I was person #100 to solve Project Euler problem 451 a while back. The problem and solution are detailed in the source code below. The solution was found in 28 seconds on a 20-core Xeon machine. As noted in the code, using the Chinese Remainder Theorem could help efficiency, but I don’t have that coded currently and so I iterated over potential C.R.T. candidates instead, which provided a quickly coded and self-contained solution.

/* * Jason B. Hill / jason@jasonbhill.com * pe451.c / solves Project Euler problem 451 * build: gcc pe451.c -fopenmp -O3 -o pe451 * execute: ./pe451 * * I logged in to see that problem 451 had around 90 solutions, and decided to * attempt to be one of the first 100 solvers to enter the proper solution. As * such, I coded this a bit quickly. A "clean" solution would make use of the * Chinese Remainder Theorem, and the timing could probably be brought down * a tiny bit. But, all things considered, this runs very fast itself. * * Problem: Consider the number 15. There are eight positive integers less than * 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14. The modular inverses * of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 because * 1*1 mod 15=1 * 2*8=16 mod 15=1 * 4*4=16 mod 15=1 * 7*13=91 mod 15=1 * 11*11=121 mod 15=1 * 14*14=196 mod 15=1 * Let I(n) be the largest positive number m smaller than n-1 such that the * modular inverse of m modulo n equals m itself. So I(15)=11. Also I(100)=51 * and I(7)=1. Find the sum of I(n) for 3<=n<=2ยท10**7. * * The solution is documented in the code below. The code is in C with OpenMP. * * The result is found in 28 seconds on a 20-core Xeon and and 2:48 on a 2-core * core i7. Here is some of the output: * prime sieve created - 0.110000 seconds * list of 1270607 primes created - 0.050000 seconds * list of prime factors created for integers <= 20000000 - 5.350000 seconds * prime factor exponents computed for all integers - 0.980000 seconds */ #include <omp.h> #include <stdio.h> #include <stdlib.h> #include <stdint.h> #include <time.h> /*****************************************************************************/ /* The following structs and routines are for integer factorization */ /*****************************************************************************/ /* * Notes on factoring: * * We need to consider the prime factorization of positive integers less than * 2*10^7. How many distinct prime factors can such numbers have? That's easy. * Since the product of the 8 smallest primes (2*3*5*7*11*13*17*19 = 9,699,690) * is less than 2*10^7 and the 9 smallest primes (9,699,690*23 = 223,092,870) * is greater than 2*10^7, we know that each positive integer less than 2*10^7 * can only have 8 distinct prime factors. Let's create a C-struct that can * keep track of this data for each factored integer. */ /* * A struct to hold factor information for an integer. * p holds a list of distinct prime factors. (at most 8) * e holds a list of exponents for the prime factors in p. * num_factors says how long the lists p and e are. */ struct factors_t { uint64_t p[8]; // list of distinct primes uint8_t e[8]; // list of exponents for the primes p[] uint64_t pp[8]; // prime power factor (p[i]**e[i]) uint8_t num_factors; // how many distinct prime factors }; /* * More notes on factoring: * * Every positive integer n has a prime factor <= sqrt(n). We use this fact to * build a prime sieve. First, we construct a function to return the square * root of a uint64_t. Then, we'll use that function to create a sieve (which * is returned as a pointer/list of uint8_ts... effectively Booleans). */ /* * Return the square root of a uint64_t */ uint64_t sqrt_uint64(uint64_t n) { uint8_t shift = 1; uint64_t res, s; while ((1<<shift) < n) shift += 1; res = (1<<((shift>>1) + 1)); while (1) { s = (n/res + res)/2; if (s >= res) return res; res = s; } } /* * Return a prime sieve identifying all primes <= limit. * This is just a list of uint8_t's where 0 means the index is composite and * 1 means the index is prime. */ uint8_t * prime_sieve(uint64_t limit) { uint8_t *sieve; // this will be the pointer that is returned uint64_t i,j; uint64_t s = sqrt_uint64(limit); // allocate memory for sieve sieve = malloc((limit + 1) * sizeof(uint8_t)); // set initial values in the sieve sieve[0] = 0; sieve[1] = 0; sieve[2] = 1; // set other initial odd values in sieve to 1, even values to 0. for (i=3;i<=limit;i++) { if (i%2==0) sieve[i] = 0; else sieve[i] = 1; } // unset composite numbers (evens are already unset) for (i=3;i<=(s+1);i=i+2) { if (sieve[i]==1) { j = i*3; // sieve[i] prime, sieve[2*i] already 0 while (j<=limit) { sieve[j] = 0; j += 2*i; } } } return sieve; } /* * Determine if a value is prime using a provided sieve. */ _Bool is_prime(uint64_t n, uint8_t * sieve) { if ((1 & sieve[n]) == 1) return 1; else return 0; } /*****************************************************************************/ /* The following functions are specific to Project Euler problem 451 */ /*****************************************************************************/ /* * Notes: * * Given a number n, we're looking for the largest m < n-1 such that m^2 is 1 * modulo n. That's a mouthful. First, field theory tells us that we're only * interested in integers that are relatively prime with n (which is why the * problem asks for m < n-1 ... otherwise it would be a lot easier). * * Now, if n=p^k is a prime power and we have x^2 = 1 mod p^k, we consider: * a) 1 is a solution for x since 1*1=1 mod p^k = 1. * b) p^k-1 is a solution since (p^k-1)^2=p^(2k)-2p^k+1 (mod p^k) = 1. But, * that's too big since we need m < n=p^k-1. :-( * c) Any other solutions? Well... that's sort of complicated. Ugh. Let's first * consider the case when p=2. For n=p^1=2^1, we only have x=1 as a * solution. When n=p^2=2^2, we only have x=1 and x=3. For n=p^k=2^k, we * also find that 2^(k-1)+1 and 2^(k-1)-1 square to 1 modulo 2^k. For other * primes, this doesn't happen and we only have 1 and p^k-1 as solutions. * * Thus, what we're going to do is as follows: * 1) Factor each integer within the given range (factoring a range of numbers * is much faster than factoring each one individually). We'll store that * factorization information using the struct defined above. * 2) Because of step 1, determining if a number is prime or a power of a prime * is easy. In case of primes and prime powers, we consider if the prime is * even or odd, returning the appropriate maximum square root of 1 directly. * 3) When the integer is composite having more than a single distinct prime * factor, we use the Chinese Remainder Theorem "in spirit" and iterate * over potential candidates (instead of computing the C.R.T. result * directly). We're only considering possible m that are both relatively * prime with n AND such that m^2 is +1 or -1 modulo the prime power * factors of n. When a candidate does not satisfy these properties, we * simply move to the next candidate. */ /* * Compute the next candidate (the next largest number that may satisfy the * given equivalence relations). This is done relative to the largest non-2 * prime power in the factorization of n. * * n is the integer for which we're computing the largest square root. * m is the largest odd prime power factor of n. * c is the current candidate (assume uninitialized when set to 0). */ uint64_t next_candidate(uint64_t n, uint64_t m, uint64_t c) { uint64_t r; // check if c is initialized if (c==0) c = n-1; // we can only consider values of \pm1 mod m r = c % m; if (r==m-1) return c-(m-2); else return c-2; } /* * Determine if the current candidate is (1) relatively prime to n and, if it * is, (2) a square root of unity modulo n. We can tweak this later for timing * as the test for being relatively prime may cut performance a bit. We'll see. * * n is the integer for which we're computing the largest square root. * cnd is the current candidate. * factors is a list of factors_t structs (prime, exponent info) */ _Bool verify_candidate(uint64_t n, uint64_t cnd, struct factors_t * factors) { uint8_t i,j; uint64_t pp; // verify that cnd is not divisible by any prime in factors[n].p[] for (i=0;i<factors[n].num_factors;i++) { if (cnd % factors[n].p[i] == 0) return 0; } // verify cnd modulo 2 when exponent of 2 in n is > 2 if (factors[n].p[0] == 2 && factors[n].e[0] > 2) { pp = factors[n].pp[0]>>1; if (cnd % pp != 1 && cnd % pp != pp-1) return 0; } // verify other primes for (i=0;i<factors[n].num_factors;i++) { if (factors[n].p[i] == 2) continue; pp = factors[n].pp[i]; if (cnd % pp != 1 && cnd % pp != pp-1) return 0; } return 1; } /* * Given a positive integer n, find the largest positive m < n-1 such that * gcd(n,m)=1 and m**2 (mod n) = 1. */ uint64_t largest_sqrt(uint64_t n, struct factors_t * factors) { uint8_t i; uint32_t j; // if n is an odd prime, or a power of an odd prime, return 1 if (factors[n].num_factors == 1 && factors[n].p[0] != 2) return 1; // if n is a power of 2 if (factors[n].num_factors == 1) { // if n is 2 or 4 if (factors[n].e[0] < 3) return 1; // if n is 2**e for e >= 3 return (factors[n].pp[0]>>1)+1; } // find the maximum odd prime power factor of n uint64_t pp = 1; uint64_t cnd; for (i=0;i<factors[n].num_factors;i++) { if (factors[n].p[i] != 2 && factors[n].pp[i] > pp) { pp = factors[n].pp[i]; } } // get the first candidate w.r.t. moppf cnd = next_candidate(n, pp, 0); while (!verify_candidate(n, cnd, factors)) cnd = next_candidate(n, pp, cnd); return cnd; } /*****************************************************************************/ /* Execute */ /*****************************************************************************/ int main() { uint64_t s = 0; // final output value uint64_t i,j,k; // iterators uint64_t limit = 20000000; // upper bound for computations uint64_t num_primes = 0; // count for primes <= limit uint8_t e; // exponents for prime factoring int tid; // thread id for openmp double time_count; // timer clock_t start, start_w; // time variables uint8_t *sieve; // prime sieve uint64_t *primes; // list of primes struct factors_t *factors; // prime factors and exponents /* start outer timer */ start_w = clock(); /* make the prime sieve */ start = clock(); sieve = prime_sieve(limit); time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("prime sieve created - %f seconds\n", time_count); /* form list of primes from sieve */ start = clock(); // compute the number of primes in sieve for (i=2;i<=limit;i++) { if (is_prime(i, sieve)) { num_primes = num_primes + 1; } } primes = malloc(num_primes * sizeof(uint64_t)); j=0; for (i=2;i<=limit;i++) { if (is_prime(i, sieve)) { primes[j] = i; j++; } } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("list of %llu primes created - %f seconds\n", num_primes, time_count); /* fill out a factors_t struct for each integer below limit */ start = clock(); // allocate memory for factors_t factors = malloc(sizeof(struct factors_t) * (limit + 1)); // set the initial number of factors for each number to 0 for (i=1;i<=limit;i++) factors[i].num_factors=0; // for each prime, add that prime as a factor to each of its multiples for (i=0;i<num_primes;i++) { j = 1; // start at 1*p for each prime p while (j*primes[i]<=limit) { k = factors[j*primes[i]].num_factors; // get proper index for p factors[j*primes[i]].p[k] = primes[i]; // add prime to p factors[j*primes[i]].num_factors++; // increase index j++; // increase multiple of prime } } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("list of prime factors created for integers <= %llu - %f seconds\n", limit, time_count); /* compute exponents for each prime in factor lists */ start = clock(); for (i=2;i<=limit;i++) { for (j=0;j<factors[i].num_factors;j++) { e=1; k=factors[i].p[j]; while (i % (k*factors[i].p[j]) == 0) { e++; k*=factors[i].p[j]; } factors[i].e[j] = e; factors[i].pp[j] = k; } } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("prime factor exponents computed for all integers - %f seconds\n", time_count); /* find largest square root of unity for each integer under limit; add to s */ #pragma omp parallel for reduction(+:s) shared(factors) schedule(dynamic,1000) for (i=3;i<=limit;i++) { s = s+largest_sqrt(i, factors); } time_count = (double)(clock() - start) / CLOCKS_PER_SEC; printf("result for 3<=i<=%llu: %llu - %f seconds\n", limit, s, time_count); time_count = (double)(clock() - start_w) / CLOCKS_PER_SEC; printf("\ntotal time: %f seconds\n\n", time_count); free(sieve); free(primes); free(factors); return 0; }