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，说明是否仍然是循环的；若有循环，循环的规律是什么等等

Based on the provided text, the question seems to be asking for an analysis of the Chandy-Lamport snapshot algorithm in certain scenarios and to provide examples of such scenarios.

In the given scenario, machine i starts the snapshot before sending message M along ch1, and machine j receives the marker after sending out message M' along ch2. The resulting snapshot state is down ∅ up M', which seems strange since this state was never reached during the computation.

One possible explanation for this is that the snapshot algorithm is designed to capture a specific moment in time, regardless of whether or not that moment was reached during the computation. In other words, the algorithm is not limited to capturing only states that were actually reached during the computation, but rather any state that could have been reached at some point.

As for examples of running scenarios, here are two possible scenarios and their characteristics:

Scenario 1:
- Process 0 sends message M1 to Process 1
- Process 1 receives M1 and sends message M2 to Process 2
- Process 2 receives M2 and sends message M3 to Process 0
- Process 0 receives M3 and terminates

In this scenario, the algorithm should record a snapshot state consisting of all messages in transit, as well as the state of each process's local variables at the moment the snapshot was taken.

Scenario 2:
- Process 0 sends message M1 to Process 1
- Process 1 receives M1 and sends message M2 to Process 2
- Process 1 receives the marker and turns red
- Process 2 receives M2 and sends message M3 to Process 0
- Process 0 receives M3 and terminates

In this scenario, the algorithm should record a snapshot state consisting of all messages in transit, as well as the state of each process's local variables at the moment the marker was received by Process 1. This scenario demonstrates that the algorithm can handle the presence of markers and still capture a consistent snapshot state.