A Solution toPOJ1811 Prime Test

STYC

Problem Description

Given an integer N which satisfies the relation 2 < N < 254, determine whether or not it is a prime number. If N is not a prime number, find out its smallest prime factor.

Key Concepts

Prime numbers: A prime number is a positive integer p > 1 that has no positive integer divisors other than 1 and p itself.

Framework of the Solution

Determine whether the given N is prime or not.If N is prime, print “Prime” and exit.Factorize N for its smallest prime factor.

The Brute-force WayTrial Division

If N is even, then 2 is its smallest prime factor.Try dividing N by every odd number k between 2 and N1/2. The smallest k by which N is divisible is the smallest prime factor of N. If such k does not exist, then N is prime.Complexity: O(N1/2) for time, O(1) for space

Modified Brute-force

Construct a table that stores all prime numbers not greater than N1/2. Try dividing N only by prime numbers.Complexity: O(N1/2logN) for time, O(N1/2) for space using Sieve of EratosthenesEstimation of space consumption: 226 bits = 223 bytes = 8,192 kilobytesMuch time is used in the process of sieving

Modified Brute-force 2

Embed a table of prime numbers smaller than Nmax

1/4 into the source. Extend the table to N1/2 by runtime calculation.Complexity: O(N3/4/logN) for time, O(N1/2/logN) for spaceEstimation of time consumption: Finding all 7,603,553 primes smaller than 227 takes approx. 1.32 x 1011 divisions or 73 minutes on a Pentium 1.5 GHz.

Brute-force with Trick

Start from N1/2 rather than 2. Do factorization recursively once a factor is found.Efficient in handling N = pq where p and q relatively close to each other.POJ accepts this! - westever’s solution

Brute-force with Trick 2Wheel Factorization

Test whether N is a multiple of 2, 3 or 5. If it is, the problem has been solved.If not, do trial division using only integers which are not multiples of 2, 3, and 5.Saves 7/15 of work.

Key Concepts (cont.)

Prime factorization algorithms: Algorithms devised for determining the prime factors of a given numberPrimality tests: Tests to determine whether or not a given number is prime, as opposed to actually decomposing the number into its constituent prime factors

Primality Tests

Deterministic: Adleman-Pomerance-Rumely Primality Test, Elliptic Curve Primality Proving…Probabilistic: Rabin-Miller Strong Pseudoprime Test…

Rabin-MillerStrong Pseudoprime Test

Given an odd integer N, let N = 2rs + 1 with s odd. Then choose a random integer a between 1 and N - 1. If as = 1 (mod N) or a2^j s = -1 (mod N) for some j between 0 and r - 1, then N passes the test. A prime will pass the test for all a.

Rabin-MillerStrong Pseudoprime Test

Requires no more than (1 + o(1))logN multiplications (mod N).A number which passes the test is not necessarily prime. But a composite number passes the test for at most 1/4 of the possible bases a.If n multiple independent tests are performed on a composite number, the probability that it passes each test is 1/4n or less.

Rabin-MillerStrong Pseudoprime Test

Smallest composite numbers passing the RMSPT using the first k primes as bases: 2,047; 1,373,653; 25,326,001; 3,215,031,751; 2,152,302,898,747; 3,474,749,660,383; 341,550,071,728,321, 341,550,071,728,321; at most 41,234,316,135,705,689,041…341,550,071,728,321 = 244.957…, 41,234,316,135,705,689,041 = 265.160…

Tests show that randomized bases may fail sometimes.

Pseudocode of RMSPT (Sprache)

function powermod(a, s, n){

p := 1b := awhile s > 0{

if s & 1 == 1 then p := p * b % nb := b * b % ns := s >> 1

}}

Pseudocode of RMSPT (cont.)

function rabin-miller(n){

if n > 2 AND powermod(2, n - 1, n) != 1 then return FALSEif n > 3 AND powermod(3, n - 1, n) != 1 then return FALSEif n > 5 AND powermod(5, n - 1, n) != 1 then return FALSEif n > 7 AND powermod(7, n - 1, n) != 1 then return FALSEif n > 11 AND powermod(11, n - 1, n) != 1 then return FALSEif n > 13 AND powermod(13, n - 1, n) != 1 then return FALSEif n > 17 AND powermod(17, n - 1, n) != 1 then return FALSEif n > 19 AND powermod(19, n - 1, n) != 1 then return FALSEif n > 23 AND powermod(23, n - 1, n) != 1 then return FALSEreturn TRUE

}

Prime Factorization Algorithms

Continued Fraction Algorithm, Lenstra Elliptic Curve Method, Number Field Sieve, Pollard Rho Method, Quadratic Sieve, Trial Division…

Pollard RhoFactorization Method

Also known as Pollard Monte Carlo factorization method.Runs at O(p1/2) where p is the largest prime factor of the number to be factored.Two aspects to this method: iteration and cycle detection.

Pollard RhoFactorization Method

Iteration:Iterate the formula xn+1 = xn

2 + a (mod N). Almost any polynomial formula (two exceptions being xn

2 and xn2 - 2) for

any initial value x0 will produce a sequence of numbers that eventually falls into a cycle.

Pollard RhoFactorization Method

Cycle detection:Keep one running copy of xi. If i is power of 2, let y = xi, and at each step, compute GCD(|xi - y|, N). If the result is neither 1 nor N, then a cycle is detected and GCD(|xi - y|, N) is a factor (not necessarily prime) of N. If the result is N, the method fails, choose another a and redo iteration.

Pseudocode of RRFM (Sprache)

function pollard-rho(n){

do{

a := random()}while a == 0 OR a == -2y := xk := 2i := 1

Pseudocode of RRFM (cont.)

while TRUE{

i := i + 1x := (x * x + a) % ne := abs(x - y)d := GCD(e, n)if d != 1 AND d != n then return delse if i == k then

{y := xk := k << 1

}}

}

Overall Efficiency Analysis

Time complexity: O(logN) for any prime; O(N1/4) for most composites under average conditions, principally decided by the factorization processSpace complexity: O(1), space demand is always independent of N

Notes on Implementation

Multiplication (mod N) in RMSPT: Requires calculation of 64bit * 64bit % 64bit. Should be computed as binary numbers using “divide and conquer” method. Use floating-point unit for (mod N) operation. Can be optimized by coding in assembly.

Notes on Implementation (cont.)

GCD(a, b) (a > b) in PRFM: Following properties of GCD helps avoiding divisions:If a = b, then GCD(a, b) = a.GCD(a, b) = 2 * GCD(a/2, b/2) with both a and b even.GCD(a, b) = GCD(a/2, b) with a even but b odd.GCD(a, b) = GCD(a - b, b) with both a and b odd.Time complexity: O(logb)

Notes on Implementation (cont.)

Combination with brute-force algorithms: Embed a prime table. Do brute-force trial division for small divisors.Minor optimizations: Use 32-bit integer division instead of the 64-bit version when possible…

Actual Timing Performance

Platform: Windows XP SP1 on a Pentium M 1.5 GHzAlgorithms tested: Adapted versions of westever’s, TN’s, lh’s and zsuyrl’s solutions and my later fixed versionTiming method: Process user time returned by the GetProcessTimes Windows API

Actual Timing Performance

Test data: Original data set on POJ, two pseudorandom data sets generated by Mathematica and one hand-made data set (all given in ascending order), new data set on POJVerification: Done with Mathematica

Actual Timing Performance

Data0(20)

Data1(10)

POJData2(24)

Data3(20)

Data5(23)

westever 7,193 117 7,937 24,462 15,935 31,375

TN 3,545 60 W/A 12,450 7,744 15,935

Wheel 2,143 40 2,453 6,932 4,182 8,929

lh 60 613 375 240(-1) 180 197

zsuyrl 100 704 484 287 203 290

STYC 60 350 312 120 67 133

Conclusions

Brute-force ways work too slow with integers that have large factors. But they are good compliments to complex methods like Pollard Rho.Original test data on POJ are too weak to have slow algorithms fail and to prove wrong solutions incorrect. New data have been much better but not yet perfect.