add example of parallelizing a min

examples/more/13_parallelize_min/example.c

@@ -0,0 +1,67 @@
+#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(n_samples * sizeof(double));
+    parallel_sampler(sampler_result, 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(quotient * sizeof(double));
+        parallel_sampler(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 = results_remainder;
+        }
+        free(results_quotient);
+        return min;
+    }
+    printf("Minimum of 1M samples of normal(5,2): %f\n", sample_n_parallel(1000000));
+
+}