/*
    debug_this.c
    Jeff Ondich, 27 March 2023

    A simple program for use with our VS Code debugging lab.

    The Nth triangular number is the sum 1 + 2 + ... + N.
    So T[1] = 1, T[2] = 3, T[3] = 6, etc.

    Yes, there is a closed-form formula (T[N] = N * (N - 1) / 2), but
    we're ignoring it so we can watch loops and recursion in action.

    This program includes an off-by-one error.
*/

#include <stdio.h>
#include <string.h>

int iterative_triangular_number(int n);
int recursive_triangular_number(int n);

int main(int argc, char *argv[]) {
    int limit = 10;

    printf("Here are the first %d triangular numbers, recursively computed:\n", limit);
    for (int k = 1; k <= limit; k++) {
        int t = recursive_triangular_number(k);
        printf("  T(%d) = %d\n", k, t);
    }
    printf("\n");

    printf("Here are the first %d triangular numbers, iteratively computed:\n", limit);
    for (int k = 1; k <= limit; k++) {
        int t = iterative_triangular_number(k);
        printf("  T(%d) = %d\n", k, t);
    }
    printf("\n");

    return 0;
}

// Returns the nth triangular number, or 0 if n <= 0.
int iterative_triangular_number(int n) {
    int total = 0;
    for (int k = 0; k < n; k++) {
        total = total + k;
    }

    return total;
}

// Same, but recursive.
int recursive_triangular_number(int n) {
    if (n < 1) {
        return 0;
    }

    int t = recursive_triangular_number(n - 1);
    t = t + n;

    return t;
}