![Algorithm:
ANALYSIS:
• Peterson's solution is restricted to two.
• The processes are numbered P0 and P1 or Pi
and Pj where j = 1-i
• Peterson's solution requires the two processes
to share two data items:
int turn;
boolean flag[2];
• The variable turn indicates whose turn it is to
enter its critical section.
• The flag array is used to indicate if a process is
ready to enter its critical section.
• To enter the critical section,
• process Pi first sets flag [i] = true
• turn = j
• If both processes try to enter at the same time,
turn will be set to both i and j at roughly the
same time.](https://image.slidesharecdn.com/osaba-240302193431-e3daca2b/85/Operatioooooooooooooooooooooooooooooooooooooooooooooo-2-320.jpg)
![To prove that solution is correct, then we need to show that
1. Mutual exclusion is preserved
• Only one process can enter the critical section.
• If Pi and Pj are the two processes that enter the critical section, the turn will be given to any one among i or j
but not both.
2. Progress requirement is satisfied
• Assume that the process is waiting continuously to enter the critical section i.e, Pi is in while loop and
flag[j]=true and turn==j
• However the process Pj in the critical section, flag[j] will become false and now Pi can enter into the critical
section to progress.
3. Bounded-waiting requirement is met
• Since each process gives the turn to another process bounded waiting is assured and each process will wait
at most one entry by another process.](https://image.slidesharecdn.com/osaba-240302193431-e3daca2b/85/Operatioooooooooooooooooooooooooooooooooooooooooooooo-3-320.jpg)
![#include <stdio.h>
#include <pthread.h>
#include <unistd.h> // for sleep()
#define TRUE 1
#define FALSE 0
int flag[2] = { FALSE, FALSE }; // Flags to indicate if process is ready to enter
critical section
int turn = 0; // Variable to indicate whose turn it is
](https://image.slidesharecdn.com/osaba-240302193431-e3daca2b/85/Operatioooooooooooooooooooooooooooooooooooooooooooooo-4-320.jpg)
![void* process(void* arg) {
int process_id = *((int*)arg);
int other = 1 - process_id;
for (int i = 0; i < 5; i++) {
flag[process_id] = TRUE;
turn = other;
while (flag[other] == TRUE && turn == other) {
// Process waits because it's not its turn or the other process is ready
}
// Process enters critical section
printf("Process %d is in the critical section.n", process_id);
sleep(1); // Simulating critical section work
// Process exits critical section
flag[process_id] = FALSE;
}
pthread_exit(NULL);
}](https://image.slidesharecdn.com/osaba-240302193431-e3daca2b/85/Operatioooooooooooooooooooooooooooooooooooooooooooooo-5-320.jpg)
![int main() {
pthread_t threads[2];
int ids[2] = { 0, 1 };
int i;
// Creating threads
for (i = 0; i < 2; i++) {
if (pthread_create(&threads[i], NULL, process, &ids[i]) != 0) {
perror("pthread_create");
return -1;
}
}
// Joining threads
for (i = 0; i < 2; i++) {
if (pthread_join(threads[i], NULL) != 0) {
perror("pthread_join");
return -1;
}
}
return 0;
}](https://image.slidesharecdn.com/osaba-240302193431-e3daca2b/85/Operatioooooooooooooooooooooooooooooooooooooooooooooo-6-320.jpg)
Operatioooooooooooooooooooooooooooooooooooooooooooooo
- 1. “Critical Section Problem Using Peterson’s Solution” By, SHREYAS V E - 4VV21CS147 PRAMUKH B V - 4VV21CS190 SHREYA CHAKRAVARTHY S - 4VV21CS145 SHANE MATHEW - 4VV21CS139 Subject: Operating Systems Subject Code: 21CS52 Faculty in-charge: Dr. Madhusudhan H S
- 2. Algorithm: ANALYSIS: • Peterson's solution is restricted to two. • The processes are numbered P0 and P1 or Pi and Pj where j = 1-i • Peterson's solution requires the two processes to share two data items: int turn; boolean flag[2]; • The variable turn indicates whose turn it is to enter its critical section. • The flag array is used to indicate if a process is ready to enter its critical section. • To enter the critical section, • process Pi first sets flag [i] = true • turn = j • If both processes try to enter at the same time, turn will be set to both i and j at roughly the same time.
- 3. To prove that solution is correct, then we need to show that 1. Mutual exclusion is preserved • Only one process can enter the critical section. • If Pi and Pj are the two processes that enter the critical section, the turn will be given to any one among i or j but not both. 2. Progress requirement is satisfied • Assume that the process is waiting continuously to enter the critical section i.e, Pi is in while loop and flag[j]=true and turn==j • However the process Pj in the critical section, flag[j] will become false and now Pi can enter into the critical section to progress. 3. Bounded-waiting requirement is met • Since each process gives the turn to another process bounded waiting is assured and each process will wait at most one entry by another process.
- 4. #include <stdio.h> #include <pthread.h> #include <unistd.h> // for sleep() #define TRUE 1 #define FALSE 0 int flag[2] = { FALSE, FALSE }; // Flags to indicate if process is ready to enter critical section int turn = 0; // Variable to indicate whose turn it is
- 5. void* process(void* arg) { int process_id = *((int*)arg); int other = 1 - process_id; for (int i = 0; i < 5; i++) { flag[process_id] = TRUE; turn = other; while (flag[other] == TRUE && turn == other) { // Process waits because it's not its turn or the other process is ready } // Process enters critical section printf("Process %d is in the critical section.n", process_id); sleep(1); // Simulating critical section work // Process exits critical section flag[process_id] = FALSE; } pthread_exit(NULL); }
- 6. int main() { pthread_t threads[2]; int ids[2] = { 0, 1 }; int i; // Creating threads for (i = 0; i < 2; i++) { if (pthread_create(&threads[i], NULL, process, &ids[i]) != 0) { perror("pthread_create"); return -1; } } // Joining threads for (i = 0; i < 2; i++) { if (pthread_join(threads[i], NULL) != 0) { perror("pthread_join"); return -1; } } return 0; }
- 8. THANK YOU