#include "../../../squiggle.h"
#include "../../../squiggle_more.h"
#include <stdio.h>
#include <stdlib.h>

int main()
{
    /* Question: can we parallelize this?
    A = normal(5,2)
    B = min(A)
    B * 20
    */

    /* Option 1: parallelize taking from n samples */
    // Question being asked: what is the distribution of sampling 1000 times and taking the min?
    double sample_min_of_n(uint64_t * seed, int n)
    {
        double min = sample_normal(5, 2, seed);
        for (int i = 0; i < (n - 2); i++) {
            double sample = sample_normal(5, 2, seed);
            if (sample < min) {
                min = sample;
            }
        }
        return min;
    }
    double sample_min_of_1000(uint64_t * seed)
    {
        return sample_min_of_n(seed, 1000);
    }

    int n_samples = 1000000, n_threads = 16;
    double* results = malloc((size_t)n_samples * sizeof(double));
    sampler_parallel(sample_min_of_1000, results, n_threads, n_samples);
    printf("Mean of the distribution of (taking the min of 1000 samples of a normal(5,2)): %f\n", array_mean(results, n_samples));
    free(results);

    /* Option 2: take the min from n samples cleverly using parallelism */
    // Question being asked: can we take the min of n samples cleverly?
    double sample_n_parallel(int n)
    {

        int n_threads = 16;
        int quotient = n / 16;
        int remainder = n % 16;

        uint64_t seed = 1000;
        double result_remainder = sample_min_of_n(&seed, remainder);

        double sample_min_of_quotient(uint64_t * seed)
        {
            return sample_min_of_n(seed, quotient);
        }
        double* results_quotient = malloc((size_t)quotient * sizeof(double));
        sampler_parallel(sample_min_of_quotient, results_quotient, n_threads, quotient);

        double min = results_quotient[0];
        for (int i = 1; i < quotient; i++) {
            if (min > results_quotient[i]) {
                min = results_quotient[i];
            }
        }
        if (min > result_remainder) {
            min = result_remainder;
        }
        free(results_quotient);
        return min;
    }
    printf("Minimum of 1M samples of normal(5,2): %f\n", sample_n_parallel(1000000));
}