Modify the following code to do the following: - Task 1 (50 pts). Eliminate the deadlock by avoiding circular waiting. (Hint: change the order to access chopsticks for some philosophers.) - Task 2 (50 pts). Eliminate the deadlock by avoiding hold and wait. In other words, if a philosopher is not able to obtain the right chopstick, he/she will release the left chopstick immediately. (Hint: Use Pthread Mutex to solve the problem.) -Task 3 (50 pts extra credit). Eliminate the deadlock by limiting that at most 3 philosophers are able to compete for the chopsticks at the same time. (Hint: Use Pthread Conditional Variable to solve the problem.)

#include
#include
#include
#include
#include

#define NUMP 5

pthread_mutex_t fork_mutex[NUMP];

int main(int argc, char* argv[])
{
int i;
pthread_t diner_thread[NUMP];
int dn[NUMP];
void *diner();
for (i=0;i pthread_mutex_init(&fork_mutex[ i], NULL);

for (i=0;i dn[i] = i;
pthread_create(&diner_thread[i] ,NULL, diner,&dn[i]);
}
for (i=0;i pthread_join(diner_thread[i],NULL);
pthread_exit(0);
}

void *diner(int *i)
{
int v;
int eating = 0;
printf("I'm diner %d\n",*i);
v = *i;
while (eating < 5) {
printf("%d is thinking\n", v);
sleep( v/2);
printf("%d is hungry\n", v);
pthread_mutex_lock(&fork_mutex[ v]);
pthread_mutex_lock(&fork_mutex[ (v+1)%NUMP]);
printf("%d is eating\n", v);
eating++;
sleep(1);
printf("%d is done eating\n", v);
pthread_mutex_unlock(&fork_mute x[(v+1)%NUMP]);
pthread_mutex_unlock(&fork_mute x[v]);
}
pthread_exit(NULL);
}