Peterson's Solution for critical section problem

#Peterson’sSolution #criticalsection #oslectures
Peterson’s Solution is a classical software based solution to the critical section problem.
In Peterson’s solution, we have two shared variables:
boolean flag[i] :Initialized to FALSE, initially no one is interested in entering the critical section
int turn : The process whose turn is to enter the critical section.
Peterson’s Solution preserves all three conditions :
Mutual Exclusion is assured as only one process can access the critical section at any time.
Progress is also assured, as a process outside the critical section does not block other processes from entering the critical section.
Bounded Waiting is preserved as every process gets a fair chance.
Disadvantages of Peterson’s Solution
It involves Busy waiting
It is limited to 2 processes.
The algorithm uses two variables, flag and turn. A flag[n] value of true indicates that the process n wants to enter the critical section. Entrance to the critical section is granted for process P0 if P1 does not want to enter its critical section or if P1 has given priority to P0 by setting turn to 0.
P0: flag[0] = true;
P0_gate: turn = 1;
while (flag[1] == true && turn == 1)
{
// busy wait
}
// critical section
...
// end of critical section
flag[0] = false;
P1: flag[1] = true;
P1_gate: turn = 0;
while (flag[0] == true && turn == 0)
{
// busy wait
}
// critical section
...
// end of critical section
flag[1] = false;

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sir, what will happen if we only put while(flag[ j ]==true); as the condition ?

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[@stviewsshorts]

Is Peterson’s Algorithm a good solution for Critical Section problem? If yes, then why? If no, then why?

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Because it satisfies me,bw,progress

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@@SudhakarAtchala can you explain more

[@gautamshenoy6170]

@@stviewsshorts petersons will satisfy only when there are 2 processes ... so it definetly wont be a "good" solution when your pc is loaded with processes