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，并要分析它们的特征，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: 2024-03-07 14:51:14 浏览: 11
第一个运行场景的例子是：Process 0 turns red; records n0=0 and sends the token along (0,1). Process 1 receives the token and turns red; records n1=0 and sends the token along (1,2). Process 2 receives the token and turns red; records n2=0 and sends the marker along (2,1). Process 1 receives the marker and turns red; records n1=0 and sends the marker along (1,0). Process 0 receives the marker and turns red; records n0=0 and sends the marker along (0,2). Process 2 receives the marker and turns red; records n2=0 and sends the token along (2,0). The algorithm then repeats from the beginning. 这个例子是一个循环，但是它的规律是进程2会在每个周期结束时将标记传递给进程0，而其他进程则按照正常的顺序传递标记。 第二个运行场景的例子是：Process 0 turns red; records n0=0 and sends the token along (0,1). Process 1 receives the token and turns red; records n1=0 and sends the token along (1,2). Process 2 receives the token and turns red; records n2=0 and sends the marker along (2,0). Process 0 receives the marker and turns red; records n0=0 and sends the marker along (0,1). Process 1 receives the marker and turns red; records n1=0 and sends the token along (1,0). Process 0 receives the token and turns red; records n0=1 and sends the marker along (0,2). Process 2 receives the marker and turns red; records n2=0 and sends the token along (2,1). Process 1 receives the token and turns red; records n1=1 and sends the marker along (1,2). The algorithm then repeats from the beginning. 这个例子也是一个循环，但是它的规律是进程0和进程1会交替地记录一个标记，而进程2则在它们之间传递标记。

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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

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1. Process O turns red ; records no =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 t ,2= l ; andsends 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 O receives the token and then receives the marker along (2,0). The algorithm terminates here .The total number of tokens recorded is as follows :( no + n ]+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 .. Let machine i start Chandy-Lamport snapshot before it has sent M along ch1. Also, let machine j receive the marker after it sends out M’ along ch2. Observe that the snapshot state is down ∅ up M’ Doesn’t this appear strange? This state was never reached during the computation!再举至少两个运行场景的例子，并分析它们的特征，比如是否仍然是循环的；若有循环，循环的规律是什么等等

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This code sets up interrupts for two buttons (SW1 and SW2) on GPIOF, and turns on/off LEDs based on the button presses. The main function continuously checks for button presses using a while loop, and...

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