Encryption and Encryption and Decryption Speaker:Tsung Ray Wang Advisor:Prof.Li-Chun Wang

Contents n MODELS,GOALS,AND EARLY CIPHER SYSTEMS n THE SECRECY OF A CIPHER SYSTEM n PRACTICAL SECURITY n STREAM ENCRYPTION n PUBLIC KEY CRYPTOSYSTEMS

Model of a cryptographic channel Plaintext M Encipher Decipher Public channel Plaintext Cryptanalyst Key Ciphertext K K Secure channel

The two primary reasons for using cryptosystems in communications n (1)privacy,to prevent unauthorized persons from exacting information from the channel n (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 n Ciphertext-Only Attack n Known-Plaintext Attack n Chosen-Text Attack

Classic Ciphers n 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 n Polybius square. Plaintext : NOWI S T H E T I M E Ciphertext: n 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 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

Trithemius progressive key

THE SECRECY OF A CIPHER SYSTEM What is Perfect Secrecy?? Entropy and Equivocation Rate of a language and Redunancy Unicity Distance and Ideal Secrecy

P(M o )=1/4M0M0 0 C0C0 1 C1C1 C2C2 2 C3C3 3 P(M 1 )=1/4 M1M1 P(M 2 )=1/4 M2M2 P(M 3 )=1/4 M3M3 PlaintextCiphertext Example of perfect secrecyKey C s =T k j (M i ) S=( ) modulo-N

PRACTICAL SECURITY n Substitution n Permutation n Product Cipher System n The Data Encryption Standard

Substitution box n=3 input 2 n = output n = input output

Permutation box input output

Individual keying capability Example of binary key

Initial Permutation (IP)

E-Table Bit Selection

P-Table Permutation

Final Permutation (IP -1 )

Key Permutation PC

Key Schedule of Left Shifts Iteration Number of left shifts i

Key Permutation PC

STREAM ENCRYPTION n Key Generation Using a Linear Feedback Shift Register n Vulnerabilities of Linear Feedback Shift Registers

Linear feedback shift register example feedback output x4x4 x3x3 x2x2 x1x1 Modulo-2 adder

PUBLIC KEY CRYPTOSYSTEMS n Signature Authentication Using a Public Key Cryptosystem n A Trapdoor One-Way Function n The Rivest-Shamir-Adelman Scheme n The Knapsack Problem n A Public Key Cryptosystem Based on a Trapdoor Knapsack

The important features of a public key cryptosystem n 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, n For each K, and are easy to compute. n For each K,the computation of from is computa-tionally intractable.

Public Key cryptosystem M Crypto machine Subscriber A Directory A- B- C-.. Crypto machine Subscriber B M

Signature authenticaton using a public key cryptosystem M Crypto machine A Date Crypto machine A Public channel Directory Crypto machine B Crypto machine B M Directory Signature storage

The Rivest-Shamir-Adelman Scheme 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 RSA

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.