Description: There are 5 philosophers sitting around a table and try to eat from the center of the table. 5 chopsticks also lay on the table. Let us call the 5 philosophers in clockwise P1, P2, ...P5. There also 5 chopsticks clockwise S1, S2, ..., S5 on the table. There are only one chopstick between every two philosophers. Every right hand chopstick will have the same index as the philosopher. For example, on the right hand side of P1, the chopstick is called S1. And every left hand side chopstick number is 1+number of the philosopher.

A philosopher spend random time to think, then he feel hungry and try to eat.

The middle dish can provide enough food for everyone at the same time.

But a philosopher only can start to eat when he picked up two chopsticks from left hand side and right hand side to form a pair of chopsticks.

If a philosopher take one chopsticks, he will try to fight with neighbours to get another one, and never back off to put down the one in his hand.

Once the philosopher is eating, you can not interrupt him.

Here comes the deadlock problem: when all the 5 philosophers start to pick up the right hand side chopsticks at the same time, they get stuck.

We use a mutex, every time whenever one philosor start to pick chopsticks, he lock the mutex, and he is the only guy can pick chopsticks. After he get two chopsticks, he unlock the mutex.

Your first task is modifying the two lines between the beginning line and end line. You can insert and edit multiple lines between special comment lines in anyways, however, as the comment indicated, do not modify the special begin and comment lines themselves!

#include <stdio.h>

#include <stdlib.h>

#include <malloc.h>

#include <time.h>

#include <unistd.h>

#include <pthread.h>

#include <semaphore.h>

#define N 5

// here we use 5 different semaphores, each chopstick use one

// semaphore. Initally, the value for each is 1. After using one

// resource (chopstick), the value -1. 0 means currently not available

sem_t chopsticks[N];

// We do not use mutex, we set up a new semaphore

// called m

// #1#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #1#END# DO NOT MODIFIE THIS COMMENT LINE!

//The id of 5 philosophers

int philosophers[N] = {0, 1, 2, 3, 4};

void delay (int len) {

inti = rand() % len;

intx;

while (i > 0) {

x = rand() % len;

while (x > 0) {

x--;

}

i--;

}

}

void *philosopher (void* arg) {

inti = *(int *)arg;

intright = i;// right hand chopstick indexed as the same id of the philospher

intleft = (i + 1) % N;// left hand chopstick index as the id of the philospher + 1

while (1) {

printf("Philosopher %d is thinking\n", i);

delay(60000);

printf("Philosopher %d is hungry \n", i);

// wait for the semaphore m first, make sure no more than 4 philoshpers

// pick their first chopstick at the same time.

// #2#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #2#END# DO NOT MODIFIE THIS COMMENT LINE!

// wait for first (right) chopstick to use

// after using it, the value will decrease 1

// THAT IS WHY use need to use &, because the semaphore need to be modified

sem_wait(&chopsticks[right]);

printf("Philosopher %d picked up Chopstick %d, can not eat with one chopstick.\n", i, right);

// wati to use the left chopstick

sem_wait(&chopsticks[left]);

printf("Philosopher %d picked up Chopstick %d, now is eating with two chopsticks.\n", i, left);

delay(60000);

// release the first chopstick, increase the semaphore 1 by sem_post

sem_post(&chopsticks[right]);

printf("Philosopher %d put down Chopstick %d\n", i, right);

// when any right chopstick been released,

// increase semaphore m to release one capacity

// #3#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #3#END# DO NOT MODIFIE THIS COMMENT LINE!

sem_post(&chopsticks[left]);

printf("Philosopher %d put down Chopstick %d\n", i, left);

}

}

int main (int argc, char **argv) {

srand(time(NULL));

pthread_tphilo[N];

inti;

// change your the following id into your banner id

// #8#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #8#END# DO NOT MODIFIE THIS COMMENT LINE!

printf("Banner id: %d\n", banner_id);

// initalize the chopstick semphore

for (i=0; i<N; i++) {

// #4#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #4#END# DO NOT MODIFIE THIS COMMENT LINE!

}

// initalize the new semaphore m

// think what should be the inital value, and why?

// #5#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #5#END# DO NOT MODIFIE THIS COMMENT LINE!

// create threads

// #6#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #6#END# DO NOT MODIFIE THIS COMMENT LINE!

// put threads into the system

for (i=0; i<N; i++) {

pthread_join(philo[i], NULL);

}

// destroy semaphores

for (i=0; i<N; i++) {

sem_destroy(&chopsticks[i]);

}

// also do not forgor to destroy the m semaphore

// #7#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #7#END# DO NOT MODIFIE THIS COMMENT LINE!

return0;

}

Question

Description: There are 5 philosophers sitting around a table and try to eat from the center of the table. 5 chopsticks also lay on the table. Let us call the 5 philosophers in clockwise P1, P2, ...P5. There also 5 chopsticks clockwise S1, S2, ..., S5 on the table. There are only one chopstick between every two philosophers. Every right hand chopstick will have the same index as the philosopher. For example, on the right hand side of P1, the chopstick is called S1. And every left hand side chopstick number is 1+number of the philosopher.

A philosopher spend random time to think, then he feel hungry and try to eat.
The middle dish can provide enough food for everyone at the same time.
But a philosopher only can start to eat when he picked up two chopsticks from left hand side and right hand side to form a pair of chopsticks.
If a philosopher take one chopsticks, he will try to fight with neighbours to get another one, and never back off to put down the one in his hand.
Once the philosopher is eating, you can not interrupt him.

Here comes the deadlock problem: when all the 5 philosophers start to pick up the right hand side chopsticks at the same time, they get stuck.

We use a mutex, every time whenever one philosor start to pick chopsticks, he lock the mutex, and he is the only guy can pick chopsticks. After he get two chopsticks, he unlock the mutex.

Your first task is modifying the two lines between the beginning line and end line. You can insert and edit multiple lines between special comment lines in anyways, however, as the comment indicated, do not modify the special begin and comment lines themselves!

#include <stdio.h>

#include <stdlib.h>

#include <malloc.h>

#include <time.h>

#include <unistd.h>

#include <pthread.h>

#include <semaphore.h>

#define N 5

// here we use 5 different semaphores, each chopstick use one

// semaphore. Initally, the value for each is 1. After using one

// resource (chopstick), the value -1. 0 means currently not available

sem_t chopsticks[N];

// We do not use mutex, we set up a new semaphore

// called m

// #1#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #1#END# DO NOT MODIFIE THIS COMMENT LINE!

//The id of 5 philosophers

int philosophers[N] = {0, 1, 2, 3, 4};

void delay (int len) {

inti = rand() % len;

intx;

while (i > 0) {

x = rand() % len;

while (x > 0) {

x--;

}

i--;

}

void *philosopher (void* arg) {

inti = *(int *)arg;

intright = i;// right hand chopstick indexed as the same id of the philospher

intleft = (i + 1) % N;// left hand chopstick index as the id of the philospher + 1

while (1) {

printf("Philosopher %d is thinking\n", i);

delay(60000);

printf("Philosopher %d is hungry \n", i);

// wait for the semaphore m first, make sure no more than 4 philoshpers

// pick their first chopstick at the same time.

// #2#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #2#END# DO NOT MODIFIE THIS COMMENT LINE!

// wait for first (right) chopstick to use

// after using it, the value will decrease 1

// THAT IS WHY use need to use &, because the semaphore need to be modified

sem_wait(&chopsticks[right]);

printf("Philosopher %d picked up Chopstick %d, can not eat with one chopstick.\n", i, right);

// wati to use the left chopstick

sem_wait(&chopsticks[left]);

printf("Philosopher %d picked up Chopstick %d, now is eating with two chopsticks.\n", i, left);

delay(60000);

// release the first chopstick, increase the semaphore 1 by sem_post

sem_post(&chopsticks[right]);

printf("Philosopher %d put down Chopstick %d\n", i, right);

// when any right chopstick been released,

// increase semaphore m to release one capacity

// #3#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #3#END# DO NOT MODIFIE THIS COMMENT LINE!

sem_post(&chopsticks[left]);

printf("Philosopher %d put down Chopstick %d\n", i, left);

}

int main (int argc, char **argv) {

srand(time(NULL));

pthread_tphilo[N];

inti;

// change your the following id into your banner id

// #8#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #8#END# DO NOT MODIFIE THIS COMMENT LINE!

printf("Banner id: %d\n", banner_id);

// initalize the chopstick semphore

for (i=0; i<N; i++) {

// #4#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #4#END# DO NOT MODIFIE THIS COMMENT LINE!

}

// initalize the new semaphore m

// think what should be the inital value, and why?

// #5#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #5#END# DO NOT MODIFIE THIS COMMENT LINE!

// create threads

// #6#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #6#END# DO NOT MODIFIE THIS COMMENT LINE!

// put threads into the system

for (i=0; i<N; i++) {

pthread_join(philo[i], NULL);

}

// destroy semaphores

for (i=0; i<N; i++) {

sem_destroy(&chopsticks[i]);

}

// also do not forgor to destroy the m semaphore

// #7#BEGIN# DO NOT MODIFIE THIS COMMENT LINE!

// #7#END# DO NOT MODIFIE THIS COMMENT LINE!

return0;

}

Accepted Answer

Answer is given below: