440 likes | 725 Views
Encryption and Decryption. Speaker:Tsung Ray Wang Advisor:Prof.Li-Chun Wang. Contents. MODELS,GOALS,AND EARLY CIPHER SYSTEMS THE SECRECY OF A CIPHER SYSTEM PRACTICAL SECURITY STREAM ENCRYPTION PUBLIC KEY CRYPTOSYSTEMS. Model of a cryptographic channel. Cryptanalyst.
E N D
Encryption and Decryption Speaker:Tsung Ray Wang Advisor:Prof.Li-Chun Wang
Contents • MODELS,GOALS,AND EARLY CIPHER SYSTEMS • THE SECRECY OF A CIPHER SYSTEM • PRACTICAL SECURITY • STREAM ENCRYPTION • PUBLIC KEY CRYPTOSYSTEMS
Model of a cryptographic channel Cryptanalyst Plaintext Plaintext Encipher Decipher Public channel M Ciphertext K K Key Secure channel
The two primary reasons for using cryptosystems in communications • (1)privacy,to prevent unauthorized persons from exacting information from the channel • (2)authentication,to prevent unauthorized persons from injecting information into the channel
System Goals The major requirements for a cryptosystem 1.To provide an easy and inexpensive means of encryption and decryption to authorized users in possession of the appropriate key 2.To ensure that the cryptanalyst’s task of producing an estimate of the plaintext without benefit of the key is made difficult and expensive
Classic Threats • Ciphertext-Only Attack • Known-Plaintext Attack • Chosen-Text Attack
Classic Ciphers • Caesar Cipher ex. Plaintext : N O W I S T H E T I M E : Ciphertext : Q R Z L V W K H W L P H • Polybius square . Plaintext : NOWI S T H E T I M E Ciphertext: 33 43 25 42 34 44 32 51 44 42 23 51 • Polyalphabetic cipher . Plaintext: NOWI S T H E T I M E Ciphertext: OQZMXZ O M CS X Q
Caesar’s alphabet with a shift of 3 Plaintext: ABCDEFGHIJKLMNOPQRSTUVWXYZ CHIPHERTEXT: DEFGHIJKLMNOPQRSTUVWXYZABC Polybius square 1 2 3 4 5 A B C D E F G H IJ K L M N O P Q R S T U V W X Y Z 1 2 3 4 5
THE SECRECY OF A CIPHER SYSTEM •What is Perfect Secrecy?? • Entropy and Equivocation • Rate of a language and Redunancy • Unicity Distance and Ideal Secrecy
Example of perfect secrecy Key 0 P(Mo)=1/4 M0 C0 1 P(M1)=1/4 2 C1 M1 P(M2)=1/4 M2 3 C2 P(M3)=1/4 M3 C3 Plaintext Ciphertext Cs=Tkj(Mi) S=( ) modulo-N
PRACTICAL SECURITY • Substitution • Permutation • Product Cipher System • The Data Encryption Standard
Substitution box 2n=8 2n=8 n=3 0 0 output 1 0 1 1 2 2 input 3 3 1 1 4 4 5 5 0 6 1 6 7 7 input 000 001 010 011 101 110 111 100 output 011 000 111 110 010 100 101 001
Permutation box 0 1 0 1 output input 0 0 1 0 0 1
Individual keying capability Example of binary key 1 0 1 0 0 0 1 0 1 1 1 1 1 0 1 1 0 1 0 1 1 1 0 1 0
Initial Permutation (IP) 58 50 42 34 26 18 10 2 60 52 44 36 28 20 12 4 62 54 46 38 30 22 14 6 64 56 48 40 32 24 16 8 57 49 41 33 25 17 9 1 59 51 43 35 27 19 11 3 61 55 45 37 29 21 13 5 63 55 47 39 31 23 15 7
E-Table Bit Selection 32 1 2 3 4 5 4 5 6 7 8 9 8 9 10 11 12 13 12 13 14 15 16 17 16 17 18 19 20 21 20 21 22 23 24 25 24 25 26 27 28 29 28 29 30 31 32 1
P-Table Permutation 16 7 20 21 29 12 28 17 1 15 23 26 5 18 31 10 2 8 24 14 32 27 3 9 19 13 30 6 22 11 4 25
Final Permutation (IP-1) 40 8 48 16 56 24 64 32 39 7 47 15 55 23 63 31 38 6 46 14 54 22 62 30 37 5 45 13 53 21 61 29 36 4 44 12 52 20 60 28 35 3 43 11 51 19 59 27 34 2 42 10 50 18 58 26 33 1 41 9 49 17 57 25
Key Permutation PC-1 57 49 41 33 25 17 9 1 58 50 42 34 26 18 10 2 59 51 43 35 27 19 11 3 60 52 44 36 63 55 47 39 31 23 15 7 62 54 46 38 30 22 14 6 61 53 45 37 29 21 13 5 28 20 12 4
Key Schedule of Left Shifts Iteration Number of left shifts i 1 1 2 1 3 2 4 2 5 2 6 2 7 2 8 2 1 9 10 2 11 2 12 2 13 2 2 14 2 15 16 1
Key Permutation PC-2 14 17 11 24 1 5 3 28 15 6 21 10 23 19 12 4 26 8 16 7 27 20 13 2 41 52 31 45 33 48 30 40 51 45 33 48 44 49 39 56 34 53 46 42 50 36 29 32
STREAM ENCRYPTION • Key Generation Using a Linear Feedback Shift Register • Vulnerabilities of Linear Feedback Shift Registers
Linear feedback shift register example output x3 x2 x1 x4 Modulo-2 adder feedback
PUBLIC KEY CRYPTOSYSTEMS • Signature Authentication Using a Public Key Cryptosystem • A Trapdoor One-Way Function • The Rivest-Shamir-Adelman Scheme • The Knapsack Problem • A Public Key Cryptosystem Based on a Trapdoor Knapsack
The important features of a public key cryptosystem • The encryption algorithm, ,and the decryption algorithm, ,are invertible transformations on the plaintext ,M,or the ciphertext ,C,defined by the key K. That is,for each K and M, • For each K, and are easy to compute. • For each K,the computation of from is computa-tionally intractable.
Public Key cryptosystem Subscriber A Subscriber B Crypto machine M Crypto machine M Directory A- B- C- . . . .
Signature authenticaton using a public key cryptosystem A A Crypto machine Crypto machine Public channel M Date Directory B B Crypto machine Crypto machine M Signature storage Directory
The Rivest-Shamir-Adelman Scheme RSA 1.Each user chooses his own value of n and another pair of positive integers (e,d) ,and n=pq, =(p-1)(q-1),gcd[ ,d]=1, ed modulo- =1,and p,q are prime numbers. 2..The user places his encryption key the number pair (n,e),in the public directory. 3. The decryption key consists of the number pair (n,d),of which d is kept secret. 4.messages are first represented as integers in the range (0,n-1) 5.Encryption: modulo-n Decryption: modulo-n
How to Compute e A variation of Euclid’s algorithm for computing the gcd of and d is to compute e 1.First,compute a series …... where = , =d ,and = modulo- ,until an =0 is found. than the gcd ( , d )= 2.For each compute numbers and such that = + 3.If =1,then is the multiplicative inverse of modulo- .If is a negative number, the solution is +
The Knapsack problem 1.Let us express the knapsack problem in terms of a knapsack vector ‘a’ and a data vector ’ x’. 2.The knapsack,S,is the sum of a subset of the components of the knapsack vector where = ax
Super-increasing and how to slove “x” 1.super-increasing is 2.When a is super-incresing,the solution of x is found by starting with if S (otherwise ) ,and continuing as follows: = where
A Public key Cryptosystem Based on a Trapdoor Knapsack -this scheme,also known as the Merkle-Hellman scheme method: 1.we form a super-increasing n-tuple a’,and select a prime number M such that ,also select a random number,W, where 1<W<M,and we form to satisfy the following relationship: W modulo -M =1,note:the vector a’ and the number M,W, are all kept hidden. 2.we form a with the elements from a’ as: modulo-M
3.When a data vector x is to be transmitted ,we multiply x by a, yielding the number S,which is sent on the public channel. 4.The authorized user receives S and converts it to S’ : = = 5.Since the authorized user knowns the secretly held super-increasing vector a’ ,he can use S’ to find x.
CONCLUSION 1.We have presented the basic models and goals of the cryptographic process,and looked at some early cipher systems. 2.We defined a system that can exhibit perfect secrecy . 3.We outlined the DES algorithm in detail,and we also considered the use of linear feedback shift registers(LFSR) for stream encryption systems. 4.RSA scheme ,based on the product of two large prime numbers, and the Merkle-Hellman scheme,based on the classical knapsack problem.