Peterson-2 互斥算法
经典 Peterson-2 算法 的无争用复杂度为 4（因为它执行 4 次读取） /对共享寄存器内存的写操作）是否有某种版本的 Peterson-2 算法，它需要较少的对共享寄存器内存的访问？ 显然1次访问是不可能的。但是2次或3次访问呢？ 谢谢
The contention-free complexity for classic Peterson-2 algorithm is 4 (because it performs 4 read/write operations to shared-registers memory)Is there some verion of Peterson-2 algorithm, which requires less accesses to shared-registers memory ?
It is obvious that 1 access is impossible.But what about 2 or 3 accesses?
Thank you
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每个临界区至少需要三个操作：进入时写入和读取（以声明获取互斥锁并验证其他进程尚未获取）、退出时写入（以释放互斥锁）。在进入时，Peterson 算法中的进程
id写入单写入器寄存器interested[id]和多写入器寄存器turn。以将有界寄存器转变为还保存无界版本号的寄存器为代价，对于两个进程，通过两个单写入器寄存器模拟多写入器寄存器，每次写入 1 次写入，每次读取 1 次读取，从而允许两者的合并写入了彼得森的算法。At least three operations per critical section are needed: a write and a read on entry (to declare acquisition of the mutex and verify that the other process has not acquired), a write on exit (to release the mutex). On entry, process
idin Peterson's algorithm writes the single-writer registerinterested[id]and the multi-writer registerturn. At the cost of turning a bounded register into one that also holds an unbounded version number, for two processes, there's a simulation of a multi-writer register by two single-writer registers that makes 1 write per write and 1 read per read, allowing the merger of the two writes in Peterson's algorithm.