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CPSC 490 Number Theory Primes, Factoring and Euler Phi-function

CPSC 490 Number Theory Primes, Factoring and Euler Phi-function. Mar.31 st , 2006 Sam Chan. Clarifications. The Euclidean Algorithm runs in logarithmic time. Finding modular inverses can be done using the extended Euclidean algorithm.

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CPSC 490 Number Theory Primes, Factoring and Euler Phi-function

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  1. CPSC 490 Number TheoryPrimes, Factoring and Euler Phi-function Mar.31st, 2006 Sam Chan

  2. Clarifications • The Euclidean Algorithm runs in logarithmic time. • Finding modular inverses can be done using the extended Euclidean algorithm. • Ex. 5x = 1 (mod 7) is equivalent to writing 5x – 7y = 1 or 5x + 7(-y) = 1. Use extended Euclidean Algorithm to find x.

  3. Primes • Def: a positive integer p > 1 is prime if and only if it has exactly two divisors, 1 and p. • Ex. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 are the fist 10 primes. • How many primes are there exactly?

  4. Euclid’s Proof • Euclid first proved that there exists an infinite number of primes. (300 BC) • Assume there are finitely many primes. • Let S be the set of all n primes, p1p2p3…pn • Let q = p1p2p3…pn + 1. • There must exist a prime factorization for q. We claim that this factorization does not contain an element of S. This would imply that q uses a prime outside the set S, thus there are in fact more than n primes. • Assume that pi, 1 <= i <= n, is in the prime factorization of q. Then it must divide p1p2p3…pn + 1. Specificially pi | 1. • But since primes are greater than 1, this is our contradiction.

  5. First approach • Consider the following code: bool isPrime( int n ) { for( int i = 2; i*i <= n; i++ ) if( n % i == 0 ) return true; return false; } • We attempt to find divisors of n. If we find one then return false. • Has a running time of O(n1/2) • In practice, primality testing comes in to varieties.

  6. Primality testing • Deterministic tests will determine with absolute certainty whether a given integer n is prime. • Probabilistic tests will determine with a small probability of error, whether a given integer n is prime.

  7. Example of a probabilistic test • Fermat’s Little Theorem • Let n be a prime and a be any positive integer such that a < n. Then a raised to the nth power is congruent to a modulo n. • an = a (mod n) or an-1-1 = 0 (mod n) • Given an integer n, all we have to do is find the remainder of an(mod n) and see if it equals a. • We know how to compute modulo powers from our previous discussion.

  8. Fermat’s Little Theorem • If an integer n satisfies this theorem, does this imply that n is prime? • NO! • If an (mod n) is not congruent to a then we are certain that n is not prime. • If it is congruent to a then there is a good chance that n is prime. • As it turns out, there are composite numbers that also satisfy Fermat’s Little Theorem.

  9. Fermat’s Little Theorem • A composite number n is a Carmichael number if n satisfies Fermat’s Little Theorem for every a that is relatively prime to n. • Smallest Carmichael number is 561. • Other examples include: 89 = 8 (mod 9) • These are called pseudoprimes in general • What can we do to avoid having our test fail?

  10. Fermat’s Little Theorem • We are free to choose any a we like, as long as a < n. • Pick many integers a to test for primality. • If n passes this test for a lot of different values of a, then it is highly probable that n is prime. • Run this test many times to reduce our probability of error.

  11. Running times • Probabilistic tests have running times around O(logn). • In 2002, Agrawal et al discovered the first algorithm for a deterministic test that is polynomial in the number of digits of n. • Running time is O(log7.5n). However, they seem to run in O(log3n) in practice. • Largest known prime currently is 230402457-1. It contains 9152052 digits.

  12. Sieve of Eratosthenes • What if we want a list of primes? • Given n, we can find all the primes up to n by using the Sieve of Eratosthenes. • Start with a list of integers from 2 to n. • At iteration i, mark the smallest integer a in the set as a prime, then cross off all multiples of a in the set. • Repeat until i2 <= n.

  13. Sieve of Eratosthenes • Ex. Let n = 10, and x denote a crossed off integer. • 2, 3, 4, 5, 6, 7, 8, 9, 10 • 2, 3, x, 5, x, 7, x, 9, x • 2, 3, x, 5, x, 7, x, x, x • Done! • So primes up to n = 10 are 2, 3, 5, and 7.

  14. Sieve of Eratosthenes • Here is an implementation: bool prime[1000000]; void sieve( int n ) { memset( prime, 1, sizeof(prime) ); prime[0] = prime[1] = 0; for( int i = 2; i <= n; i++ ) if( prime[i] ) { for( int j = i+i; j <= n; j += i ) prime[j] = 0; } }

  15. Sieve of Eratosthenes • At the end of this algorithm, prime[i] = 1 if and only if i is prime, 0 otherwise. • Outer loop starts at the smallest position that hasn’t been crossed off. • Inner loop repeatedly crosses of multiples of i. • With some analysis, we can show that this has a running time of O(nlogn). • Can we improve this algorithm?

  16. Sieve of Eratosthenes • Consider after having done the first iteration in the outer loop. So i = 3. • We start at i + i = 6. • However, we already crossed off all multiples of 2 (all the even numbers) in the first iteration. So why are we looking here? • Similarly, when we have crossed off all multiples of 3, we should not look at any multiples of 3 in any future iteration.

  17. Sieve of Eratosthenes • Improved implementation bool prime[1000000]; void sieve( int n ) { memset( prime, 1, sizeof(prime) ); prime[0] = prime[1] = 0; for( int i = 2; i*i <= n; i++ ) if( prime[i] ) { for( int j = i*i; j <= n; j += i ) prime[j] = 0; } } • Changes are in bold.

  18. Sieve of Eratosthenes • Outer loop only goes up to the square root of n. • Ex. n = 10, there is no need to check any multiples of a where a > 3. Surely they will all have been crossed off by some previous combination. (ie. 2x5 = 5x2) • Inner loop starts at i*i. This prevents looking at positions that obviously is crossed off. • Improved running time is still O(nlogn), but in practice the speed usually speeds it up by 50%.

  19. Sieve of Eratosthenes • For anyone interested: • There are other implementations that exploit the fact that our prime[] array is just an array of 1’s and 0’s. • Use bits to lower memory usage, perhaps speed things up. • Other optimizations that can improve running time in practice. • Sieve of Atkin, optimized version of the Sieve of Eratosthenes.

  20. Factoring • Given n, we are not interested in finding positive integers a and b such that a*b = n. • Factoring is hard, when n contains 100 or more digits. • Many cryptographic systems depend heavily on this assumption.

  21. Is factoring really that hard? • Consider n = 2193 (n is given as such because my calculator doesn’t like numbers that have more than 10 digits, so pretend we do not know the prime factorization of n ahead of time). How long would it take to factor this number? • It turns out that we can factor this number in 193 division operations. We just divide our number by 2 193 times. • It was trivial to factor this number, so why is factoring considered a hard problem?

  22. First attempt at factoring • Recall our first approach in primality testing. bool isPrime( int n ) { for( int i = 2; i*i <= n; i++ ) if( n % i == 0 ) return true; return false; } • We can do exactly the same thing, except instead of returning true when we have found a factor, we simply return that number that divided n.

  23. Further optimizations of our first attempt • To speed things up even further, we can only divide by primes between 2 and square root of n. • How well does this work in practice? • What is the worst case input?

  24. Worst case for factoring • What if we are given two 100 digit prime numbers a and b. Let n = a*b. • To factor this number with our approach, we will have to check all primes between 2 and the minimum of a and b. • As an exmaple, consider n = 2193 – 1. • The first prime divisor is 13,821,503. • Assuming that our computer can perform a billion division instructions per second, it will take more than 35,000 years to find the second largest factor of n!

  25. Worst case for factoring • As it turns out, the full factorization of n is: • n = 13,821,503 * 61,654,440,233,248,340,616,559 * 14,732, 265,321,145,317,331,353,282,383. • Conclusion: Factoring is hard for most cases.

  26. RSA Factoring Challenge • http://www.rsasecurity.com/rsalabs/node.asp?id=2093 • Monetary reward for successful factoring of very large integers. • Latest factored number: RSA-640. An integer that is 640 bits long. • To quote the website: “The effort took approximately 30 2.2 GHz- Opteron-CPU years according to the submitters, over five months of calendar time. (This is about half the effort for RSA-20)

  27. Factoring and the Sieve • We know we can’t factor general numbers very efficiently. • Here is an implementation of an augmented Sieve of Eratosthenes that will factor for us: int factors[1000000]; void fsieve( int n ) { for( int i = 1; i <= n; i++ ) factors[i] = i; for( int i = 2; i <= n; i++ ) if( factors[i] == I ) { for( int j = i*i; j <= n; j += i ) if( factors[j] == j ) f[j] = i; } }

  28. Factoring Sieve • Again we loop over the numbers from 2 to n. • The inner loop will set factor[j] = i if i is a multiple of j. Since we loop in ascending order, it is the smallest multiple of j. • We can retrieve the list o factors for any arbitrary position. • Ex. n = 6 • Before the first iteration, factors[]: 1, 2, 3, 4, 5, 6 • After the last iteration, factors[]: 1, 2, 3, 2, 5, 2 • To retrieve the factors of 6, go to factors[6]. Divide that 6 by that value and get 3, go to factors[3], divide 6 by that value and get 2, go to factors[2], divide 6 by 2 to get 1. Stop. • Factors are 1, 2, 3 and 6.

  29. Euler’s Phi Function • Def: “Phi of n”, denoted as Φ(n), counts the number of integers in the range of [1,n-1] that are relatively prime to n. • Def: Two positive integers a and b are said to be relatively prime if gcd(a,b) = 1. • Note that the definition does not require either a or b to be prime themselves. • Ex. a = 6, b = 25. (6,25) = 1

  30. Euler Phi Function • Ex. Φ(6) = 2 since 1 and 5 are relatively prime to 6. • Ex. Φ(2000) = 800. How do we know this? • Let’s first establish some ways we can calculate Φ(n).

  31. Case when n is prime • If n is prime, then by definition, it contains no factors other than itself and 1. Since (n,1) = 1, the number integers less than n that are relatively prime to n is just n – 1. • Ex. Φ(7) = 6 since 1, 2, 3, 4, 5, 6 are all relatively prime to 7.

  32. Case when n = pr, r > 1 • If n is a power of a prime number p, can we derive a formula for Φ(n)? • Let’s consider what pr is not relatively prime to. • Obviously, (p,pr) = p. In fact, all integers ap <= pr where a is some positive integer are not relatively prime to p. • Also, all powers of p, up to pr-1 are not relatively prime to p.

  33. Case when n = pr, r > 1 • Thus the numbers p, 2p, 3p, … (pr-1-1)p are all not relatively prime to p. • There are pr-1 numbers less than pr. There are pr-1-1 numbers that are not relatively prime to pr. • So Φ(n) = (pr-1)-(pr-1-1) = pr - pr-1.

  34. General Case • Given n and it’s prime factorization, can we derive a general formula for n? In general:

  35. Looking ahead • Next class, Andrew will be talking about cryptography. • One of the more famous algorithms for public-key encryption is RSA. • Each step of the algorithm involves something that we talked about in each of the 3 (including today) discussions. • We will need to find two large prime numbers. (primality) • We will need to solve linear equations such as ax + by = 1. (extended Euclidean algorithm) • We will need to compute Φ(n) • Try to be at least a little familiar with all 3 or otherwise, the next discussion may be hard to follow.

  36. References • http://mathworld.wolfram.com/PrimalityTest.html • http://www.fortunecity.com/emachines/e11/86/tourist2d.html • http://www.rsasecurity.com/rsalabs/node.asp?id=2093 • http://en.wikipedia.org/wiki/RSA • Last Year’s notes.

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