Petersons algorithm doubt

[Rahul Shah]

Hi professor,

I did not understand why the Peterson's algorithm cannot be used for n
processes.

Suppose I consider 3 processes and the

flag[0]=1;
flag[1]=1;
flag[2]=1;
turn=1;
While((flag[1]==1&&turn==1)OR
While((flag[1]==1&&turn==0)OR
While((flag[1]==1&&turn==1)OR
(flag[2]==1&&turn==2))
(flag[2]==1&&turn==2))
(flag[0]==1&&turn==0))


CRITICAL SECTION CRITICAL
SECTION CRITICAL
SECTION

flag[0]=0;
flag[1]=0;
flag[2]=0;


Can you just tell me the scenario where this will not work. I am not
able to figure it out.

Thanks,
Rahul

[Partha Dasgupta]

There is something wrong, semantically and/or typo in this section. Fix it, indent it properly so that I can interpret it and I will try to see why it does not work.

While((flag[1]==1&&turn==1)OR
While((flag[1]==1&&turn==0)OR
While((flag[1]==1&&turn==1)OR
(flag[2]==1&&turn==2))
(flag[2]==1&&turn==2))
(flag[0]==1&&turn==0))

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[Rahul Shah]

oh. I did not see this. Sorry for that. Here's my question


//Process 0
flag[0]=1;
turn=1;
While((flag[1]==1&&turn==1)OR (flag[2]==1&&turn==2))

CRITICAL SECTION

flag[0]=0;

//Process 1
flag[1]=1;
turn = 2;

While((flag[0]==1&&turn==0)OR (flag[2]==1&&turn==2))

CRITICAL SECTION
flag[1]=0;

//Process 2
flag[2]=1;
turn = 0;

While((flag[0]==1&&turn==0)OR (flag[1]==1&&turn==1))

CRITICAL SECTION
flag[2]=0;

Regards,
Rahul Shah

[Partha]

Does not work. Should I tell you why?

OK a short description and you should try to work out the details.

0 and 1 try to get in, but 1 gets in 0 gets stuck. Then 2 tries to get
in, enabling 0 and letting 0 go through, and now we have 2 processes
in the CS.

[Rahul Shah]

Ok. That means any other process can change the turn variable when
some process is in the critical section. It is not locked.