11384_Diffie Hellman

Transposition Techniques Transposition Techniques (Class-L7) (Class-L7)

Transposition Techniques Transposition Techniques (Class-L7) (Class-L7)

Transposition Ciphers• now consider classical

transposition or permutation ciphers

• these hide the message by rearranging the letter order without altering the actual letters used

• can recognise these since have the same frequency distribution as the original text

Types of Transposition Ciphers Techniques

• Rail Fence Technique

• Columnar Transposition Techniques– Simple Columnar Transposition

Techniques– Simple Columnar Transposition

Techniques with multiple rounds

Rail Fence cipher Technique

• write message letters out diagonally over a number of rows

• then read off cipher row by row

1. Write down the plain text message as a sequence of diagonals

2. Read the plain text written in Step1 as a sequence of rows.

Example1. Given Plain text – Hello EveryoneH l o v r o e e l E e y n

2. Cipher Text : HlovroeelEeyn

Example:

• Plain text : CAPSTONE PROJECT

Simple Columnar Transposition Techniques

1. Write the plain text message row by row in a rectangle of a pre-defined size

2. Read the message column by column. It can be any random order such as 2,3,1 etc

3. Thus, The message obtained is the cipher text message.

Example• 1. Given Plain text– Hello Everyone

• Consider a Rectangle with four columns.

H e l l

o E v e

r y o n

e

Example• Decide a random order : 2 4 1 3• Read the text in the order of these

columns• Cipher Text : eEylenHorelvo

Simple Columnar Transposition Techniques

with multiple rounds• Given Text – Hello Everyone• After Round 1, Cipher text is

eEylenHorelvo• Perform Step 1 to 3 as many times

as desired

Example• Round 2:• eEylenHorelvo

• Choose same order of columns : 2 4 1 3• Read the text in the order of these columns

Cipher Text: EneloveeroyHl

e E y l

e n H o

r e l v

o

ExamplePlain Text : CAPSTONE PROJECT1. CPTNPOETASOERJC2. SRAECOJPPTCNETO3. ETRPTONAPOSJECC

Review Questions1. Sita meets Ram and says Rjjy rj ts ymj xfggfym bj bnqq inxhzxx ymj uqfs. If she is using modified Caesar Cipher,What does she wants to convey ???

2. What would be the transformation of a message ‘HAPPY BIRTH DAY TO YOU’ using Rail Fence Technique

3. The following message was received by Bob :hs yis ls. eftstof n^ TyymrieraseMr^ e ho ec^etose Dole^. If the message is encrypted by using The Simple ColumnarTransposition method, with the key as 24153? Find the original Plain text ?

Review Questions4. Consider a scheme involving the replacement

of alphabets as follows:Original : A B C ………. X Y ZChanged to : Z Y X ………. C B A

If Sita sends a message HSLDNVGSVNLMVB, What

should Ram get from this ?

Encryption and Decryption

• As we know the process to transform plain text to cipher text or vice versa – encryption and decryption

Encryption and Decryption… Cont

• In computer communication – sender send the encrypt message through the network.

• Receiver received the message and decrypt it to plain text.

• To encrypt and decrypt the message – encryption and decryption algorithm

• Usage both must be same each other – otherwise the decryption cannot success

• Others method using key = one time pad in Vernam Technique

• Algorithm – know to everyone– to made the message secure – use the key.


• Every Encryption and Decryption process has two aspects :

– Algorithms– Key (used for encryption and

decryption)

• There are 2 cryptography mechanism– Symmetric key Cryptography = use

same key to encrypt and decrypt the message

– Asymmetric key Cryptography – use different key in encrypt and decrypt the message


Symmetric and Asymmetric Key Cryptography

• Symmetric Key Cryptography and the problem key problem

Symmetric Key Cryptography and key distribution problem… Cont

– Problem in transmission. Create the same problem– By courier may improve the situation – same

problem happened– Another option by hand-delivery mechanism– Others idea – put the envelope in box and locked it –

difficult to receiver to opened it – Another KEY?– Send key by another way– No solution completely acceptable – either not fully

proof or not practically possible – called key distribution problem / key exchange problem

– Same key to encrypt and decrypt – symmetric key operation

– Let say A want to send different message to 2 person B and C. so need 2 different pair of key



– How about involve more than 5 person?

•10 pairs of key and locked needed– So we can write in Mathematic

•Person involve (PI)– PI * (PI – 1) / 2

•Let say 1000 person involve – 1000 * (1000 – 1) / 2 = 499,500 lock and key

pairs

– Remember that locked and key pair must be maintained by somebody

Diffie-Hellman Key Exchange / Agreement Algorithm

– Introduction

•Solution to the problem of key agreement or key exchange

•Both parties can agree on a symmetric key – used in encryption / decryption

•Based on mathematical principle – describe the step in algorithm, illustrate by example and discuss mathematic basic

Diffie-Hellman Key Exchange / Agreement Algorithm… Cont

– Description of the algorithm

– Let say Alice and Bob agree upon a key that used in encrypt/ decrypt. The step by using Diffie-Helman algorithm shown in the figure.

• Base on the step involve, actually the keys is similar; K1=K2=K

Diffie-Hellman Key Exchange / Agreement Algorithm

Alice

A = g^x mod n

A

K1=B^x modn

Bob

B = g^y mod n

B

K2=A^y modn

Alice and Bob agree on two prime numbers, n and g

1

2

3

6

4

5

7

As it turns out, K1=K2=K. Thus K becomes the shared Symmetric key

between Alice and Bob

– Example of the algorithm•Let take small exam to prove the Difie-

Helman Key Exchange. This have been shown in Fig. 3.2



– Mathematical Theory behind the algorithm•1st lets look at technical description of the

complexity of the algorithm

•What is actually means– Take a look what Alice does in step 6.

Here the compute»K1 = Bx mod n»What is B? from step 4.»B = gy mod n»There for, substitute this value of B

in step 6 we have following equation»K1 = (gy)x mod n = gyx mod n


– What Bob does in step 7. Here Bob compute;

» K2 = Ay mod n » What is A? from step 2 we have;» A = gx mod n» There for, substitute this value of A in step 7 we have

following equation» K2 = (gx)y mod n = gxy mod n

– Now basic mathematic say» Kyx=Kxy

– So we get K1=k2=k. Hence the Proof– Obviously question, if Alice and Bob can generate key

separately – so can attacker. – Solution: exchange n,g,A,B; base on x and y that cannot

easily be calculated – mathematically


Why Diffie-Hellman Works ???

Alice Bob

g gOne-Third of the key(g) is public

g g

x y

g

y

x

g

x

y

Alice sends the key to Bob

Bob sends the key to Alice

Alice fillup

Another one-third

Of the key using

her secret random number

Bob fillsup

Another one-third

Of the key using

his secret random number

Alice Completes

The keyBy addingThe last

PartReceived

FromBob

BobCompletes

The keyBy addingThe last

PartReceived

FromAlice

Both keys are the same because the order of Completion (x First or y First) does not matter

– Problem with algorithm• Diffie Helman Key Exchange – not solve all the

problem associated with key exchange• Can fall pray to the man-in-the-middle attack

that also called bucket brigade attack.• This happened as follow

– As usual Alice send n and g to Bob, let say n=11 and g=7 (those code is basic in calculate the key K1=K2=K)

– She not realize that the hacker (Tom) listening for their conversation. Tom copy all the value.


– Lets say all select random value for x and y as shown in fig.3.24

– All the three person calculate A and B with x and y that have been selected. Note that Alice calculate for A and Bob calculate for B but Tom calculate both, A and B. this can be look at fig. 3.25


Fig. 3.24

Fig. 3.25

– The real Drama:»Alice send A(e.i: 2) to Bob, Tom

intercept it and send his A (e.i: 9)to Bob. Bob not realized it.

»Bob send his B (e.i: 8) to Alice, Tom intercept it and send his B (e.i: 4) Alice. Alice accept it and not realized what had happened

»At this juncture, Alice, Bob, and Tom have a value A and B as shown in 3.27




» Base the key, those three person generate the key as shown in 3.28

» Why Tom generate 2 keys?

Fig. 3.27

Fig. 3.28

Asymmetric Key Operation

• Asymmetric Key Operation

– Have 2 keys ; encrypt and decrypt the message.

– Let say A want to send a message to B, so B will send K1 to A to be used in encrypting the message

– B will open / decrypt the message by using K2

– K1 != K2– K1 everyone know… but not K2; only B

knows.– K1 known as public Key and K2 as private

key



– What if B want to received a message from C; C may use the same key (K1) to encrypt the message and B can use K2 (same as used to decrypt message A)

– Only a pair of key is needed to execute cryptography process (if B want to receive from 1000 person)

– But if they want to communicate with each other they will need 1000 lock, 1000 K1, 1000 K2 – not as symmetric (499, 500)

• Technique that hiding the message inside other message

• Historically, the sender use invisible ink, tiny pin puncher on specific character, etc

• Of late – hiding behind the picture

Steganography

Fig. 3.30

• The cryptanalyst is armed with the following information– The encryption/ decryption algorithm– The encrypted message– Knowledge about the key size

• Keys – challenge for the attackers• Attackers can develop programming to

solve the keys – depend to size of key

Key Range and Key Size

Key Range and Key Size… Cont

• How attackers determine either the message that he/she decrypt is the plain text or the right keys?

• How to prevent?– Size of the keys (40, 56, 120 etc)– In bit– Bigger the key size; long time to

crack



– Can also represent the possible values in the key range using hexadecimal

– Complexity to the attackers– The keys size chasing by the technology

• today- 56 bit not safe; • Tomorrow – 128bit may not safe• Another day – 256-

– But impossible to 512 bit; why?• Suppose that every atom in the universe is

actually a computer• In the world – 2300; if each computer can check

2300 keys in one second (which cannot happen)• 2162 millennia to search 1% of 512 bit key;



• 3 possibility Attacks can be occurs– Cipher Text only Attack

• The attacker doesn’t know about the plain text. Has some and all the cipher text. Guess the meaning of the message

– Know Plain Text Attack• Know some pair of plain text and corresponding

the cipher text – tries to find other pairs of plain text

– Chosen Plain Text Attack• Chosen the plain text block – try to look for

encryption of the same in cipher text

Possible Types of Attacks

Question• Alice and Bob want to establish a

secret key using the Diffie-Hellman Key Exchange protocol.

• Assume n=11, g=5, x=2, y=3. • Find out the values of A, B and the

secret Key (K1, K2)

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11384_Diffie Hellman