Step by Step solution about How to resolve the algorithm Blum integer step by step in the C programming language

This C program generates and analyzes Blum integers, which are a class of integers with special properties in number theory. Here's a detailed explanation of the code:

Header Inclusions:

• <stdio.h>: For input and output operations.
• <stdbool.h>: For the bool data type (boolean values).
• <locale.h>: For setting the locale to use localized numeric formatting (e.g., commas as thousand separators).

Global Variables:

• inc[8]: An array to help in the efficient generation of prime numbers (used in firstPrimeFactor).

Function Declarations:

• isPrime(int n): Checks if the given integer n is prime.
• firstPrimeFactor(int n): Finds and returns the smallest prime factor of the given odd integer n.

Main Function:

• i: Main loop counter, starts from 1.
• j: Loop counter for various purposes.
• bc: Counter for Blum integers found.
• p, q: Variables to hold prime factors.
• blum[50]: Array to store the first 50 Blum integers.
• counts[4]: Array to count the frequency of Blum integers ending in 1, 3, 5, and 7.
• digits[4]: Array of digits 1, 3, 5, and 7 to correspond with counts.

Main Loop:

• The program enters an infinite loop (while (true)) to continuously search for Blum integers.
• It calls firstPrimeFactor on the current i to find its smallest prime factor p.
• If p is congruent to 3 modulo 4 (p % 4 == 3), it calculates q as i / p.
• If both q and i are congruent to 3 modulo 4 and q is prime, it has found a Blum integer.
• It stores the Blum integer in the blum array and increments the corresponding count in counts.
• The program prints various status messages and updates based on certain milestones:
  • When 50 Blum integers have been found, it prints the first 50.
  • When the 26,828th or any subsequent 100,000th Blum integer is found, it prints the number of Blum integers found so far.
  • When the 400,000th Blum integer is found, it prints the percentage distribution of the first 400,000 Blum integers ending in 1, 3, 5, or 7.
• The loop continues indefinitely, searching for and analyzing Blum integers.
• The program runs indefinitely until you terminate it.

#include <stdio.h>
#include <stdbool.h>
#include <locale.h>

int inc[8] = {4, 2, 4, 2, 4, 6, 2, 6};

bool isPrime(int n) {
    if (n < 2) return false;
    if (n%2 == 0) return n == 2;
    if (n%3 == 0) return n == 3;
    int d = 5;
    while (d*d <= n) {
        if (n%d == 0) return false;
        d += 2;
        if (n%d == 0) return false;
        d += 4;
    }
    return true;
}

// Assumes n is odd.
int firstPrimeFactor(int n) {
    if (n == 1) return 1;
    if (!(n%3)) return 3;
    if (!(n%5)) return 5;
    for (int k = 7, i = 0; k*k <= n; ) {
        if (!(n%k)) {
            return k;
        } else {
            k += inc[i];
            i = (i + 1) % 8;
        }
    }
    return n;
}

int main() {
    int i = 1, j, bc = 0, p, q;
    int blum[50], counts[4] = {0}, digits[4] = {1, 3, 5, 7};
    setlocale(LC_NUMERIC, "");
    while (true) {
        p = firstPrimeFactor(i);
        if (p % 4 == 3) {
            q = i / p;
            if (q != p && q % 4 == 3 && isPrime(q)) {
                if (bc < 50) blum[bc] = i;
                ++counts[i % 10 / 3];
                ++bc;
                if (bc == 50) {
                    printf("First 50 Blum integers:\n");
                    for (j = 0; j < 50; ++j) {
                        printf("%3d ", blum[j]);
                        if (!((j+1) % 10)) printf("\n");
                    }
                    printf("\n");
                } else if (bc == 26828 || !(bc % 100000)) {
                    printf("The %'7dth Blum integer is: %'9d\n", bc, i);
                    if (bc == 400000) {
                        printf("\n%% distribution of the first 400,000 Blum integers:\n");
                        for (j = 0; j < 4; ++j) {
                            printf("  %6.3f%% end in %d\n", counts[j]/4000.0, digits[j]);
                        }
                        break;
                    }
                }
            }
        }
        i += (i % 5 == 3) ? 4 : 2;
    }
    return 0;
}