A STUDY OF DATA ENCRYPTION STANDARD USING GRAPH THEORY

A STUDY OF DATA ENCRYPTION STANDARD USING GRAPH THEORY

Femina. P

, Antony Xavier.D

1Student, ²Associate Professor, PG and Research Dept. of Mathematics, Loyola College, Chennai (India)

ABSTRACT

The purpose of any creation is to attain a secure condition. To achieve the state of being secure many rules and processes are abided and followed, in this paper, the DES creation is a 16 subkeys block ciphertext applying the rules of Hamilton’s Path and the process of Anti-Magic Graph Labeling on a Cube to arrive at the ultimate secure condition, which makes any attack tedious.

Keywords: Anti-Magic Labeling, Ciphertext, DES, Hamilton Path, Cube

I. INTRODUCTION

Cryptography provides a number of security goals to ensure the privacy of the data [1,2]. The goals of the cryptography are Confidentiality, Integrity, Availability, Authenticity, Non-Reputation, Access control [1]. In cryptography the encryption algorithms can be classified into two broad categories- Symmetric key and asymmetric key encryption[2]. In Symmetric key cryptography, the key used for encryption and decryption is same. Thus, the key must be distributed through the secure channel before transmission started.

A magic square is an arrangement of numbers into a square such that the sum of each row, column and diagonal are equal [3]. The term “anti-magic” then comes from being the opposite of magic / arranging numbers in a way such that no two sums are equal.

A Hamiltonian path is a graph that visits each vertex exactly once, whichis applied on the cube graph [4].

The DES (Data Encryption Standard) algorithm[5] is the most widely used encryption algorithm in the world.

For many years, and among many people, "secret code making” and DES have been synonymous. DES is a block cipher--meaning it operates on plaintext blocks of a given size (64-bits) and returns ciphertext blocks of the same size. Thus DES results in a permutation among the 2^64 (read this as: "2 to the 64th power") possible arrangements of 64 bits, each of which may be either 0 or 1 [4]. Each block of 64 bits is divided into two blocks of 32 bits each, a left half block L and a right half R.The 16-rounds of the block cipher is the core of DES. Here at first 64-bit block plain text is divided into two 32-bit words, LPT and RPT [6]. In each iteration, the second word RPT is fed into a function “f‟ and the result is XORed with the left word LPT. Then the both words are swapped, and the algorithm continues. The function “f ‟ of DES algorithm has following steps: 64-bit key is the input of the algorithm. Every eighth-bit position of the key is ignored to produce 56 bit key[5]. This key is first subjected to a permutation by permuted choice one and the resultant 56 bit key divided into two 28 bit words.

They are circularly left-shifted according to the number of left shift in the particular round. The resulting shift

value serves as the input to permuted choice two to produce 48-bit outputs. Encryption is the process of an obscuring message to make them unreadable in the absence special knowledge.

In this paper, a study of DES by using graph theory is presented. It is achieved by using Hamilton path and anti- magic labeling on a cube graph. Here a Hamilton circuit of a cube graph is applied to arrive at a secret key and by obtaining the process of anti-magic labeling we arrive at sixteen un-identical subkeys on the cube graph. To optimize and minimize any threats in identifying plain text, an authentication process of encryption and decryption is commuted between the sender and receiver to confirm, if the authenticator is genuine.

Therefore, with these analytical concepts, a block ciphertext code has been presented proposing minimum code attack possibilities.

II. KEY TRANSFORMATION

In the cube graph, the number of directions using Hamiltonian path with a starting node is given. Hence, the Hamilton path for the given cube graph is 26-33-21-20-8-13-23-12.

Fig 1: Anti-Magic Labeling and Hamilton Path on Cube Graph

2.1 Create 16 Subkeys, Each of Which is 48-Bits Long

The input is an arbitrary Hamiltonian path of the cube graph in 64-bit binary format. This is the secret key.

Altogether 16 arbitrary mappings are stored in the secretmapping table for producing subkeys of 16 rounds. The key transformation algorithm for the first round is given below

Algorithm – Key Transformation

Step 1: 64-bit secret key is permuted using the first entry of the secret mapping table.

Step 2: Every eighth-bit position of the sub key is discarded to produce 56-bit subkey.

Step 3: 56-bit subkey is permuted using the permuted choice two table in the standard DES algorithm and the subkey for round 1 is presented.

Illustration

From the original 64-bit key

K = 00011010 00100001 00010101 00010100 00001000 00001110 00010111 00001011 we get the 56-bit permutation

K+ = 0001101 0010000 0001010 0001010 0000100 0000111 0001011 0000101.

The permuted choice two table is given below.

PC-2

14 17 11 24 1 5 3 28 15 6 21 10 23 19 12 4 26 8 16 7 27 20 13 2 41 52 31 37 47 55 30 40 51 45 33 48 44 49 39 56 34 53 46 42 50 36 29 32 After we apply the permutation, PC-2, we will arrive,

K₁ = 000001 000001 000100 011100 100000 010011 010100 110000.

By following the same method, we will get the 16 sub keys.

Table 1: Secrete Mapping table

Round Mapping

1 1<->2 2 2<->5 3 3<->9 4 6<->8 5 6<->12 6 7<->11 7 3<->7

8 1->2 , 2->5 , 5->1 9 3->9 , 9->4 , 4->3 10 1<->8

11 5<->10 , 9<->8 12 1<->11 , 5<->4 , 8<->3

13 9<->8 , 10<->11 , 1<->4 , 7<->5 14 9<->12 , 2<->7 , 8<->3 , 5<->11 15 1->2 , 2->5 , 9<->8 , 10<->11 16 1->2 , 2->8 , 8->10 , 10->6 , 6->1

For the other keys we have,

K₂ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₃ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₄ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₅ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₆ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₇ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₈= 000001 000001 000100 011100 100000 010011 010100 110000.

K₉ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₁₀ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₁₁ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₁₂= 000001 000001 000100 011100 100000 010011 010100 110000.

K₁₃= 000001 000001 000100 011100 100000 010011 010100 110000.

K₁₄ = 000001 000001 000100 011100 100000 010011 010100 110000.

K₁₅= 000001 000001 000100 011100 100000 010011 010100 110000.

K₁₆= 000001 000001 000100 011100 100000 010011 010100 110000.

Table 2: Hexadecimal format for the above subkeys Round Keys

1 04111c813530

2 041114853520

3 043518852020

4 051114853520

5 94031c813560 6 240396853420

7 041114043620

8 041114853530

9 043518852030

10 00119c8436a0 11 2503968521a0 12 a1418a8d0221 13 0431188130b0 14 202714880661 15 043514853520 16 25431e8524b0

III. DETAILED WORK OF DES

DES is a block cipher-meaning it operates on plaintext blocks of a given size (64-bits) and returns ciphertext blocks of the same size.

Thus,DES results in a permutation among the 2⁶⁴ (read this as: "2 to the 64th power") possible arrangements of 64 bits, each of which may be either 0 or 1.Each block of 64 bits is divided into two blocks of 32 bits each, a left half block L, and a right half R.

Illustration:

Suppose Mbe the plain text message say, M = 0123456789ABCDEFWhere, M is in hexadecimal (base 16) format.

Rewriting M in binary format, we get the 64-bit block of text:

M = 0000 0001 0010 0011 0100 0101 0110 01111000 1001 1010 1011 1100 1101 1110 1111 L = 0000 0001 0010 0011 0100 0101 0110 0111

R = 1000 1001 1010 1011 1100 1101 1110 1111

The first bit of M is "0". The last bit is "1". We read from left to right.DES operates on the 64-bit blocks using key sizes of 56- bits. The keys are actually stored as being 64 bits long, but every 8th bit in the key is not used (i.e. bits numbered 8, 16, 24, 32, 40, 48, 56, and 64).

3.1. ENCODE EACH 64-BIT BLOCK OF DATA.

There is an initial permutation IP of the 64 bits of the message data M.

This rearranges the bits according to the following table, where the entries in the table show the new arrangement of the bits from their initial order. The 58th bit of M becomes the first bit of IP.

IP

58 50 42 34 26 18 10 2 60 52 44 36 28 20 12 4 62 54 46 38 30 22 14 6 64 56 48 40 32 24 16 8 57 49 41 33 25 17 9 1 59 51 43 35 27 19 11 3 61 53 45 37 29 21 13 5 63 55 47 39 31 23 15 7

Illustration:

Applying the initial permutation to the block of text M, given previously, we get

M = 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 IP = 1100 1100 0000 0000 1100 1100 1111 1111 1111 0000 1010 1010 1111 0000 1010 1010 Next divide the permuted block IP into a left half L0 of 32 bits, and a right half R0 of 32 bits.

Illustration:

From IP, we get L₀ and R₀

L₀ = 1100 1100 0000 0000 1100 1100 1111 1111 R0 = 1111 0000 1010 1010 1111 0000 1010 1010

We now proceed through 16 iterations, for 1<=n<=16, using a function f which operates on two blocks--a data block of 32 bits and a key Kn of 48 bits to produce a block of 32 bits.

Let + denote XOR addition, (bit-by-bit addition modulo 2).

Then for n going from 1 to 16 we calculate L_n = R_n-1

R_n = L_n-1 + f(R_n-1,K_n)

This results in a final block, for n = 16, of L₁₆R_16.

Illustration:

K₁ = 000001 000001 000100 011100 100000 010011 010100 110000.

L₁ = R₀ = 1111 0000 1010 1010 1111 0000 1010 1010 R₁ = L₀ + f(R₀,K₁)

It remains to explain how the function f works. To calculate f, we first expand each block R_n-1 from 32 bits to 48 bits. This is done by using a selection table that repeats some of the bits in R_n-1 . We'll call the use of this selection table the function E. Thus E(R_n-1) has a 32 bit input block, and a 48 bit output block.

Thus the first three bits of E(R_n-1) are the bits in positions 32, 1 and 2 of R_n-1 while the last 2 bits of E(R_n-1) are the bits in positions 32 and 1.

E BIT-SELECTION TABLE

32 1 2 3 4 5 4 5 6 7 8 9 8 9 10 11 12 13 12 13 14 15 16 17

16 17 18 19 20 21 20 21 22 23 24 25 24 25 26 27 28 29 28 29 30 31 32 1

Illustration:

R₀ = 1111 0000 1010 1010 1111 0000 1010 1010

E(R₀) = 011110 100001 010101 010101 011110 100001 010101 010101

Next in the f calculation, we XOR the output E(R_n-1) with the key K_n (i.e.,K_n + E(R_n-1)).

Illustration:

K₁ = 000001 000001 000100 011100 100000 010011 010100 110000 E(R₀) = 011110 100001 010101 010101 011110 100001 010101 010101 K₁+E(R₀) = 011111 100000 010001 001001 111110 110010 000001 100101.

We have not yet finished calculating the function f. To this point, we have expanded R_n-1 from 32 bits to 48 bits, using the selection table, and XORed the result with the key K_n. We now have 48 bits / eight groups of six bits.We now do something strange with each group of six bits: we use them as addresses in tables called “S- boxes.” Each group of six bits will give us an address in a different S-box. Write the previous result, which is 48 bits, in the form:

K_n + E(R_n-1) =B₁B₂B₃B₄B₅B₆B₇B₈

K1+E(R0) = 011111 100000 010001 001001 111110 110010 000001 100101

where each B_i is a group of six bits. We now calculateS₁(B₁)S₂(B₂)S₃(B₃)S₄(B₄)S₅(B₅)S₆(B₆)S₇(B₇)S₈(B₈) where S_i(B_i) refers to the output of the ith S- box.

S1

Column Number

Row No. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 14 4 13 1 2 5 11 8 3 10 6 12 5 9 0 7 1 0 15 7 4 14 2 13 1 10 6 12 11 9 5 3 8 2 4 1 14 8 13 6 2 11 15 12 9 7 3 10 5 0 3 15 12 8 2 4 9 1 7 5 11 3 14 10 0 6 13 The remaining tables defining the functions S₂,...,S₈ are the following:

S2

15 1 8 14 6 11 3 4 9 7 2 13 12 0 5 10 3 13 4 7 15 2 8 14 12 0 1 10 6 9 11 5 0 14 7 11 10 4 13 1 5 8 12 6 9 3 2 15 13 8 10 1 3 15 4 2 11 6 7 12 0 5 14 9

S3

10 0 9 14 6 3 15 5 1 13 12 7 11 4 2 8 13 7 0 9 3 4 6 10 2 8 5 14 12 11 15 1

13 6 4 9 8 15 3 0 11 1 2 12 5 10 14 7 1 10 13 0 6 9 8 7 4 15 14 3 11 5 2 12

S4

7 13 14 3 0 6 9 10 1 2 8 5 11 12 4 15 13 8 11 5 6 15 0 3 4 7 2 12 1 10 14 9 10 6 9 0 12 11 7 13 15 1 3 14 5 2 8 4 3 15 0 6 10 1 13 8 9 4 5 11 12 7 2 14

S5

2 12 4 1 7 10 11 6 8 5 3 15 13 0 14 9 14 11 2 12 4 7 13 1 5 0 15 10 3 9 8 6 4 2 1 11 10 13 7 8 15 9 12 5 6 3 0 14 11 8 12 7 1 14 2 13 6 15 0 9 10 4 5 3

S6

12 1 10 15 9 2 6 8 0 13 3 4 14 7 5 11 10 15 4 2 7 12 9 5 6 1 13 14 0 11 3 8 9 14 15 5 2 8 12 3 7 0 4 10 1 13 11 6 4 3 2 12 9 5 15 10 11 14 1 7 6 0 8 13

S7

4 11 2 14 15 0 8 13 3 12 9 7 5 10 6 1 13 0 11 7 4 9 1 10 14 3 5 12 2 15 8 6

1 4 11 13 12 3 7 14 10 15 6 8 0 5 9 2 6 11 13 8 1 4 10 7 9 5 0 15 14 2 3 12

S8

13 2 8 4 6 15 11 1 10 9 3 14 5 0 12 7 1 15 13 8 10 3 7 4 12 5 6 11 0 14 9 2 7 11 4 1 9 12 14 2 0 6 10 13 15 3 5 8 2 1 14 7 4 10 8 13 15 12 9 0 3 5 6 11

To repeat, each of the functions S1, S2,..., S8, takes a 6-bit block as input and yields a 4-bit block as output. The table to determine S1 is shown and explained below:

B1 = 011111 B2 = 100000 B3 = 010001 B4 = 001001 FL = 01 → 1 FL = 10 → 2 FL = 01 → 1 FL = 01 → 1 M = 1111 → 15M = 0000 → 0 M = 1000 → 8 M = 0100 → 4 S₁ = 8 S₂ = 0S₃ = 2 S₄ = 6

S₁(B₁) = 1000 S₂(B₂) = 0000S₃(B₃) = 0010 S₄(B₄) = 0110 B₅ = 111110 B₆ = 110010 B₇ = 000001 B₈= 100101 FL = 10 →2 FL = 10 → 2 FL = 10 → 1 FL = 11 → 3 M = 1111 →15 M = 1001 → 9 M = 0000 → 0 M = 0010 → 2 S5 = 14S6 = 0S7 = 13 S8 = 14

S₅(B₅) = 1110 S₆(B₆) = 0000S₇(B₇) = 1101 S₈(B₈) = 1110

The net result is that the eight groups of 6 bits are transformed into eight groups of 4 bits (the 4-bit outputs from the S boxes) for 32 bits total.

Illustration:

For the first round, we obtain as the output of the eight S boxes:

K₁ + E(R₀) = 011111 100000 010001 001001 111110 110010 000001 100101.

S₁(B₁)S₂(B₂)S₃(B₃)S₄(B₄)S₅(B₅)S₆(B₆)S₇(B₇)S₈(B₈) = 1000 0000 0010 0110 1110 0000 1101 0000.

The final stage in the calculation of “f” is to do a permutation P of the S-box output to obtain the final value of f:

f = P(S₁(B₁)S₂(B₂)...S₈(B₈))

The permutation P is defined in the following table. P yields a 32-bit output from a 32-bit input by permuting the bits of the block.

P

16 7 20 21 29 12 28 17 1 15 23 26

5 18 31 10 2 8 24 14

32 27 3 9 19 13 30 6 22 11 4 25

Illustration:

S₁(B₁)S₂(B₂)S₃(B₃)S₄(B₄)S₅(B₅)S₆(B₆)S₇(B₇)S₈(B₈) = 1000 0000 0010 0110 1110 0000 1101 0000.

we get f = 0000 1011 1101 0110 0001 0000 1010 0101

R₁ = L₀ + f(R₀ , K₁ ) = 1100 1100 0000 0000 1100 1100 1111 1111 + 0000 1011 1101 0110 0001 0000 1010 0101 = 1100 0111 1101 0110 1101 1100 0101 1010

In the next round, we will have L₂ = R₁, which is the block we just calculated, and then we must calculate R₂

=L₁ + f(R₁, K₂), and so on for 16 rounds. At the end of the sixteenth round, we have the blocks L₁₆ and R₁₆.

We then reverse the order of the two blocks into the 64-bit block

R₁₆L₁₆ . and apply a final permutation IP^-1 as defined by the following table:

Round Key LPT_i RPT_i

IP cc00ccee e0aae0aa

1 04111c813530 e0aae0aa c7dbdc5a

2 041114853520 c7dbdc5a d038c554

3 043518852020 d038c554 dcbc5990

4 051114853520 dcbc5990 23c4d856

5 94031c813560 23c4d856 94ca3586

6 240396853420 94ca3586 c9c12990

7 041114043620 c9c12990 76cbe890

8 041114853530 76cbe890 c4de4c47

9 043518852030 c4de4c47 9d71a755

10 00119c8436a0 9d71a755 f5183394 11 2503968521a0 f5183394 f5e7c0d7 12 a1418a8d0221 f5e7c0d7 54e0fad9

13 0431188130b0 54e0fad9 5165e549

14 202714880661 5165e549 F9c0ffbe

15 043514853520 f9c0ffbe 4e212bd6

16 25431e8524b0 4e212bd6 B3d5eb83

IP^-1

40 8 48 16 56 24 64 32 39 7 47 15 55 23 63 31 38 6 46 14 54 22 62 30 37 5 45 13 53 21 61 29 36 4 44 12 52 20 60 28 35 3 43 11 51 19 59 27 34 2 42 10 50 18 58 26 33 1 41 9 49 17 57 25

Illustration:

If we process all 16 blocks using the method defined previously, we get, on the 16th round, L₁₆ = 0100 1110 0010 0001 0010 1011 1101 0110

R₁₆ = 1011 0011 1101 0101 1110 1011 1000 0011

We reverse the order of these two blocks and apply the final permutation to

R₁₆L₁₆ = 10110011 11010101 11101011 10000011 01001110 00100001 00101011 11010110 IP^-1 = 01111101 11001111 10010010 10001100 01010010 0110110 10010110 01010111 Therefore the hexadecimal format is 7dcf928c526c9657.

This is the encrypted form of M = 0123456789ABCDEF: namely, C = 7dcf928c526c9657.

Decryption is simply the inverse of encryption, following the same steps as above, but reversing the order in which the subkeys are applied.

IV. CONCLUSION

The entire document deals with security considerations. Still, it makes sense to summarize a few key points here. It should be clear by now that the DES algorithm offers little deterrence for a determined adversary. DES is a 64-bit block cipher having a key size of 56-bits.

Security is a very complex topic. Here a graph based modified DES (GMDES) algorithm is discussed which is more secure than the classical DES algorithm. A graph based modified algorithm using an anti-magic labeling process on a cube to arrive at a secret cipher key that reduces the probability of various attacks is under investigation.Besides this, an unauthorized user cannot be able to decrypt the message in a feasible amount of time in spite of knowing the secret key. The future depends on detailed simulation and comparison with other encryption algorithm, as well as complete statistical survey and analysis with respect to other attacks.

Furthermore, this proposed technique can be implemented in the embedded system, smart card security, cloud computer security, etc.,Thus, the amount of time taken and the number of elementary operations performed by the algorithm differ by at most a constant factor.

REFERENCES

[1] Kahate, and Atul, Cryptography & Network Security (Tata Mc Graw-Hill Education, 2011).

[2] Debajit Sensarma, Samar Sen Sarma. A, Graph Based Modified Data Encryption Standard Algorithm with Enhanced Security, International Journal of Research in Engineering, 3(3), 2014.

[3] J. A. Macdougall, M. Miller, Slamin, and W. D. Wallis, Vertex-magic total labeling of graph, submitted.

[4] A. Dougals, Introduction to Graph Theory Upper Saddle River (Prentice Hall inc, 1996).

[5] The DES algorithm illustrated by j.orlin grabbe

[6] Stallings, and William, Cryptography and Network Security Principles and Practice(Prentice Hall, 2011).