Skip to main content

Basic Cryptographic Operation

Basic Cryptographic Operation


In this post, We will learn some terminologies that used in computer security.

Basic Terms in Cryptography

  • Encryption: Encryption is a process to translate a plaintext to a cipher.
  • Decryption: Decryption is a process to translate a cipher to plaintext as it was before encryption.
  • Plaintext: This is a message, a string that we want send securely like an email, a text message etc.
  • Cipher: Cipher is a secure message that produces as a result of encryption.
  • Initialization vector (IV): IV provides semantic security to the cipher. It means that same message should have a different cipher.
  • Padding: Padding is a way to take data that may or may not be a multiple of the block size for a cipher and extend it out so that it is. Padding required of may many cipher mode because data to be encrypted to the exact multiple of block size.
  • Key: In cryptography, a key is a value that is applied using an algorithm to a string or block of unencrypted plaintext to produce encrypted text, or to decrypt encrypted text. There are two type of key
    • Public Key: The public keys are shared over the network. it uses to encrypt plain text.
    • Private Key: The private keys are stored securely and only respective know his private key. A key encrypted by the public key is decrypted by the private key
  • Mode: Direct use of cipher is not advisable because an enemy can build a codebook equivalent to plaintext or ciphertext. There are five standard modes of operation.
    • Electronic Code Book (ECB)
    • Cipher Block Chaining (CBC)
    • Cipher Feedback  (CFB)
    • Output Feedback (OFB)
    • Counter (CTR) 

Cryptography Model

  • There are two type of encryption and decryption models that widely used in the security domain to secure sensitive information
    • Symmetric Key Model
    • Asymmetric Key Model

Symmetric Key Model

In symmetric key encryption, we use the same key to encrypt as well as decrypt.

Above image representing an example of symmetric encryption. You can clearly see all those in some scenarios a hacker able to get cipher but he/she can not read data because he does not have the key. But there is a problem here because both parties are using same how do they share this key? We use key wrapping and unwrapping technique to protect key.

Key Wrapping and Unwrapping

  • Key Wrapping: Key wrapping is a technique to encrypt a symmetric key material. It is used to protect keys at rest or transmit them over insecure networks
  • Key Unwrapping: Key unwrapping is a technique used to decrypt a key material.

Asymmetric Key Model

 Asymmetric cryptographic systems use one key to encrypt data and the different key to decrypt. These keys are known as public and private key

In the asymmetric algorithm, the public key of the recipient is share across the channel. The sender encrypts the message with the public key of recipient and recipient decrypt with his private key which is securely stored.

Symmetric Key Algorithms 

There is a set of the symmetric key algorithm which use to encrypt a plaintext. We will discuss some of them here
  1. Advanced Encryption Standard (AES)
  2. Data Encryption Standard (DES)
  3. Triple Data Encryption Algorithm (3DES)

Asymmetric Key Algorithms

There is a set of an asymmetric key algorithm which use to encrypt a plaintext. We will discuss some of them here
  1.  Rivest Shamir Adleman (RSA)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Comments

  1. I gathered many useful informations about this topic. Really very useful for learning the skills and will continue your blog reading in the future.
    Blue Prism Training in Anna nagar
    Blue Prism Training in Chennai
    Blue Prism training chennai
    RPA Training in Anna nagar
    RPA Training in Adyar
    Data Science course in Anna nagar

    ReplyDelete

Post a Comment

Popular posts from this blog

SSC Practice Problem Solution

SSC Practice Problem Solution In this post, we will discuss solution of  SSC Practice Problem . Question 1  The efficiency of (A + B) = 100/24 = (25/6)% The efficiency of (A + B + C) = 100/8 = (25/2)% The efficiency of C = 25/2 – 25/6 = (50/6)% Hence, C can alone finish this job in = 100/(50/6) = 12 days Answer: 12 days option (1) Question 2  Original price = Rs. 6000; Price after discount = 6000 – 1200 = Rs. 4800; Price after raising service contract = 4800 + 480 = Rs. 5280 Answer: Rs. 5280 option (2) Question 3  Suppose the received money by A, B, and C is respectively 5x, 6x, and 9x. 5x = 450; => x = 90; Hence, the total money = 20x = 20*90 = Rs. 1800; Answer: Rs. 1800 option (2) Question 4 Suppose the Cost price of the bag= Rs. x; Hence, x + 0.15x = 230; => x = Rs. 200; Selling price after selling on 20% = 200 + 20% of 200 = Rs. 240 Answer: Rs. 240 option (3) Question 5  The required percentage ...

Average

Average In this post, We will learn about average.  Average of any collection of data is the sum of data divide by number of data. Average = (Sum of observations /  Number of observations) Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.  problems The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is (CAT 2018) As specified in the question mean of x and y and z is 80 x+y+z=3*80 mean of x, y, z,u,v is 75 x+y+z+u+v=75*5 2u=x+y 2v=y+z x+y+z+((x+y+y+z)/2)=75*5 y+3(x+y+z)=75*10 y=750-720 y=30 since x+y+z=240 y=30 so x+z=210 since x>=z so minimum value of x=210/2 = 105 Problem2: In first 5 inings sehwag has scored at average of 40. In sixth ining He had scored 52 what is His average after 6th ining. one simple approach is getting total score divided by no of inin...

Complex Number

Complex Number While solving the equation of the form  a x 2 + b x + c = 0 , the roots of the equations can take three forms which are as follows: Two Distinct Real Roots Similar Root No Real roots (Complex Roots) The introduction of complex numbers in the 16th century made it possible to solve the equation  x 2 + 1 = 0 . The roots of the equation are of the form  x = ± − 1 ‾ ‾ ‾ √  and no real roots exist. Thus, with the introduction of complex numbers, we have Imaginary roots. We denote  − 1 ‾ ‾ ‾ √  with the symbol i, where i denotes Iota (Imaginary number). An equation of the form z= a+ib, where a and b are real numbers, is defined to be a complex number. The real part is denoted by Re z = a and the imaginary part is denoted by Im z = b. Algebraic Operation on Complex numbers: Addition of two complex numbers Subtraction of two complex number Multiplication of two complex number Division of two complex number Power of Iota (i) ...

Logarithm

Logarithm When a x   = N , then we say that x = logarithm of N to the base a and write it as x = log a N . In simple words, it represents the power to which a number must be raised. Let me expand on that by giving a simpler example. If we are asked, what would be the result if ‘a’ is multiplied with itself ‘b’ times; then your answer would be x = a*a*a*…. b (times). This can also be written as a^b. This is also known as ‘a raised to the power of b’ If we are asked, which number multiplied with itself ‘b’ times, will result in a; then you are asked for the value of x such that x*x*x*… b(times) = a => x^b = a => x = a^(1/b) This is also known as ‘bth root of a’ If swe are asked, how many times should you multiply ‘a’ with itself to get ‘b’, that is where the concept of logarithm comes into the picture. You are asked for the value of ‘x’ such that a*a*a…. x (times) = b => x = Log  a  b This is also known as ‘ Log b to the base a’ ...