Old and new questions about prisoners

From http://www.iqstar.net

Source: http://www.oursci.org/bbs/oursci/showthread.php? S = & threadid = 2494.

There is an old question:

There were 100 prisoners in one cell. There was a lamp in the wind, and a switch was used to control the light. Pull the light and turn it off.

One day, the Director gathered the prisoners and announced that from now on, whenever I was happy, I would choose a prisoner to let the wind go. If at some time, one of you would report to me, all the prisoners have been picked up by me, and you are all released. However, if the reported person reports an error, all of you will be shot.

The prison director added that the lights and switches in the Windy place were not moved by any other prisoner, and they were of good quality. In addition, there is no guarantee. Selecting a suitable prisoner is not necessarily equal to everyone's opportunities, but he can ensure that for any prisoner, for any large natural number N, as long as the time is long enough, he will be released n times.

Before returning to their cell, prisoners may sometimes get together to discuss a plan that is absolutely reliable and can be released. They knew that once they returned to the cell, there would be no contact information except the lamp. They could not see or hear any other prisoner (and of course they could not see the Prison Director), nor could they see the place of the wind in the cell. The worst thing is that they don't even know whether the light is on or off. In addition, the Director is moody. Sometimes he may be happy for 100 times a day, and sometimes he may be unhappy for several days. The new Director is also very smart. He is very good at observation. If you want to make other marks in a place where the wind shines, for example, put something, draw something, and so on, it is likely that he will secretly put something similar or draw something, or simply clean it up there. So it is best not to be smart in addition to the lamp.

If you are one of the prisoners, how can you provide an absolutely reliable solution when everyone is together?

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The following question is from me (here, it refers to Mr. Sany's boss:

If the new director says that, after everyone goes back to his prison, he will pick a person and release it, let the remaining 99 people play the game (the remaining 99 people do not know who has been picked); and he told everyone that the light is off now. If you think there is no absolutely reliable solution, prove this point of view. If you think there is still an absolutely reliable solution, propose it.

The answer below is provided by path2math of the three ideas forum, which is prepared by a cold meal.

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1) First, define the state of a prisoner when the wind is released: if the light is in the same status as when the last wind is over (both the light and the light are on ), the prisoner's status is "normal". (For example, if the light is on when the prisoner blew the wind and the light is off, the prisoner will see that the light is still off, if the light is on, It is abnormal. Otherwise, it is an "exception ". If the prisoner is in the first windfall state because there is no previous light, the first windfall state of the prisoner is always "abnormal" by definition (which can also be defined as part of the prisoner's state ). There are three possible actions for prisoners to perform each exposure: to change the light status by clicking the button, to do nothing, or to report that everyone has come out ("report" for short ").

 

(2)-5) Normalize the possible policies of prisoners.

(2) the prisoner has no information except the light state that he sees when the wind blows and the information about his operations when the wind blows, therefore, the actions required by the prisoner in the policy for the nth release (press the light, do not press, or report) are only dependent on the status of the lamp that was previously seen n-1 times and his operations. If, in the prisoner's strategy, even if the light is "normal" during the nth blow, the prisoner still needs to click (or report ), then, the director can immediately ask the prisoner to release the prisoner again after N-1 occurrence. In this way, the prisoner's policy is equivalent to the act opposite to his original act at the n-1 th. If the first n-th policy is to press the light, otherwise, press (or n-1 time, if the original policy is to report for the nth time ). In this way, we prove that if there is a policy that allows prisoners to be free, there must be a strategy that does not require "Although the lights are 'normal', but still need to press the lights or report, prisoners can also be given freedom.

3) Likewise, the prisoner's strategy does not need to distinguish the number of consecutive normal operations. In the course of wind exposure, if the prisoner's policy is to detect exceptions during the nth wind release, he will do some operations (by, not by or report) -- For any k> = 0, if the prisoner is normal from the nth wind to the nth + k wind, and the next N + k + 1 is abnormal, then, he should perform the same operations when the nth + k + 1 Wind is released and when the nth wind is released, the exception is discovered. Because in the same logic as II), the prison director can allow the prisoner to release k + 1 consecutive times after the nth stroke of N-1 prisoner, this makes it useless for the prisoner's policy of "Discovering exceptions during the nth release.

4) according to the above discussion, the prisoner's policy description can be roughly described as follows: "1st exceptions, some operations; 2nd exceptions, perform an operation; perform an operation for 3rd exceptions ;......", There is no need to consider whether a certain risk is normal. First, follow the instructions in step 2. The normal state must not do anything, and then follow the instructions in step 3. It is not important to know how many normal exceptions occur between the two.

5) there is an inaccuracy in the statement in section 4, that is, there may be two situations in the case of 1st exceptions (that is, when each person is brought out for the first time, that is, the light is on or the light is off. In the future, the status of the abnormal lights will be determined by 1st exceptions: if the light is turned on for 1st exceptions, the light is turned off for 2nd times, the light is turned on for 3rd Times, and the light is turned off for 4th times ......; If the light is off for 1st exceptions, the opposite is true. So correctly, each prisoner has two sets of strategies:
Initial Opening Strategy: One set is "If 1st lights are turned on, perform some operations; 2nd exceptions, perform some operations; 3rd exceptions, perform some operations ;......", This policy is called S1;
Initial elimination policy: the other set is "If 1st lights are off, perform an operation; 2nd exceptions, perform an operation; 3rd exceptions, and perform an operation; ......", This is an S2 policy.
Of course, in actual use, each prisoner only uses one of them, which is determined by the light state seen by the prisoner during the first windfall.

 

The next step is the core of path2math proof.

6) for each prisoner, in his S1, the actions taken against the first exception are switched, the exception A + 1 is not (not pressed or reported); in S2, the action taken for the first B is switched, the exception in Section B + 1 is not. Therefore, the number of S1 of the prisoner is a and the number of S2 is B. A and B can be 0, any natural number greater than 0, or infinity. It determines the number of S1 and S2 of a prisoner. In fact, it only determines the part of his strategy (the previous part). However, we will see that this part of strategy already contains all the difficulties. That is to say, either the prison director does not need to consider anything other than this part of the strategy to make the prisoners fail, or only this part of the strategy can make the prisoners succeed.

7) A simple, interesting and important conclusion is that when a prisoner adopts the S1 strategy (that is to say, the light is on when he first releases him ), after an exception occurs for a consecutive time (a is his S1 number), if the next exception occurs, he does not decide to report it (if not all of them have been released, he will not decide to report it ), he will have a "no-sensitive moment", that is, the next time when the wind goes off, no matter whether the light is on or off, he will do nothing, and determines that the light is normal at this time. The S2 policy has a similar conclusion.

8) each prisoner belongs to and belongs to only one of the following four situations:
A) S1> 0, S2 = 0;
B) S1 = 0, S2> 0;
C) S1> 0, S2> 0;
D) S1 = 0, S2 = 0;

It is impossible for all prisoners to have S2 = 0: In that case, no one will open the light when the first wind is put on, and everyone will think that the light is normal after the first wind is put on, so no one will turn on the light. It is impossible for all prisoners to have S1 = 0: according to the previous statement, there must be a person S2> 0. The Director first asks this person to open the light. This person will think that the light is normal, none of the others will turn off the lights when the first wind is set, and after the first wind is set, all people will think that the lights are normal, so no one will turn off the lights.

We want to prove that we can divide a) B) c) three prisoners into Groups A and B so that the following conditions are met: (we will see later, if the prisoners want a successful strategy, d) the number of prisoners in the case must be 0)
1) All prisoners meeting a) are in;
2) All prisoners meeting B) are in B;
3) make AA = {A1, A2, A3 ,......} This is a set of S1 numbers for each prisoner in group A. BB = {B1, B2, B3 ,......} This is a set of S2 numbers for each prisoner in Group B (note that these AI and Bi are strictly greater than 0). | A | indicates the number of prisoners in group, | B | Number of people in group B, number of members in HA (AA) and number of members in HA (which may be infinite), and number of members in Hb (which may be infinite ). We have ha> = | B | and Hb> = | A |.

9) Suppose 8) the Group in can't be true, but because ha >=| A |, Hb >=| B |, SO 3) The two inequalities in cannot all be true. Let's assume that there is at least one group so that "Ha> = | B |" is true ("at least one group exists, so that Hb> = | A |" is similar ). In this way, we can take a group so that "Ha> = | B |" is true, and HB-| A | is the largest (but it is assumed that it will be smaller than 0 ), take this group now.

"HB-| A | <0" means to use steps such as "bring one person from B and bring one person from A to turn off the light, each person in B is taken out of the corresponding Bi times (Bi is the number of S2 of that person), and a different person is selected from each time in, eventually there are still prisoners that have never been taken out of. At this time, all the prisoners in Group B entered the "insensitive moment" (see 7), and all the prisoners in group A who had been taken out considered that the light was normal. If all of the people in a who have not been brought out are S2 = 0, all (including D) prisoners will not do anything in the next wind, and think that the light is normal, so no one will turn on the light. So among all the people in a that have not been taken out, there must be a person's S2> 0 (that is, it belongs to C ).

Consider a special situation: if all the people in group A enter the "insensitive time" (that is to say, all their S1 numbers are 1 and they are taken out once, (7). Then, take this person out once (he will go to S2 policy and turn on the light). Then he will think that the light is on normally, while other people (including D) will think that the light is on normally) they are all in the "no-sensitive moment". When anyone comes out of the storm, they will be in a normal state, which will be in an infinite loop.

So there must be another person in group A whose S1 number is "not used up" (or he hasn't been out yet, or his S1 is greater than 1 ). Transfer the S2> 0 person in group A to Group B. We find that HA> = | B | is not affected, because the number of people in group A that have been released earlier is equal to the number of people in group B. Even if group B has a new member, in addition, the number of S1 s in the group A described above is enough to offset the number of S1 s in use. | B | increase by 1. But now | A | 1 is missing, and HB increases the S2 number of the person transferred from Group A, which is in conflict with the previous "HB-| A | largest. This means that we assume that the Group in "8) cannot establish a conflict ". Therefore, we can certainly find a group that meets the conditions of (8) 1) 2) 3.

 

Now, set the group A and group B that meet the conditions of (8) medium, (1), (2), and (3.

10) If both ha and HB are infinite (that is to say, there is an infinite number of S1 for one person m in a, and the number of S2 for one person n in B is infinite), then bring n first and then m, in this case, the light is off, and N will use its S2 policy in the future, and m will use its S1 policy in the future. Once again, all prisoners are taken out (no one will report it for the first time). Assuming that the light is off (the reasoning behind the light is similar), then the following cycle is followed: all prisoners who think that the light is normal, n all those who think that the light is normal, M ,....... No one will report this infinite process.

11) so we can only assume that at least one of HA and Hb is limited. Assuming that Hb is limited and HA> = HB (HA is limited and Hb> = ha is similar in case of reasoning, but there is an asymmetry in the conclusions of 12), we will discuss it in section 13 ).

Ha> = HB >=| A |. This means that, by switching between B and A in a similar way as in 9), the governor can achieve the number of times that all B's people are taken out of their S2, all members of a are taken out at least once, and the light is off. We note that if d) the number of people in the case is not equal to 0, then according to the conclusion in 7), all people in B must be in the "no-sensitive moment". At this time, anyone (including D) when the light is put out, everything will not work and the light is regarded as normal, so as to go into an infinite loop.

So d) the number of people in the case must be 0, and there must be one person in B. The next time an exception occurs, a report "All people have released" will be made. This person is called "S2 privileged person ". According to the definition, the number of such S2 privileged characters may not have only one, but we will prove that he can only have one. (Similarly, if we assume that HA is limited and Hb> = ha, we will find that there must be an "S1 privileged character" in group ").

Assume that the S2 number of a S2 privileged character X is B, and the HA> = Hb> = B According to the Hb definition, if there are other people in group B who are different from X (their S2 count is greater than 0, so Hb> B), as long as the Director deducts this person, only the prisoners in group X and group A are put out in turn. So, Because ha> B can cause group B to encounter B + 1 exception when group B is not released, therefore, reporting is impossible. So there can only be one prisoner in group B, whose S2 number is equal to HB.

This proves that under the assumption that there was only one S2 privileged character TB at the beginning of the 11 S, and group B was composed of him.

12) in addition to the S2 privileged character TB, assume that the number of S2 of prisoner y in a is greater than 0. Now we hold y and take the TB and other people in group A in turn, so that all people in group A except group y enter the "insensitive time" (because y is deducted, so no one will report it next time.) At this time, the light is off, and TB considers the light to be on normally. If y is taken out now, he will turn on the light (because of its S2> 0). Now no one will turn off the light and enter the infinite loop.

Therefore, apart from the unique S2 privileged figures, the S2 numbers of all other prisoners are 0.

13) Now let's assume that HA is limited and Hb> = ha. We can also prove that a group is composed of only one S1 vip ta.

In addition to the S1 privilege Ta, the S1 number of a prisoner Z in B is greater than 0. Now we hold down Z and take the rest of group B and TA out in turn, so that all others except Z in Group B enter the "insensitive time" (Because Z is deducted, so no one will report it again), but in the last "Bring B group members to turn on the light, and then bring TA to turn off the light" process, do not bring ta out, but replace it with Z. In this case, both TA and Z think that the light is normal. Now no one turns on the light and enters an infinite loop.

Note that the above actions to replace TA with Z are valid only when TA was indeed taken out once before (in this case, we can assert that "ta considers the light to be normal "). If the sum of the S2 numbers of all people except Z in Group B is 1, this argument cannot be established. This happens in Group B where only Z and another prisoner W are involved, in addition, the S2 number of w is 1, which is the case of the three people mentioned in the 20100 post. If the total number of prisoners is greater than or equal to 4, more than two persons are in group B. This argument is always valid.

Therefore, when the total number of prisoners is greater than or equal to 4, the S1 count of all other prisoners is 0 except for the unique S1 privilege.

 

Now let's look at the situation where the Director will pick a person to release.

14) assume that the number of prisoners is greater than 4. After a prisoner is released, if the policy of the remaining prisoner gives them freedom, 11)-13) The conclusion must be true. Note 11) that, prisoners whose S1 and S2 are both 0 should not exist.

However, the director can calculate the S1 and S2 numbers for each person. If the S1 count of only one prisoner is not 0, the S2 count of other prisoners cannot be 0. Otherwise, "S1 and S2 count are both 0, so the number of prisoners whose S1 number is not 0 is released. If the number of S2 of only one prisoner is not 0, the number of S1 of other prisoners cannot be 0, so the prisoner whose S2 number is not 0 is released. In this way, no matter how many prisoners are divided into group A and group B, their strategies cannot meet the conclusions in group 11)-13.

 

Answers to new questions by path2math

Proof of New Problems

First, for each prisoner, define two states of the lamp: "normal" and "abnormal ". That is to say, if the light was on before the last time the wind came back, it would be "abnormal" and "normal" if the light was turned off when it was called out for the wind ". If the light is turned off before the last time the wind came back, it turns on as abnormal and turned off as normal.
In this way, prisoners can be divided into two categories: group A is considered abnormal when the light is turned on, and group B is considered abnormal when the light is turned off. Since the prisoners knew from the very beginning that the light was originally turned off, all of them were in group.
Next we will discuss what a strategy should be like if there is a policy to prevent prisoners from being put to death or permanently shut down.
We can think that the prison director knows the prisoner's plan and will try to prevent them from being released.
Assume that the prison director takes such a behavior: every time he decides to bring a person out of the air, he will not stop bringing it out until he does nothing. In the prisoner's policy, we can eliminate this situation: even if the light is "normal", the switch will still be clicked from time to time. For example, if a prisoner uses this policy in some circumstances and someone has not been taken out of the system, the prison director can use this policy to control the switch of the light at will, let other prisoners mistakenly think that everyone has been taken out, and if the prisoner takes this policy, others will have been taken out, then he can directly ask the prison director for release without using this policy. Therefore, when the possibility of such a policy is ruled out, the prison director can adopt the above behavior.
On the contrary, if a strategy succeeds under the aforementioned behavior mode of the prison director, a slight change to it will inevitably be able to adapt to all the behavior methods of the prison director: you only need to turn on or off the light as soon as possible. If the light remains "normal" in the future, you can do nothing.
Therefore, it may be assumed that the prison director has taken the above actions. In this way, as long as the prisoner is taken out, he will not do anything when he sees the "normal" lamp. The following discussions follow this premise.
First, there must be a person who can turn on the light by bringing it out. Otherwise, everyone will be shut down. So the prison director first took the man out and lit the light. In this way, the person is changed to Group B.
Next, the prison director held a person in group A and did not consider taking him out for the moment. Unless otherwise stated, this person is outside the scope of the discussion. Apparently, before this person is taken out, no one can report to the prison director that the whole person has been taken out.
Consider the remaining persons in group. For everyone, if they bring it out and see the "abnormal" lamp, there are only two types of results: X. Keep the light on and turn it into Group B; Y. Turn off the lights and leave them in group. Bring out a prisoner (if any) who takes policy y, the light is turned off. If there is still one person in the rest who can turn on the light by bringing it out, then bring it out and turn it on. So we will start the process again.
The above loop will end in a limited number of times, because the number of people in group A is reduced by one each time. There may be two possible results: 1. Group A is the person who adopts Strategy X, and the light is on; 2. The light is off, and no one in group A can turn the light on by bringing it out. For the case where the lights are turned off and the situation in Group B is discussed, each prisoner in Group B may see the "abnormal" light. There are two possible strategies:. Keep the lights closed and turn them into group A; group B. Turn on the light and leave it in group B. If group B has a prisoner who adopts policy B, bring the prisoner out and turn on the light. If group A still has a policy y, bring the prisoner out again and turn off the light; if group B still has a prisoner who adopts policy B, bring him out again and turn on the light ...... If both group A and group B have always taken policy y and always adopt policy B, the cycle will continue; otherwise, the final situation will be either: r1. Group A is the person who adopts Strategy X, and the light is on; or: R2. The light is off, and no one in group A can turn the light on by bringing it out, and group B is also a prisoner of Strategy.
In the case of circulation, the prison director can use this to control the switch of any lamp. Therefore, there is always a way for the prisoner to make a wrong report or keep off, so this should not happen.
In the case of R2, the first person to be held out will not turn on the light if it is taken out, and the prisoners will only be shut down forever. So he will turn on the light and turn him into group B, and then he must always adopt the B strategy until a person in group A reports to the prison director that everyone has been taken out. If this is not the case, he will always bring him out after he enters group B. Before a person in group A reports to the prison director, he will either turn into a strategy, in this way, everyone will be shut down forever; or he will report to the prison director, but if he is in such a strategy, the prison director only needs to hold another person in group B in advance to replace him, you can report the error when a person is held down and not taken out. Therefore, in group A, a series of Y actions will report to the prison director that everyone has been taken out, and the detained person will use enough B action to force him to respond with y action until he reports it. If there are other people in group A, we can see from the above discussion that there is no difference between these people bring out and not bring out when the lights are off, so we can let the person in group A make an error report without bringing them out. Therefore, there must be only one person with the privilege to report to the prison director.
In the case of R1, if someone else has not been taken out of the detained person at this time, you can think that the person who was held in the first place has opened the person, repeat the above process. The final possibility is still loop, R1 or R2. We have already discussed the loop and R2. in the case of R1, releasing the held person will not be of any effect (because it is known that he adopts the policy X ), so the prisoner will be locked forever.
If all the people outside the detained person are taken out of R1, then the situation is similar to R2, but the AB group has changed its position. So we can discuss it in the same way.

To sum up, there is only one way to live in this step: there must be only one person who has the privilege to report to the prison director. Such a strategy was originally available, but since the prison director promulgated new regulations ......
No way to survive ......

[Comments from Mr. sensan]

For me, it seems that some points are not too clear. The final proof is vague and needs to be enhanced.

Proof of New Problems
First, for each prisoner, define two states of the lamp: "normal" and "abnormal ". That is to say, if the light was on before the last time the wind came back, it would be "abnormal" and "normal" if the light was turned off when it was called out for the wind ". If the light is turned off before the last time the wind came back, it turns on as abnormal and turned off as normal.

In this way, prisoners can be divided into two categories: group A is considered abnormal when the light is turned on, and group B is considered abnormal when the light is turned off. Since the prisoners knew from the very beginning that the light was originally turned off, all of them were in group.

According to the following discussion, a prisoner who has not been released at one time should not belong to Group A, because although they know that the light is "normal ", however, when you see the "normal" light, you will press it. For the sake of rigor, their status should be classified as a special "initial" group. My reasoning below is based on three states.

Next we will discuss what a strategy should be like if there is a policy to prevent prisoners from being put to death or permanently shut down.

We can think that the prison director knows the prisoner's plan and will try to prevent them from being released.

Assume that the Prison Director adopts the following behavior: every time I decided to take a man, Yi, O, O, NE, baking, and sending Huang, 5, yi, yi, Ba, O it's time to restore the wisdom of Hangzhou and Hangzhou! Bao Yun for the generation of Huang Ba Yu huai qihang Yun Xiao Yu Yue mystery Yao Yun I badgine Yan frequency eye Wu cap that Yi Fu Yi XU Xu Yu su Huai qihang playing Huan Yu sorry, the fan was covered, and the haze was pushed to the silk gap, and the Huai Yi was knocked on this tip. k Yu Zhi Bai yuanjing value Shu zookeeper fresh twist is more than luyun free trade R knocking on the window when about an acre of growth shoot allows the visual equipment male twist sword brown gray Huang Ben TU you can fight for the canal worm gun on the wall! Is it a pure umbrella? /Font>

There is a loophole in the above reasoning, but it is easy to make up for it. Suppose a prisoner only clicks the light from time to time when the light is on and "normal, then the prison director cannot use him to control the lights at Will-he can use him to turn off the lights at most.

However, it is not necessary for the Governor to be able to use his arbitrary control lights to invalidate the policy of such a prisoner, as long as he simply invalidates the policy of the prisoner.

The reasoning may be written like this: the prisoner has no information except the light state when the prisoner is released and the information about his operations, so the actions that the prisoner requires in the n + 1 release strategy (by lighting, not by pressing, or reporting that the prisoner has all been put off the wind ("Report ")), it only depends on the status and operation of the lamp that you saw N times ago. If, in the prisoner's strategy, even if the light is "normal" during the nth blow, the prisoner still needs to click (or report ), then, the director can immediately ask the prisoner to release the prisoner again after N-1 occurrence. In this way, the prisoner's policy is equivalent to the act opposite to his original act at the n-1 th. If the first n-th policy is to press the light, otherwise, press (or n-1 time, if the original policy is to report for the nth time ). In this way, we can prove that if there is a policy that allows prisoners to be free, there must be a strategy that does not require "Although the lights are 'normal', but still need to press the lights, prisoners can also be given freedom. Here we can see that it is necessary to distinguish between "normal" and "initial". For the "initial" status, there is no n-1 time, and the above reasoning cannot be performed.

Therefore, there are two types of prisoners who think that the light is on abnormally or in the initial group. This is group A, and that the light is switched OFF abnormally is group B.

Therefore, it may be assumed that the prison director has taken the above actions. In this way, as long as the prisoner is taken out, he will not do anything when he sees the "normal" lamp. The following discussions follow this premise.

Now, we can omit the phrase "prisoner is taken out" because we distinguish "normal" from "initial". Of course, there are some troubles, therefore, such a distinction may not simplify the description, but it may prevent you from wondering whether there is any reasoning vulnerability.

First, there must be a person who can turn on the light by bringing it out. Otherwise, everyone will be shut down. So the prison director first took the man out and lit the light. In this way, the person is changed to Group B.

Next, the prison director held a person in group A (in the initial group) and did not consider taking him out for the moment. Unless otherwise stated, this person is outside the scope of the discussion. Apparently, before this person is taken out, no one can report to the prison director that the whole person has been taken out.

Consider the remaining persons in group. For everyone, if they bring it out and see the "abnormal" lamp, there are only two types of results: X. Keep the light on and turn it into Group B; Y. Turn off the lights and leave them in group. Bring out a prisoner (if any) who takes policy y, the light is turned off. If there is still one person in the rest who can turn on the light by bringing it out, then bring it out and turn it on. So we will start the process again.

The above loop will end in a limited number of times, because the number of people in group A is reduced by one each time. There may be two possible results: 1. Group A is the person who adopts Strategy X, and the light is on; 2. The light is off, and no one in group A can turn the light on by bringing it out.

I will slightly change the description:

Consider the remaining persons in group. For everyone, if they bring him out and see the "abnormal" light, he has only two possible strategies: X) Keep the light on, and change himself into Group B; y) turn off the lights and leave them in group. Bring out a prisoner (if any) who takes policy y, the light is turned off. If there is still one person in group A who can turn on the light by bringing it out, then bring it out and turn it on. So we will start the process again.

The above loop will end in a limited number of times, because the number of people in group A is reduced by one each time. There may be two possible results:
1) The light is on, but no one in group A can choose to turn off the light using the y strategy. That is to say, now all the people in group A will be on, changed to Group B;
2) The lights are switched off, but the lights cannot be switched on in group A. That is to say, if all the people in group A are taken out, they will keep the lights off and stay in group.

Here we need to add a description of the initial group: whether it is 1) or 2). Next we will take the people in the initial group of the remaining group A (except the person who is deducted) if it is the case 1), these people will not switch, and they will switch to Group B. If it is 2), these people will not switch, remain in group. In this case, it will still be 1) or 2), but there will be no other person in the initial group except the person who was previously detained-all others have been released.

For the case where the lights are turned off and the situation in Group B is discussed, each prisoner in Group B may see the "abnormal" light. There are two possible strategies:. Keep the lights closed and turn them into group A; group B. Turn on the light and leave it in group B. If group B has a prisoner who adopts policy B, bring the prisoner out and turn on the light. If group A still has a policy y, bring the prisoner out again and turn off the light; if group B still has a prisoner who adopts policy B, bring him out again and turn on the light ...... If both group A and group B have always taken policy y and always adopt policy B, the cycle will continue; otherwise, the final situation will be either: r1. Group A is the person who adopts Strategy X, and the light is on; or: R2. The light is off, and no one in group A can turn the light on by bringing it out, and group B is also a prisoner of Strategy.

In the case of circulation, the prison director can use this to control the switch of any lamp. Therefore, there is always a way for the prisoner to make a wrong report or keep off, so this should not happen.

In the case of R2, the first person to be held out will not turn on the light if it is taken out, and the prisoners will only be shut down forever. So he will turn on the light and turn him into group B, and then he must always adopt the B strategy until a person in group A reports to the prison director that everyone has been taken out. If this is not the case, he will always bring him out after he enters group B. Before a person in group A reports to the prison director, he will either turn into a strategy, in this way, everyone will be shut down forever; or he will report to the prison director, but if he is in such a strategy, the prison director only needs to hold another person in group B in advance to replace him, you can report the error when a person is held down and not taken out. Therefore, in group A, a series of Y actions will report to the prison director that everyone has been taken out, and the detained person will use enough B action to force him to respond with y action until he reports it. If there are other people in group A, we can see from the above discussion that there is no difference between these people bring out and not bring out when the lights are off, so we can let the person in group A make an error report without bringing them out. Therefore, there must be only one person with the privilege to report to the prison director.

In the case of R1, if someone else has not been taken out of the detained person at this time, you can think that the person who was held in the first place has opened the person, repeat the above process. The final possibility is still loop, R1 or R2. We have already discussed the loop and R2. in the case of R1, releasing the held person will not be of any effect (because it is known that he adopts the policy X ), so the prisoner will be locked forever.
If all the people outside the detained person are taken out of R1, then the situation is similar to R2, but the AB group has changed its position. So we can discuss it in the same way.

To sum up, there is only one way to live in this step: there must be only one person who has the privilege to report to the prison director. Such a strategy was originally available, but since the prison director promulgated new regulations ......
No way to survive ......

This paragraph is too vague. First, it seems that only the first two cases are demonstrated, but not the case 1) "the light is on ". In addition, "in the case of circulation, the prison director can use this to control the switch of the lamp at will, so there is always a way for the prisoner to make a wrong report or keep turning it off, so this should not happen ." It is also very vague.

It should be noted that there is an absolutely reliable strategy if there are four people at first (and then one person leaves three people, therefore, the argument will certainly require more than three people.