Symmetric cipher model Model of Symmetric Cryptosystem cryptography substitution technique

3.  Plaintext:
This is often the initial intelligible message
or information that's fed into the algorithmic program as
input.
 Encryption algorithm:
The cryptography algorithmic
program performs numerous substitutions and
transformations on the plaintext.

4.  Secret key:
 The key key's conjointly input to
the coding algorithmic program.
 The key's a worth freelance of the plaintext and of
the algorithmic program.
 The algorithmic program can turn out a
special output betting on the precise key getting
used at the time.
 The precise substitutions and transformations
performed by the algorithmic program rely on the
key.

5.  Cipher text:
 It depends on the plaintext and also the secret
key.
 For a given message, 2 totally different
completely different} keys can turn
out 2 different cipher texts.
 The cipher text is Associate in
Nursing apparently random stream of
knowledge and, because it stands, is
unintelligible.

6.  Decryption algorithm:
 This is primarily the cryptography rule run in
reverse.It takes the ciphertext and also the secret
key and produces the first plaintext.

8.  This is primarily the cryptography rule run in
reverse.
 It takes the ciphertext and also the secret key
and produces the first plaintext.
 Cryptographic systems area
unit characterized 3 freelance dimensions.
The type of operation is
employed for remodeling plaintext to cipher text.

9.  All encoding algorithmic rule area
unit supported 2 general principles: substitution.
 within which every component within
the plaintext(bit, letter, cluster of bits or letter) is
mapped into another component and transposition.
 within which components within the plaintext area
unit rearranged.
 The basic demand is that no info be lost(i.e., that
each one operations area unit reversible).
 Most system, referred to as product systems,
involve multiple stages of substitution and
transpositions.

10.  Cryptanalysis:
cryptology attacks have faith in the character of
the algorithmic Program and maybe
some information of the overall characteristics
of the plaintext or maybe some sample plaintext-
cipher text pairs.
 this sort of attacks exploits the characteristics of
the algorithmic program to aim to deduce a
particular plaintext or to deduce the key getting
used.

11.  Brute-Force attack:
 The attacker tries every possible key on a piece
of cipher-text until an intelligible translation into
plaintext is obtained.
 On average, half of all possible keys must be
tried to achieve success.

13.  Is one during which the letters of
plaintext are replaced by alternative letters or by
numbers or symbols.
 If the plaintext is viewed as a sequence of bits,
then substitution involves exchange plaintext bit
patterns with ciphertext bit patterns

14.  Simplest and earliest best-known use of a
substitution cipher used by general.
 Involves replacement every letter of the alphabet
with the letter standing 3 places additional down the
alphabet
 Alphabet is wrapped around in order that the letter
following Z may be a
 plain: meet me after the toga party
 cipher: PHHW PH DIWHU WKH WRJD
SDUWB

15.  Can define transformation as:
a b c d e f g h i j k l m n o p q r s t u v w x y z
D E F G H I J K L M N O P Q R S T U V W X Y Z A B C
 Mathematically give each letter a number
a b c d e f g h i j k l m n o p q r s t u v w x y z
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
 Algorithm can be expressed as: c = E(3, p) = (p + 3) mod
(26)
 A shift may be of any amount, so that the general Caesar
algorithm is:
C = E(k , p ) = (p + k ) mod 26
 Where k takes on a value in the range 1 to 25;
the decryption algorithm is simply:
 p = D(k , C ) = (C - k ) mod 26

17.  With only 25 possible keys, the caeser cipher is so
far from secure.
 A dramatic increase in the key space can achieved
by allowing arbitrary substitution.
 Before proceeding, we define the term permutation.
 A Permutation of finite set of elements S is an
ordered sequence of all the elements of S, with each
element appearing exactly once.
 For example, if S={a, b, c}, there are six
permutations of S.
abc, acb, bac, bca, cab, cba,

18.  If the “cipher” line may be any permutation of
the twenty six alphabetic characters, then
there square measure 26!
or larger than 4x1026 potential keys. this is
often ten orders of magnitude larger than the
key house for DES
 Approach is named as a monoalphabetic
substitution cipher as a result of one cipher
alphabet is employed per message

19.  The known multiple letter cryptography cipher is
that the playfair, that treats diagrams within
the plaintext as single units and interprets these
units into ciphertext diagrams.
 The playfair algorithm is based on the use of 5x5
matrix of letters constructed using a keyword.

20.  Hill cipher is developed by the man of
science Lester Hill in 1929. Strength is that
it fully hides single-letter frequencies.
 The use of a bigger matrix hides a lot
of frequency info
 A 3x3 Hill cipher hides not solely single-
letter however additionally two-letter
frequency info try other relevant Tools

21.  This example will rely on some linear algebra
and some number theory. The key for a hill
cipher is a matrix
 e.g.

22.  In the higher than case, we've got taken the
dimensions to be 3×3, but it will be any size (as
long because it is square).
 Assume we wish to inscribe the message ATTACK
AT DAWN.
 To inscribe this, we'd like to interrupt the message
into chunks of three.
 we tend to currently take the
primary three characters from our plaintext, ATT
and produce a vector that corresponds to the letters
(replace A with 0, B with 1 ... Z with 25 etc.) to get:
[0 nineteen 19] (this is ['A' 'T' 'T']).

23.  To get our ciphertext we perform a matrix
multiplication (you may need to revise matrix
multiplication if this doesn't make sense):

24.  This method is performed for all three letter
blocks within the plaintext.
 The plaintext might have to be compelled
tube soft with some further letters to
create positive that there's a full range of
blocks.
 Now for the tricky part, the decryption.
 We need to find an inverse matrix modulo 26
to use as our 'decryption key.

25.  we want something that will take 'PFO'
back to 'ATT'. If our 3 by 3 key matrix is
called K, our decryption key will be the 3 by 3
matrix K-1, which is the inverse of K.
 To find K-1 we have to use a bit of maths.
 It turns out that K-1 above can be calculated
from our key