How to resolve the algorithm Giuga numbers step by step in the C++ programming language

Published on 7 June 2024 03:52 AM
#cpp

How to resolve the algorithm Giuga numbers step by step in the C++ programming language

Table of Contents

Problem Statement

A Giuga number is a composite number n which is such that each of its distinct prime factors f divide (n/f - 1) exactly. All known Giuga numbers are even though it is not known for certain that there are no odd examples. 30 is a Giuga number because its distinct prime factors are 2, 3 and 5 and:

Determine and show here the first four Giuga numbers. Determine the fifth Giuga number and any more you have the patience for.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Giuga numbers step by step in the C++ programming language

C++ Implementation of Giuga Sequence Generators:

Approach 1: Fast Check for Giuga Numbers:

The first C++ code efficiently checks whether a positive even integer n is a Giuga number, defined as an even number with exactly one factor of 2 and all its prime factors are of the form 6k+1 or 6k-1. The function is_giuga performs this check using several optimizations:

  • It assumes n is even with one factor of 2.
  • It quickly eliminates non-Giugas by checking for factors of 3 and 5.
  • It iterates through prime numbers of the form 6k+1 to check further divisibility.
  • If no factor of the form 6k+1 is found, it checks for a special case where n itself or half of n minus 1 is divisible by n.

If all these checks pass, the function returns true, indicating that n is a Giuga number.

Approach 2: Generating Giuga Numbers Using Rational Arithmetic:

The second C++ code generates Giuga numbers for a given n. It uses rational arithmetic to efficiently compute the sum of reciprocals of prime factors of integers.

  • The core of this approach is the giuga_numbers function, which generates and prints Giuga numbers up to a specified n.
  • It initializes vectors p and s to track prime factors p[t] and their reciprocals s[t].
  • The algorithm iteratively constructs the Giuga sequence by alternating between increasing t and choosing primes p[t] to ensure that the sum s[t] remains less than 1 or equals 1.
  • When t is less than n-2, it optimizes the selection of p[t] based on the previous sum s[t-1].
  • When t equals n-2, it analyzes potential divisors of a specific polynomial to generate Giuga numbers.

Overall, these C++ implementations provide efficient ways to check for Giuga numbers and generate them for small values of n.

Source code in the cpp programming language

#include <iostream>

// Assumes n is even with exactly one factor of 2.
bool is_giuga(unsigned int n) {
    unsigned int m = n / 2;
    auto test_factor = [&m, n](unsigned int p) -> bool {
        if (m % p != 0)
            return true;
        m /= p;
        return m % p != 0 && (n / p - 1) % p == 0;
    };
    if (!test_factor(3) || !test_factor(5))
        return false;
    static constexpr unsigned int wheel[] = {4, 2, 4, 2, 4, 6, 2, 6};
    for (unsigned int p = 7, i = 0; p * p <= m; ++i) {
        if (!test_factor(p))
            return false;
        p += wheel[i & 7];
    }
    return m == 1 || (n / m - 1) % m == 0;
}

int main() {
    std::cout << "First 5 Giuga numbers:\n";
    // n can't be 2 or divisible by 4
    for (unsigned int i = 0, n = 6; i < 5; n += 4) {
        if (is_giuga(n)) {
            std::cout << n << '\n';
            ++i;
        }
    }
}


#include <boost/rational.hpp>

#include <algorithm>
#include <cstdint>
#include <iostream>
#include <vector>

using rational = boost::rational<uint64_t>;

bool is_prime(uint64_t n) {
    if (n < 2)
        return false;
    if (n % 2 == 0)
        return n == 2;
    if (n % 3 == 0)
        return n == 3;
    for (uint64_t p = 5; p * p <= n; p += 4) {
        if (n % p == 0)
            return false;
        p += 2;
        if (n % p == 0)
            return false;
    }
    return true;
}

uint64_t next_prime(uint64_t n) {
    while (!is_prime(n))
        ++n;
    return n;
}

std::vector<uint64_t> divisors(uint64_t n) {
    std::vector<uint64_t> div{1};
    for (uint64_t i = 2; i * i <= n; ++i) {
        if (n % i == 0) {
            div.push_back(i);
            if (i * i != n)
                div.push_back(n / i);
        }
    }
    div.push_back(n);
    sort(div.begin(), div.end());
    return div;
}

void giuga_numbers(uint64_t n) {
    std::cout << "n = " << n << ":";
    std::vector<uint64_t> p(n, 0);
    std::vector<rational> s(n, 0);
    p[2] = 2;
    p[1] = 2;
    s[1] = rational(1, 2);
    for (uint64_t t = 2; t > 1;) {
        p[t] = next_prime(p[t] + 1);
        s[t] = s[t - 1] + rational(1, p[t]);
        if (s[t] == 1 || s[t] + rational(n - t, p[t]) <= rational(1)) {
            --t;
        } else if (t < n - 2) {
            ++t;
            uint64_t c = s[t - 1].numerator();
            uint64_t d = s[t - 1].denominator();
            p[t] = std::max(p[t - 1], c / (d - c));
        } else {
            uint64_t c = s[n - 2].numerator();
            uint64_t d = s[n - 2].denominator();
            uint64_t k = d * d + c - d;
            auto div = divisors(k);
            uint64_t count = (div.size() + 1) / 2;
            for (uint64_t i = 0; i < count; ++i) {
                uint64_t h = div[i];
                if ((h + d) % (d - c) == 0 && (k / h + d) % (d - c) == 0) {
                    uint64_t r1 = (h + d) / (d - c);
                    uint64_t r2 = (k / h + d) / (d - c);
                    if (r1 > p[n - 2] && r2 > p[n - 2] && r1 != r2 &&
                        is_prime(r1) && is_prime(r2)) {
                        std::cout << ' ' << d * r1 * r2;
                    }
                }
            }
        }
    }
    std::cout << '\n';
}

int main() {
    for (uint64_t n = 3; n < 7; ++n)
        giuga_numbers(n);
}

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