Cloud Computing and securityWith Homomorphic Encryption

The Conclusion

Session 1-Intro To Cloud

What is Cloud? Types of Cloud Components Advantages Disadvantage

Session 2-Basics of Security

Principles of Security Types of attacks Cryptography techniques Types of Cryptography Hashing-Physical & Digital Data level Security

Session 3-Homorphic Encryption

Various Algorithms RSA An Example (sort of!) Conclusion

Papers

Kutub Thakur, “An Investigation on Cyber Security Threats and Security Models”, 2015 IEEE

Maha TEBAA and EL HAJII, “Secure Cloud Computing through Homomorphic Encryption", University Mohammed V–Agdal

Darko Hrestak and Stjepan Picek, “Homomorphic Encryption in the cloud", University of Zagreb, Croatia.

Some Algorithms

RSA – Digital Signatures, iBanking, Credit Card Transactions Paillier Algorithm – E Voting systems Enhanced Homomorphic Cryptosystem(EHC)-Efficient Secure

Message transmission in mobile AdHoc Networks BGV Encryption Scheme - Encrypt bits, Security of integer

polynomials

Insight into RSA

Key GenerationStep 1 : Each user generates a public-private key pair by selecting two large primes at random – say p & q. Step 2 : Computing RSA-modulus N = p . q & ø(N) = (p-1) (q-1). Step 3 : Selecting at random, the encryption key “e” where, 1<e<ø(N) s.t. GCD(e, ø(N)) = 1Step 4 : Publish their public encryption key: KU= { e , N } & keep the secret private decryption key: KR= {d , N}.

Insight into RSA contd..

Encryption

Step 1: Obtains public key of recipient KU={e , N} Step 2: Computes: Ci = Msg^e mod N, where 0≤M<N.

Insight into RSA contd..

Decryption

Step 1 : Uses their private key KR={d , N}

Step 2 : Computes: M=C.d mod N

An Example : Area_rect[7,3]

Step 1 : p = 11 & q = 13Step 2 : System modulus N = 11 * 13 = 143 and ø(143) = (11-1) (13-1) = 120. Step 3 : Random Encryption key e, where, 1 < e < 120 : GCD (e , 120) = 1 ; e = 23Step 4 : Public Encryption Key: KU = {23,143}

Secret Private Decryption Key: KR= { d ,N }. To Calculate : 23 . d ≡ 1 mod ø(143)

Now, 23 . d + k . 120 = 1 = GCD(e,120) . Hence from the above equation, d = 47 & k = -9

KU= { e , N } & keep the secret private decryption key: KR= {d , N}.

Area_rect[7,3] contd..

Encrypt(message) ≡ Message^pub.key mod RSA-modulus

OR

Encrypt(message) ≡ Message^23 mod 143

Property is : Encrypt(m) * Encrypt(n) = Encrpyt(m*n)

Area_rect[7,3] contd..

Width = 7 private key : (47,143)Height = 3 public key : (23,143) E(w) = width^e mod N = 7^23 mod 143 = 2E(h) = height^e mod N = 3^23 mod 143 = 126

E(Area) = 252

Area_rect[7,3] contd..

D(E(Area)) = cipher^d mod N = 252^47 mod 143

= 21

VOILA!

Area = 21

Conclusion

Encryption is the only way FHE is still impractical for real-world Applications It makes it an interesting problem [ Academic + Industrial ] This will lead to design of more efficient algos in future Since there are applications where SWHE would be powerful

enough, it seems we are closer to that goal than it may look like

Fin.

Thank You!