DES | Assymetric Key

DES Attack à man in the middle attack

Sender …DES encrypted message + key by hacker ….Receiver

IDEA àInternational Data Encryption Standard (Commercial Version)

Input   : 128 bits

Key     : 128 bits

Output: 128 bits

To get down encryption it take 8 rounds

For doing email encryption is using PGPà Pretty good privacy

RC-5 à Rivest Cipher / Smart Card

AES à Advanced Encryption Standard

Input : 128 bits or 256 bits

Key: 128 bits or 256 bits

Output : 128 bits or 256 bits

Week 4 Day 5 – ITEC Programme

AXCRYPTà Encryptionà EAS

 For Securing key exchange there is save way to do it using Diffie Hellman Key Exchange or Agreement.

Sender ------à------ Receiver

Step 1 : 1^st sender & receiver agree upon two large prime numbers

Step 2 : Sender choses another large random number.

Ex : it is X

Such that sender calculate A= g ^x mod n

Step 3 : This A is sent to receiver

Step 4 : At the receiver end, receiver also chooses large random number that is suppose Y

And calculate B such that B= g ^y mod n

Step 5 : Receiver sends B to sender

Step 6 : Sender calculates the key as K₁= B ^x mod n

Step 7 : Receiver calculates the key as K₂=A ^y mod n so K=K₁=K₂

Example

Step 1 : Let n=11 g=7

Step 2 : ender chooses x  suppose x =3

Step 3 : A=g ^x mod n

                 =7 ³ mod 11

                 = 2

Step 4 : suppose y= 6 and B =4

              B = g ^y mod n

                  = 7 ⁶ mod 11

                  = 4

Step 5 : B will be sent to sender

Step 6 : Sender calculates key value

             K₁ = B ^x mod n

                  = 4 ³ mod 11

                  = 9

Step 7 : Receiver end calculate key value

             K₂ = A ^y mod n

                  = 2⁶ mod 11

                  = 9

            K = k ₁ = k ₂

Assymetric Key

Assymetric Keyà Same Key (Encryption & Decryption)

Assymetric Key à Public Key (Known to all)  and Private key (Remind private or an individual)

Public Key à Encryption, Private Key à Decryption

But if the aim is authentication

Private Key à Encryption, Public Key à Decryption

RSA Algorithm

Working of RSA

Step 1 : Change two large prime numbers , suppose P & Q

Step 2 : Calculate N= P.Q

Step 3 : Select the public key E such that it shared not be the factor of (p-1) (q-1)

Step 4 : Encryption is given by

              C= (PT) ^E mod n

              Where  C =Cipher text

                           PT = Plain text

                            E = Public key

                            N = P.Q

Step 5 : The private key D should be chosen by receiver such that

              (D.E) mod (P-1).(Q-1) = 1

Step 6 : Decryption

             PT = C ^D mod n

              D = Private key

              C = Cipher text

              PT= Plain text

Example :

Step 1 : suppose P=7 Q = 17

Step 2 : n = p. q

                = 7 . 17

                = 119

Step 3 : E such that (p-1) (q-1)

               (p-1) (q-1) = 6.16 = 96

               Suppose E= 5 bcoz 6=1,2,3,6 (6 can be devided by numbers written beside)

               And 16 (1,2,4,8,16) /prime #. So choose which is not there.

Step 4 : C=(PT) E mod N

                = (10) 5 mod 119

                = 40

Step 5 : (D.E) mod (P-1) . (Q-1) =1

              (D.5) mod 96 = 1

              To figure out D à 96 . 4 = 384 + 1 = 385 / 5 = 77

              (D.5) mod 96 = 1

              (77.5) mod 97 = 1

Step 6 : Decryption

             PT = C ^D mod n

                  = 40 ⁷⁷ mod 119

                  = 10

Introduction to Cryptography

History of Cryptography

•Cryptography has roots that began around 2000 B.C. in Egypt

•Encryption methods evolved from being mainly for show into practical applications used to decorate tombto tell the story of the life of the deceased. The practice was not as much to hide the messages themselves, but to make them seem more noble, ceremonial, and majestic used to hide information from others.

•A Hebrew cryptographic method required the alphabet to be flipped so that each letter in the original alphabet is mapped to a different letter in the flipped alphabet. The encryption method was called atbash. An example of an encryption key used in the atbash encryption scheme is shown in following:

    ABCDEFGHI JK LMNOPQ R STU VW XYZ

    ZYXWVUTSR QP ONMLKJ I HGF ED CBA

•Around 400 B.C., the Spartans used a system of encrypting information by writing a message on a sheet of papyrus, which was wrapped around a staff. (This would look like a piece of paper wrapped around a stick or wooden rod.) The message was only readable if it was around the correct staff, which allowed the letters to properly match

What is Cryptography

•From Greek word Kryptos + graphy which means secret writing

•Art and science of achieving security by encoding messages to make them non-readable.

Basic Of Encryption :

Plain text -> Encryption -> Cipher Text -> Decryption -> Plain Text

Conceptual cryptography is to make readable message unreadable and so the opposite.

Goals of cryptosystem

•Cryptosystems can provide confidentiality, authenticity, integrity, and nonrepudiation services. It does not provide availability of data or systems. Confidentiality means that unauthorized parties cannot access information. Authenticity refers to validating the source of the message to ensure the sender is properly identified.  Integrity provide assurance that the message was not modified during transmission, accidentally or intentionally. Nonrepudiation means that a sender cannot deny sending the message at a later date, and the receiver cannot deny receiving it.

What is Cryptanalysis?

Technique of decoding messages from non-readable format back to readable format without knowing how they were initially converted from readable format to non-readable format.

Cryptology

•Combination of Cryptography and Cryptanalysis

•Cryptography + Cryptanalysis= Cryptology

Cryptography

•Plain text

    The original message or data that is fed into the algorithm as input.

    e.g -  This is the plain text.

• Encryption Algorithm

     The algorithm which converts the plain text

     into cipher text by various transpositions and substitutions.

•Secret Key

    The Key is a value independent of the plain text. The algorithm will produce a different output depending on the specific key being used at that time.

•Cipher Text

    The scrambled message produced as the output while changing the plain text as cipher text using encryption algorithm.

•Decryption Algorithm

    It takes the cipher text and the secret key and produces the original plain text.

Requirements for Secure Use of Encryption 

There are two requirements for secure use of encryption:

•Strong Encryption Algorithm

•

•Sender and Receiver must have copies of the secret key and should keep the key secure.

Cryptographic Systems are characterized  by three independent dimensions:

•Type of operations used for transforming plaintext to ciphertext.

•The number of keys used

The way in which the plaintext is processed

•Cryptanalysis

    Cryptanalytic attacks rely on nature of the algorithm

•Brute Force Attacks --> only possible combination are checked for

•Passwords : dictionary and brute force  

DES | DATA ENCRYPTION STANDARD

DES (DATA ENCRYPTION STANDARD)

This is a symmetric key algorithm.

Input is taken à 64 bits

Key is taken à 56 bits

Output is taken à 64 bits

In case of this symmetric key algorithm, the size of the output is either equal or lesser than the size of the input. Generally the size of the output is longer than the input.

16 rounds processing to get it done.

In 1976, a Professor from Santa Clara University Edward found Simplified DES to make a symmetric more simple.

Input à 5 bits

Key à10 bits

Output à 8 bits

P10 : Permutation applied on 10 bits

 P8 : Permutation applied on 8 bits

IP : Initial Permutation

Fk : Complex function

SW : Switch function

Ip^-1 : Inverse of initial Permutation

Let’s try the exercise !

10 bit keyàp10àShiftàp8àShiftàp8

10 bit key = 1010000010

Step 1 à Apply  P10 on the 10 bit key

Given p10 =  3 5 2 7 4 10 1 9 8 6

#      = 1 2 3 4 5 6 7 8 9 10

Key = 1 0 1 0 0 0 0 0 1  0

Answer P10 à 10000 01100

Step 2 : Apply left circular shift separately on second 5 bit

1 0 0 0  0         0 1 1 0 0

1 2 3 4  5         1 2 3 4 5

Start from #2  which is 0 and 3^rd which is 0 (reverse) and 4^th is 0  and 5 which is 0 and 1 which 1

And the result will be 00001 and the rest also start from #2 so the result is 11000

Answer Shift à 0000111000

Step 3 à Apply p8 on the output step 2

Given P8 : 6 3 7 4 8 5 10 9

#      = 1 2 3 4 5 6 7 8 9 10

Key = 0 0 0 0 1 1 1 0 0 0

Answer P8 à 1 0 1 0 0  1 0 0  (k1 Subkey)

Step 4 à Apply left cecular  by 2 bits on output of step 2

0 0 0 0 1               1 1 0 0 0

1 2 3 4 5               1 2 3 4 5

Start from #2  which is 0 and 3^rd which is 0 (reverse) and 4^th is 0  and 5 which is 0 and 1 which 1

And the result will be 00100 and the rest also start from #2 so the result is 00011
Answe Shift à 00100 00011 (k2)

Step 5 à Apply p8 on output of step 4

Given p8 : 6 3 7 4 8 5 10 9

#      = 1 2 3 4 5 6 7 8 9 10

Key = 0 0 1 0 0 0 0 0 1  1

Final Cipher text à 0 1 0 0 0 0 1

Encryption

8 bits Plain text à IPà FKàSWàFKàIP à 8 bits Cipher text

Suppose : PT : 1011101

Step 1 : Initial Permutation

Given à 2 6 3 1 4 8 5 7

#      = 1 2 3 4 5 6 7 8

Key = 1 0 1 1 1 1 0 1

Answer p8 à 0 1 1 1 1 1 1 0

Step 2 : Complex function is combination of substitution and permutation.

0 1 1 1                   1 1 1 0

    L                            R

F_k (L₁R) = L + F (R,SK),R)

Where + is Exclusive OR

F  is Mapping Function

R is Right side

SK is Subkey à K₁

L is left side

Let’s see the result k1 and k2 in previous steps

K1 : 10100100

K2 : 01000011

Suppose output of F (1110,10100100) = 1101

L = 0111

Fk (0111,1110)=(0111+1101,1110)

XOR

0 1 1 1

1 1 0 1

1 0 1 0   à output

Fk = ( 1010,  1110)

             L         R

Step 3 à Switch function

 ( the left L is shifted to R and R is shifted to L / interchanging of L&R)

( 1010,  1110)

    R         L

Step 4 : Complec funcion

Fk (1110,1010)= (1110+F (1010,01000011),1010)

Fk(1010,01000011)= 1100

Fk(1110,1010)=(1110+1100,1010)

                       = (0010,1010)

XOR

1 1 1 0

1 1 0 0

0 0 1 0

Step 5 : Apply IP^-1 to output of step 4

Geven : 4 1  3 5 7 2 8 6

#      = 1 2 3 4 5 6 7 8

Key = 0 0 1 0 1 0 1 0

Final Cipher text à 0 0 1 1 1 0 0 0

TRANSPOSITION

Is the position of the characters of the plaintext

1. Rail Fence Technique

In this case will be given plaintext and dept =2 or depth=3

Step 1 : Arrange the alphabets of the plain text diagonally

Step 2 : The cipher text will be read as row –wise

Ex : Dept = 2 , Given text à Come home 

C                             M                            H                             M                                                                                 

                O                             E                              O                             E                                                        

From this case, the cipher text will be C M H M     O E O E 

Depth= 3

C                                                             H                                                                                                                        

                O                             E                              O                             E                                                        

                                M                                                            M                                                         

From this case, the cipher text will be CH OEOE  MM

2. Simple Columnar

In this step will be given Plaintext (PT) and Random number of columns

Step 1 : Arrange the alphabets of the plaintext row-wise in the number of columns decided by the user

Step 2 : Specify the random order of columns.

Step 3 : Write the cipher text alphabets according to the order in step 2

Ex:

Given à Number of columns 1-6 , Plain Text à Come Home Tomorrow

1	2	3	4	5	6
C	O	M	E	H	O
M	E	T	O	M	O
R	R	O	w

So the Chiper text wil be  6,4,3,1, 2,5 oo,eow,mto,cmr,oer,hm or 6,5,4,3,2,1 its random.

In this case the Cipher text can be applied for second time so the plain text can’t be easily read.

1	2	3	4	5	6
O	O	E	O	W	M
T	O	C	M	R	O
E	R	M	H

3. One-time Pad or VERNAM CIPHER

In this case will be given PT and One-time Pad and A=0

Step 1 : Convert the alphabets of the plaintext into their corresponding numerical equivalents.

Step 2 : Convert the alphabets of the one-time pad into their corresponding numerical equivalents.

Step 3 : Add the corresponding numerical equivalents with the numerical equivalents of one-time pad.

Step 4 : Subtract 26 if  the sum is greater than 25

Ex:

PT à  H O W ARE YOU

One-time Pad à N C B T Z Q A R X

Change it into numerical code & add them each

PT	H	O	W	A	R	E	Y	O	U
	7	14	22	0	17	4	24	14	20
OTP	N	C	B	T	Z	Q	A	R	X
Step 3	13	2	1	19	25	16	0	17	23
SUM	20	16	23	19	42	20	24	31	43
Step 4					26			26	26
	20	16	23	19	16	20	24	5	17
Ciphtxt	U	Q	X	T	Q	U	Y	F	R

Decryption :

In this case will be given One-time pad

Step 1 : Convert the alphabets of the plaintext into their corresponding numerical equivalents.

Step 2 : Convert the alphabets of the one-time pad into their corresponding numerical equivalents.

Step 3 : Subtract one-time pad from cipher text

Step 4 : If the different is negative than add 26 to it.

CT	U	Q	X	T	Q	U	Y	F	R
	20	16	23	19	16	20	24	5	17
OTP	N	C	B	T	Z	Q	A	R	X
Step 3	13	2	1	19	25	16	0	17	23
SUM	7	14	22	0	-9	4	24	-12	-6
step 4					26			26	26
	7	14	22	0	17	4	24	14	20
PlainText	H	O	W	A	R	E	Y	O	U