摘要：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. We prove a lower bound of m >= n^2, as well as a tight lower bound of m >= 2^n for two restricted versions of the problem, where the processes are oblivious or where the signaller always resets the blackboard to the same fixed value. We also consider a one-shot version of the problem, where each reader takes at most two steps. In this case, we prove that it is necessary and sufficient that the blackboard can store m=n+1 values.
关键词：Signal detection; ABA problem; space complexity; lower bound
