Process 0 turns red ; records n0 =0,| sent (0,1)|=1, and | received (2,0)|=0; and sends the marker along (0,1). 2. Process 1 receives the token and forwards it along (1,2) before receiving the marker . Then , it receives the marker ; turns red ; records n1=0,| received (0,1)|=1, and |sent(1,2)|=1 ; and sends the marker along (1,2). 3. Process 2 receives the token and forwards it along (2,0) before receiving the marker . Then , it receives the marker ; turns red ; records n2=0,| received (1,2)|=1, and | sent (2,0)|=1; and forwards the marker to process 0. 4. Process 0 receives the token and then receives the marker along (2,0). The algorithm terminates here .The total number of tokens recorded is as follows :( n0 + n1+n2)+(| sent (0,1)|- |received (0,1)|)+(| sent (1,2)|-| received (1,2)|)+(| sent (2,0)|-| received (2,0)|)=1. This is consistent with the expected outcome .再举至少两个运行场景的例子,并分析它们的特征,说明是否仍然是循环的;若有循环,循环的规律是什么等等,用中文回答
时间: 2024-02-03 14:03:05 浏览: 25
第一个运行场景的例子已经给出了,这里给出另外两个:
第二个运行场景的例子是:Process 0 turns red; records n0=0, |sent(0,1)|=1, and |received(2,0)|=0; and sends the marker along (0,1). 2. Process 1 receives the marker and turns red; records n1=0, |received(0,1)|=1, and |sent(1,2)|=0; and terminates. 3. Process 2 receives the marker and turns red; records n2=0, |received(1,2)|=0, and |sent(2,0)|=0; and sends the marker along (2,0). 4. Process 0 receives the marker and turns red; records n0=0, |received(2,0)|=1, and |sent(0,1)|=0; and terminates. In this scenario, the loop is broken because two processes terminate, leading to the termination of the algorithm. The characteristic of this scenario is that the algorithm terminates due to the termination of two processes.
第三个运行场景的例子是:Process 0 turns red; records n0=0, |sent(0,1)|=1, and |received(2,0)|=0; and sends the marker along (0,1). 2. Process 1 receives the marker and turns red; records n1=0, |received(0,1)|=1, and |sent(1,2)|=0; and sends the marker along (1,2). 3. Process 2 receives the marker and turns red; records n2=0, |received(1,2)|=1, and |sent(2,0)|=0; and terminates. In this scenario, the loop is broken because one process terminates, leading to the termination of the algorithm. The characteristic of this scenario is that the algorithm terminates due to the termination of one process.