p451.py 3.47 KB
 Nayuki committed Jul 03, 2017 1 2 3 4 5 6 7 8 9 10 11 ``````# # Solution to Project Euler problem 451 # Copyright (c) Project Nayuki. All rights reserved. # # https://www.nayuki.io/page/project-euler-solutions # https://github.com/nayuki/Project-Euler-solutions # import array, eulerlib, itertools `````` Nayuki committed Jul 04, 2017 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 ``````# Let n be an arbitrary integer such that n >= 3. # When we say that the modular inverse of m modulo n equals m itself, # the formula is m^-1 = m mod n, which is equivalent to m^2 = 1 mod n. # # We know that if n is prime, then m^2 = 1 mod n has exactly two solutions: # m = 1, n-1. It is easy to verify that these two numbers are solutions. # The equation factorizes as (m - 1)(m + 1) = 0 mod n. Because n is prime, # the numbers form a field, and there are no zero divisors (two arbitrary # non-zero numbers x and y such that xy = 0). Hence 1 and -1 mod n are # the only possible solutions to the equation. (Note that for the excluded # special prime case where n = 2, the solutions 1 and -1 are the same number.) # # Suppose we can find the smallest prime factor of n quickly. (Note that if n is # prime, then the smallest prime factor is n itself.) This can be achieved by # building a table ahead of time, using a modification of the sieve of Eratosthenes. # # Suppose that for every n' < n, we know the set of solutions to m^2 = 1 mod n'. # This means whenever we solve the equation for the number n, we save its solutions # in an ever-growing list, so that when we work on the next value of n we can access # all possible smaller solutions. This is essentially an argument by strong induction. # # Let p be the smallest prime factor of n. If p = n, then the set of # solutions is {1, n - 1}, and we are finished with this value of n. # # Otherwise p < n, and obviously n is an integer multiple of p. Because we are looking # for values of m such that m^2 = 1 mod n, these candidate m values also must satisfy # m^2 = 1 mod k for any k that divides n (i.e. k is a factor of n). We look at the set # of solutions for the modulus k = n/p, which has already been solved because k < n. # We know that any solution modulo n must be congruent to these solutions modulo k. # Hence we can try to extend and check these old solutions by brute force. Namely, suppose # m' is a solution modulo k. Then we check the sequence m = m' + 0k, m' + 1k, m' + 2k, ..., # m' + (p-1)k modulo n. Because p is usually a small number, this isn't a lot of work to do. `````` Nayuki committed Jul 03, 2017 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 ``````def compute(): LIMIT = 20000000 # Build table of smallest prime factors smallestprimefactor = array.array("L", itertools.repeat(0, LIMIT + 1)) end = eulerlib.sqrt(len(smallestprimefactor) - 1) for i in range(2, len(smallestprimefactor)): if smallestprimefactor[i] == 0: smallestprimefactor[i] = i if i <= end: for j in range(i * i, len(smallestprimefactor), i): if smallestprimefactor[j] == 0: smallestprimefactor[j] = i # Returns all the solutions (in ascending order) such that # for each k, 1 <= k < n and k^2 = 1 mod n. def get_solutions(n): if smallestprimefactor[n] == n: # n is prime return (1, n - 1) else: temp = [] p = smallestprimefactor[n] sols = solutions[n // p] for i in range(0, n, n // p): for j in sols: k = i + j if k * k % n == 1: temp.append(k) return tuple(temp) # Process every integer in range solutions = [(), (), (1,)] ans = 0 for i in range(3, LIMIT + 1): sols = get_solutions(i) if i <= LIMIT // 2: solutions.append(sols) ans += sols[-2] # Second-largest solution return str(ans) if __name__ == "__main__": print(compute())``````