Question

//In this assignment, we use multiple threads to calculate the frequencies of the first digits
//of the numbers in an array of long integers
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <assert.h>

//#define NUM_THREADS 2
#define DIGITS 10
#define MAX 100000000
unsigned long a[MAX];

struct thread_data
{
int thread_num;
int i, j;                           //the staring and ending index
unsigned long freq[DIGITS];           // results
};

//initialize the array
void init_array(unsigned long a[], int n)
{   
a[0] = 1;
for(int i = 1; i < n; i++)
{   
a[i] = a[i-1] + i + 1;
}
}

//get the first digit of the long integer k
int first_digit(unsigned long k)
{
while(k >= DIGITS)
   {
k = k/DIGITS;
}
return k;
}
//frequencies of the first digits among the numbers in a[],
//there are total n numbers
void frequencies(unsigned long a[], int n, unsigned long freq[DIGITS])
{
for(int k=0; k<DIGITS; k++)
freq[k] = 0;

for(int i = 0; i<n; i++)
{
int m = first_digit(a[i]);
freq[m] ++;
}
}

void* thread_freq(void* threadarg)
{
struct thread_data* my_data = (struct thread_data*) threadarg;
   //fill in code below, only line is needed
  
   pthread_exit(NULL);
}

int main(int argc, char* argv[])
{
   if(argc!=3)
   {
       printf("Usage: %s n NUM_THREADS\n", argv[0]);
       exit(-1);
   }
   int n = atoi(argv[1]);
   assert(n>=1 && n<=MAX);
   int NUM_THREADS = atoi(argv[2]);
   assert(NUM_THREADS >=1 && NUM_THREADS <=10);

   init_array(a, n);
  
pthread_t threads[NUM_THREADS];
struct thread_data thread_data_array[NUM_THREADS];
int rc, t;
   for( t=0; t<NUM_THREADS; t++ ) {
thread_data_array[t].thread_num = t;
thread_data_array[t].i = t*n/NUM_THREADS;
       thread_data_array[t].j = (t+1)*n/NUM_THREADS - 1;
rc = pthread_create(&threads[t], NULL, thread_freq, &thread_data_array[t]);
if (rc) {
printf("ERROR; return code from pthread_create() is %d\n", rc);
exit(-1);
}
}
unsigned long freq[DIGITS] ={0};
for( t=0; t<NUM_THREADS; t++ ) {
rc = pthread_join( threads[t], NULL );
if( rc ){
printf("ERROR; return code from pthread_join() is %d\n", rc);
exit(-1);
}
       //fill in code below
       //no more than two lines of code expected

}
for(int i=1; i<DIGITS; i++)
printf("%d: %8lu\n", i, freq[i]);
return 0;
   //Run this code using different number of threads and see whether multiple threads
   //lead to time saving
   //Compare the running times with using just 1 thread
}

Answer 1

Please find the following program to find the frequceny of 1st digit in the array.

Note:

I have attached the screenshots and comments inline for better understanding.

Program:

#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <assert.h>

//Remove the following lines if not needed.
#include <sys/time.h>

//#define NUM_THREADS 2
#define DIGITS 10
#define MAX 100000000
unsigned long a[MAX];

struct thread_data
{
   int thread_num;
   int i, j; //the staring and ending index
   unsigned long freq[DIGITS]; // results
};

//initialize the array
void init_array(unsigned long a[], int n)
{   
   a[0] = 1;
   for(int i = 1; i < n; i++)
   {   
       a[i] = a[i-1] + i + 1;
   }
}

//get the first digit of the long integer k
int first_digit(unsigned long k)
{
   while(k >= DIGITS)
   {
       k = k/DIGITS;
   }
   return k;
}
//frequencies of the first digits among the numbers in a[],
//there are total n numbers
//COMMENT: We have to pass the start/end index for each thread to calculate the
//frequency of first digit for numbers in a array
void frequencies(unsigned long a[], int start, int n, unsigned long freq[DIGITS])
{
   for(int k=0; k<DIGITS; k++)
       freq[k] = 0;

   for(int i = start; i<=n; i++)
   {
       int m = first_digit(a[i]);
       freq[m] ++;
   }
}

void* thread_freq(void* threadarg)
{
   struct thread_data* my_data = (struct thread_data*) threadarg;
   //fill in code below, only line is needed
   //call the thread function to find the frequency of 1st digit from
   //a array
   frequencies(a, my_data->i, my_data->j, my_data->freq);

   pthread_exit(NULL);
}

int main(int argc, char* argv[])
{
   if(argc!=3)
   {
       printf("Usage: %s n NUM_THREADS\n", argv[0]);
       exit(-1);
   }
   //Remove the next line if not needed
   struct timeval tv1, tv2;
   int n = atoi(argv[1]);
   assert(n>=1 && n<=MAX);
   int NUM_THREADS = atoi(argv[2]);
   assert(NUM_THREADS >=1 && NUM_THREADS <=10);

   init_array(a, n);

unsigned long freq[DIGITS] ={0,0,0,0,0,0,0,0,0,0};
   pthread_t threads[NUM_THREADS];
   struct thread_data thread_data_array[NUM_THREADS];
   int rc, t;
   //Remove the next line if not needed
gettimeofday(&tv1, NULL);
   for( t=0; t<NUM_THREADS; t++ ) {
       thread_data_array[t].thread_num = t;
       thread_data_array[t].i = t*n/NUM_THREADS;
       thread_data_array[t].j = (t+1)*n/NUM_THREADS - 1;
       //uncomment the following line to find the start/end index for each thread
       //printf("thread=%d start=%d end=%d\n", t, thread_data_array[t].i, thread_data_array[t].j);
       rc = pthread_create(&threads[t], NULL, thread_freq, &thread_data_array[t]);
       if (rc) {
           printf("ERROR; return code from pthread_create() is %d\n", rc);
           exit(-1);
       }
   }
   for( t=0; t<NUM_THREADS; t++ ) {
       rc = pthread_join( threads[t], NULL );
       if( rc ){
           printf("ERROR; return code from pthread_join() is %d\n", rc);
           exit(-1);
       }
       //fill in code below
       //no more than two lines of code expected
       //once a thread is completed its work, take the frequency values from freq array in
       //thread data structure
       for (int j=1; j<DIGITS; j++)
freq[j]=freq[j] + thread_data_array[t].freq[j];

   }
   //Remove the next if not needed
   gettimeofday(&tv2, NULL);

   for(int i=1; i<DIGITS; i++)
       printf("%d: %8lu\n", i, freq[i]);

   //Remove the next printf if not needed
   printf ("Total time with %d THREADS = %f seconds\n", NUM_THREADS,
           (double) (tv2.tv_usec - tv1.tv_usec) / 1000000 +
           (double) (tv2.tv_sec - tv1.tv_sec));
   return 0;
   //Run this code using different number of threads and see whether multiple threads
   //lead to time saving
   //Compare the running times with using just 1 thread
}

Output:

osboxes@osboxes:~/Chegg/C$ ./a.out 10 3
1: 3
2: 2
3: 2
4: 1
5: 1
6: 1
7: 0
8: 0
9: 0
Total time with 3 THREADS = 0.000470 seconds
osboxes@osboxes:~/Chegg/C$ ./a.out 10 1
1: 3
2: 2
3: 2
4: 1
5: 1
6: 1
7: 0
8: 0
9: 0
Total time with 1 THREADS = 0.004904 seconds

Screen Shots:

Sample Output Screen Shot: