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Public Key Algorithms

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Modular Arithmetic

Public key algorithms are based on modular arithmetic.

Modular addition. Modular multiplication. Modular exponentiation.

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Modular Addition

Addition modulo (mod) K Poor cipher with (dk+dm) mod K, e.g.,

if K=10 and dk is the key.

Additive inverse: addition mod K yields 0.

“Decrypt” by adding inverse.

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Modular Multiplication Multiplication modulo K Multiplicative inverse: multiplication

mod K yields 1 Only some numbers have inverse Use Euclid’s algorithm to find inverse

Given x, n, it finds y such that xy mod n = 1

All number relatively prime to n will have mod n multiplicative inverse

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Totient Function x, m relative prime: no other

common factor than 1 Totient function ø(n): number of

integers less than n relatively prime to n if n is prime, ø(n)=n-1 if n=pq, and p, q are primes,

ø(n)=(p-1)(q-1)

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Modular Exponentiation

xy mod n = xy mod ø(n) mod n

if y = 1 mod ø(n) then xy mod n = x mod n

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RSA (Rivest, Shamir, Adleman)

The most popular one. Support both public key encryption and

digital signature. Assumption/theoretical basis:

Factoring a big number is hard. Variable key length (usually 512 bits). Variable plaintext block size.

Plaintext must be “smaller” than the key. Ciphertext block size is the same as the key

length.

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What Is RSA? To generate key pair:

Pick large primes (>= 256 bits each) p and q

Let n = p*q, keep your p and q to yourself!

For public key, choose e that is relatively prime to ø(n) =(p-1)(q-1), let pub = <e,n>

For private key, find d that is the multiplicative inverse of e mod ø(n), i.e., e*d = 1 mod ø(n), let priv = <d,n>

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How Does RSA Work?

Given pub = <e, n> and priv = <d, n> encryption: c = me mod n, m < n decryption: m = cd mod n signature: s = md mod n, m < n verification: m = se mod n

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Why Does RSA Work?

Given pub = <e, n> and priv = <d, n> n =p*q, ø(n) =(p-1)(q-1) e*d = 1 mod ø(n) xed = x mod n encryption: c = me mod n decryption: m = cd mod n = med mod

n = m mod n = m Why????????

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e*d = 1 mod ø(n) So e*d = 1 + kø(n)

…med mod n = m 1 + kø(n)

m (m kø(n) mod n) = m ????

What is Fermat’s theorem???

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Why Is RSA Secure? Factoring 512-bit number is very

hard! But if you can factor big number n

then given public key <e,n>, you can find d, hence the private key by: Knowing factors p, q, such that, n =

p*q Then ø(n) =(p-1)(q-1) Then d such that e*d = 1 mod ø(n)

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Attacks on RSA

• Brute force key search• Mathematical attacks• Timing attacks

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Math-Based Attacks

Three possible approaches: – Factor n = pq – Determine F(n) – Find the private key d directly • All the above are equivalent to factoring n

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Brute Force

An adversary just tries all possiblekeys and keeps his fingers crossed

thatthe right key is not the last key he

willtry !

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Timing Attacks By measuring the time required to

perform decryption (exponentiation with the

private key as exponent), an attacker can figure out the private key

Possible countermeasures: – use constant exponentiation time – add random delays – blind values used in calculations

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Other Attacks on RSA Small encryption exponent e E=3, Alice sends the message m to three people (public keys (e, n1), (e, n2), (e,n3)) An attacker can compute a solution to the

followingsystem x = c1 mod n1 x = c2 mod n2 x = c3 mod n3 Then, compute m from x = m3 Countermeasure: padding required

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Forward Search Attack If message space is small, the attacker cancreate a dictionary of encrypted messages(public key known, encrypt all possiblemessages and store them) When the attacker ‘sees’ a message on thenetwork, compares the encryptedmessages,

so he finds out what particular message was encrypted

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Small decryption exponent d

Choosing a small exponent helps efficiency BUT

If size of d is 1/4 size of n (in bits) and gcd(p-1,q-1) is small, there is a way to compute d only from e and n.

Countermeasure: d should be about the same size as n.

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Common modulus attack Each entity must choose its own modulus Assume Alice and Bob generated keys

using the same modulus n, ((e1, n ), d1)) and ((e2, n), d2))

C1 = Me1 mod n, C2 = Me2 mod n (e1)a + (e2) b = 1 if gcd(e1,e2)=1 M = C1a C2 b mod n

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Cycling attack Intercepted ciphertext: C C1 = Ce mod n C2 = C1

e mod n C3 = C2

e mod n

Ck = Ck-1e mod n

If Ck = C then stop P = Ck-1

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Attacker Goals Total break: the attacker finds the key (thesymmetric key for ciphers or the private key for publickey cryptosystems); after that all ciphertexts can bedecrypted. Partial break: with some probability , the adversaryis able to decrypt previously unseen ciphertexts,without knowing the key. Or the adversary can findout info about the plaintext, just by looking at theciphertext. Distinguishability: with probability > 0.5, theadversary can distinguish between encryption of twodifferent plaintexts, or between an encryption and arandom string.

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Diffie-Hellman Key Exchange

Shared key, public communication No authentication of partners What’s involved?

P is a prime (about 512 bits), and g < p

P and g are publicly known

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Diffie-Hellman Key Exchange

ProcedureAlice Bobpick secret Sa randomly pick secret Sb

randomlycompute TA=gSa mod p compute TB=gSb

mod psend TA to Bob send TB to Alice

compute TBSa mod p compute TA

Sb mod p

Alice and Bob reached the same secret gSaSb

mod p, which is then used as the shared key.

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DH Security - Discrete Logarithm Is Hard

T = gs mod p Conjecture: given T, g, p, it is

extremely hard to compute the value of s (discrete logarithm)

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Diffie-Hellman Scheme

Security factors Discrete logarithm very difficult. Shared key (the secret) itself never

transmitted.

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Disadvantages: Expensive exponential operation

DoS possible. The scheme itself cannot be used to

encrypt anything – it is for secret key establishment.

No authentication, so you can not sign anything …

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Bucket Brigade Attack...Man In The Middle

Alice Trudy BobgSa=123 gSx =654 gSb =255 123 --> 654 -->

<--654 <--255 654Sa=123Sx 255Sx=654Sb

Trudy plays Bob to Alice and Alice to Bob

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Diffie-Hellman in Phone Book Mode

DH was subject to active man-in-the-middle attack because their public key-component was intercepted and substituted

Phone book mode allows everyone to generate the public key-component in advance and publish them through other reliable means, e.g. <TB> for bob

All communicating parties agree on their common <g, p>

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Encryption With Diffie-Hellman

Everyone computes and publishes <p, g, T> T=gS mod p

Alice communicates with Bob: Alice

Picks a random secret Sa

Computes gbSa mod pb

Use Kab = TbSa mod pb to encrypt message

Send encrypted message along with gbSa

mod pb

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Bob (gb

Sa)Sb mod pb = (gbSb)Sa mod pb = Tb

Sa mod pb = Kab

Use Kab to decrypt