Task 1: Converting a Numerical Score to a Letter Grade

The function grade_converter maps an integer score to a corresponding letter grade ('A' through 'E'). The functon takes an int parameter and returns a char value. A corrected implementation of a similar switch-case structure is shown below:

char grade_converter(int score_val) {
    const char* grade;
    switch(score_val / 10) {
        case 10:
        case 9:
            grade = "A";
            break;
        case 8:
            grade = "B";
            break;
        case 7:
            grade = "C";
            break;
        case 6:
            grade = "D";
            break;
        default:
            grade = "E";
    }
    return grade[0];
}

Key issues in a flawed implementation include type mismatch (assigning a const char* to a char variable) and missing break statements, which cause fall-through errors in the switch block.

Task 2: Summing the Digits of an Integer

The objective is to calculate the sum of all digits in a given integer. This can be achieved through both iterative and recursive methods.

// Iterative approach
int sum_digits_iterative(int num) {
    int total = 0;
    while(num > 0) {
        total += num % 10;
        num /= 10;
    }
    return total;
}

// Recursive approach
int sum_digits_recursive(int num) {
    if(num == 0) return 0;
    return (num % 10) + sum_digits_recursive(num / 10);
}

Both methods produce identical results. The iterattive version uses a loop, while the recursive version relies on self-calls to break down the problem.

Task 3: Computing Exponential Values

The function compute_power calculates x raised to the power of n. It is implemented recursively.

double compute_power(double base, int exp) {
    if(exp == 0) return 1.0;
    return base * compute_power(base, exp - 1);
}

Task 4: Identifying Twin Prime Numbers

This program finds and counts all twin prime pairs (primes differing by 2) within 100.

#include <stdio.h>
#include <math.h>

int check_prime(int value);

int main() {
    int pair_count = 0;
    printf("Twin primes under 100:\n");
    for(int i = 3; i + 2 <= 100; i++) {
        if(check_prime(i) && check_prime(i + 2)) {
            printf("%d %d\n", i, i + 2);
            pair_count++;
        }
    }
    printf("Total twin prime pairs: %d\n", pair_count);
    return 0;
}

int check_prime(int value) {
    if(value <= 1) return 0;
    for(int divisor = 2; divisor <= sqrt(value); divisor++) {
        if(value % divisor == 0) return 0;
    }
    return 1;
}

Task 5: Solving the Towers of Hanoi Puzzle

This program simulates the Towers of Hanoi puzzle, calculating and displaying the sequence and total number of disk moves.

#include <stdio.h>

int solve_hanoi(unsigned int disks, char src, char aux, char dest);
void log_move(unsigned int disk_id, char from, char to);

int main() {
    unsigned int disk_count;
    while(scanf("%u", &disk_count) != EOF) {
        int moves = solve_hanoi(disk_count, 'A', 'B', 'C');
        printf("Total moves: %d\n", moves);
    }
    return 0;
}

int solve_hanoi(unsigned int disks, char src, char aux, char dest) {
    int move_count = 0;
    if(disks == 1) {
        log_move(disks, src, dest);
        move_count++;
    } else {
        move_count += solve_hanoi(disks - 1, src, dest, aux);
        log_move(disks, src, dest);
        move_count++;
        move_count += solve_hanoi(disks - 1, aux, src, dest);
    }
    return move_count;
}

void log_move(unsigned int disk_id, char from, char to) {
    printf("%u: %c -> %c\n", disk_id, from, to);
}

Task 6: Calculating Binomial Coefficients

This task involves computing the binomial coefficient C(n, m) using both recursive and iterative methods.

// Recursive calculation
int binomial_recursive(int n_val, int m_val) {
    if(m_val == 0 || m_val == n_val) return 1;
    return binomial_recursive(n_val - 1, m_val) + binomial_recursive(n_val - 1, m_val - 1);
}

// Iterative calculation
int binomial_iterative(int n_val, int m_val) {
    if(m_val > n_val) return 0;
    if(m_val == 0 || m_val == n_val) return 1;
    int coeff = 1;
    for(int i = 1; i <= m_val; i++) {
        coeff = coeff * (n_val - i + 1) / i;
    }
    return coeff;
}

Task 7: Computing the Greatest Common Divisor of Three Numbers

This program finds the GCD of three integers by successively applying the GCD function.

#include <stdio.h>

int calculate_gcd(int val_a, int val_b);

int main() {
    int a, b, c, result;
    while(scanf("%d %d %d", &a, &b, &c) != EOF) {
        result = calculate_gcd(calculate_gcd(a, b), c);
        printf("GCD: %d\n", result);
    }
    return 0;
}

int calculate_gcd(int val_a, int val_b) {
    while(val_b != 0) {
        int remainder = val_a % val_b;
        val_a = val_b;
        val_b = remainder;
    }
    return val_a;
}