Mutual Exclusion Problem - mutex

Please take a look on the following pseudo-code:
boolean blocked[2];
int turn;
void P(int id) {
while(true) {
blocked[id] = true;
while(turn != id) {
while(blocked[1-id])
/* do nothing */;
turn = id;
}
/* critical section */
blocked[id] = false;
/* remainder */
}
}
void main() {
blocked[0] = false;
blocked[1] = false;
turn = 0;
parbegin(P(0), P(1)); //RUN P0 and P1 parallel
}
I thought that a could implement a simple Mutual - Exclution solution using the code above. But it's not working. Has anyone got an idea why?
Any help would really be appreciated!

Mutual Exclusion is in this exemple not guaranteed because of the following:
We begin with the following situation:
blocked = {false, false};
turn = 0;
P1 is now executes, and skips
blocked[id] = false; // Not yet executed.
The situation is now:
blocked {false, true}
turn = 0;
Now P0 executes. It passes the second while loop, ready to execute the critical section. And when P1 executes, it sets turn to 1, and is also ready to execute the critical section.
Btw, this method was originally invented by Hyman. He sent it to Communications of the Acm in 1966

Mutual Exclusion is in this exemple not guaranteed because of the following:
We begin with the following situation:
turn= 1;
blocked = {false, false};
The execution runs as follows:
P0: while (true) {
P0: blocked[0] = true;
P0: while (turn != 0) {
P0: while (blocked[1]) {
P0: }
P1: while (true) {
P1: blocked[1] = true;
P1: while (turn != 1) {
P1: }
P1: criticalSection(P1);
P0: turn = 0;
P0: while (turn != 0)
P0: }
P0: critcalSection(P0);

Is this homework, or some embedded platform? Is there any reason why you can't use pthreads or Win32 (as relevant) synchronisation primitives?

Maybe you need to declare blocked and turn as volatile, but without specifying the programming language there is no way to know.

Concurrency can not be implemented like this, especially in a multi-processor (or multi-core) environment: different cores/processors have different caches. Those caches may not be coherent. The pseudo-code below could execute in the order shown, with the results shown:
get blocked[0] -> false // cpu 0
set blocked[0] = true // cpu 1 (stored in CPU 1's L1 cache)
get blocked[0] -> false // cpu 0 (retrieved from CPU 0's L1 cache)
get glocked[0] -> false // cpu 2 (retrieved from main memory)
You need hardware knowledge to implement concurrency.

Compiler might have optimized out the "empty" while loop. Declaring variables as volatile might help, but is not guaranteed to be sufficient on multiprocessor systems.

Related

Backtracking results in same repeating course

I am trying to solve a puzzle, and it has been suggested that I use backtracking - I did not know the term so did some investigation, and found the following in Wikipedia:
In order to apply backtracking to a specific class of problems, one must provide the data P for the particular instance of the problem that is to be solved, and six procedural parameters, root, reject, accept, first, next, and output. These procedures should take the instance data P as a parameter and should do the following:
root(P): return the partial candidate at the root of the search tree.
reject(P,c): return true only if the partial candidate c is not worth completing.
accept(P,c): return true if c is a solution of P, and false otherwise.
first(P,c): generate the first extension of candidate c.
next(P,s): generate the next alternative extension of a candidate, after the extension s.
output(P,c): use the solution c of P, as appropriate to the application.
The backtracking algorithm reduces the problem to the call backtrack(root(P)), where backtrack is the following recursive procedure:
procedure backtrack(c) is
if reject(P, c) then return
if accept(P, c) then output(P, c)
s ← first(P, c)
while s ≠ NULL do
backtrack(s)
s ← next(P, s)
I have attempted to use this method for my solution, but after the method finds a rejected candidate it just starts again and finds the same route, rather than the next possible one.
I now don't think I have used the next(P,s) correctly, because I don't really understand the wording 'after the extension s'.
I've tried 2 methods:
(a) in the first() function, generating all possible extensions, storing them in a list, then using the first. The next() function then uses the other extensions from the list in turn. But this maybe can't work because of the calls to backtrack() in between the calls to next().
(b) adding a counter to the data (i.e. the class that includes all the grid info) and incrementing this for each call of next(). But can't work out where to reset this counter to zero.
Here's the relevant bit of code for method (a):
private PotentialSolution tryFirstTrack(PotentialSolution ps)
{
possibleTracks = new List<PotentialSolution>();
for (Track trytrack = Track.Empty + 1; trytrack < Track.MaxVal; trytrack++)
{
if (validMove(ps.nextSide, trytrack))
{
ps.SetCell(trytrack);
possibleTracks.Add(ps);
}
}
return tryNextTrack(ps);
}
private PotentialSolution tryNextTrack(PotentialSolution ps)
{
if (possibleTracks.Count == 0)
{
ps.SetCell(Track.Empty);
return null;
}
ps = possibleTracks.First();
// don't use same one again
possibleTracks.Remove(ps);
return ps;
}
private bool backtrackTracks(PotentialSolution ps)
{
if (canExit)
{
return true;
}
if (checkOccupiedCells(ps))
{
ps = tryFirstTrack(ps);
while (ps != null)
{
// 'testCells' is a copy of the grid for use with graphics - no need to include graphics in the backtrack stack
testCells[ps.h, ps.w].DrawTrack(g, ps.GetCell());
if (ps.TestForExit(endColumn, ref canExit) != Track.MaxVal)
{
drawRowColTotals(ps);
return true;
}
ps.nextSide = findNextSide(ps.nextSide, ps.GetCell(), ref ps.h, ref ps.w);
if (ps.h >= 0 && ps.h < cellsPerSide && ps.w >= 0 && ps.w < cellsPerSide)
{
backtrackTracks(ps);
ps = tryNextTrack(ps);
}
else
return false;
}
return false;
}
return false;
}
and here's some code using random choices. This works fine, so I conclude that the methods checkOccupiedCells() and findNextSide() are working correctly.
private bool backtrackTracks(PotentialSolution ps)
{
if (canExit)
{
return true;
}
if (checkOccupiedCells(ps))
{
Track track = createRandomTrack(ps);
if (canExit)
return true;
if (track == Track.MaxVal)
return false;
ps.SetCell(track);
ps.nextSide = findNextSide(ps.nextSide, track, ref ps.h, ref ps.w);
if (ps.h >= 0 && ps.h < cellsPerSide && ps.w >= 0 && ps.w < cellsPerSide)
backtrackTracks(ps);
else
return false;
}
}
If it helps, there's more background info in the puzzle itself here

order of execution of forked processes

#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
#include<sys/sem.h>
#include<sys/ipc.h>
int sem_id;
void update_file(int number)
{
struct sembuf sem_op;
FILE* file;
printf("Inside Update Process\n");
/* wait on the semaphore, unless it's value is non-negative. */
sem_op.sem_num = 0;
sem_op.sem_op = -1; /* <-- Amount by which the value of the semaphore is to be decreased */
sem_op.sem_flg = 0;
semop(sem_id, &sem_op, 1);
/* we "locked" the semaphore, and are assured exclusive access to file. */
/* manipulate the file in some way. for example, write a number into it. */
file = fopen("file.txt", "a+");
if (file) {
fprintf(file, " \n%d\n", number);
fclose(file);
}
/* finally, signal the semaphore - increase its value by one. */
sem_op.sem_num = 0;
sem_op.sem_op = 1;
sem_op.sem_flg = 0;
semop( sem_id, &sem_op, 1);
}
void write_file(char* contents)
{
printf("Inside Write Process\n");
struct sembuf sem_op;
sem_op.sem_num = 0;
sem_op.sem_op = -1;
sem_op.sem_flg = 0;
semop( sem_id, &sem_op, 1);
FILE *file = fopen("file.txt","w");
if(file)
{
fprintf(file,contents);
fclose(file);
}
sem_op.sem_num = 0;
sem_op.sem_op = 1;
sem_op.sem_flg = 0;
semop( sem_id, &sem_op, 1);
}
int main()
{
//key_t key = ftok("file.txt",'E');
sem_id = semget( IPC_PRIVATE, 1, 0600 | IPC_CREAT);
/*here 100 is any arbit number to be assigned as the key of the
semaphore,1 is the number of semaphores in the semaphore set, */
if(sem_id == -1)
{
perror("main : semget");
exit(1);
}
int rc = semctl( sem_id, 0, SETVAL, 1);
pid_t u = fork();
if(u == 0)
{
update_file(100);
exit(0);
}
else
{
wait();
}
pid_t w = fork();
if(w == 0)
{
write_file("Hello!!");
exit(0);
}
else
{
wait();
}
}
If I run the above code as a c code, the write_file() function is called after the update_file () function
Whereas if I run the same code as a c++ code, the order of execution is reverse... why is it so??
Just some suggestions, but it looks to me like it could be caused by a combination of things:
The wait() call is supposed to take a pointer argument (that can
be NULL). Compiler should have caught this, but you must be picking
up another definition somewhere that permits your syntax. You are
also missing an include for sys/wait.h. This might be why the
compiler isn't complaining as I'd expect it to.
Depending on your machine/OS configuration the fork'd process may
not get to run until after the parent yields. Assuming the "wait()"
you are calling isn't working the way we would be expecting, it is
possible for the parent to execute completely before the children
get to run.
Unfortunately, I wasn't able to duplicate the same temporal behavior. However, when I generated assembly files for each of the two cases (C & C++), I noticed that the C++ version is missing the "wait" system call, but the C version is as I would expect. To me, this suggests that somewhere in the C++ headers this special version without an argument is being #defined out of the code. This difference could be the reason behind the behavior you are seeing.
In a nutshell... add the #include, and change your wait calls to "wait(0)"

Atomically setting a variable without comparing first

I've been reading up on and experimenting with atomic memory access for synchronization, mainly for educational purposes. Specifically, I'm looking at Mac OS X's OSAtomic* family of functions. Here's what I don't understand: Why is there no way to atomically set a variable instead of modifying it (adding, incrementing, etc.)? OSAtomicCompareAndSwap* is as close as it gets -- but only the swap is atomic, not the whole function itself. This leads to code such as the following not working:
const int N = 100000;
void* threadFunc(void *data) {
int *num = (int *)data;
// Wait for main thread to start us so all spawned threads start
// at the same time.
while (0 == num) { }
for (int i = 0; i < N; ++i) {
OSAtomicCompareAndSwapInt(*num, *num+1, num);
}
}
// called from main thread
void test() {
int num = 0;
pthread_t threads[5];
for (int i = 0; i < 5; ++i) {
pthread_create(&threads[i], NULL, threadFunc, &num);
}
num = 1;
for (int i = 0; i < 5; ++i) {
pthread_join(threads[i], NULL);
}
printf("final value: %d\n", num);
}
When run, this example would ideally produce 500,001 as the final value. However, it doesn't; even when the comparison in OSAtomicCompareAndSwapInt in thread X succeeds, another thread Y can come in set the variable first before X has a chance to change it.
I am aware that in this trivial example I could (and should!) simply use OSAtomicAdd32, in which case the code works. But, what if, for example, I wanted to set a pointer atomically so it points to a new object that another thread can then work with?
I've looked at other APIs, and they seem to be missing this feature as well, which leads me to believe that there is a good reason for it and my confusion is just based on lack of knowledge. If somebody could enlighten me, I'd appreciate it.
I think that you have to check the OSAtomicCompareAndSwapInt result to guarantee that the int was actually set.

Bounded Buffers (Producer Consumer)

In the shared buffer memory problem , why is it that we can have at most (n-1) items in the buffer at the same time.
Where 'n' is the buffer's size .
Thanks!
In an OS development class in college, I had an adjunct teacher that claimed it was impossible to have a software-only solution that could use all N elements in the buffer.
I proved him wrong with something I decided to call the race track solution (inspired by the fact that I like to run track).
On a race track, you are not limited to a 400 meter race; a race can consist of more than one lap. What happens if two runners are neck and neck
in a race? How do you know whether they are tied, or whether one runner has lapped the other? The answer is simple: in a race, we don't monitor a runner's position
on the track; we monitor the distance each runner has traversed. Thus, when two runners are neck and neck, we can disambiguafy between a tie and when one runner has
lapped the other.
So, our algorithm has an N-element array, and manages a 2N race. We don't restart the producer/consumer's counter back to zero until they finish their respective 2N race.
We don't allow the producer to be more than one lap ahead of the consumer, and we don't allow the consumer to be ahead of the producer.
Actually, we only have to monitor the distance between the producer and consumer.
The code is as follows:
Item track[LAP];
int consIdx = 0;
int prodIdx = 0;
void consumer()
{ while(true)
{ int diff = abs(prodIdx - consIdx);
if(0 < diff) //If the consumer isn't tied
{ track[consIdx%LAP] = null;
consIdx = (consIdx + 1) % (2*LAP);
}
}
}
void producer()
{ while(true)
{ int diff = (prodIdx - consIdx);
if(diff < LAP) //If prod hasn't lapped cons
{ track[prodIdx%LAP] = Item(); //Advance on the 1-lap track.
prodIdx = (prodIdx + 1) % (2*LAP);//Advance in the 2-lap race.
}
}
}
It's been a while since I originally solved the problem, so this is according to my best recollection. Hopefully I didn't overlook any bugs.
Hope this helps!
Oops, here's a bug fix:
Item track[LAP];
int consIdx = 0;
int prodIdx = 0;
void consumer()
{ while(true)
{ int diff = prodIdx - consIdx; //When prodIdx wraps to 0 before consIdx,
diff = 0<=diff? diff: diff + (2*LAP); //think in 3 Laps until consIdx wraps to 0.
if(0 < diff) //If the consumer isn't tied
{ track[consIdx%LAP] = null;
consIdx = (consIdx + 1) % (2*LAP);
}
}
}
void producer()
{ while(true)
{ int diff = prodIdx - consIdx;
diff = 0<=diff? diff: diff + (2*LAP);
if(diff < LAP) //If prod hasn't lapped cons
{ track[prodIdx%LAP] = Item(); //Advance on the 1-lap track.
prodIdx = (prodIdx + 1) % (2*LAP);//Advance in the 2-lap race.
}
}
}
Well, theoretically a bounded buffer can hold elements upto its size. But what you are saying could be related to certain implementation quirks like a clean way of figuring out when the buffer is empty/full. This question -> Empty element in array-based bounded buffer deals with a similar thing. See if it helps.
However you can of course have implementations that have all n slots filled up. That's how the bounded buffer problem is defined anyway.

Why threads give different number on my program using ThreadPool?

why Number has different value?
Thx
class Program
{
static DateTime dt1;
static DateTime dt2;
static Int64 number = 0;
public static void Main()
{
dt1 = DateTime.Now;
for (int i = 0; i < 10; i++)
{
ThreadPool.QueueUserWorkItem(new WaitCallback(WorkThread), DateTime.Now);
}
dt2 = DateTime.Now;
Console.WriteLine("***");
Console.ReadLine();
}
public static void WorkThread(object queuedAt)
{
number = 0;
for (Int64 i = 0; i < 2000000; i++)
{
number += i;
}
Console.WriteLine("number is:{0} and time:{1}",number,DateTime.Now - dt1);
}
}
number is being shared between all of your threads, and you're not doing anything to synchronize access to it from each thread. So one thread might not have even started it's i loop (it may or may not have reset number to 0 at this point), while another can be half way through, and another might have finished it's loop completely and be at the Console.WriteLine part.
Here you have 10 threads acting on the static variable number at indeterminate times. One thread could on its 10000 iteration while another could just be beginning execution. And your routine begins by resetting number to 0. This logic would produce interesting results but nothing predictable.
If multiple threads access the same variable all at once, there is a risk of race conditions. A race condition is basically when the operations of the two threads are interwoven such that they interfere with eachother. To add a value to "number", the old value must be read, the sum computed, and the new value set. If those steps are being done by many threads at the same time, the value-setting can overwrite work done by previous threads, and the final result can change. You must use a lock (also called a critical section, mutex, or monitor) to protect the variable so this can't happen.