Operating Systems: Peterson's solution

Question

I am trying to understand the peterson's solution for synchronisation. For reference, I am attaching the source of reading:

This is from the wikipedia page. Now, Let's say P1 wants to enter the critical section. It sets flag1 = true and turn=0. If P0 is already in its critical section, P1 will continuously wait in its while loop while(flag[0] == true && turn ==0) as flag[0] = true. My doubt is:

1. What will happen in the case: P1 is executing its while loop and when P0 just tries to enter the critical section and executed the line flag[0] = true . Due to some reason, it couldnt't execute the next line and terminated. In this, flag[0] is also true and turn is also 0. P1 unnecessarily waits while no process is inside the critical section.
2. Why it needs to check for the turn variable. Why not only this: while(flag[0]== true) As soon as P0 will leave from critical section, flag[0] = false and P1 will be able to enter.

I am little confused in this synchronisation problem, any help will be appreciated.

Answer 1

2. Why it needs to check for the turn variable. Why not only this: while(flag[0]== true) As soon as P0 will leave from critical section, flag[0] = false and P1 will be able to enter

If you were to use just flag[0] and flag[1] it would lead to a deadlock. Let us consider P0 sets the flag[0] = true and goes ahead to check the condition in while loop. Before P0 can load the flag[1] value in a register for comparison, P1 sets flag[1] = true. So now P0 loads the value true and will keep waiting and since the flag[0] too is true and P0 is stuck in the while loop, P1 too will wait on the while loop. This is a deadlock condition since both are waiting for the other to change the flag value to false.

1. What will happen in the case: P1 is executing its while loop and when P0 just tries to enter the critical section and executed the line flag[0] = true . Due to some reason, it couldnt't execute the next line and terminated. In this, flag[0] is also true and turn is also 0. P1 unnecessarily waits while no process is inside the critical section.

For this question I don't think P0 terminating abruptly proves the implementation of the Peterson's solution for synchronization to be wrong. It is flaw in the program and any buggy code could lead to such a deadlock condition.