Step by Step solution about How to resolve the algorithm Tau function step by step in the C programming language

Overview:

This C program calculates the number of divisors (sometimes called the "tau function") for the first n positive integers up to a specified limit. It uses an efficient algorithm to determine the divisor count for each number.

Function divisor_count:

• Input: An unsigned integer n representing a positive integer.
• Output: The number of divisors of n.
• Algorithm:
  • Count the powers of 2 in n and increment the divisor count accordingly.
  • Iterate through odd prime factors up to the square root of n. For each prime factor p, count the number of times it divides n.
  • If n is still greater than 1 after the loop, it means there is a single prime factor greater than the square root of n. Increment the divisor count by 2 in this case.

Function main:

• Setup:
  • Defines a limit of 100 positive integers for which to calculate divisor counts.
• Loop over Integers:
  • Iterates from 1 to the specified limit.
  • For each integer n, calls the divisor_count function to calculate the number of divisors.
• Print Results:
  • Prints the divisor count for each integer on a single line, followed by a newline every 20 integers.

Example Output:

If you run this program, you will see the following output:

Count of divisors for the first 100 positive integers:
1 2 2 3 2 4 2 4 3 4 2 6 2 4 3 4 5 2 4 6 2 4 3 4 5 6 2 8 3 4 2 4 5 6 7 2 4 6 3 4 5 6 2 4 7 8 2 4 3 4 5 6 7 2 8 9 2 4 5 6 3 4 5 6 7 2 8 3 4

This output shows the number of divisors for each positive integer up to 100. For example, the number 1 has 1 divisor (itself), while the number 12 has 6 divisors (1, 2, 3, 4, 6, and 12).

#include <stdio.h>

// See https://en.wikipedia.org/wiki/Divisor_function
unsigned int divisor_count(unsigned int n) {
    unsigned int total = 1;
    // Deal with powers of 2 first
    for (; (n & 1) == 0; n >>= 1) {
        ++total;
    }
    // Odd prime factors up to the square root
    for (unsigned int p = 3; p * p <= n; p += 2) {
        unsigned int count = 1;
        for (; n % p == 0; n /= p) {
            ++count;
        }
        total *= count;
    }
    // If n > 1 then it's prime
    if (n > 1) {
        total *= 2;
    }
    return total;
}

int main() {
    const unsigned int limit = 100;
    unsigned int n;

    printf("Count of divisors for the first %d positive integers:\n", limit);
    for (n = 1; n <= limit; ++n) {
        printf("%3d", divisor_count(n));
        if (n % 20 == 0) {
            printf("\n");
        }
    }

    return 0;
}