#### Transcript Asynchronous Exclusive Selection

Asynchronous Exclusive Selection Bogdan Chlebus, U. Colorado Darek Kowalski, U. Liverpool Model and Problem 2 9 17 12 1 11 3 28 27 4 8 READ WRITE 5 14 7 15 6 13 1 processes registers Model Details PROCESSES: REGISTERS: • n processes • r registers • unique ids from {1,...,N} • unique ids from {1,...,r} • asynchronous • read/write • prone to crash failures • store O(log N) bits each PROBLEMS: Assigning exclusive integers to processes Plan Example 1: Renaming Example 2: Store&Collect Example 3: Unbounded Naming Example 1: Renaming Problem definition: • k ≤ n processes contend to acquire unique integers as new names in a smaller range {1,...,M} Complexity measures: – Max number of local steps per process – Length of the new name interval M – Number r of registers used Competing for Register Pair of registers is more useful than single ones Splitter(R): Compete-for-Register(R): • If one process contends • If one process contends then it wins R then it wins R • R can be won by at most • R can be won by at most one contending process one contending process • If there is more than one contending process then at least two different outputs are achieved . . . Main Technique 8 28 7 27 6 17 5 13 4 3 5 2 1 1 processes registers Main Technique: Majority Renaming (l,N)-Majority-Renaming (MR) Problem: at least half of l contending processes acquire new unique names in the interval {1,...,M}. We implement MR for any l, N, and for M=12e7l log(N/l) We need: • Bipartite graph G between set of processes and set of registers such that: – Input degree is 4 log(N/l) – A majority of contending processes have unique neighbours (i.e., no other contending process has it as a neighbour) From Majority Renaming to Renaming • Assume known k known N • Let Si, for i=0,1,...,lg k, be mutually disjoint sets of registers, s.t., |Si|=12e7(k/2i) log(2iN/k) • Solve (k/2i,N)-Majority-Renaming for i=0,1,...,lg k – For ith execution, use set Si of registers COMPLEXITY of algorithm Basic-Rename(k,N): • Local steps: O(log k log N) • M, r = O(k log(N/k)), i.e., M = 24e7k log(N/k) Cascade Renaming • Assume known k known N • Execute Basic-Rename(k,Nj), for j=0,1,...,j* – N0 := N – Nj := 24e7klog(Nj-1/k), for j>0, until Nj ≤ e14k COMPLEXITY of algorithm Cascade-Rename(k,N): • Local steps: O(log k (log N + log k loglog N)) • M ≤ e14k • r = O(k log(N/k)) Relaxing parameters k and N • Known k unknown N – Local steps: O(k) – M = 2k-1 – r = O(k2) • Unknown k known N – Local steps: O(log2k (log N + log k loglog N)) – M = O(k) – r = O(n log(N/n)) • Unknown k, N – Local steps: – M= – r= MA O(k) 8k – lg k -1 O(n2) AF AM O(k) O(k log k) O(k2) O(k) O(n2) O(n2) O(k2) 2k-1 O(n2) Example 2: Store&Collect • Problem definition: some arbitrary k processes repeatedly execute operations Store and Collect – Store: updates the value that the process wants to be collected by others – Collect: learns all the values proposed most recently by other processes, one value per process • Complexity Measures: – Max number of local steps per process – Number r of registers used From Renaming to Store&Collect • Organize registers into consecutive intervals of lengths 2,4,8,... • Associate a control register with every interval First Store operation (of process p): • Set control regs of intervals of size smaller than p into used • Acquire a new name i using renaming algorithm and deposit the value in the register with the name i located in the shortest interval of length not smaller than p Collect operation: • Collect all values from consecutive intervals until the first empty control register occurs Store&Collect results • N,k – known: – Store time: O(log k (log N + log k loglog N)) – Collect time: O(k) – Number of registers: O(k log(N/k)) ... • N,k – unknown: – Store/Collect time: O(k) – Number of registers: O(n2) (different approach: Afek, De Levie, DC 2007) Lower Bounds Any wait-free solution of Renaming requires 1 + min{k-1, log2r (N/2M)} local steps in the worst case. Space-efficient solution: O(k) registers Any wait-free space-efficient solution of Store operation in Store&Collect requires (min{k, logr (N/k)}) local steps in the worst case. Example 3: Unbounded Naming Problem: • Infinite number of registers dedicated to depositing • Fairly distributed deposit requests (each process eventually receives a new value to deposit) • Deposit operation must be acknowledged Measure: • Number of registers never used for depositing Naive solution (infinite number of unused registers): • Process p deposits in consecutive registers congruent to p modulo N Repository Repository: concurrent data structure, s.t., each process can deposit consecutive values in dedicated registers to guarantee: Persistence – for any register R dedicated for depositing, after an ack(R) event, no value is ever written to R Non-blocking – each time at least one non-faulty process wants to deposit a value, then eventually the value gets deposited Implementing a Repository Minimizing the number of unused registers (n-1) • Each process p maintains (locally): – list Lp of 2n-1 register numbers (available for deposits) – next possibly empty register Ap • Atomic snapshot object of n SWMR registers • Verification procedure: process p verifies list Lp by reading consecutive registers, removing non-empty ones and searching for a new one (starting from Ap) • Choosing by rank: a process of rank k among other acquiring processes selects kth register from its list and verifies, using snapshot, if it is unique Implementing a Repository cont. Wait free implementation with n(n-1) deposit-free registers • Array Help[1..n,1..n] of shared registers • Two parallel threads intertwined: – Verification: process p keeps reading Help[p,*] in a cyclic fashion, every Help[p,q] = null is replaced by new name obtained from the previous algorithm – Depositing: process p keeps reading Help[*,p] in a cyclic fashion, until finding q s.t. Help[q,p] = x ≠ null, deposits in Rx and writes null to Help[q,p] Conclusion • We presented a variety of selection techniques and their applications • There is enough evidence that new methods and applications are needed • How to reduce the number of registers: – Quadratic number of registers is used in most of the presented applications