 Mod 26 15 18 10 17 16 16 12 25 19 1 3 13 Second decrypted

message

P S K R Q Q M Z T B D N

Step-4: Bob use Reverse offset rule with Pell number to P2 get plaintext P

Reverse offset rule with Pell number

P S K R Q Q M Z T B D N

15 - 0

18 - 1

10 - 2

17 5

16 - 12

16 - 29

12 - 70

25 - 169

19 - 408

1 - 985

3 - 2378

13 - 5741 15 17 8 12 -13 -58 -144 -389 -984 -2375 -5728

Mod 26 15 17 8 12 4 13 20 12 1 4 17 18

Third decrypted message

P R I M E N U M B E R S

7. CONCLUSIONS

For triple encryption only two keys are employed. In the process the plain text is encrypted with the first key K1 then encrypted with the inverse of the second key K2 and then encrypted with the K1.The triple encryption is

performed using with two keys only instead of three independent keys.

8. REFERENCES

[1] A. ChandraSekhar, D. Chaya Kumari, S. Ashok Kumar "Symmetric Key Cryptosystem for Multiple Encryptions",International Journal of Mathematics Trends and Technology (IJMTT). V29 (2):140-144 January 2016. ISSN:2231-5373.

[2] A. Chandra Sekhar, Prasad Reddy. P.V.G.D, A.S.N.Murty, B.Krishna Gandhi "Self-Encrypting Data Streams Using Graph Structures" IETECH International Journal Of Advanced Computations PP 007-009, 2008, vol 2.

[3] A.P.Stakhov “ The Golden matrices and a new kind of cryptography” chaos, solutions and Fractals 32(2007) pp1138-1146.

[4] A.P.Stakhov “ The Golden section and modern harmony mathematics. Applications of Fibonacci numbers” ,kluwer Academic publishers (1998). pp393-399

[5] “E.H.Lock Wood, A single-light on pascal's triangle, Math, Gazette 51(1967), PP 243-244.

[6] Fibonacci, Lucas and Pell numbers andpascal's triangle, Thomas Khoshy, Applied Probability Trust, PP 125-132.

[7] Linear independent spanning sets and linear transformations for multi-level encryption, A.ChandraSekhar, V.Anusha, B.Ravi Kumar, S.Ashok Kumar Vol36(2015) , No.4, PP;385-392.

[8] International journal on cryptography and information security(IJCI") “Image encryption using Fibonacci-Lucas transformation” Vol.2,No3,September 2012.

[9] On the security of multiple encryption, Ralphe.merle, Elxs, Inti Martin E.Hellmon Standford University, Communication of the ACM July 1981, Vol 24 No 7.

[10] Branstad,D.K., Gait,J., and Katzke,S.Report of the workshop on cryptography in support of computer security, National Bureau of Standards Rep.NBSIR 77-1291(Sept.21-22,1976).

[11] Diffie,W., and Hellman, M.Exhaustive cryptanalysis of the NBS data encryption Standard. Computer (June 1977),74-84. [12] Hellman, M.E., An extension of the Shannon theory approach to cryptography,IEEE Trans.Info.IT-23, (May 1977),289-294. [13] Kolata,G.B Computer encryption and the national security agency,Science 1977(July 29,1977)438-440.