forked from personal/squiggle.c
71 lines
2.2 KiB
C
71 lines
2.2 KiB
C
#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));
|
|
}
|