Lecture 3.2: Public Key Cryptography II CS 436/636/736 Spring 2014 Nitesh Saxena.

Lecture 3.2: Public Key Cryptography II

CS 436/636/736 Spring 2014

Nitesh Saxena

Course Administration

• HW1 solution has been posted• Grades posted• We will distribute after lecture today• Any questions, or help needed?

• We need to something to make-up the missed material– Waiting on UAB for guidelines

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Outline of Today’s Lecture

• The RSA Cryptosystem (Encryption)

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“Textbook” RSA: KeyGen• Alice wants people to be able to send her encrypted

messages.• She chooses two (large) prime numbers, p and q and

computes n=pq and . [“large” = 1024 bits +]• She chooses a number e such that e is relatively prime to

and computes d, the inverse of e in , i.e., ed =1 mod • She publicizes the pair (e,n) as her public key. (e is called RSA

exponent, n is called RSA modulus). She keeps d secret and destroys p, q, and

• Plaintext and ciphertext messages are elements of Zn and e is the encryption key.

4

)(n)(n

)(nZ

)(n

)(n

RSA: Encryption

• Bob wants to send a message x (an element of Zn

*) to Alice.

• He looks up her encryption key, (e,n), in a directory.

• The encrypted message is

• Bob sends y to Alice.5

nxxEy e mod)(

RSA: Decryption

• To decrypt the message

she’s received from Bob, Alice computes

Claim: D(y) = x6

nyyD d mod)(

nxxEy e mod)(

RSA: why does it all work• Need to show

D[E[x]] = x E[x] and D[y] can be computed efficiently if keys

are known E-1[y] cannot be computed efficiently without

knowledge of the (private) decryption key d.

• Also, it should be possible to select keys reasonably efficiently This does not have to be done too often, so

efficiency requirements are less stringent.7

E and D are Inverses

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nxnx

nxx

nx

nx

nx

nnx

nyyD

t

tn

nt

ed

de

de

d

modmod1

mod)(

mod

mod

mod)(

mod)mod(

mod)(

)(

1)(

Because

From Euler’s Theorem

)(mod1 ned

Tiny RSA example.

• Let p = 7, q = 11. Then n = 77 and

• Choose e = 13. Then d = 13-1 mod 60 = 37.• Let message = 2.• E(2) = 213 mod 77 = 30.• D(30) = 3037 mod 77=2

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60)( n

Slightly Larger RSA example.

• Let p = 47, q = 71. Then n = 3337 and

• Choose e = 79. Then d = 79-1 mod 3220 = 1019.• Let message = 688232… Break it into 3 digit

blocks to encrypt.• E(688) = 68879 mod 3337 = 1570. E(232) = 23279 mod 3337 = 2756• D(1570) = 15701019 mod 3337 = 688. D(2756) = 27561019 mod 3337 = 232.

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322070*46)( pq

Security of RSA: RSA assumption• Suppose Oscar intercepts the encrypted

message y that Bob has sent to Alice.• Oscar can look up (e,n) in the public directory

(just as Bob did when he encrypted the message)

• If Oscar can compute d = e-1 mod then he can use to recover the plaintext x.

• If Oscar can compute , he can compute d (the same way Alice did). 11

xnyyD d mod)(

)(n

)(n

Security of RSA: factoring

• Oscar knows that n is the product of two primes

• If he can factor n, he can compute • But factoring large numbers is very difficult:– Grade school method takes divisions.– Prohibitive for large n, such as 160 bits– Better factorization algorithms exist, but they are

still too slow for large n– Lower bound for factorization is an open problem

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)(n

)( nO

How big should n be?

• Today we need n to be at least 1024-bits– This is equivalent to security provided by 80-bit

long keys in private-key crypto

• No other attack on RSA known– Except some side channel attacks, based on

timing, power analysis, etc. But, these exploit certain physical charactesistics, not a theoretical weakness in the cryptosystem!

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Key selection

• To select keys we need efficient algorithms to– Select large primes• Primes are dense so choose randomly.• Probabilistic primality testing methods known. Work in

logarithmic time.

– Compute multiplicative inverses• Extended Euclidean algorithm

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RSA in Practice

• Textbook RSA is insecure– Known-plaintext?– CPA?– CCA?

• In practice, we use a “randomized” version of RSA, called RSA-OAEP– Use PKCS#1 standard for RSA encryptionhttp://www.rsa.com/rsalabs/node.asp?id=2125– Interested in details of OAEP: refer to (section 3.1 of)

http://isis.poly.edu/courses/cs6903/Lectures/lecture13.pdf

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http://www.rsa.com/rsalabs/node.asp?id=2125




Some questions

• c1 = RSA_Enc(m1), c2 = RSA_Enc(m2). – What is RSA_Enc(m1m2)?

• Homomorphic property

– What is RSA_Enc(2m1)?• Malleability (not a good property!)

• Is it possible to find inverses mod n (RSA modulus)?

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Some Questions

• RSA stands for Robust Security Algorithm, right?• If e is small (such as 3)

– Encryption is faster than decryption or the other way round?

• Private key crypto has key distribution problem and Public key crypto is slow– How about a hybrid approach?– Do you know how ssl/ssh works?

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Some Questions

• I encrypt m with Alice’s RSA PK, I get c– I encryt m again, I get --?– What does this mean?

• What if I do the above with DES?

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Further Reading

• Section 8.2 of HAC• Section 9 of Stallings

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Outline of Today’s Lecture

• Discrete Logarithm System• El Gamal Encryption• Digital Signatures

Lecture 3.4: Public Key Cryptography IV

Discrete Logarithm Assumption• Work with a cyclic group G with generator g• Let |G| = m• G = {g0, g1, g2,…,gm-1}

• Given any y = gx in G (where x belongs to Zm), g and and m, it is not possible to compute x

• This is known as the DL assumption• Of course, x should be fairly large – at least

160-bits in length • This suggests that one can possibly use x as

the secret key, and y (and other parameters) as the public key

El Gamal Encryption -- KeyGen• p, q primes such that q|p-1• g is an element of order q and generates a group Gq of

order q – g = g’(p-1)/q (were g’ is the generator of Zp*)

• x in Zq, y = gx mod p• DL assumption -- given (p, q, g, y), it is computationally

hard to compute x– No polynomial time algorithm known– p should be 1024-bits and q be 160-bits

• x becomes the private key and y becomes the public key

ElGamal Encryption/Decryption

• Encryption (of m in Gq):– Choose random r in Zq

– k = gr mod p– c = myr mod p– Output (k,c)

• Decryption of (k,c)– M = ck-x mod p

• Secure under (a variant of) the discrete logarithm assumption


ElGamal Example: dummy

• Let’s construct an example• KeyGen:

– p = 11, q = 2 or 5; let’s say q = 5– g’ = 2 is a generator of Z11*

– g = 22 = 4– x = 2; y = 42 mod 11 = 5

• Enc(3):– r = 4 k = 44 mod 11 = 3– c = 3*54 mod 11 = 5

• Dec(3,5):– m = 5*3-2 mod 11 = 3


El Gamal Security

• Secure against CPA attacks assuming that discrete logarithm is hard

• Not secure against CCA attacks; why?– It is possible to massage the ciphertext in a

meaningful way– Given a ciphertext (k, c), compute k’ = kgr’and c’ =

cyr’ (r’ is picked by the adversary)– Query the decryption oracle on (k’,c’); it decrypts

and returns the response -- m


CCA Security

• Like in the case of symmetric key encryption, we can derive CCA secure encryption using CPA secure encryption

• Just prevent any massaging of the ciphertext• Integrity protection mechanism is needed– But, now a public-key based mechanism is needed

• Digital signatures -- next


Digital Signatures

• Message Integrity– Detect if message is tampered with while in the

transit

• Source/Sender Authentication– No forgery possible

• Non-repudiation– If I sign something, I can not deny later– A trusted third party (court) can resolve dispute

• Many applications – signed email, e-contracts, e-transactions…

Public Key Signatures

• Signer has public key, private key pair• Signer signs using its private key• Verifier verifies using public key of the signer


Security Notion/Model for Signatures

• Existential Forgery under (adaptively) chosen message attack (CMA)– Adversary (adaptively) chooses messages mi of its

choice– Obtains the signature si on each mi

– Outputs any message m (≠ mi) and a signature s on m


RSA Signatures

• Key Generation: same as in encryption• Sign(m): s = md mod N• Verify(m,s): (se == m mod N)

• The above text-book version is insecure; why?• In practice, we use a randomized version of

RSA (implemented in PKCS#1)– Hash the message and then sign the hash


Digital Signature Standard (DSS)

• Adopted as standard in 1994• Security based on hardness of the discrete

logarithm problem


DSS – KeyGen; Signing; Verification• KeyGen: the same way as El Gamal

– p, q primes such that q|p-1– g is an element of order q and generates a group Gq of order q

• g = g’(p-1)/q (were g’ is the generator of Zp*)

– x in Zq, y = gx mod p

• Sign: – Pick random r from Z*q

– k = (gr mod p) mod q; c = (m + xk)r-1 mod q– Output (k,c) and also the message m

• Verify: kc == gm.yk mod p


DSS Example

• Refer to 11.57 of HAC


Some Questions

• I encrypt m with Alice’s ElGamal PK, I get c– I encrypt m again, I get --?– What does this mean?

• Is RSA-OAEP CCA secure?• Is El Gamal CCA secure?


Further Reading

• Stalling Chapter 10• HAC Chapter 8 and Chapter 11


Lecture 3.2: Public Key Cryptography II CS 436/636/736 Spring 2014 Nitesh Saxena.

Documents

computes n

security of rsa

textbook rsa

rsa assumption

rsa exponent

rsa modulus

pair e

inverse of e