An introduction to the sick dog god question, a classic logic question about the sick dog

3 mad dogs, using the hypothetical method, starting from 1 and so on.

I think there's one answer, who told you that one shot can go out of business with a dog.

3 (there were several sick dogs on the first few days of gunfire).

To put it into an inductive way: if there is only one sick dog, then the family with the sick dog sees 49 good dogs, and there is a sick dog in the village, it must be his own, and he will kill his dog on the first day A gunshot goes off.

Suppose there are two sick dogs, and the family of the sick dog is considered A and B on the first day, A sees 48 good dogs and B's sick dog, and A thinks that if his dog is a good dog, then B should see 49 good dogs, and he should kill B's dog on the first day, but B doesn't shoot on the first day, which means that A's dog is also a sick dog, so A kills his own dog the next day, and in the same way, B thinks the same thing, so B also kills his own dog, and two gunshots go out.

The inference is that if a person sees n sick dogs, then if he doesn't hear a gunshot on day n, then his dog must be sick too, and he is going to shoot his own dog on day n+1.

On the third day, the gunshot rang out, indicating that the family that proved that there was a sick dog saw two sick dogs from another family, but waited for two days and did not hear a gunshot, indicating that their own dog was also a sick dog, so there were three sick dogs in total.

3 Analysis:

On the first day, if everyone sees the other 49 dogs jumping around, then he doesn't have to look at his own dog, according to premise 1, everyone understands that if there is at least one sick dog in the yard, it must be their own dog who is sick, then according to condition 2, the gun will go off on the first day. Since there were no gunshots on the first day, the assumption is incorrect, i.e., it is not right for everyone to see the other 49 dogs alive, and conversely, everyone can see at least one sick dog.

Now the question is: everyone has seen at least one sick dog, but why hasn't the gun been fired? Because the owner of a sick dog sees someone else's sick dog, if the owner of the sick dog he sees shoots his own dog, he thinks:

Thankfully, my dog was not sick, and on the first day, he did not dare to kill his own dog rashly until the owner of the sick dog he saw shot his dog, and if he did, he would mislead the owner of the sick dog he saw, and he would think that his dog was not sick. There were no gunshots on the first day, then everyone knew that there were at least two sick dogs in the yard.

By the next day, if everyone sees 48 live dogs and one sick dog, then don't hesitate, your dog is sick and killed. But the next day there was still no gunshots, then everyone knew that there were at least three sick dogs in the yard, and that could only wait until the third day.

By the third day, if everyone sees 47 dogs jumping around alive and two sick dogs, then don't hesitate to kill your own dog too. On the third day, there was a burst of gunfire, indicating that three dogs had been killed.

1 Three died (there were a few gunshots fired on the first few days).

Simple analysis: Starting with an abnormal dog, it is clear that a gunshot will be heard on the first day. Here's the gist.

You just have to put yourself in the shoes of the owner of the abnormal dog.

If there are two words, the train of thought continues, only consider people who have two abnormal dogs, and the rest do not need to think about it. Through the first day.

They learned about each other. Kill your own dog the next day. In other words, everyone needs a day to prove themselves.

The dog is normal. If there are three articles, again, only those three people are considered, and each of them will take two days to prove themselves.

Your dog is a normal dog.

3. If there is only one sick dog, then only the owner of the sick dog cannot see the other sick dogs on the first day, then he can be sure that his dog must be a sick dog on the first day, and the gunshot will sound on the first day.

If there are two sick dogs, then the owners of the two sick dogs only see one of each other's sick dogs on the first day, they will assume that there is only one sick dog and wait for the other person to shoot, but they do not shoot on the first day, indicating that their assumption is wrong, there is more than one sick dog, but they only see one and the other homes do not, then there is another and only one is in their own home. The next day both came to this conclusion, and both opened fire.

If there are three sick dogs, the owners of the three sick dogs on the first day see two sick dogs, waiting for the other two to shoot, but not on the first day, the three owners will assume that there are only two sick dogs, according to the above logic, the other two will shoot the next day, but the other two will also wait for him to shoot the next day, so there will be no movement the next day, at this time the owners of the three sick dogs can judge that the sick dogs are not only two but three, one of which is in their own home, and the three will shoot on the third day.

By analogy, the owner of the sick dog assumes that he is not the owner of the sick dog, and the number of sick dogs is the true number minus 1, and waits for the other sick dog owners to shoot every day until he expects all the other sick dog owners to deduce that the sick dog is in his house on the day and finds that none of them have fired, indicating that there is another sick dog in his house. All owners of sick dogs follow this logic and find their sick dogs on the same day and shoot them on the same day.

The necessary condition for this question to be true: all 50 protagonists in the question understand the logic of the solution to this question.

Sick dogs have already been mentioned in the title, assuming that the number of sick dogs is n. (In the following analysis, a: indicates the owner of the sick dog, and b: indicates the owner of the non-sick dog).

Day 1. If n = 1, there is no doubt that A cannot see a sick dog, so you can be sure that your dog is sick. That's when A shoots.

B sees A shooting and understands that A does not see any other sick dogs. for if a sees it, he cannot be sure that his dog is a sick dog; Therefore b knows that his dog is not a sick dog.

The answer emerges: n=1.

If n 2, a must have seen the sick dog, will not shoot on the same day, enter let roll the next day. Good stool.

The next day. From the first day, it can be concluded that n 2;If A sees only one sick dog on the first day, there is no doubt that his dog is sick. Two aces shoot, answer n=2. If A sees sick dog 2, there will be no answer at this time, and it will continue to enter the third day. and gives n 3

Day 3. Friends have come up with n 3, if A saw two sick dogs, it proves that his dog is sick. Three A shots, answer n = 3. If A sees sick dog 3, he will also not be able to judge the result and enter the fourth day. and gives n 4

Until one day.

B will always see one more sick dog than A sees, so B won't shoot).

Day p. From the above conclusions, when N P,A saw P-1 sick dog, all A understood: his dog was a sick dog. Answer: n=(p-1)+1;i.e. n=p

In conclusion: the first few days of gunfire indicates that there are several sick dogs.