Announcements: Questions? This week: Digital signatures, DSA Secret sharing

Announcements:Announcements:

Questions? Questions?

This week:This week: Digital signatures, Digital signatures, DSADSA Secret sharingSecret sharing

DTTF/NB479: DszquphsbqizDTTF/NB479: Dszquphsbqiz Day 30Day 30

RSA Signatures allow you to recover the message RSA Signatures allow you to recover the message from the signature; ElGamal signatures don’tfrom the signature; ElGamal signatures don’t

Sig = f(user, message) Sig = f(user, message)

RSARSAAlice chooses: Alice chooses:

p,q, n=pq, p,q, n=pq, e: gcd(n, (p-1)(q-1))=1, e: gcd(n, (p-1)(q-1))=1, d: ed ≡ 1(mod ((p-1)(q-1))d: ed ≡ 1(mod ((p-1)(q-1))

Publishes n, ePublishes n, eAlice’s signature:Alice’s signature:

y ≡ my ≡ mdd(mod n). Delivers (m, y)(mod n). Delivers (m, y)

Bob’s verification:Bob’s verification: Does m ≡ yDoes m ≡ yee (mod n)? (mod n)?

ElGamalElGamalAlice chooses: Alice chooses:

p,primitive root p,primitive root , secret , secret aa, , and and ≡ ≡ a a (mod p)(mod p)

Publishes (p, Publishes (p, ), keeps ), keeps aa secretsecret

Alice’s signature:Alice’s signature: Chooses k: random, Chooses k: random,

gcd(k, p-1)=1gcd(k, p-1)=1 Sends m, (r,s), where:Sends m, (r,s), where:

r ≡ r ≡ kk (mod p) (mod p)s ≡ ks ≡ k-1-1(m – ar) (mod p-1)(m – ar) (mod p-1)

Bob’s verification:Bob’s verification: Does Does rrrrss ≡ ≡ mm (mod p)? (mod p)?

It’s quicker to sign a short digest than to sign a long It’s quicker to sign a short digest than to sign a long messagemessage

Note that we need to choose Note that we need to choose n n > m> m in RSA, p in RSA, p > m> m in ElGamal in ElGamal Problem: m could be long!Problem: m could be long! But h(m) is short!But h(m) is short!

So Alice sends (m, sig(h(m)))So Alice sends (m, sig(h(m)))

Eve intercepts this, wants to sign m’ with Alice’s Eve intercepts this, wants to sign m’ with Alice’s signature, so needs sig(h(m’)) = sig(h(m)), and signature, so needs sig(h(m’)) = sig(h(m)), and thus h(m)=h(m’)thus h(m)=h(m’) Why can’t she do this?Why can’t she do this?

Birthday attacks can be successful on signatures that Birthday attacks can be successful on signatures that are too shortare too short

Slightly different paradigm: two rooms with r Slightly different paradigm: two rooms with r people each. What’s the probability that someone people each. What’s the probability that someone in this room has the same birthday as someone in in this room has the same birthday as someone in the other room.the other room.

Approximation: Approximation: We divide by N, not 2N.We divide by N, not 2N. But setting the probability = 0.5 and solving for r, we But setting the probability = 0.5 and solving for r, we

get get r=c*sqrt(n) r=c*sqrt(n) again again (where c=sqrt(ln 2)~.83)(where c=sqrt(ln 2)~.83)

Consider a 50-bit hash. Only need O(2^25) documentsConsider a 50-bit hash. Only need O(2^25) documents These are relatively easy to generate, actually.These are relatively easy to generate, actually.

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2

1

Birthday attacks on signatures that are too shortBirthday attacks on signatures that are too short

MalloryMallory generates 2 groups of documents: generates 2 groups of documents:

She takes a match (mShe takes a match (m11, m, m22) between them such that h(m) between them such that h(m11) = ) = h(mh(m22))MalloryMallory sends (m sends (m11, h(m, h(m11)) to Alice, who returns signed copy: )) to Alice, who returns signed copy: (m(m11, sig(h(m, sig(h(m11)).)).MalloryMallory replaces m replaces m11 with m with m22 and uses sig(h(m and uses sig(h(m11) as the ) as the signature. signature.

The pair (mThe pair (m22, sig(h(m, sig(h(m11)) looks like Alice’s valid signature!)) looks like Alice’s valid signature!

Alice’s defense? What can she do to defend herself?Alice’s defense? What can she do to defend herself?

r “good docs” r “fraudulent docs”

Alice’s defenseAlice’s defense

She changes a random bit herself!She changes a random bit herself!Note this changes her signature: Note this changes her signature: (m(m11’, sig(h(m’, sig(h(m11’))’)) Mallory is forced to generate another message with the Mallory is forced to generate another message with the

same hash as this new document. same hash as this new document. Good luck!Good luck!

Lessons:Lessons: Birthday attacks essentially halve the number of bits of Birthday attacks essentially halve the number of bits of

security. security. So SHA-1 is still secure against themSo SHA-1 is still secure against them

Make a minor change to the document you sign!Make a minor change to the document you sign!

Code-talkers?Code-talkers?

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As far as I can tell, Navajo doesn’t have a word for As far as I can tell, Navajo doesn’t have a word for zero. Do-neh-lini means zero. Do-neh-lini means neutralneutral..Seriously, the Navajo code talkers created effective Seriously, the Navajo code talkers created effective codes used in WWII: codes used in WWII: http://en.wikipedia.org/wiki/Code_talker

DSA: Digital Signature AlgorithmDSA: Digital Signature Algorithm

19941994

Similar to ElGamalSimilar to ElGamal signature with appendixsignature with appendix But verification is fasterBut verification is faster And it’s guaranteed to be more secureAnd it’s guaranteed to be more secure

Assume m is already hashed using SHA: Assume m is already hashed using SHA: so we are signing a 160-bit message, m.so we are signing a 160-bit message, m.

DSA: Digital Signature AlgorithmDSA: Digital Signature Algorithm

Alice’s Setup:Alice’s Setup: m: 160-bit messagem: 160-bit message q: 160-bit primeq: 160-bit prime p: 512-bit prime, such that q is a factor of (p-1)p: 512-bit prime, such that q is a factor of (p-1) g: a primitive root of p. g: a primitive root of p. ≡≡gg(p-1)/q(p-1)/q (mod p) (mod p)

Then Then q q ≡ 1 (mod p). (Why?)≡ 1 (mod p). (Why?) ≡ ≡ aa. Secret a, 0 < a < q-1. Secret a, 0 < a < q-1 Publishes: (p,q,Publishes: (p,q,))

Sig = (r,s)Sig = (r,s) random k, 0 < k < q-1random k, 0 < k < q-1 r ≡ r ≡ kk (mod q) (mod q) s = ks = k-1-1(m + ar) (mod q)(m + ar) (mod q)

Verify:Verify: Compute u1 ≡ sCompute u1 ≡ s-1-1m (mod q), u2 ≡ sm (mod q), u2 ≡ s-1-1r (mod q)r (mod q) Does (Does (u1u1u2u2 (mod p))(mod q) = r? (mod p))(mod q) = r?

q=17

p=103g=2

=?

1-3

DSA: Digital Signature AlgorithmDSA: Digital Signature Algorithm

Alice’s Setup:Alice’s Setup: m: 160-bit messagem: 160-bit message q: 160-bit primeq: 160-bit prime p: 512-bit prime, such that q is a factor of (p-1)p: 512-bit prime, such that q is a factor of (p-1) g: a primitive root of p. g: a primitive root of p. ≡≡gg(p-1)/q(p-1)/q (mod p) (mod p)

Then Then q q ≡ 1 (mod p). (Why?)≡ 1 (mod p). (Why?) ≡ ≡ aa. Secret a, 0 < a < q-1. Secret a, 0 < a < q-1 Publishes: (p,q,Publishes: (p,q,))

Sig = (r,s)Sig = (r,s) random k, 0 < k < q-1random k, 0 < k < q-1 r ≡ r ≡ kk (mod q) (mod q) s = ks = k-1-1(m + ar) (mod q)(m + ar) (mod q)

Verify:Verify: Compute u1 ≡ sCompute u1 ≡ s-1-1m (mod q), u2 ≡ sm (mod q), u2 ≡ s-1-1r (mod q)r (mod q) Does (Does (u1u1u2u2 (mod p))(mod q) = r? (mod p))(mod q) = r?

q=17

p=103g=2

=64

Advantages over Advantages over ElGamal?ElGamal?

In ElGamal, if you could In ElGamal, if you could solve r = solve r = kk (mod p) by (mod p) by Pollig-Hellman, you’d have Pollig-Hellman, you’d have k.k.

In DSA, (p-1) has a large In DSA, (p-1) has a large factor, q.factor, q.

If you could solve the non-q If you could solve the non-q factors, there would still be factors, there would still be q possibilities for k.q possibilities for k.

How many ints (mod p) How many ints (mod p) give a specific int (mod q)?give a specific int (mod q)?

4

DSA: Digital Signature AlgorithmDSA: Digital Signature Algorithm

Alice’s Setup:Alice’s Setup: m: 160-bit messagem: 160-bit message q: 160-bit primeq: 160-bit prime p: 512-bit prime, such that q is a factor of (p-1)p: 512-bit prime, such that q is a factor of (p-1) g: a primitive root of p. g: a primitive root of p. ≡≡gg(p-1)/q(p-1)/q (mod p) (mod p)

Then Then q q ≡ 1 (mod p). (Why?)≡ 1 (mod p). (Why?) ≡ ≡ aa. Secret a, 0 < a < q-1. Secret a, 0 < a < q-1 Publishes: (p,q,Publishes: (p,q,))

Sig = (r,s)Sig = (r,s) random k, 0 < k < q-1random k, 0 < k < q-1 r ≡ r ≡ kk (mod q) (mod q) s = ks = k-1-1(m + ar) (mod q)(m + ar) (mod q)

Verify:Verify: Compute u1 ≡ sCompute u1 ≡ s-1-1m (mod q), u2 ≡ sm (mod q), u2 ≡ s-1-1r (mod q)r (mod q) Does (Does (u1u1u2u2 (mod p))(mod q) = r? (mod p))(mod q) = r?

q=17

p=103g=2

=64

How hard is it to search for a How hard is it to search for a 512-bit prime p = kq + 1 for 512-bit prime p = kq + 1 for some even number k?some even number k?

How do we search for primes?How do we search for primes? 1/115 of odd 100-digit 1/115 of odd 100-digit

numbers are prime.numbers are prime. What fraction of odd 512-bit What fraction of odd 512-bit

integers are prime?integers are prime? Recall our discussion of the Recall our discussion of the

density of primesdensity of primes

(Day 21) Using within a primality testing (Day 21) Using within a primality testing schemescheme

Finding large probable primesFinding large probable primes

#primes < x = #primes < x =

Density of primes: ~1/ln(x)Density of primes: ~1/ln(x)

For 100-digit numbers, ~1/230.For 100-digit numbers, ~1/230.

So ~1/115 of odd 100-digit So ~1/115 of odd 100-digit numbers are primenumbers are prime

Can start with a random large odd Can start with a random large odd number and iterate, applying number and iterate, applying M-R to remove composites. M-R to remove composites. We’ll soon find one that is a We’ll soon find one that is a likely prime.likely prime.

Odd?

div by other small primes?

Prime by Factoring/advanced techn.?

n

no

no

yes

yes

prime

Pass M-R?

)ln()(

x

xx

DSA: Digital Signature AlgorithmDSA: Digital Signature Algorithm

Alice’s Setup:Alice’s Setup: m: 160-bit messagem: 160-bit message q: 160-bit primeq: 160-bit prime p: 512-bit prime, such that q is a p: 512-bit prime, such that q is a

factor of (p-1)factor of (p-1) g: a primitive root of p. g: a primitive root of p. =g=g(p-1)/q(p-1)/q (mod p) (mod p)

Then Then q q = 1 (mod p). (Why?)= 1 (mod p). (Why?) = = aa. Secret a, 0 < a < q-1. Secret a, 0 < a < q-1 Publishes: (p,q,Publishes: (p,q,))

Sig = (r,s)Sig = (r,s) random k, 0 < k < q-1random k, 0 < k < q-1 r = r = kk (mod p) (mod p) s = ks = k-1-1(m + ar) (mod q)(m + ar) (mod q)

Verify:Verify: Compute u1 = sCompute u1 = s-1-1m, u2 = sm, u2 = s-1-1rr Does (aDoes (au1u1bbu2u2 (mod p))(mod q) = r? (mod p))(mod q) = r?

Show that order of ops matters:Show that order of ops matters:((kk (mod p))(mod q) ≠ ((mod p))(mod q) ≠ (kk (mod q))(mod p) (mod q))(mod p)

Easier: find Easier: find (a(a(mod p))(mod q) ≠ (a(mod q))(mod p)(mod p))(mod q) ≠ (a(mod q))(mod p)

5

Latest versionsLatest versions

Recommended:Recommended: SHA-224/256/384/512 as the hash functionSHA-224/256/384/512 as the hash function qq of size 224 and 256 bits of size 224 and 256 bits p p of size 2048 and 3072. of size 2048 and 3072.

http://csrc.nist.gov/publications/drafts/fips_186-3/Draft_FIPS-186-3%20_November2008.pdfhttp://csrc.nist.gov/publications/drafts/fips_186-3/Draft_FIPS-186-3%20_November2008.pdf