1. Max Registers, Counters, and Monotone Circuits James Aspnes, Yale University Hagit Attiya, Technion Keren Censor, Technion PODC 2009

2. Counting  Counting is critical for some programs in multiprocessing systems  Example: Algorithms for randomized consensus  Required: Counters with sub-linear (in the number of processes n) step complexity per operation PODC 2009 2 012

3. Counter Model:  System of n processes  Asynchronous system: no timing assumptions  Implement using shared Read/Write registers  Crash failures: require wait-free implementations Can be implemented using snapshots in linear time (in n) PODC 2009 3 Counter increment ok readCounter v +1

4. Related work  Lower bound of Ω (n) for time complexity by Jayanti, Tan, and Toueg [PODC 1996] and similar lower bounds by Ellen, Hendler, and Shavit [FOCS 2005]  Motivated work on approximate counting [Aspnes and C, SODA 2009] PODC 2009 4

5. Exact counting  Give up on sub-linear exact counting?  Or inspect lower bound more carefully:  Based on executions with many increments  But some applications use a small number of increments We show an implementation of a bounded counter where each operation takes sub-linear time PODC 2009 5 long operation

6. A tree-based counter s1s1 s1s1 s2s2 s2s2 s3s3 s3s3 snsn snsn s4s4 s4s4 … s 1 +s 2 s 3 +s 4 s 1 +...+s 4 ∑s i … … s n-1 +s n ReadCounter : return value at root Increment : recursively increment from leaf to root +1 p 1 increments update p k reads PODC 2009 6 O(log n) steps to increment O(1) steps to read counter

7. Seems nice, but…  If each node is a multi-writer register, then even for 2 processes and 2 increments this does not work s1s1 s1s1 s2s2 s2s2 s 1 +s 2 +1 p 1 increments PODC 2009 p 2 increments +1 update 1 update 2 Counter is incorrect 7

8. Max register  Replace multi-writer registers with Max Registers  In this case the tree-based counter works  If max registers are linearizable then so is counter PODC 2009 8 Max Register WriteMax( v ) ok ReadMax v Maximal value previously written

9. A tree-based counter s1s1 s1s1 s2s2 s2s2 s3s3 s3s3 snsn snsn s4s4 s4s4 … s 1 +s 2 s 3 +s 4 s 1 +...+s 4 ∑s i … … s n-1 +s n ReadCounter : return value at root Increment : recursively increment from leaf to root PODC 2009 9

10. Max register – recursive construction  MaxReg 0 : Max register that supports only the value 0  WriteMax is a no-op, and ReadMax returns 0  MaxReg 1 supports values in {0,1}  Built from two MaxReg 0 objects  and one additional multi-writer register “switch” PODC 2009 MaxReg 0 10 MaxReg 0 switch WriteMax 0 0 1 1 =1 ReadMax = ? switch=0 : return 0 switch=1 : return 1

11. Max register – recursive construction  MaxReg k supports values in {0,…,2 k -1}  Built from two MaxReg k-1 objects with values in {0,…,2 k-1 -1}  and one additional multi-writer register “switch” PODC 2009 MaxReg k-1 MaxReg k switch WriteMax t t t t < 2 k-1 ? t t t -2 k-1 ReadMax =1 = ? t t t t switch=0 : return t switch=1 : return t +2 k-1 11

12. MaxReg k unfolded PODC 2009 switch MaxReg 0 … switch MaxReg 0 … … Complexity does not depend on n: WriteMax and ReadMax in O(k) steps MaxReg k 12

13. A tree-based counter s1s1 s1s1 s2s2 s2s2 s3s3 s3s3 snsn snsn s4s4 s4s4 … s 1 +s 2 s 3 +s 4 s 1 +...+s 4 ∑s i … … s n-1 +s n ReadCounter : return value at root Increment : recursively increment from leaf to root PODC 2009 13 m-valued counter: ReadCounter : O(log m) steps Increment : O(log n log m) steps

14. Analysis  Inductive linearizability proof No contradiction with lower bound of JTT because of bounded size of max register and counter  Extension to unbounded max registers (and counters) with complexity according to value written or read  Both WriteMax and ReadMax of value v take O(min(log v, n)) steps PODC 2009 14

15. Lower bound of min(log m, n-1)  S m = {executions with WriteMax operations up to value m by p 1 …,p n-1, followed by one ReadMax operation by p n }  T(m,n) = worst case cost of ReadMax in S m PODC 2009 15 p n reads

16. Lower bound of min(log m, n-1)  No process takes steps after p n so p n does not write  Reads a fixed register R. Did anyone write to R?  k = minimal such that there is a write to R in S k  No one in S k-1 writes to R so T(m,n)≥T(k-1,n)+1 PODC 2009 16 p n reads R

17. Lower bound of min(log m, n-1)  In addition, consider a run in S k that writes to R PODC 2009 17 write to R by p i Finish writes except by p i Non-concurrent writes in {k,…,m} write to R by p i T(m,n) ≥ T(m-k+1,n-1)+1 p n returns maximal value from {k,…,m} p n reads Solve recurrence: T(m,n) ≥ 1+ min k {max(T(k-1,n), T(m-k+1,n-1))}, we had T(m,n)≥T(k-1,n)+1 R R p n reads

18. Summary  Implementation of max registers with O(min(log v, n)) steps per operation writing or reading the value v  Sub-linear implementation of counters  Extension of counters to any monotone circuit with monotone consistency instead of linearizability PODC 2009 18

19. Summary  Lower bounds  An alternative proof for JTT  Tight lower bound for max registers  Same lower bound proof for counters Further research: close gap between upper and lower bounds on counters  Randomized lower bound Further research: randomized algorithm? Take-home message: Lower bounds do not always have the final say PODC 2009 19

20. Thank you PODC 2009 20

21. Unbalanced tree PODC 2009 MaxReg 0 … switch MaxReg 0 21 Bentley and Yao [1976] switch MaxReg 0 Leaf i is at depth O(log i)

22. Unbounded max register PODC 2009 MaxReg 0 … switch Snapshot-based counter switch MaxReg 0 WriteMax and ReadMax of v in O(min(log v, n)) steps 22