CMSC 414 Computer and Network Security Lecture 6 Jonathan Katz.

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CMSC 414 Computer and Network Security Lecture 6 Jonathan Katz
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Transcript of CMSC 414 Computer and Network Security Lecture 6 Jonathan Katz.

CMSC 414Computer and Network Security

Lecture 6

Jonathan Katz

Administrative announcements

Midterm I– March 6

GRACE accounts set up

Read the essays and papers linked from the course webpage– Will discuss next Tuesday and/or Thursday

Public-key cryptography

The public-key setting A party (Alice) generates a public key along with

a matching secret key (aka private key)

The public key is widely distributed, and is assumed to be known to anyone (Bob) who wants to communicate with Alice– We will discuss later how this can be ensured

Alice’s public key is also known to the attacker!

Alice’s secret key remains secret

Bob may or may not have a public key of his own

The public-key setting

c = Encpk(m)

pk

c = Encpk(m)

pk

Private- vs. public-key I

Disadvantages of private-key cryptography– Need to securely share keys

• What if this is not possible?

• Need to know in advance the parties with whom you will communicate

• Can be difficult to distribute/manage keys in a large organization

– O(n2) keys needed for person-to-person communication in an n-party network

• All these keys need to be stored securely

– Inapplicable in open systems (think: e-commerce)

Private- vs. public-key II

Why study private-key at all?– Private-key is orders of magnitude more efficient

– Private-key still has domains of applicability• Military settings, disk encryption, …

– Public-key crypto is “harder” to get right• Needs stronger assumptions, more math

– Can combine private-key primitives with public-key techniques to get the best of both (for encryption)

• Still need to understand the private-key setting!

– Can distribute keys using trusted entities (KDCs)

Private- vs. public-key III

Public-key cryptography is not a cure-all– Still requires secure distribution of public keys

• May (sometimes) be just as hard as sharing a key

• Technically speaking, requires only an authenticated channel instead of an authenticated + private channel

– Not clear with whom you are communicating (for public-key encryption)

– Can be too inefficient for certain applications

Cryptographic primitives

Private-key setting Public-key setting

ConfidentialityPrivate-key encryption

Public-key encryption

IntegrityMessage

authentication codesDigital signature

schemes

Public-key encryption

Functional definition

Key generation algorithm: randomized algorithm that outputs (pk, sk)

Encryption algorithm:– Takes a public key and a message (plaintext), and

outputs a ciphertext; c Epk(m)

Decryption algorithm:– Takes a private key and a ciphertext, and outputs a

message (or perhaps an error); m = Dsk(c)

Correctness: for all (pk, sk), Dsk(Epk(m)) = m

Security? Just as in the case of private-key encryption, but

the attacker gets to see the public key pk

That is:– For all m0, m1, no adversary running in time T, given pk

and an encryption of m0 or m1 can determine the encrypted message with probability better than 1/2 +

Public-key encryption must be randomized (even to achieve security against ciphertext-only attacks)

Security against ciphertext-only attacks implies security against chosen-plaintext attacks

El Gamal encryption

We have already (essentially) seen one encryption scheme:

p, g

hA = gx mod p

hB = gy mod p

KAB = (hB)x KBA = (hA)y

p, g, hA = gx

Receiver Sender

c = (KBA . m) mod phB, c

Security

If the DDH assumption holds, the El Gamal encryption scheme is secure against chosen-plaintext attacks

RSA: background

N=pq, p and q distinct, odd primes

(N) = (p-1)(q-1)– Easy to compute (N) given the factorization of N

– Hard to compute (N) without the factorization of N

Fact: for all x ZN*, it holds that x(N) = 1 mod N

– Proof: take CMSC 456!

If ed=1 mod (N), then for all x it holds that (xe)d = x mod N

RSA key generation Generate random p, q of sufficient length

Compute N=pq and (N) = (p-1)(q-1)

Compute e and d such that ed = 1 mod (N) – e must be relatively prime to (N) – Typical choice: e = 3; other choices possible

Public key = (N, e); private key = (N, d)

We have an asymmetry!– Given c=xe mod N, receiver can compute x=cd mod N– No apparent way for anyone else to recover x

Hardness of the RSA problem?

The RSA problem:– Compute x given N, e, and xe mod N

If factoring is easy, then the RSA problem is easy

We know of no other way to solve the RSA problem besides factoring N– But we do not know how to prove that the RSA

problem is as hard as factoring

The upshot: we believe factoring is hard, and we believe the RSA problem is hard

“Textbook RSA” encryption

Public key (N, e); private key (N, d)

To encrypt a message m ZN*, compute

c = me mod N

To decrypt a ciphertext c, compute m = cd mod N

Correctness clearly holds…

…what about security?

Textbook RSA is insecure!

It is deterministic!

Furthermore, it can be shown that the ciphertext leaks specific information about the plaintext

Padded RSA Public key (N, e); private key (N, d)

– Say |N| = 1024 bits

To encrypt m {0,1}895, – Choose random r {0,1}128

– Compute c = (r | m)e mod N

Decryption done in the natural way…

Essentially this idea has been standardized as RSA PKCS #1 v1.5

Hybrid encryption

Public-key encryption is “slow”

Encrypting “block-by-block” would be inefficient for long messages

Hybrid encryption gives the functionality of public-key encryption at the (asymptotic) efficiency of private-key encryption!

Hybrid encryption

Enc’

message

Enck

pk

“encapsulated key”

“encrypted message”

ciphertext

Enc = public-key encryption schemeEnc’ = private-key encryption scheme

Security

If public-key component and private-key component are secure against chosen-plaintext attacks, then hybrid encryption is secure against chosen-plaintext attacks

Malleability/chosen-ciphertext security All the public-key encryption schemes we have

seen so far are malleable– Given a ciphertext c that encrypts an (unknown)

message m, may be possible to generate a ciphertext c’ that encrypts a related message m’

In many scenarios, this is problematic– E.g., auction example; password example

Note: the problem is not integrity (there is no integrity in public-key encryption, anyway), but malleability