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Fence Complexity in Concurrent Algorithms. Petr Kuznetsov TU Berlin/DT-Labs. STM is about ease-of-programming and efficiency. What is “efficient“ in a concurrent system?. Cost metrics. Space: used memory Cheap Advanced garbage-collection Time: - PowerPoint PPT Presentation
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Fence Complexity in Concurrent Algorithms
Petr KuznetsovTU Berlin/DT-Labs
STM is about ease-of-programmingand efficiency
What is “efficient“ in a concurrent system?
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Cost metrics
Space: used memoryCheapAdvanced garbage-collection
Time: the number of reads and writes (per operation)the number of stalls
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Relaxed memory modelsMemory is much slower than CPURead: check the cache -> read the memoryWrite: invalidate the caches -> update the memoryTo overcome “stalled writes” – reorder operations
Reordering may result in inconsistency
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What is inconsistency?
Process P:
Write(X,1)
Read(Y)
Process Q:
Write(Y,1)
Read(X)
P
QW(Y,1)
R(Y)W(X,1)
R(X)
W(X,1)
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Possible outcomes
P Q
P reads before Q writes
P reads after Q writes
Q reads after P writes
Q reads before P writes
Out-of-order
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Fixing out-of-order Memory fences: read-after-write (RAW)
write(X,1)
fence() // enforce the order
read(Y)
P
QW(Y,1)
R(Y)W(X,1)
R(X)
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Fixing out-of-order Atomic operations: atomic-write-after-read atomic{
read(Y)
…
write(X,1)
}E.g., CAS, TAS, Fetch&Add,…
RAW/AWAR fences take ~60 RMRs
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Our result
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Any concurrent program in a certain class must use RAW/AWARs
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What programs?
Concurrent data types:queues, counters, hash tables, trees,…Non-commutative operationsLinearizable solo-terminating implementations
Mutual exclusion
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Non-commutative operations
Operation A is non-commutative if there exists operation B where (applied to some state):
A influences Band
B influences A
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Example: Queue enq(v) – add v to the end of the queue deq() – dequeues the item at the head of the queue
Q=1;2
Q.deq():1;Q.deq():2 vs. Q.deq():2;Q.deq():1deq() influence each other
Q.enq(3):ok;Q.deq():1 vs. Q.deq():1;Q.enq(3):okenq() is commutative
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Proof sketch A non-commutative operation must write Suppose not
deq():1 deq():11;2
there must be a write!
w
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Proof sketch Let w be the first write Suppose there are no AWAR
deq():11;2
A(w) - the longest atomic construct containing w
w
w must be the first base-object event in A(w)!
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Proof sketch Suppose there are no RAWs
deq():11;2
No RAW - no difference for deq()!
deq():1
A(w)
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Mutual exclusionLock() – acquire the lockUnlock() – release the lock (Mutex) No two process holds the lock at the
same time (Deadlock-freedom) If at least one process
executes Lock() and no active process fails, at least one process acquires the lock
Two Lock() operations influence each other!
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Our result
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In any implementation of mutual exclusion or a concurrent data type with a non-
commutative operation op, a complete execution of op or lock() contains a
RAW or AWAR
Every successful lock acquire incurs a RAW/AWAR fence
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Why do we care?
Hardware design: what primitives must be optimized?
API design: returned values matterSet with add returning fail vs. returning ok
Verification – early catch of obviously incorrect algorithm
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What’s next? Weaker primitives?
Idempotent Work Stealing [Michael et al,PPoPP’09 ] Tight lower bounds?
How many RAW/AWAR fences are incurred? Other patterns
Read-after-readWrite-after-writeMulti-RAW:
write(Xi,1)
collect(X1,..,Xn)
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References H. Attiya, R. Guerraoui, D. Hendler, P. Kuznetsov,
M. Michael, M. VechevLaws of Order: Expensive Synchronization in Concurrent Algorithms Cannot be EliminatedIn POPL 2011
Srivatsan’s talk on STM fence complexity, TR on the way
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QUESTIONS?
