1. DISTRIBUTED ALGORITHMS AND SYSTEMS Spring 2014 Prof. Jennifer Welch CSCE 668 1

2. Shared Memory Model  Processors communicate via a set of shared variables, instead of passing messages.  Each shared variable has a type, defining a set of operations that can be performed atomically. 2

3. Shared Memory Model Example 3 p0p0 p1p1 p2p2 X Y readwrite read

4. Shared Memory Model  Changes to the model from the message-passing case:  no inbuf and outbuf state components  configuration includes a value for each shared variable  only event type is a computation step by a processor  An execution is admissible if every processor takes an infinite number of steps 4

5. Computation Step in Shared Memory Model  When processor p i takes a step:  pi 's state in old configuration specifies which shared variable is to be accessed and with which operation  operation is done: shared variable's value in the new configuration changes according to the operation's semantics  p i 's state in new configuration changes according to its old state and the result of the operation 5

6. Observations on SM Model  Accesses to the shared variables are modeled as occurring instantaneously (atomically) during a computation step, one access per step  Definition of admissible execution implies  asynchronous  no failures 6

7. Mutual Exclusion (Mutex) Problem  Each processor's code is divided into four sections:  entry: synchronize with others to ensure mutually exclusive access to the …  critical: use some resource; when done, enter the…  exit: clean up; when done, enter the…  remainder: not interested in using the resource 7 entrycriticalexitremainder

8. Mutual Exclusion Algorithms  A mutual exclusion algorithm specifies code for entry and exit sections to ensure:  mutual exclusion: at most one processor is in its critical section at any time, and  some kind of "liveness" or "progress" condition. There are three commonly considered ones… 8

9. Mutex Progress Conditions  no deadlock: if a processor is in its entry section at some time, then later some processor is in its critical section  no lockout: if a processor is in its entry section at some time, then later the same processor is in its critical section  bounded waiting: no lockout + while a processor is in its entry section, other processors enter the critical section no more than a certain number of times.  These conditions are increasingly strong. 9

10. Mutual Exclusion Algorithms  The code for the entry and exit sections is allowed to assume that  no processor stays in its critical section forever  shared variables used in the entry and exit sections are not accessed during the critical and remainder sections 10

11. Complexity Measure for Mutex  An important complexity measure for shared memory mutex algorithms is amount of shared space needed.  Space complexity is affected by:  how powerful is the type of the shared variables  how strong is the progress property to be satisfied (no deadlock vs. no lockout vs. bounded waiting) 11

12. Test-and-Set Shared Variable  A test-and-set variable V holds two values, 0 or 1, and supports two (atomic) operations:  test&set(V): temp := V V := 1 return temp  reset(V): V := 0 12

13. Mutex Algorithm Using Test&Set  code for entry section: repeat t := test&set(V) until (t = 0) An alternative syntactic construction is: wait until test&set(V) = 0  code for exit section: reset(V) 13

14. Mutual Exclusion is Ensured  Suppose not. Consider first violation, when some p i enters CS but another p j is already in CS 14 p j enters CS: sees V = 0, sets V to 1 p i enters CS: sees V = 0, sets V to 1 no node leaves CS so V stays 1 impossible!

15. No Deadlock  Claim: V = 0 iff no processor is in CS.  Proof is by induction on events in execution, and relies on fact that mutual exclusion holds.  Suppose there is a time after which a processor p is in its entry section but no processor ever enters CS. 15 p is in entry but no processor enters CS p is still in entry, no processor is in CS V always equals 0, next t&s by p returns 0 p enters CS, contradiction!

16. What About No Lockout?  One processor could always grab V (i.e., win the test&set competition) and starve the others.  No Lockout does not hold.  Thus Bounded Waiting does not hold. 16

17. Read-Modify-Write Shared Variable  The state of this kind of variable can be anything and of any size.  Variable V supports the (atomic) operation  rmw(V,f ), where f is any function temp := V V := f(V) return temp  This variable type is so strong there is no point in having multiple variables (from a theoretical perspective). 17

18. Mutex Algorithm Using RMW  Conceptually, the list of waiting processors is stored in a shared circular queue of length n  Each waiting processor remembers in its local state its location in the queue (instead of keeping this info in the shared variable)  Shared RMW variable V keeps track of active part of the queue with first and last pointers, which are indices into the queue (between 0 and n - 1)  so V has two components, first and last 18

19. Conceptual Data Structure 19 The RMW shared object just contains these two "pointers" 12 3 4 5 6 7 8 910 11 12 13 14 15 0 first last

20. Mutex Algorithm Using RMW  Code for entry section: // increment last to enqueue self position := rmw(V,(V.first,V.last+1)) // wait until first equals this value repeat queue := rmw(V,V) until (queue.first = position.last)  Code for exit section: // increment first to dequeue self rmw(V,(V.first+1,V.last)) 20

21. Correctness Sketch  Mutual Exclusion:  Only the processor at the head of the queue (V.first) can enter the CS, and only one processor is at the head at any time.  n-Bounded Waiting:  FIFO order of enqueueing, and fact that no processor stays in CS forever, give this result. 21

22. Space Complexity  The shared RMW variable V has two components in its state, first and last.  Both are integers that take on values from 0 to n - 1, n different values.  The total number of different states of V thus is n 2.  And thus the required size of V in bits is 2*log 2 n. 22

23. Spinning  A drawback of the RMW queue algorithm is that processors in entry section repeatedly access the same shared variable  called spinning  Having multiple processors spinning on the same shared variable can be very time-inefficient in certain multiprocessor architectures  Alter the queue algorithm so that each waiting processor spins on a different shared variable 23

24. RMW Mutex Algorithm With Separate Spinning Shared RMW variables:  Last : corresponds to last "pointer" from previous algorithm  cycles through 0 to n - 1  keeps track of index to be given to the next processor that starts waiting  initially 0 24

25. RMW Mutex Algorithm With Separate Spinning Shared RMW variables (continued):  Flags[0..n-1] : array of binary variables  these are the variables that processors spin on  make sure no two processors spin on the same variable at the same time  initially Flags[0] = 1 (proc "has lock") and Flags[i] = 0 (proc "must wait") for i > 0 25

26. Overview of Algorithm  entry section:  get next index from Last and store in a local variable myPlace increment Last (with wrap-around)  spin on Flags[myPlace] until it equals 1 (means proc "has lock" and can enter CS)  set Flags[myPlace] to 0 ("doesn't have lock")  exit section:  set Flags[myPlace+1] to 1 (i.e., give the priority to the next proc) use modular arithmetic to wrap around 26

27. Question  Do the shared variables Last and Flags have to be RMW variables?  Answer: The RMW semantics (atomically reading and updating a variable) are needed for Last, to make sure two processors don't get the same index at overlapping times. 27

28. Invariants of the Algorithm 1. At most one element of Flags has value 1 ("has lock") 2. If no element of Flags has value 1, then some processor is in the CS. 3. If Flags[k] = 1, then exactly (Last - k) mod n processors are in the entry section, spinning on Flags[i], for i = k, (k+1) mod n, …, (Last-1) mod n. 28

29. Example of Invariant 29 00100000 0 1 2 3 4 5 6 7 Flags 5 Last k = 2 and Last = 5. So 5 - 2 = 3 procs are in entry, spinning on Flags[2], Flags[3], Flags[4]

30. Correctness  Those three invariants can be used to prove:  Mutual exclusion is satisfied  n-Bounded Waiting is satisfied. 30

31. Lower Bound on Number of Memory States Theorem (4.4): Any mutex algorithm with k-bounded waiting (and no-deadlock) uses at least n states of shared memory. Proof: Assume in contradiction there is an algorithm using less than n states of shared memory. 31

32. Lower Bound on Number of Memory States  Consider this execution of the algorithm:  There exist i and j such that C i and C j have the same state of shared memory. 32 p 0 p 0 p 0 …p1p1 p2p2 p n-1 CC0C0 C2C2 C n-1 C1C1 …… p 0 in CS by ND p 1 in entry sec. p 2 in entry sec. p n-1 in entry sec. initial config., all in rem.

33. Lower Bound on Number of Memory States Shared memory state is same in C i as in C j 33 CiCi CjCj p 0 in CS, p 1 -p i in entry, rest in rem. p 0 in CS, p 1 -p j in entry, rest in rem. p i+1, p i+2, …, p j  = sched. in which p 0 -p i take steps in round robin by ND, some p h has entered CS k+1 times  p h enters CS k+1 times while p i+1 is in entry

34. Lower Bound on Number of Memory States  But why does p h do the same thing when executing the sequence of steps in  when starting from C j as when starting from C i ?  All the processors p 0,…,p i do the same thing because:  they are in same states in the two configs  shared memory state is same in the two configs  only differences between C i and C j are (potentially) the states of p i+1, …,p j and those processors don't take any steps in  34