Refactor

Author: isoyuki

challenges/parallel-computing/MonteCarlo.c

@@ -1,36 +1,14 @@
 #include <stdio.h>
 #include <stdlib.h>
-//add code here
-
-//
 #include <time.h>
 
-int main ()
-    {
-    time_t start;
-    time_t end;
+// Step 1: Implement a non-parallel version of this program
+// Step 2: Measure the time it takes to run on large arrays
+// Step 3: Implement a parallel version of this program
+// Step 4: Measure the time again and observe the difference
 
-    int Ncirc = 0;
-    int num_trials = 100000000;
-    double pi, x, y;
-    double r = 1.0;  
+int main() {
 
-    start = time(NULL);
-    //add code here
-    
-    //
-    for(int i=0; i<num_trials; i++)
-    {
-        x = ((double) rand() / (RAND_MAX));
-        y = ((double) rand() / (RAND_MAX));
-        if ((x*x + y*y) <= r*r) 
-        {
-            Ncirc++;
-        }    
-    }
-    end = time(NULL); 
+    // Implement a program where it calculates pi using the Monte Carlo method
 
-    printf("took %ld seconds to process\n", end - start); 
-    pi = 4.0 * ((double)Ncirc/(double)num_trials);
-    printf("\n %d trials, pi is %f \n",num_trials, pi);
-    }
+}

challenges/parallel-computing/array.c

@@ -1,28 +1,15 @@
 #include <stdio.h>
-//add code here
-
-//
 #include <unistd.h>
 #include <time.h>
 
-int main() {
+// Step 1: Implement a non-parallel version of this program
+// Step 2: Measure the time it takes to run on large arrays
+// Step 3: Implement a parallel version of this program
+// Step 4: Measure the time again and observe the difference
 
-    int array[20];
-    time_t start;
-    time_t end;
+int main() {
 
-    start = time(NULL);
-    // add code here
-        
-    //          
-    for(int i=0; i<sizeof(array)/sizeof(int); ++i) 
-    {
-        array[i] = i;
-        printf("%d\n", array[i]);
-        sleep(1);
-    }  
-    end = time(NULL); 
+    // Implement a program where it creates an array of 1000000 (or any large number) of elements
 
-    printf("took %ld seconds to process\n", end - start);    
     return 0;
 }

challenges/parallel-computing/jacobi.c

@@ -35,27 +35,27 @@ int main(int argc, char** argv)
     const int n = 2000;
     const int m = 2000;
     const int iter_max = 500;
-    
+
     const double tol = 1.0e-6;
     double error = 1.0;
 
     double *restrict A    = (double*)malloc(sizeof(double)*n*m);
     double *restrict Anew = (double*)malloc(sizeof(double)*n*m);
-    
+
     initialize(A, Anew, m, n);
 
     printf("Jacobi relaxation Calculation: %d x %d mesh\n", n, m);
-    
+
     double start = omp_get_wtime();
     int iter = 0;
-   
+
     while ( error > tol && iter < iter_max )
     {
         error = calcNext(A, Anew, m, n);
         swap(A, Anew, m, n);
 
         if(iter % 100 == 0) printf("%5d, %0.6f\n", iter, error);
-        
+
         iter++;
 
     }

challenges/parallel-computing/laplace2d.c

@@ -1,30 +1,3 @@
-/* Copyright (c) 2012, NVIDIA CORPORATION. All rights reserved.
- *
- * Redistribution and use in source and binary forms, with or without
- * modification, are permitted provided that the following conditions
- * are met:
- *  * Redistributions of source code must retain the above copyright
- *    notice, this list of conditions and the following disclaimer.
- *  * Redistributions in binary form must reproduce the above copyright
- *    notice, this list of conditions and the following disclaimer in the
- *    documentation and/or other materials provided with the distribution.
- *  * Neither the name of NVIDIA CORPORATION nor the names of its
- *    contributors may be used to endorse or promote products derived
- *    from this software without specific prior written permission.
- *
- * THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AS IS'' AND ANY
- * EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
- * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
- * PURPOSE ARE DISCLAIMED.  IN NO EVENT SHALL THE COPYRIGHT OWNER OR
- * CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL,
- * EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO,
- * PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
- * PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY
- * OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
- * (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
- * OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
- */
-
 #include <math.h>
 #include <stdlib.h>
 
@@ -35,7 +8,7 @@ void initialize(double *restrict A, double *restrict Anew, int m, int n)
     memset(A, 0, n * m * sizeof(double));
     memset(Anew, 0, n * m * sizeof(double));
 
-    
+    // Can you parallelize this loop?
     for(int i = 0; i < m; i++){
         A[i] = 100.0;
         Anew[i] = 100.0;
@@ -45,29 +18,15 @@ void initialize(double *restrict A, double *restrict Anew, int m, int n)
 double calcNext(double *restrict A, double *restrict Anew, int m, int n)
 {
     double error = 0.0;
-    
-    for( int j = 1; j < n-1; j++)
-    {
-        for( int i = 1; i < m-1; i++ )
-        {
-            Anew[OFFSET(j, i, m)] = 0.25 * ( A[OFFSET(j, i+1, m)] + A[OFFSET(j, i-1, m)]
-                                           + A[OFFSET(j-1, i, m)] + A[OFFSET(j+1, i, m)]);
-            error = fmax( error, fabs(Anew[OFFSET(j, i, m)] - A[OFFSET(j, i , m)]));
-        }
-    }
+
+    // Implementation here
+
     return error;
 }
 
 void swap(double *restrict A, double *restrict Anew, int m, int n)
 {
-    
-    for( int j = 1; j < n-1; j++)
-    {
-        for( int i = 1; i < m-1; i++ )
-        {
-            A[OFFSET(j, i, m)] = Anew[OFFSET(j, i, m)];    
-        }
-    }
+    // Implementation here
 }
 
 void deallocate(double *restrict A, double *restrict Anew)

challenges/parallel-computing/private.c

@@ -1,45 +1,15 @@
 #include <stdio.h>
-//add code here
-
-
-//
 #include <unistd.h>
 #include <time.h>
 
-int main() {
+// Step 1: Implement a non-parallel version of this program
+// Step 2: Measure the time it takes to run on large arrays
+// Step 3: Implement a parallel version of this program
+// Step 4: Measure the time again and observe the difference
 
-    int array[20];
-    time_t start;
-    time_t end;
-    int x;
-    int sum = 0;
-
-    start = time(NULL);
-    //add code here
-    
-    //              
-    for(int i=0; i<sizeof(array)/sizeof(int); ++i) 
-    {
-        array[i] = i;
-        sleep(1);
+int main() {
 
-        //add code here
-        x = pow(array[i],2);
-        
-        //
-    }  
-    printf("sum of the array is %d\n", sum); 
-    if (sum == 2470)
-    {
-        printf("correct\n");
-    }
-    else
-    {
-        printf("wrong answer\n");
-    }
-    
-    end = time(NULL); 
+    // Implement a program where it squares each element in the array and outputs the sum of them
 
-    printf("took %ld seconds to process\n", end - start);    
     return 0;
 }

challenges/parallel-computing/reduction.c

@@ -7,32 +7,6 @@
 
 int main() {
 
-    int array[20];
-    time_t start;
-    time_t end;
+    // Implement a program where it sums all the elements in the array
 
-    //add code here
-    
-    //    
-    start = time(NULL);
-
-    //add code here
-    
-    //              
-    for(int i=0; i<sizeof(array)/sizeof(int); ++i) 
-    {
-        array[i] = i;
-        sleep(1);
-
-        //add code here
-        
-        //
-    }  
-    
-    printf("sum of the array is %d\n", sum); 
-    
-    end = time(NULL); 
-
-    printf("took %ld seconds to process\n", end - start);    
-    return 0;
 }