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Indian Journal of Science and Technology

Vol.2 No. 8 (Aug 2009) ISSN: 0974- 6846

Research article

Ιndian Society for Education and Environment (iSee) http://www.indjst.org Indian J.Sci.Technol.

“Cipher system” Sokouti et al.

9

An approach in improving transposition cipher system

Massoud Sokouti 1, Babak Sokouti 2 and Saeid Pashazadeh 1

1 Faculty of Electrical and Computer Engineering, University of Tabriz, Tabriz, Iran

2 Faculty of Engineering, Islamic Azad University -Tabriz Branch,Tabriz, Iran

m_sokouti@yahoo.com

An approach in improving transposition cipher system

Abstract: Abstract: Transposition ciphers are stronger than simple

substitution ciphers. However, if the key is short and the

message is long, then various cryptanalysis techniques

can be applied to break such ciphers. By adding 8 bits

(one byte) for each byte using a function and another

mathematical function to position the bits in a binary tree

and using its in-order tour, this cipher can be made

protected. Using an in-order tour of binary tree can diffuse

the eight bits (includes 7 bits produced by the function

and 1 random bit) and eight bits of the plaintext. This can

highly protect the cipher. However, if the key

management processes are not secured the strongest

ciphers can easily be broken.

Keywords: Cryptography, binary tree, cipher protection,

transposition cipher.

Introduction Introduction

With the growth of Internet,

hackers and spies make big issues

for army forces, organizations and

companies. Transposition ciphers are

still the most important

techniques in the construction of

modern symmetric encryption

algorithms. We will clearly see

combinations of substitution and

transposition ciphers in two important

modern symmetric encryption

algorithms: DES and

(Mao Hewlett W-Packard,

DES is not very secure because of

the limitation of the space of the key

is 56 bits. On the other hand AES is a

new cipher. The benefit of these 2

ciphers is that they have two factors

of cryptology and security, diffusion

and confusion (Schinier, 1996). In

part 3 we will completely clarify these

two factors. In this paper we present

a new algorithm that mostly uses

transposition cipher, diffusion and

modification of block cipher.

Transposition ciphers Transposition ciphers

Transposition ciphers are an

important family of classical ciphers,

in additional substitution ciphers,

which are widely used in the

constructions of modern

ciphers (Mao Hewlett W-Packard,

2003).

In a transposition cipher the

kernel

AES

2003).

block

plaintext remains the same, but the order of characters is

shuffled around. In a simple columnar transposition

cipher, the plaintext is written horizontally onto a piece of

graph paper of fixed width and the cipher text is read off

vertically. Decryption is a matter of writing the cipher text

vertically onto a piece of graph paper of identical width

and then reading the plaintext off horizontally (Schinier,

1996).

Cryptanalysis of these ciphers is discussed in

(Sinkov, 1966; Gaines, 1956). Since the letters of the

cipher text are the same as those of the plaintext, a

frequency analysis on the cipher text would reveal that

each letter has approximately the same likelihood as in

English. This gives a better clue to a cryptanalyst, who

can apply a variety of techniques to determine the right

ordering of the letters to obtain the

plaintext (Schinier, 1996)

Putting the cipher text through a

second transposition, cipher greatly

enhances security. There are even

more complicated

ciphers, but computers can break

almost all of them. The German

ADFGVX cipher, used during World

War I, is a transposition cipher

combined with a simple substitution. It

was a very complex algorithm for its

day but was broken by Georges

Painvin, a French cryptanalyst (Kahn,

1967; Schinier, 1996).

We can give an example for

transposition cipher. In our example

the key is a small number for

example 5. In order to encrypt a

message using this key, we write

the key in rows of 5 letters and

encrypt by writing the letters of the

first column first, then the second

column, etc. If the length of the plain

text is not multiple of 5 then we add

the appropriate number of “z”s at

the end before we encrypt (Table 1,

2).

Transpositions of this type are

easy to break. Since the key must

be a advisor of the cryptogram

length, an attacker has only to count

the length of the cryptogram and try

each divisor in turn (Piper &

Murphy, 2002).

iitosesetteepnfgccadidatgelznhlkrersrhrnaz

transposition

Teble 1

Plain text:

ifnightclocksaredesiredstartthegreenplan

Key: 5

i i

h h

c c

e e

r r

a a

e e

n n

f f

t t

k k

d d

e e

r r

g g

p p

n n

c c

s s

e e

d d

t t

r r

l l

i i

l l

a a

s s

s s

t t

e e

a a

g g

o o

r r

i i

t t

h h

e e

n n

Cipher text:

ihceraenftkdergpncsedtrlilassteagorithen

Table 2

Plain text:

ifnightclocksaredesiredstartthegreenplan

Key: 3

i f i f

i g i g

t c t c

o c o c

s a s a

e d e d

s i s i

e d e d

t a t a

t t t t

e g e g

e e e e

p l p l

n z n z

Cipher text:

n n

h h

l l

k k

r r

e e

r r

s s

r r

h h

r r

n n

a a

z z

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Indian Journal of Science and Technology

Rotor machines Rotor machines

In the 1920s, various mechanical encryption devices

were invented to automate the process of encryption.

Most were based on the concept of a rotor, a mechanical

wheel wired to perform a general substitution.

Go!

Keyword

-----------

Go!

A rotor machine has a keyboard and a series of

rotors, and implements a version of the Vigenère cipher.

Each rotor is an arbitrary permutation of the alphabet, has

26 positions, and performs a simple substitution. For

example, a rotor might be wired to substitute “F” for “A,”

“U” for “B,” “L” for “C,” and so on. And the output pins of

one rotor are connected to the input pins of the next. For

an example, in a 4-rotor machine the first rotor might

substitute “F” for “A,” the second might substitute “Y” for

“F,” the third might substitute “E” for “Y,” and the fourth

might substitute “C” for “E”; “C” would be the output

cipher text. Then some of the rotors shift, so next time the

substitutions will be different.

It is the combination of several rotors and the gears

moving them that make the machine secure. Because the

rotors all move at different rates, the period for an n-rotor

machine is 26n. Some rotor machines can also have a

different number of positions on each rotor, further

frustrating cryptanalysis.

The best-known rotor device is the Enigma. The

Enigma was used by the Germans during World War II.

The idea was invented by Arthur Scherbius and Arvid

Gerhard Damm in Europe. It was patented in the United

States by Arthur Scherbius (Schinier, 1996; Scherbius,

1928). The Germans beefed up the basic design

considerably for wartime use.

The German Enigma had three rotors, chosen from a

set of five, a plug board that slightly permuted the

plaintext, and a reflecting rotor that caused each rotor to

operate on each plaintext letter twice. As complicated as

the Enigma was, it was broken during World War II. First,

a team of Polish cryptographers broke the German

Enigma and explained their attack to the British. The

Germans modified their Enigma as the war progressed,

and the British continued to cryptanalyze the new

versions. Explanations of how rotor ciphers work and how

they were broken have already been reported (Kahn,

1967; Barker, 1977; Deavours & Kruh,1985; Diffie &

Hellman, 1979; Deavours,1980; Konheim, 1981; Rivest,

1981; Welchman, 1982; Hagelin,1994). Two fascinating

accounts of how the Enigma was broken were provided

by Hodges (1983) and Kahn (1991). For the enigma

simulator vide Carlson, Andy, Enigma simulator, http://

homepages.tesco.net/~andycarlson/enigma/eigma_i.html

Confusion and diffusion Confusion and diffusion

The two basic techniques for obscuring the

redundancies in a plaintext message are confusion and

diffusion (Shannon, 1949). Confusion obscures the

Vol.2 No. 8 (Aug 2009) ISSN: 0974- 6846

Research article

Indian Society for Education and Environment (iSee) http://www.indjst.org Indian J.Sci.Technol.

“Cipher system” Sokouti et al.

10

relationship between the plaintext and the cipher text.

This frustrates attempts to study the cipher text looking

for redundancies and statistical patterns. The easiest way

to do this is through substitution. A simple substitution

cipher, like the Caesar Cipher, is one in which every

identical letter of plaintext is substituted for a single letter

of cipher text. Modern substitution ciphers are more

complex: A long block of plaintext is substituted for a

different block of cipher text, and the mechanics of the

substitution change with each bit in the plaintext or key.

This type of substitution is not necessarily enough; the

German Enigma is a complex substitution algorithm that

was broken during World War II.

Diffusion dissipates the redundancy of the plaintext

by spreading it out over the cipher text. A cryptanalyst

looking for those redundancies will have a harder time

finding them. The simplest way to cause diffusion is

through transposition (also called permutation). A simple

transposition cipher, like columnar transposition, simply

rearranges the letters of the plaintext. Modern ciphers do

this type of permutation, but they also employ other forms

of diffusion that can diffuse parts of the message

throughout the entire message. Stream ciphers rely on

confusion alone, although some feedback schemes add

diffusion. Block algorithms use both confusion and

diffusion. As a general rule, diffusion alone is easily

cracked (although double transposition ciphers hold up

better than many other pencil-and-paper systems)

(Schinier, 1996).

Block cipher Block cipher

In a block cipher, the bit-string is divided into blocks

of a given size and the encryption algorithm acts on that

block to produce a cryptogram block that, for most

symmetric ciphers, has the same size.

Block ciphers have many applications. They can be

used to provide confidentiality, data integrity, or user

authentication, and even be used to provide the key

stream generator for stream ciphers. With stream ciphers,

it is very difficult to give a precise assessment of their

security. Clearly the key size provides an upper bound of

an algorithm’s cryptographic strength. However with the

Simple Substitution Ciphers, having a large number of

keys is no guarantee of strength. A symmetric algorithm

is said to be well designed if an exhaustive key search is

the simplest form of attack. Of course, an algorithm can

be well designed but, if the number of keys is too small,

also be easy to break.

Designing strong encryption

specialized skill. Nevertheless there are a few obvious

properties that a strong block cipher should possess and

which are easy to explain. If an attacker has obtained a

known plaintext pair for an unknown key, then that should

not enable them deduce easily the cipher text

corresponding to any other plaintext block. For example,

an algorithm in which changing the plaintext block in a

known way produces a predictable change in the cipher

text, would not have this property. This is just one reason

algorithms is a

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Indian Journal of Science and Technology

for requiring a block cipher to satisfy the diffusion property

which is that a small change in the plaintext, maybe for

example in one or two positions, should produce an

unpredictable change in the cipher text.

During the threat posed by exhaustive key searches it

is likely that the attacker tries a key that differs from the

correct value in only a small number of positions. If there

were any indications that the attacker had, for instance,

tried a key which disagreed with the correct key in only

one position, then the attacker could stop his search and

merely change each position of that specific wrong key in

turn. This would significantly reduce the time needed to

find the key and is another undesirable property. Thus

block ciphers should satisfy the confusion property which

is, in essence, that if an attacker is conducting an

exhaustive key search then there should be no indication

that they are ‘near’ to the correct key (Singh, Simon,

2000; Piper & Murphy, 2002).

Since block ciphers are used in CBC mode, the

plaintext must be converted into an integral sequence of

blocks. For this we append padding to the plaintext and a

padding length of 1 byte. The padding length must be

equal to all bytes of the padding, and the total length (the

plaintext, the padding, and the padding length) must be a

multiple of the block size. When the cipher text is

decrypted, the last byte specifies the length of the

padding to be removed. The padding structure is also

checked and an error is issued if it is not valid.

Note that the padding does not need to be the

shortest one. It can actually be longer in order to hide the

real size of the plaintext to a potential adversary

(Vaudenay, 2006).

Method Method

We represent an algorithm in turbo C++ programming

language. The benefit of this algorithm is that it uses

diffusion of 15 bits for every one byte using a binary tree

and doubling the size into two bytes. Also the block cipher

is used for every byte. We use a function to diffuse the 15

bits between the bits of plain text, F(x), consists of a

Vol.2 No. 8 (Aug 2009) ISSN: 0974- 6846

Research article

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“Cipher system” Sokouti et al.

11

prime number more than the length of the plaintext (p),

the key of the transposition cipher (k), a (g) number

generator less than p, a (c) positive constant number less

than eight and x that represents the (n-1)th character of

the plaintext which starts from 0. The equation (1) shows

the F(x) function. To reduce the result of F(x) in the range

of 0 to 127 (to make the binary tree), F(x) will be reduced

by performing F(x) mod 128.

F(x) = (k.gx + c) mod p (1)

Now suppose that the outputs of F(x) in bits are f [0], f

[1], f [2],…,f [6] and the bits of plain text are p [0], p [1], p

[2], … ,p [7]. We can diffuse these 15 bits in the binary tree

shown in Fig 1 that can be written as follows if we name

the binary tree nest as T(0…14) and will be substituted by

f(0..6) and p(0..7) (Fig.1):

In this paper, the messages can be represented as

M=m0||m1||…||mn (n: 0 … infinite) in which mi is a character

that is consists of 8 binary bits while converted from the

ASCII codes mi=b0||b1||…||b8 ( i:1..n ). We used another

function for filling the binary tree in a manner that is not

predictable, the function Y(j) is as follows in equation (2):

Y(j)=(gk+c) mod 15 j:0..14 , Y(j): 0 .. 14 (2)

In which the key of the transposition cipher (k), a (g)

number generator less than p, a (c) positive constant

number less than eight. Notice that in this function we will

repeat the calculations until all the numbers of nests in

the binary tree (i.e. 0..14) are produced since some of the

produced numbers will be produced two times. In the

random number generation of 0..14 that are produced in

a loop, c and k are increased plus one in each loop which

is repeated.

Now in this stage the new tree will be constructed

from the normal tree with T(0..14) components in a

manner that the first component of the array Y(j) is the

position of the first component of the array T and will be

continued until the new tree is filled that is illustrated in

Fig 2.

Recall that in an in-order tour

in-order tour of a binary tree, codes

are in Appendix 1, we traverse the tree by first visiting all

the nodes in the left subtree using an in-order tour, then

visiting the root, and finally visiting all the nodes in the

right subtree using an in-order tour. To use the

transposition cipher with k key the in-order tour of the tree

will be written and after converting to the particular ASCII

p [0]

f [0]

f [2]

f [1]

f [5]

f [6]

f [4]

f [3]

p [1]

p [2]

p [3]

p [4]

p [5]

p [6]

p [7]

Fig1. Diffusion of 15 bits, f[ ] are the F(x) bits and p [ ] are

the plain text bits in level-order tour of a binary tree

T(0)=f[0]; T(1)=f[1]; T(2)=f[2]; T(3)=f[3]; T(4)=f[4];

T(5)=f[5]; T(6)=f[6]; T(7)=p[0];

T(8)=p[1]; T(9)=p[2]; T(10)=p[3]; T(11)=p[4]; T(12)=p[5];

T(13)=p[6]; T(14)=p[7];

T(4)

T(3)

T(10)

T(14)

T(0)

T(6)

T(5)

T(12)

T(13)

T(9)

T(11)

T(8)

T(1)

T(7)

T(2)

Fig.2. . A typical sample of the new tree which is used to

make the in-order tour of the binary tree

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Indian Journal of Science and Technology

code of the 16 bits in which the last bit is randomly

produced, the transposition cipher algorithm is applied to

the ASCII characters.

But in a manner that “z” character in not used for

filling the empty positions of the transposition cipher,

instead we use random characters from 256 characters of

ASCII code.

Encryption algorithm

• Input Message (M)

• Input Keys (k,p,g,c)

• Convert Message Characters to Bits

• Calculate F(x)=((kgx+c) mod p) mod 128) for every 8

bits (one character)

• Construct primary level-order tour of binary tree using

extra bits (f[0],…,f[6]),

(p[0],…,p[7])

• Calculate Y(j)=(gk+c) mode 15 to find the new positions

of the tree bits Y(j),j:0..14

• Construct the secondary binary tree according to the

new situations calculated by Y(j)

• In-order tour of the secondary binary tree is calculated

• Calculate the 16th bit of the double character randomly

• Convert the 16bits in two 2 ASCII characters

concatenated to each other to form the new plaintext

• Encrypt the new plaintext using transposional cipher

algorithm and k key and a random character instead of

“z” to from the ciphertext

Decryption Algorithm

• Input Cipher text

• Input keys (k,g,c)

• Decrypt the cipher text using transposition cipher

algorithm and k key

• Convert the ASCII characters to bits

• Every 16 bits (left to right) will be presented as one

secondary binary tree

• Reverse in-order tour of the decrypted cipher text is

calculated to form the secondary binary tree

• Reverse level-order tour of the secondary binary tree.

• Calculate Y(j)=(gk+c) mode 15 to find the positions of

the tree bits Y(j),j:0..14

• Construct the primary binary tree using extra bits

(f[0],…,f[6]), plaintext character bits (p[0],…,p[7]) by using

Y(j)

• Plaintext will be resolved from converting the resulted

bits into ASCII characters

Discussion Discussion

Using this algorithm with this kind of transposition

cipher can make the cipher unbreakable. It doubles the

length of the plaintext and diffuses the bits of plain text by

using the in-order tour of binary tree.

Since we have used the in-order tour of binary tree, no

one can guess the original plaintext. This algorithm has

another advantage that the results of F(x) bits are

diffused by the in-order tour of binary tree too. This cipher

uses 256 characters of ASCII code. The key which is

needed for encryption is:

Vol.2 No. 8 (Aug 2009) ISSN: 0974- 6846

Research article

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“Cipher system” Sokouti et al.

12

plaintext character bits

Key: (k,p, g, c)

And the key which is needed for decryption is:

Key: (k,g,c)

When determining the order of an algorithm, we are

only concerned with its category, not a detailed analysis

of the number of steps. We obviously want to use the

most efficient algorithm in our programs. Whenever

possible, choose an algorithm that requires the fewest

number of steps to process data. We want to show that

this method can be strongly stable against the attacks of

transposition cipher which are listed as below:

1. If the attacker checks all divisors of the length of the

cipher text as a key of transposition cipher he/she will

find the plaintext. And the order of finding the plaintext

is as follows:

2. We had to use just the character z at the end of the

cipher text according to the key.

3. If we give a key more than length of the key the

plaintext will not be encrypted. For example if key is

more than plaintext length then: Plaintext = flower

Key = 7 7

Cipher text = flowerz flowerz

4. The order of encryption of transposition cipher

according to the following algorithm is O(2n):

Input(text, key) Input(text, key)

j ? -1;

forfor i? 0 to length of (texttext)-1 do

if if (i mod keymod key = 0) then then ++j;

ciphercipher[j][i modmod keykey] ? text

//construct the column model using k key with the

order of O(n)O(n)

if if(length of (text text) mod key mod key != 0) then

for for z?length of (texttext)mod keymod key) to

ciphercipher[(length of (text

//adding ‘z’ characters where ever needed with the

order of Θ Θ(k) =O (n) k<=n (k) =O (n) k<=n

Return

5. The order of decryption of transposition cipher

according to the following algorithm is O(n):

Input(cipher, key) Input(cipher, key)

a? 0;

for for i?0 to to length of (cipher) (cipher)/key

for for j?0 to key 0 to key-1 do do

texttext[a] ? ciphercipher[(l/key

++a;

//decrypt the ciphertext using transposition cipher

and k key with the order of O(n)

6. The complexity of brute force attack against

transposition cipher according to the following algorithm

is O(n2):

Input(cipher) Input(cipher)

forfor key?1 to to length of (cipher cipher) do

if if (length of (ciphercipher) mod mod key = 0)

//finds the divisors of cipher text and takes it as a

key with the order of O(n)O(n)

for for i?0 to to (length of (ciphercipher) / key ) – 1 do

for for j?0 to to j<=key-1 do

flower

do

text[i];

then

to (keykey-1) do

key)][z] = 'z';

do

text) / key

key-1 dodo

key)*j+i];

O(n)

do

do

do

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Indian Journal of Science and Technology

texttext[a] ? cipher

* j + i];

++a;

//checks the divisor keys in the cipher text to reveal

the plaintext

−

==

k

k

10i0j

Our method characteristics:

1. If the attacker can break the transposition part, again

the brute force is the only way to break the cipher and

its order is calculated as below.

2. Instead of using the character z we use a random

character from the 256 characters of ASCII code. This

makes the cipher safer.

3. Because we use 2 functions to fill the binary tree and

in-order tour so we can give a key more than the length

of the plaintext.

Plaintext = flower flower

Key = 7 7 p=19 19 g=5 5 c=2 2

Cipher text = 'rC6j?'rC6j?‼ ‼#r` #r`╥¼6 ¼6τ τ

The order of encryption method according to the following

algorithm is O(3n2):

Input(plaintext,key,p,g,c) Input(plaintext,key,p,g,c)

v ? 0, z ? 1, y ? 1, flag ? 1, l ? 0, c1

key key;

while while (l != 15) dodo

for for i?1 to key to key dodo

y ? g g * y;

form1[l] ? ( y + c c) mod mod 15;

if if (l >= 1) thenthen

forfor j?0 to to l-1 do do

if if (form1[j] = form1[l]) then then

flag? 0;

if if (flag = 1) then then

++l;

++keykey, ++c c, flag ? 1, y ? 1;

//creates 15 numbers between 0 to 14 with

the order of O(2n O(2n2 2) )

forfor i?0 to to length of (plaintextplaintext) – 1 do

forfor j?0 to to 7 do

bit_text[j] ? j'th bit of plaintext plaintext[i];

x ? i;

y? 1;

for for j?0 to to x-1 do do

y ? g g * y;

//order of the power according to the last for is O(n

form2 ? ((k k * y + c1 c1) mod mod p p) mod

y? 1;

for for j?0 to to 7 dodo

bit_form2[j] ? j'th bit of form2[i];

forfor j?1 to to 7 do do

tree1[j - 1] ? bit_form2[j];

for for j?7 to to 14 dodo

tree1[j] ? bit_text[j - 7];

forfor j?0 to to 99 do do

tree2[j] ? ' ';

Vol.2 No. 8 (Aug 2009) ISSN: 0974- 6846

Research article

Indian Society for Education and Environment (iSee) http://www.indjst.org Indian J.Sci.Technol.

“Cipher system” Sokouti et al.

13

cipher[(length of (cipher cipher / key)

∑∑

=

∑

=

k

∑∑

=

1

∑

==

−

=

===

n

k

l

nn

k

l

i

n nllkOrder

1

2

1

1

0

1 -k

1

c1 ? c c, k k ?

do

do

O(n2 2) )

mod 128;

//the order of in_order tour is O(n)

the shape of tree it is O(1)

tree4[15] ? a random bit;

cipher1[2 * i] ? character of tree4[0...7];

cipher1[2 * i + 1] ? character of tree4[8...15];

v ? 0;

j ? -1;

for for i?0 to to length of (cipher1) - 1 do

if if (i mod mod k k = 0) then

++j;

cipher2cipher2[j][i mod k k] ? cipher1[i];

//starts the transposition cipher with the order of O(n)

if if (length of (cipher1) mod mod k k != 0) then

for u?length of (cipher1) mod

cipher2 cipher2[length of (cipher1) / k k][u] ? a random character;

//puts the random characters according the key with the

order of O(n) O(n)

return cipher2cipher2;

4. The order of decryption method according to the

following algorithm is O(2n2):

Input(cipher,key,g,c) Input(cipher,key,g,c)

a ? 0 , k , k ? key key , v ? 0 , m ? 0;

l? length of (cipher cipher);

while while (t != 15) do do

for for i? 1 to key do to key do

y ? g g * y;

form1[t] ? ( y + c c) mod mod 15;

if if (t >= 1) thenthen

for for j? 0 to to t-1 do do

if if (form1[j] = form1[t]) then

flag ? 0;

if if (flag = 1) then then

++t;

++key key, ++c c, flag ? 1, y ? 1;

//creates 15 numbers between 0 to 14 with the

order of O(2n O(2n2 2) )

for for i? 0 to to l/k k-1 do do

for for j? 0 to to k k-1 do do

ciph[a] ? ciphercipher[(l / k k) * j + i];

++a;

//opens the transposition cipher with the order of O(n)

for for j? 0 to to 98 dodo

tree1[j]? ' ';

for for j? 0 to to 14 do do

tree1[j]? 2;

invinor ? invinorder(tree1,tree1[0],0);

//finds the positions of a default tree in in_order tour

form with the order of O(1) O(1)

forfor i? 0 to to l-1 do do

for for j ? 0 to to 7 do do

bit_ciph[j] ? j'th bit of ciph[i]

for for j? 8 to to 15 dodo

bit_ciph[j] ? (j-8)'th bit of ciph[i+1]

forfor j? 0 to to 14 do do

for for j?0 to to 14 do

tree2[form1[j]] ? tree1[j];

tree4 ? inorder inorder(tree2,tree2[0],0);

do

O(n) but because we know

O(1)

do

then

O(n)

mod k tok to k k-1 do do

then

(n)

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Indian Journal of Science and Technology

inorder[j] ? bit_ciph[j];

for for j? 0 to

tree1[invinor[j]] ? inorder[j];

forfor j? 0 to to 14 do

tree2[form1[j]]? tree1[j];

for for j? 7 to to 14 do

bit_text[j-7]? tree2[j];

texttext[m]? character of bit_text[0…7];

++i, ++m;

//decrypts the tree cipher with the order of O(n)

return texttext;

5. The only way to break this cipher system is to brute

force and make exhausted searches with the

complexity of ((28)2)n in which it represents the double

sized cipher text and n is the plaintext length.

This problem has the complexity of 512n, where n is

the plaintext length. The plaintext of eight or more

characters would provide more resistance to a brute force

attack in comparison to 128-bit AES and has the same

resistance to brute force attack as 256-bit AES.

This methodology does have some potential

drawbacks. The most important drawback is that the size

of the encrypted plaintext will be doubled. For areas with

low bandwidth or limited storage capacity this cipher is

not useful. However for most communication channels

where encryption is required, an increase in the plaintext

size will not have a significant impact. Also using a good

random generator in function F(x) can make the cipher

effective. All of the bits are important because if one of

the bits is lost the remainder of the message will be

useless.

Conclusion Conclusion

Although the classical transposition cipher is not

secure when it is used to encrypt unpadded messages,

this and other stream ciphers can be made as secure as

most block ciphers by using two functions, one (F(x)) for

producing the 7 extra bits of padding to each byte to

diffuse the language characteristics and the second

function (Y(j)) for positioning the bits in the binary tree to

use its in-order tour in the encryption and decryption

process that will make it as secure as 256-bit AES cipher

system. This method can resolve all the limits of the

classical transposition cipher. The methodology can also

be used for a 64 bit plaintext resulting in 128 bit cipher

text including of 127-nest binary tree plus 1 random bit

that will produce the whole 128 bits. Considering that our

method was used for 8 bit plaintext and was as secure as

the 256-bit AES and if it is used for a 64-bit plaintext its

security will be increased dramatically.

It is worth mentioning that if the key management is

not well-provided, all strong cipher will be compromised

easily. Our investigations find that there is a paucity of

data on improvements of transposition ciphers. When

designing solutions to programming problems, we are

concerned with the most efficient solutions regarding time

and space. We will consider memory requirements at a

later time. Speed issues are resolved based on the

Vol.2 No. 8 (Aug 2009) ISSN: 0974- 6846

Research article

Indian Society for Education and Environment (iSee) http://www.indjst.org Indian J.Sci.Technol.

“Cipher system” Sokouti et al.

14

to 14 dodo

do

do

O(n)

number of steps required by algorithms. Finally we

mention that if in this method, the number of Child is not

the same (we did it with maxChild=2), it will add to its

strength dramatically that our future work will be done on

this subject.

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Cryptologia. 18 (3), 204–242.

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Indian Journal of Science and Technology

Appendix Appendix

In-order tour function In-order tour function:

With mch = 2; With mch = 2;

func inorder inorder (str1[],st,po)

if (st != ' ') then

inorder inorder(str1,leftMostChild(str1,po,

mch),(mch*po)+1);

str2[v]= st;

++v;

inorder inorder(str1,rightSibling(str1,(mch * po) + 1,mch),

(mch * po) + mch);

return str2;

func leftMostChildleftMostChild(str1,pos,maxChild)

x = 1;

for i<--0 to 98 do

str[i] = str1[i];

while (x <= maxChild)

i = maxChild * pos + x;

if (str[i] != ' ')

return str[i];

++x;

return ' ';

func rightSibling(str1,pos,maxChild)

for i<--0 to 98 do

str[i] = str1[i];

p = pos mod maxChild;

if (p = 0) then

return ' ';

if (str[pos] != ' ')

while (p < maxChild) do

if (str[pos + 1] != ' ') then

return str[pos + 1];

++p , ++pos;

return ' ';

Reverse in-order tour function: Reverse in-order tour function:

With mch = 2; With mch = 2;

func invinorder invinorder(str1[],st,po)

if (st != ' ') then

invinorder invinorder(str1,leftMostChild1(str1,po,

mch),(mch*po)+1);

str2[v]= st;

++v;

invinorder invinorder(str1,rightSibling1(str1,(mch

1,mch), (mch * po) + mch);

return str2;

func leftMostChild1 leftMostChild1(str1,pos,maxChild)

x = 1;

for i<--0 to 98 do

str[i] = str1[i];

while (x <= maxChild)

i = maxChild * pos + x;

if (str[i] != ' ')

return str[i];

++x;

return ' ';

func rightSibling1(str1,pos,maxChild)

Vol.2 No. 8 (Aug 2009) ISSN: 0974- 6846

Research article

Indian Society for Education and Environment (iSee) http://www.indjst.org Indian J.Sci.Technol.

“Cipher system” Sokouti et al.

15

:

* po) +

for i<--0 to 98 do

str[i] = str1[i];

p = pos mod maxChild;

if (p = 0) then

return ' ';

if (str[pos] != ' ')

while (p < maxChild) do

if (str[pos + 1] != ' ') then

return str[pos + 1];

++p , ++pos;

return ' ';

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