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 ≥ 2ⁿ. But a proof of this conjecture remains elusive. For the general case, we prove a lower bound of m ≥ n². 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 ≥ 2ⁿ. 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.

许多共享内存算法必须处理确定共享对象的值是否在进程的两次连续访问之间更改的问题，当来自两者的响应相同时。受这个问题的启发，我们定义了信号检测问题，可以在纯粹的组合水平上进行研究。考虑一个有n + 1 个进程的系统，它由n 个读取器和一个信号器组成。进程通过共享黑板进行通信，该黑板可以存储大小为m的域中的值. 进程由对手安排。调度时，进程读取黑板，任意修改其内容，并且如果它是读取器，则返回一个布尔值。如果信号者自阅读者的前一步之后采取了一步，则阅读者必须返回真；否则它必须返回false。直观地说，在一个有n 个进程的系统中，信号检测至少需要n位的共享信息，即m ≥ 2 ⁿ。但这一猜想的证明仍然难以捉摸。对于一般情况，我们证明m ≥ n ²的下界. 对于问题的受限版本，其中进程是无意识的，或者信号器必须写入固定的值序列，我们证明了m ≥ 2 ⁿ的严格下限。我们还考虑了每个读者最多两个步骤的问题版本。在这种情况下，我们证明m = n + 1 黑板值是必要且充分的。