forked from personal/squiggle.c
68 lines
2.1 KiB
C
68 lines
2.1 KiB
C
#include "../../../squiggle.h"
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#include "../../../squiggle_more.h"
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#include <stdio.h>
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#include <stdlib.h>
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int main()
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{
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/* Question: can we parallelize this?
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A = normal(5,2)
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B = min(A)
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B * 20
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*/
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/* Option 1: parallelize taking from n samples */
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// Question being asked: what is the distribution of sampling 1000 times and taking the min?
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double sample_min_of_n(uint64_t* seed, int n){
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double min = sample_normal(5, 2, seed);
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for(int i=0; i<(n-2); i++){
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double sample = sample_normal(5, 2, seed);
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if(sample < min){
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min = sample;
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}
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}
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return min;
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}
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double sampler_min_of_1000(uint64_t* seed) {
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return sample_min_of_n(seed, 1000);
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}
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int n_samples = 1000000, n_threads = 16;
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double* results = malloc(n_samples * sizeof(double));
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parallel_sampler(sampler_min_of_1000, results, n_threads, n_samples);
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printf("Mean of the distribution of (taking the min of 1000 samples of a normal(5,2)): %f\n", array_mean(results, n_samples));
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free(results);
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/* Option 2: take the min from n samples cleverly using parallelism */
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// Question being asked: can we take the min of n samples cleverly?
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double sample_n_parallel(int n){
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int n_threads = 16;
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int quotient = n / 16;
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int remainder = n % 16;
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uint64_t seed = 1000;
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double result_remainder = sample_min_of_n(&seed, remainder);
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double sample_min_of_quotient(uint64_t* seed) {
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return sample_min_of_n(seed, quotient);
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}
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double* results_quotient = malloc(quotient * sizeof(double));
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parallel_sampler(sample_min_of_quotient, results_quotient, n_threads, quotient);
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double min = results_quotient[0];
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for(int i=1; i<quotient; i++){
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if(min > results_quotient[i]){
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min = results_quotient[i];
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}
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}
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if(min > result_remainder){
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min = result_remainder;
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}
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free(results_quotient);
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return min;
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}
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printf("Minimum of 1M samples of normal(5,2): %f\n", sample_n_parallel(1000000));
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}
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