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Public key cryptography and
RSA algorithm
Name:R.POONGUZHALI
Class:II Msc Computer Science
Batch:2017-2019
Incharge Staff:Ms.M.Florence Dayana
PUBLIC KEY
CRYPTOGRAPHY
• It is computationally infeasible to determine the decryption key given only
knowledge of the cryptographic algorithm and the encryption key
• Either of the two related keys can be used for encryption, with the other used for
decryption
• A public key encryption scheme has six ingredients
• Plaintext : This is the readable message or data that is fed into the algorithmas
input.
• Encryption algorithm : The encryption algorithm Perform various
transformations on the Plaintext.
• Public and Private keys : This is a pair of keys that have been selected so
that if one is used for encryption the other used for decryption
• Decryption algorithm : This algorithm accepts the ciphertext and thematching
key and Produces the original plaintext.
Key distribution
The communications already shares a key or someonce has been distributed
key
How to secure communications in general without having to trust a KDC
with yours key
Conventional Encryption Public key Encryption
1. The same algorithm with the same keys used for
encryption and decryption
2. The sender and receiver must share the algorithm
and the key
Need for security
1. The key must be kept secret
2. It must be impossible or at least impractical
decipher a message if the key is kept secret.
1. One algorithms used for encryption and related
algorithm for decryption with a pair or key one or
encryption and to decryption
2. The sender and receiver must each have one of the
matched key
Need for security
1. One of the two must be kept secret
2. It must be impossible or at least impractical to
decipher a message if one of the key secret.
Application for public
key cryptosystems
• Encryption/decryption : The sender encrypts a message with the recipients
public key
• Digital signature : The sender “signs” a message with private key
• Key exchange : Two sides cooperate to exchange a session key.
THE RSA ALGORITHM
RSA
ALGORITH
M
• Ron Rivest , Adi shamir and Len Adleman of MIT1977
• Best know & widely used public key scheme
• Based on exponentials in a finiate field over integers modulo a prime
• Uses the large integers (ex:1024 bits)
• Security due to cost of factoring large number
Each user generates a public/private key pair By :
1. Selecting two large primes at random –P,q
2. Computing their system modulus N=p.q
RSA
use
To encrypt a message M the sender:
• Obtains Public Key of recipient KU={e,N}
• Computes:C=Me mod N, where 0≤M<N
To decrypt the ciphertext C the Owner:
• Uses their private KR={pd,p,q}
• Computes : M= Cd mod N
That the message M must be smaller than the modulus N
Algorithm Requirement
1. It is possible to find values of e,d,f
Such that Med mod n = M for all M<n
It is relatively easy to calculate M mod n
and Cd mod n for all values of M < n
It is infeasible to determine d given e
and n
RSA
SECURIT
Y
Three approaches attacking RSA:
• Brute force key search(infeasible given size of numbers)
• Mathematical attacks (based on difficulty of computing)
• Timing attacks(on running of decryption)
Optimizing Private key operation
1.Cd mod n =cd mod pq
Compute cd mod p and cd mod q
Use Chinese remainder theorem to compute cd mod pq
Chines remainder theorem requires p-1mod q and q-1mod p.
Since d is much bigger than p,cd mod p=cr mod p where r=d mod(p-1)
1. D=k(p-1)+r
2. Modp : ad=ak(p-1)+r=ak
3. (p)ar=ar[Euler’s theorem]

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Pooguzhali

  • 1. Public key cryptography and RSA algorithm Name:R.POONGUZHALI Class:II Msc Computer Science Batch:2017-2019 Incharge Staff:Ms.M.Florence Dayana
  • 2. PUBLIC KEY CRYPTOGRAPHY • It is computationally infeasible to determine the decryption key given only knowledge of the cryptographic algorithm and the encryption key • Either of the two related keys can be used for encryption, with the other used for decryption • A public key encryption scheme has six ingredients • Plaintext : This is the readable message or data that is fed into the algorithmas input. • Encryption algorithm : The encryption algorithm Perform various transformations on the Plaintext.
  • 3. • Public and Private keys : This is a pair of keys that have been selected so that if one is used for encryption the other used for decryption • Decryption algorithm : This algorithm accepts the ciphertext and thematching key and Produces the original plaintext. Key distribution The communications already shares a key or someonce has been distributed key How to secure communications in general without having to trust a KDC with yours key
  • 4.
  • 5.
  • 6. Conventional Encryption Public key Encryption 1. The same algorithm with the same keys used for encryption and decryption 2. The sender and receiver must share the algorithm and the key Need for security 1. The key must be kept secret 2. It must be impossible or at least impractical decipher a message if the key is kept secret. 1. One algorithms used for encryption and related algorithm for decryption with a pair or key one or encryption and to decryption 2. The sender and receiver must each have one of the matched key Need for security 1. One of the two must be kept secret 2. It must be impossible or at least impractical to decipher a message if one of the key secret.
  • 7. Application for public key cryptosystems • Encryption/decryption : The sender encrypts a message with the recipients public key • Digital signature : The sender “signs” a message with private key • Key exchange : Two sides cooperate to exchange a session key.
  • 9. RSA ALGORITH M • Ron Rivest , Adi shamir and Len Adleman of MIT1977 • Best know & widely used public key scheme • Based on exponentials in a finiate field over integers modulo a prime • Uses the large integers (ex:1024 bits) • Security due to cost of factoring large number Each user generates a public/private key pair By : 1. Selecting two large primes at random –P,q 2. Computing their system modulus N=p.q
  • 10. RSA use To encrypt a message M the sender: • Obtains Public Key of recipient KU={e,N} • Computes:C=Me mod N, where 0≤M<N To decrypt the ciphertext C the Owner: • Uses their private KR={pd,p,q} • Computes : M= Cd mod N That the message M must be smaller than the modulus N
  • 11. Algorithm Requirement 1. It is possible to find values of e,d,f Such that Med mod n = M for all M<n It is relatively easy to calculate M mod n and Cd mod n for all values of M < n It is infeasible to determine d given e and n
  • 12.
  • 13. RSA SECURIT Y Three approaches attacking RSA: • Brute force key search(infeasible given size of numbers) • Mathematical attacks (based on difficulty of computing) • Timing attacks(on running of decryption)
  • 14.
  • 15. Optimizing Private key operation 1.Cd mod n =cd mod pq Compute cd mod p and cd mod q Use Chinese remainder theorem to compute cd mod pq Chines remainder theorem requires p-1mod q and q-1mod p. Since d is much bigger than p,cd mod p=cr mod p where r=d mod(p-1) 1. D=k(p-1)+r 2. Modp : ad=ak(p-1)+r=ak 3. (p)ar=ar[Euler’s theorem]