Many shared memory algorithms have to deal with the problem of determining whether the value of a shared object has changed in between two successive accesses of that object by a process when the responses from both are the same. Motivated by this problem, we define the signal detection problem, which can be studied on a purely combinatorial level. Consider a system with n + 1 processes consisting of n readers and one signaller. The processes communicate through a shared blackboard that can store a value from a domain of size m. Processes are scheduled by an adversary. When scheduled, a process reads the blackboard, modifies its contents arbitrarily, and, provided it is a reader, returns a Boolean value. A reader must return true if the signaller has taken a step since the reader's preceding step; otherwise it must return false. Intuitively, in a system with n processes, signal detection should require at least n bits of shared information, i.e., m ≥ 2 n . But a proof of this conjecture remains elusive. For the general case, we prove a lower bound of m ≥ n 2. For restricted versions of the problem, where the processes are oblivious or where the signaller must write a fixed sequence of values, we prove a tight lower bound of m ≥ 2 n . We also consider a version of the problem where each reader takes at most two steps. In this case, we prove that m = n + 1 blackboard values are necessary and sufficient.
Keywords: ABA problem; Lower bounds; Signal detection; Space complexity.
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