Public-Key Cryptanalysis 1: Introduction and Factoring

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1 Public-Key Cryptanalysis 1: Introduction and Factoring Nadia Heninger University of Pennsylvania July 21, 2013

2 Adventures in Cryptanalysis Part 1: Introduction and Factoring. What is public-key crypto really based on? How to break RSA from the comfort of your very own home. Part 2: Generating keys. How not to generate your crypto keys. And how many people who should know better have screwed it up. Part 3: Breaking keys with lattices. How fragile is RSA? And why not to post your private keys to Pastebin.

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5 Textbook Diffie-Hellman [Diffie Hellman 1976] Public Parameters G a group (e.g. F p, or an elliptic curve) g group generator Key Exchange g a g b g ab g ab

6 Computational problems Discrete Log Problem: Given g a, compute a. Solving this problem permits attacker to compute shared key by computing a and raising (g b ) a. Discrete log is in NP and conp not NP-complete (unless P=NP or similar).

7 Computational problems Diffie-Hellman problem Problem: Given g a, g b, compute g ab. Exactly problem of computing shared key from public information. Reduces to discrete log in some cases: Diffie-Hellman is as strong as discrete log for certain primes [den Boer 1988] both problems are (probabilistically) polynomial-time equivalent if the totient of p 1 has only small prime factors Towards the equivalence of breaking the Diffie-Hellman protocol and computing discrete logarithms [Maurer 1994] if... an elliptic curve with smooth order can be construted efficiently, then... [the discrete log] can be reduced efficiently to breakingthe Diffie-Hellman protocol (Computational) Diffie-Hellman assumption: This problem is hard in general.

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11 Textbook RSA [Rivest Shamir Adleman 1977] Public Key N = pq modulus e encryption exponent Private Key p, q primes d decryption exponent (d = e 1 mod (p 1)(q 1)) Encryption public key = (N, e) ciphertext = message e mod N message = ciphertext d mod N

12 Textbook RSA [Rivest Shamir Adleman 1977] Public Key N = pq modulus e encryption exponent Private Key p, q primes d decryption exponent (d = e 1 mod (p 1)(q 1)) Signing public key = (N, e) signature = message d mod N message = signature e mod N

13 Computational problems Factoring Problem: Given N, compute its prime factors. Computationally equivalent to computing private key d. Factoring is in NP and conp not NP-complete (unless P=NP or similar).

14 Computational problems eth roots mod N Problem: Given N, e, and c, compute x such that x e c mod N. Equivalent to decrypting an RSA-encrypted ciphertext. Equivalent to selective forgery of RSA signatures. Conflicting results about whether it reduces to factoring: Breaking RSA may not be equivalent to factoring [Boneh Venkatesan 1998] an algebraic reduction from factoring to breaking low-exponent RSA can be converted into an efficient factoring algorithm Breaking RSA generically is equivalent to factoring [Aggarwal Maurer 2009] a generic ring algorithm for breaking RSA in Z N can be converted into an algorithm for factoring RSA assumption : This problem is hard.

15 Public service announcement: Textbook RSA is insecure RSA encryption is homomorphic under multiplication. This lets an attacker do all sorts of fun things: Attack: Malleability Given a ciphertext c = m e mod N, ca e mod N is an encryption of ma for any a chosen by attacker. Attack: Chosen ciphertext attack Given a ciphertext c, attacker asks for decryption of ca e mod N and divides by a to obtain m. Attack: Signature forgery Attacker convinces a signer to sign z = xy e mod N and computes a valid signature of x as z d /y mod N. So in practice we always use padding on messages.

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17 The DSA Algorithm DSA Public Key p prime q prime, divides (p 1) g generator of subgroup of order q mod p y = g x mod p Private Key x private key Verify u 1 = H(m)s 1 mod q u 2 = rs 1 mod q r? = g u 1 y u 2 mod p mod q Sign Generate random k. r = g k mod p mod q s = k 1 (H(m) + xr) mod q

18 Computational problems Discrete Log Breaking DSA is equivalent to computing discrete logs in the random oracle model. [Pointcheval, Vaudenay 96]

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20 The rest of this lecture.

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22 Not ethical research.

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26 The other two lectures.

27 Factoring (and discrete log) the hard way Some material borrowed from Dan Bernstein and Tanja Lange

28 Preliminaries: Using Sage Working code examples will be given in Sage. Sage is free open source mathematics software. Download from Sage is based on Python sage: 2*3 6

29 Preliminaries: Using Sage Working code examples will be given in Sage. Sage is free open source mathematics software. Download from Sage is based on Python, but there are a few differences: sage: 2^3 8 ˆ is exponentiation, not xor

30 Preliminaries: Using Sage Working code examples will be given in Sage. Sage is free open source mathematics software. Download from Sage is based on Python, but there are a few differences: sage: 2^3 8 ˆ is exponentiation, not xor It has lots of useful libraries: sage: factor(15) 3 * 5

31 Preliminaries: Using Sage Working code examples will be given in Sage. Sage is free open source mathematics software. Download from Sage is based on Python, but there are a few differences: sage: 2^3 8 ˆ is exponentiation, not xor It has lots of useful libraries: sage: factor(15) 3 * 5 sage: factor(x^2-1) (x - 1) * (x + 1)

32 Practicing Sage and Textbook RSA Key generation: sage: p = random_prime(2^512); q = random_prime(2^512) sage: N = p*q sage: e = sage: d = inverse_mod(e,(p-1)*(q-1)) Encryption: sage: m = Integer( helloworld,base=35) sage: c = pow(m,65537,n) Decryption: sage: Integer(pow(c,d,N)).str(base=35) helloworld

33 So how hard is factoring?

34 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32))

35 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32)) * Time: CPU 0.01 s, Wall: 0.01 s

36 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32)) * Time: CPU 0.01 s, Wall: 0.01 s sage: time factor(random_prime(2^64)*random_prime(2^64))

37 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32)) * Time: CPU 0.01 s, Wall: 0.01 s sage: time factor(random_prime(2^64)*random_prime(2^64)) * Time: CPU 0.13 s, Wall: 0.15 s

38 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32)) * Time: CPU 0.01 s, Wall: 0.01 s sage: time factor(random_prime(2^64)*random_prime(2^64)) * Time: CPU 0.13 s, Wall: 0.15 s sage: time factor(random_prime(2^96)*random_prime(2^96))

39 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32)) * Time: CPU 0.01 s, Wall: 0.01 s sage: time factor(random_prime(2^64)*random_prime(2^64)) * Time: CPU 0.13 s, Wall: 0.15 s sage: time factor(random_prime(2^96)*random_prime(2^96)) * Time: CPU 4.64 s, Wall: 4.76 s

40 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32)) * Time: CPU 0.01 s, Wall: 0.01 s sage: time factor(random_prime(2^64)*random_prime(2^64)) * Time: CPU 0.13 s, Wall: 0.15 s sage: time factor(random_prime(2^96)*random_prime(2^96)) * Time: CPU 4.64 s, Wall: 4.76 s sage: time factor(random_prime(2^128)*random_prime(2^128))

41 So how hard is factoring? sage: time factor(random_prime(2^32)*random_prime(2^32)) * Time: CPU 0.01 s, Wall: 0.01 s sage: time factor(random_prime(2^64)*random_prime(2^64)) * Time: CPU 0.13 s, Wall: 0.15 s sage: time factor(random_prime(2^96)*random_prime(2^96)) * Time: CPU 4.64 s, Wall: 4.76 s sage: time factor(random_prime(2^128)*random_prime(2^128)) * Time: CPU s, Wall: s

42 Factoring in practice Two kinds of factoring algorithms: 1. Algorithms that have running time dependent on the size of the factor. Good for factoring small numbers, and finding small factors of big numbers. 2. General-purpose algorithms that depend on the size of the number to be factored. State-of-the-art algorithms for factoring really big numbers.

43 Trial Division Good for finding very small factors Takes p/ log p trial divisions to find a prime factor p.

44 Pollard rho Good for finding slightly larger prime factors Intuition Try to take a random walk among elements modn. If p divides n, there will be a cycle of length p.

45 Pollard rho Good for finding slightly larger prime factors Intuition Try to take a random walk among elements modn. If p divides n, there will be a cycle of length p. Details Expect a collision after searching about p random elements. To avoid having to store and search every element, use Floyd s cycle-finding algorithm. (Store two pointers, and move one twice as fast as the other until they coincide.)

46 Why is it the rho algorithm?

47 Pollard rho impementation def rho(n): a = ; c=10 # some random values f = lambda x: (x^2+c)%n a1 = f(a) ; a2 = f(a1) while gcd(n, a2-a1)==1: a1 = f(a1); a2 = f(f(a2)) return gcd(n, a2-a1) sage: N = sage: rho(n) 2053

48 Pollard s p 1 method Good for finding special small factors Intuition If a r 1 mod p then p gcd(a r 1, N). Don t know p, pick very smooth number r, hoping for ord(a) p to divide it. ( Smooth means has only small prime factors.)

49 Pollard s p 1 method Good for finding special small factors Intuition If a r 1 mod p then p gcd(a r 1, N). Don t know p, pick very smooth number r, hoping for ord(a) p to divide it. ( Smooth means has only small prime factors.) N= r=lcm(range(1,2^22)) # this takes a while... s=integer(pow(2,r,n)) sage: gcd(s-1,n)

50 Pollard p 1 method This method finds larger factors than the rho method (in the same time)...but only works for special primes. In the previous example, p 1 = has only small factors (aka. p 1 is smooth). Many crypto standards require using only strong primes a.k.a primes where p 1 is not smooth. This recommendation is outdated. The elliptic curve method (next slide) works even for strong primes.

51 Lenstra s Elliptic Curve Method Good for finding medium-sized factors Intuition Pollard s p 1 method works in the multiplicative group of integers modulo p. The elliptic curve method is exactly the p 1 method, but over the group of points on an elliptic curve modulo p: Multiplication of group elements becomes addition of points on the curve. All arithmetic is still done modulo N.

52 Lenstra s Elliptic Curve Method Good for finding medium-sized factors Intuition Pollard s p 1 method works in the multiplicative group of integers modulo p. The elliptic curve method is exactly the p 1 method, but over the group of points on an elliptic curve modulo p: Multiplication of group elements becomes addition of points on the curve. All arithmetic is still done modulo N. Theorem (Hasse) The order of an elliptic curve modulo p is in [p p, p p]. There are lots of smooth numbers in this interval. If one elliptic curve doesn t work, try more until you find a smooth order.

53 Elliptic Curves def curve(d): frac_n = type(d) class P(object): def init (self,x,y): self.x,self.y = frac_n(x),frac_n(y)... def add (a,b): return P((a.x*b.y + b.x*a.y)/(1 + d*a.x*a.y*b.x*b.y) (a.y*b.y - a.x*b.x)/(1 - d*a.x*b.x*a.y*b.y) def mul (self, m): return double_and_add(self,m,p(0,1))

54 Elliptic Curve Factorization def ecm(n,y,t): # Choose a curve and a point on the curve. frac_n = Q(n) P = curve(frac_n(1,3)) p = P(2,3) q = p * lcm(xrange(1,y)) return gcd(q.x.t,n) Method runs very well on GPUs. Still an active research area. ECM is very efficient at factoring random numbers, once small factors are removed. Heuristic running time L p (1/2) = O(e c ln p ln ln p ).

55 Quadratic Sieve Intuition: Fermat factorization Main insight: If we can find two squares a 2 and b 2 such that Then a 2 b 2 mod N a 2 b 2 = (a + b)(a b) 0 mod N and we might hope that one of a + b or a b shares a nontrivial common factor with N.

56 Quadratic Sieve Intuition: Fermat factorization Main insight: If we can find two squares a 2 and b 2 such that a 2 b 2 mod N Then a 2 b 2 = (a + b)(a b) 0 mod N and we might hope that one of a + b or a b shares a nontrivial common factor with N. First try: 1. Start at c = N 2. Check c 2 N, (c + 1) 2 N,... until we find a square. This is Fermat factorization, which could take up to p steps.

57 Quadratic Sieve General-purpose factoring Intuition We might not find a square outright, but we can construct a square as a product of numbers we look through.

58 Quadratic Sieve General-purpose factoring Intuition We might not find a square outright, but we can construct a square as a product of numbers we look through. 1. Sieving Try to factor each of c 2 N, (c + 1) 2 N, Only save a d i = c 2 i N if all of its prime factors are less than some bound B. (If it is B-smooth.) 3. Store each d i by its exponent vector d i = 2 e 2 3 e 3... B e B. 4. If i d i is a square, then its exponent vector contains only even entries.

59 Quadratic Sieve General-purpose factoring Intuition We might not find a square outright, but we can construct a square as a product of numbers we look through. 1. Sieving Try to factor each of c 2 N, (c + 1) 2 N, Only save a d i = c 2 i N if all of its prime factors are less than some bound B. (If it is B-smooth.) 3. Store each d i by its exponent vector d i = 2 e 2 3 e 3... B e B. 4. If i d i is a square, then its exponent vector contains only even entries. 5. Linear Algebra Once enough factorizations have been collected, can use linear algebra to find a linear dependence mod2.

60 Quadratic Sieve General-purpose factoring Intuition We might not find a square outright, but we can construct a square as a product of numbers we look through. 1. Sieving Try to factor each of c 2 N, (c + 1) 2 N, Only save a d i = c 2 i N if all of its prime factors are less than some bound B. (If it is B-smooth.) 3. Store each d i by its exponent vector d i = 2 e 2 3 e 3... B e B. 4. If i d i is a square, then its exponent vector contains only even entries. 5. Linear Algebra Once enough factorizations have been collected, can use linear algebra to find a linear dependence mod2. 6. Square roots Take square roots and hope for a nontrivial factorization. Math exercise: Square product has 50% chance of factoring pq.

61 An example of the quadratic sieve Let s try to factor N = 2759.

62 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b).

63 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 =

64 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 = = 157.

65 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 = = = 266.

66 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 = = = = 377.

67 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 = = = = = 490 =

68 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 = = = = = 490 = = 605 =

69 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 = = = = = 490 = = 605 = The product is a square:

70 An example of the quadratic sieve Let s try to factor N = Recall idea: If a 2 N is a square b 2 then N = (a b)(a + b) = 50 = = = = = 490 = = 605 = The product is a square: QS computes gcd{2759, } = 31.

71 Quadratic Sieve running time How do we choose B? How many numbers do we have to try to factor? Depends on (heuristic) probability that a randomly chosen number is B-smooth. Running time: L n (1/2) = e ln n ln ln n.

72 Number field sieve Best running time for general purpose factoring Insight Replace relationship a 2 = b 2 mod N with a homomorphism between ring of integers O K in a specially chosen number field and Z N. Algorithm 1. Polynomial selection Find a good choice of number field K. 2. Relation finding Factor elements over O K and over Z. 3. Linear algebra Find a square in O K and a square in Z 4. Square roots Take square roots, map into Z, and hope we find a factor. Running time: L n (1/3) = e (ln n ln ln n)1/3.

73 Discrete log: State of the art Discrete log: Compute g l = t in some group. (Recall Diffie-Hellman.) For decades, progress in discrete log closely followed progress in factoring. Best running time was L(1/3) using number field or function field sieve.

74 Discrete log: State of the art Discrete log: Compute g l = t in some group. (Recall Diffie-Hellman.) For decades, progress in discrete log closely followed progress in factoring. Best running time was L(1/3) using number field or function field sieve.... until this year... A new index calculus algorithm with complexity L(1/4 + o(1)) in very small characteristic. Antoine Joux, February 2013 A quasi-polynomial algorithm for discrete logarithm in finite fields of small characteristic. Razvan Barbulescu, Pierrick Gaudry, Antoine Joux, and Emmanuel Thomé, June 2013

75 Index calculus for discrete log. Goal: Solve g l t mod p. 1. Relation finding: Collect g r p r pr k k mod p for many choices of r. This gives us a congruence of discrete logarithms for the p i : r r 1 log g p r k log g p k mod (p 1) 2. Linear algebra Once we have enough relations, use linear algebra to solve for the log g p i. 3. Smooth relation Search through values of R until we find one that results in a smooth value for g R t: 4. Output discrete log: g R t p τ pτ k k mod p log g t R + τ 1 log g p τ k log g p k mod (p 1)

76 Recent discrete log progress The new discrete log algorithms have resulted in very impressive new computational results. We still need to verify whether heuristic assumptions line up in practice. Open problem: Do these new results mean anything for discrete logs in large characteristic (working mod p?) Open problem: Do these new results lead to new results in factoring?

77 Factoring in practice 1977 Rivest estimates factoring a 125-digit number would require 40 quadrillion years. (Letter to Martin Gardner) 1978 An 80-digit n provides moderate security against an attack using current technology; using 200 digits provides a margin of safety against future developments. A Method for Obtaining Digital Signatures and Public-Key Cryptosystems 1994 Atkins, Graff, Lenstra, Leyland factor RSA-129 challenge from Martin Gardner s 1977 column. We conclude that commonly-used 512-bit RSA moduli are vulnerable to any organiation prepared to spend a few million dollars and wait a few months.

78 Factoring in practice 1999 te Riele, Cavallar, Dodson, Lenstra, Lioen, Montgomery, Murphy factor (512-bit) RSA-155 challenge in 4 months. Spent 35.7 CPU years Benjamin Moody factors 512-bit RSA signing key for TI-83+ calculator in 73 days on a 1.9 GHz dual-core processor Zachary Harris factors Google s 512-bit domain authentication key in 72 hours for $75 using Amazon ec2.

79 An amateur can use off the shelf software to factor a 512-bit RSA modulus in about a day for a few hundred dollars.

80 Factoring in the cloud Amazon Elastic Compute Cloud + Cluster computing toolkit for ec2 + Number field sieve implementation

81 My first attempt: Running time details 30 hours to factor a 512-bit modulus, $234 in computation time. Polynomial selection: 7 minutes (parallel) Sieving: 14 hours (parallel) Generating matrix: 1 hour 30 minutes (single core) Linear algebra: 14 hours (partially parallel) Square root: 23 minutes (single core) 21 8x large compute optimized cluster compute instances = 20 spot instances ($0.27/hour) + 1 on demand instance control node ($2.40/hour)

82 The dirtier details of factoring +14 hours sieving first attempt + 6 hours dealing with running out of disk space and StarCluster crashing when I tried to increase size of volume, gave up and restarted + 30 minutes figuring out why CADO-NFS crashed parallelizing linear algebra across 21 nodes until I figured out it wouldn t crash if run with 16 nodes $550 total Amazon bill over four days including time to figure out installation, configuration, test smaller experiments and false starts

83 Suggestions for future experiments and optimizations Likely improvements Experiment with using more (cheaper) instances for sieving (used about 75% CPU on each instance) Experiment with effect of oversieving more to make linear algebra easier Optimize parallelization for linear algebra step. Linear algebra is only partially parallelized, and the parts that were parallelized only used 15% of available CPU on each instance.

84 How hard is it to factor longer keys? key size ECU years estimated cost $ $422, $38 million $560 million $ Using the cloud to determine key strengths T. Kleinjung, A.K. Lenstra, D. Page, and N.P. Smart. 2012

85 Summary: It costs $560 million to factor a 1024-bit key. Next Lecture: Factoring 60, bit keys for $5.

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