Mutual Exclusion By Shiran Mizrahi. Critical Section class Counter { private int value = 1;...

Mutual Exclusion

By Shiran Mizrahi

Critical Section

class Counter {private int value = 1; //counter starts at one

public Counter(int c) { //constructor initializes countervalue = c;}

public int inc() { //increment value & return prior valueint temp = value; //start of danger zonevalue = temp+1; //end of danger zonereturn temp;}

Critical Section

The problem occurs if two threads both read the value field at the line marked “start of danger zone”, and then both update that field at the line marked “end of danger zone”.

int temp = value; value = temp+1;

Critical Section

Value

read 1

read 1

write 2

read 2

write 3

write 2

2 3 2

time

int temp = value; value = temp+1;

The mutual exclusion problem

remainder coderemainder code

entry codeentry code

critical sectioncritical section

exit codeexit code

The problem is to design the entry and exit code in a way that guarantees that the mutual exclusion and deadlock-freedom properties are satisfied.

Good properties

Mutual Exclusion: No two threads are in their critical sections at the same time.

Deadlock-freedom: If a thread is trying to enter its critical section, then some thread, not necessarily the same one, eventually enters its critical section.

Starvation-freedom: If a thread is trying to enter its critical section, then this thread must eventually enter its critical section.

Starvation-freedom is a stronger property than Deadlock-freedom.

Mutual Exclusion: No two threads are in their critical sections at the same time.

Deadlock-freedom: If a thread is trying to enter its critical section, then some thread, not necessarily the same one, eventually enters its critical section.

Starvation-freedom: If a thread is trying to enter its critical section, then this thread must eventually enter its critical section.

Starvation-freedom is a stronger property than Deadlock-freedom.

Discussion Topics

The mutual exclusion problem and proposed algorithms

Peterson’s algorithm Kessels’ single-writer algorithm Tournament algorithms The Filter algorithm The Bakery algorithm

The mutual exclusion problem and proposed algorithms

Peterson’s algorithm Kessels’ single-writer algorithm Tournament algorithms The Filter algorithm The Bakery algorithm

Proposed solutions for two threads

We begin with two inadequate

but interesting algorithms

Some notations

A B

event A precedes event B CSA

thread A is in the critical section writeA(x=v)

the event in which thread A writes to x readA(x==v)

the event in which thread A reads from x

Algorithm 1

Thread 0

flag[0] = true

while (flag[1]) {}

critical section

flag[0]=false

Thread 1

flag[1] = true

while (flag[0]) {}

critical section

flag[1]=false

Mutual Exclusion

Algorithm 1 satisfies

mutual exclusion

Proof

Assume in the contrary that two threads can be in their critical section at the same time.

From the code we can see:

write0(flag[0]=true) read0(flag[1]==false) CS0

write1(flag[1]=true) read1(flag[0]==false) CS1

From the assumption:

read0(flag[1]==false) write1(flag[1]=true)Thread 0

flag[0] = true

while (flag[1]) {}

critical section

flag[0]=false

Thread 1

flag[1] = true

while (flag[0]) {}

critical section

flag[1]=false

Proof

We get:

write0(flag[0]=true) read0(flag[1]==false) write1(flag[1]=true) read1(flag[0]==false)

That means that thread 0 writes (flag[0]=true) and then thread 1 reads that (flag[0]==false), a contradiction.

Thread 0

flag[0] = true

while (flag[1]) {}

critical section

flag[0]=false

Thread 1

flag[1] = true

while (flag[0]) {}

critical section

flag[1]=false

Deadlock freedom

Thread 0

flag[0] = true

while (flag[1]) {}

critical section

flag[0]=false

Thread 1

flag[1] = true

while (flag[0]) {}

critical section

flag[1]=false

Algorithm 1 fails dead-lock freedom: Concurrent execution can deadlock. If both threads write flag[0]=true and flag[1]=true

before reading (flag[0]) and (flag[1]) then both threads wait forever.

Algorithm 2

Thread 0

victim = 0;

while (victim == 0) ;{}

critical section

Thread 1

victim = 1;

while (victim == 1) ;{}

critical section

Mutual Exclusion

Algorithm 2 satisfies

mutual exclusion

Proof

Assume in the contrary that two threads can be in their critical section at the same time.

From the code we can see:

write0(victim=0) read0(victim==1) CS0

write1(victim=1) read1(victim==0) CS1

Thread 0

victim = 0;

while (victim == 0) ;{}

critical section

Thread 1

victim = 1;

while (victim == 1) ;{}

critical section

Proof

Since thread 1 must assign 1 to victim between the events write0(victim=0) and read0(victim==1), and since this assignment is the last, we get:

write0(victim=0) write1(victim=1) read0(victim==1)

Once victim is set to 1, it does not change, so every read will return 1, and this is a contradiction to the former equation:

write1(victim=1) read1(victim==0) CS1

Thread 0

victim = 0;

while (victim == 0) ;{}

critical section

Thread 1

victim = 1;

while (victim == 1) ;{}

critical section

Deadlock freedom

Algorithm 2 also fails deadlock freedom. It deadlocks if one thread runs completely before

the other.

Thread 0

victim = 0;

while (victim == 0) ;{}

critical section

Thread 1

victim = 1;

while (victim == 1) ;{}

critical section

Algorithms for Two Threads

We’ll describe two algorithms that solve the mutual exclusion problem for two Threads. They are also deadlock-free and starvation free.

Peterson’s Algorithm

Thread 0

flag[0] = true

victim = 0

while (flag[1] and victim == 0)

{skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and victim == 1)

{skip}

critical section

flag[1] = false

Peterson’s Algorithm

0/1 indicates that the thread is contending for the critical section by setting flag[0]/flag[1] to true.

victim shows who got last Then if the value of flag[i] is true then there is no contending

by other thread and the thread can start executing the critical section. Otherwise the first who writes to victim is also the first to get into the critical section

Thread 0

flag[0] = true

victim = 0

while (flag[1] and

victim == 0) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and

victim == 1) {skip}

critical section

flag[1] = false

Schematic for Peterson’s mutual exclusion algorithmSchematic for Peterson’s mutual exclusion algorithm

Indicate contendingflag[i] := true

Indicate contendingflag[i] := true

Barriervictim := i

Barriervictim := i

Contention?flag[j] = true ?Contention?

flag[j] = true ?

critical sectioncritical section

exit codeflag[i] = false

exit codeflag[i] = false

First to cross the barrier?victim = j ?

First to cross the barrier?victim = j ?yes

yes

no / maybe

no

The structure shows that thefirst thread to cross the barrier isthe one which gets to enter thecritical section. When there is nocontention a thread can enter thecritical section immediately.

Mutual Exclusion

Peterson’s algorithm

satisfies mutual exclusion

Proof

Assume in the contrary that two threads can be in their critical section at the same time.

From the code we see:

(*) write0(flag[0]=true) write0(victim=0) read0(flag[1]) read0(victim) CS0

write1(flag[1]=true) write1(victim=1) read1(flag[0]) read1(victim) CS1

Thread 0

flag[0] = true

victim = 0

while (flag[1] and

victim == 0) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and

victim == 1) {skip}

critical section

flag[1] = false

Proof

Assume that the last thread to write to victim was 0. Then:

write1(victim=1) write0(victim=0) This implies that thread 0 read that victim=0 in equation (*) Since thread 0 is in the critical section, it must have read

flag[1] as false, so:

write0(victim=0) read0(flag[1]==false)

Thread 0

flag[0] = true

victim = 0

while (flag[1] and

victim == 0) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and

victim == 1) {skip}

critical section

flag[1] = false

Proof

Then, we get:write1(flag[1]=true) write1(victim=1)

write0(victim=0) read0(flag[1]==false)

Thus:write1(flag[1]=true) read0(flag[1]==false)

There was no other write to flag[1] before the critical section execution and this yields a contradiction.

Thread 0

flag[0] = true

victim = 0

while (flag[1] and

victim == 0) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and

victim == 1) {skip}

critical section

flag[1] = false

Starvation freedom

Peterson’s algorithm

is starvation-free

Proof

Assume to the contrary that the algorithm is not starvation-free

Then one of the threads, say thread 0, is forced to remain in its entry code forever

Thread 0

flag[0] = true

victim = 0

while (flag[1] and

victim == 0) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and

victim == 1) {skip}

critical section

flag[1] = false

Proof

This implies that at some later point thread 1 will do one of the following three things:1. Stay in its remainder forever2. Stay in its entry code forever, not succeeding and

proceeding into its critical section3. Repeatedly enter and exit its critical section

Thread 0

flag[0] = true

victim = 0

while (flag[1] and

victim == 0) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and

victim == 1) {skip}

critical section

flag[1] = false

We’ll show that each of the three possible cases leads to a contradiction.

Proof

In the first case flag[1] is false, and hence thread 0 can proceed.

The second case is impossible since victim is either 0 or 1, and hence it always enables at least one of the threads to proceed.

In the third case, when thread 1 exit its critical section and tries to enter its critical section again, it will set victim to 1 and will never change it back to 0, enabling thread 0 to proceed.

Thread 0

flag[0] = true

victim = 0

while (flag[1] and

victim == 0) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

victim = 1

while (flag[0] and

victim == 1) {skip}

critical section

flag[1] = false

Kessels’ single-writer Algorithm

What if we replace the multi-writer register victim with two single-

writer registers. What is new algorithm?

Answer (Kessels’ Alg.)victim = 0 victim[0]

=victim[1]victim = 1 victim[0]

victim[1]

Kessels’ single-writer Algorithm

Thread 0

flag[0] = true

local[0] = victim[1]

victim[0] = local[0]

while (flag[1] and

local[0]=victim[1]) {skip}

critical section

flag[0] = false

Thread 1

flag[1] = true

local[1]=1-victim[0]

victim[1] = local[1]

while (flag[0] and

local[1] victim[0])) {skip}

critical section

flag[1] = false

Thread 0 can write the registers victim[0] and flag[0] and read the registers victim[1] and flag[1]Thread 1 can write the registers victim[1] and flag[1] and read the registers victim[0] and flag[0]

Solutions for Many Threads

How can we use a two-thread algorithm to construct an algorithm for many threads? How can we use a two-thread algorithm to construct an algorithm for many threads?

Tournament Algorithms

1 2 3 4 5 6 7 8

Tournament Algorithms

A simple method which enables the construction an algorithm for n threads from any given algorithm for two threads.

Each thread is progressing from the leaf to the root, where at each level of the tree it participates in a two thread mutual exclusion algorithm.

As a thread advanced towards the root, it plays the role of thread 0 when it arrives from the left subtree, or of thread 1 when it arrives from the right subtree.

The Filter Algorithm for n Threads

A direct generalization of Peterson’s algorithm to multiple threads.

The Peterson algorithm used a two-element boolean flag array to indicate whether a thread is interested in entering the critical section. The filter algorithm generalizes this idea with an N-element integer level array, where the value of level[i] indicates the latest level that thread i is interested in entering.

ncs

cslevel n-1

Filter

There are n-1 “waiting rooms” called levels At each level

– At least one enters level– At least one blocked if

many try

Only one thread makes it throughncs

cs

level 0

level n-1

The Filter Algorithm

Thread i

for (int L = 1; L < n; L++) { level[i] = L; victim[L] = i; while (( k != i level[k] >= L) and victim[L] == i ) {} }

critical section

level[i] = 0;

Thread i

for (int L = 1; L < n; L++) { level[i] = L; victim[L] = i; while (( k != i level[k] >= L) and victim[L] == i ) {} }

critical section

level[i] = 0;

Filter

One level at a time

Filter

Thread i

for (int L = 1; L < n; L++) { level[i] = L; victim[L] = i; while (( k != i level[k] >= L) and victim[L] == i ) {} }

critical section

level[i] = 0;

Announce intention to enter level L

Filter

Thread i

for (int L = 1; L < n; L++) { level[i] = L; victim[L] = i; while (( k != i level[k] >= L) and victim[L] == i ) {} }

critical section

level[i] = 0;

Give priority to anyone but me (at every level)

Filter

Thread i

for (int L = 1; L < n; L++) { level[i] = L; victim[L] = i; while (( k != i level[k] >= L) and victim[L] == i ) {} }

critical section

level[i] = 0;

Wait as long as someone else is at same or higher level, and I’m designated victim.

Thread enters level L when it completes the loop.

Claim

There are at most n-L threads enter level L Proof: by induction on L and by contradiction At L=0 – trivial Assume that there are at most n-L+1 threads at level

L-1. Assume that there are n-L+1 threads at level L Let A be the last thread to write victim[L] and B any

other thread at level L

Proof structure

ncs

cs

Assumed to enter L-1

By way of contradictionall enter L

n-L+1 = 4

n-L+1 = 4

A B

Last to writevictim[L]

Show that A must have seen B at level L and since victim[L] == Acould not have entered

Proof

From the code we get:

From the assumption:

writeB(level[B]=L)writeB(victim[L]=B)

writeA(victim[L]=A)readA(level[B])

writeB(victim[L]=B)writeA(victim[L]=A)

for (int L = 1; L < n; L++) { level[i] = L; victim[L] = i; while (( k != i level[k] >= L) and victim[L] == i ) {} }

critical section

level[i] = 0;

Proof

When combining all we get:

Since B is at level L, when A reads level[B], it reads a value greater than or equal L and so A couldn’t completed its loop and still waiting (remember that victim=A), a contradiction.

writeB(level[B]=L) readA(level[B])

for (int L = 1; L < n; L++) { level[i] = L; victim[L] = i; while (( k != i level[k] >= L) and victim[L] == i ) {} }

critical section

level[i] = 0;

A conclusion

The filter algorithm satisfies

mutual exclusion

At level n-1 there are at most n-(n-1)=1 threads, which means at most one thread in the critical section

Starvation-freedom

Filter Lock satisfies properties:– Just like Peterson algorithm at any level– So no one starves

Fairness

Starvation freedom guarantees that if a thread is trying to enter its critical section, it will eventually do so

There is no guarantee about how long it will take

We wish for fairness: if thread A enters the entry code before thread B, then A should enter the critical section first

Bounded waiting

We divide our method into two parts:Doorway interval:

- Written DA

- always finishes in finite steps

Waiting interval:- Written WA

- may take unbounded steps

entry code

exit code

criticalsection

remainder

doorway

waiting

The mutual exclusion problem

Mutual Exclusion

Deadlock-freedom

Starvation-freedom

FIFO

r-Bounded Waiting

For threads A and B:– If DA

k DB j

A’s k-th doorway precedes B’s j-th doorway

– Then CSAk CSB

j+r

A’s k-th critical section precedes B’s (j+r)-th critical section

B cannot overtake A by more than r times

First-come-first-served means r = 0.

Fairness in Filter Algorithm

Filter satisfies properties:– No one starves– But very weak fairness

Not r-bounded for any r!– That’s pretty lame…

Bakery Algorithm

The idea is similar to a line at the bakery A customer takes a number greater than

numbers of other customers Each of the threads gets a unique identifier

Bakery Algorithm

Thread i

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section flag[i] = false;

Bakery Algorithm

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1;

while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section flag[i] = false;

Doorway

Bakery Algorithm

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section flag[i] = false;

I’m interested

Bakery Algorithm

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section

flag[i] = false;

Take an number numbers are always increasing!

Bakery Algorithm

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section

flag[i] = false;

Someone is interested

Bakery Algorithm

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section

flag[i] = false;

There is someone with a lowernumber and identifier.

pair (a,b) > (c,d) if a>c, or a=c and b>d )lexicographic order(

Deadlock freedom

The bakery algorithm is deadlock free Some waiting thread A has a unique least

(number[A],A) pair, and that thread can enter the critical section

FIFO

The bakery algorithm is first-come-first-served

If DA DB then A’s number is earlier– writeA(number[A]) readB(number[A])

writeB(number[B]) readB(flag[A])

So B is locked out while flag[A] is true

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section

flag[i] = false;

Starvation freedom

The bakery algorithm satisfies deadlock freedom and first-come-first-served and those properties implies starvation freedom

Mutual Exclusion

Suppose A and B in CS together Suppose A has an earlier number When B entered, it must have seen

– flag[A] is false, or– number[A] > number[B]

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section

flag[i] = false;

Mutual Exclusion

numbers are strictly increasing so

B must have seen (flag[A] == false) numberingB readB(flag[A]) writeA(flag[A])

numberingA

Which contradicts the assumption that A has an earlier number

flag[i]=true; number[i] = max(number[0], …,number[n-1])+1; while (k!= i flag[k] && (number[i],i) > (number[k],k)) {}; critical section

flag[i] = false;

TheEnd