IBM interview questions

There are 50 people in the village, each having a dog. There are diseased dogs in the 50 dogs (this disease will not be contagious ). So people need to find the sick dog. Everyone can observe the other 49 dogs to determine if they are ill. Only their dogs can't watch them. The results obtained after observation cannot be exchanged, or the owner of the sick dog cannot be notified. Once the host determines that his or her family is a sick dog, he or she has to shoot his or her dog. The first day and the second day were not shot. On the third day, a gunshot was reported, asking how many sick dogs were there and how to calculate them?

First inference:

A. Suppose there is a sick dog. The host of the sick dog will see that the other dogs are not ill, so he will know that his dog is ill, so there will be a shot on the first night. Because there is no shot sound, it indicates that the number of dogs is greater than 1.

B. Suppose there are two sick dogs, and the host of the sick dog will see one sick dog, because the first day did not hear a gun. The number of sick dogs is greater than 1, so the host of the sick dog will know that his dog is a sick dog, so there will be a shot the next day. Since the second day is also shot, it indicates that the number of dogs is greater than 2.

From this reasoning, if the third day is shot, there will be three sick dogs.

Second Inference

1 if it is set to 1, the dog will die on the first day, because the dog owner does not see the sick dog, but the sick dog exists.

If 2 is used, the master of the diseased dog is a and B. A sees a sick dog, B also sees a sick dog, but a sees B's sick dog not dead. The number of unknown dogs is not 1, and others are not ill, so his dog must be a sick dog, but the idea of B is the same as that of.

Therefore, when the number is 2, the last two dogs will die on the first day.

3. If there are three, set the dog owner to A, B, and C. A sees two sick dogs on the first day. If a does not set his own dog, reasoning 2. If a does not set his own dog on the second day, the two dogs will not die, so the number of dogs is definitely not 2, other people are not ill, so their dogs must be ill, so they should be shot. B and C have the same idea as a, so they also shot.

Therefore, three dogs will die the next day.

4. If there are four, set the dog owner to A, B, C, and D. A saw three sick dogs on the first day. If a set his own dog instead of a, reasoning 3. When he watched the third day, the three dogs were not dead, so the number of dogs was definitely not 3, other people are not ill, so their dogs must be ill, so they should be shot. B and C and D have the same idea as a, so they should also shot.

Therefore, four dogs will die on the third day.

5. The remainder is recursive, and N is introduced from n-1 of the year.

Answer: N is 4. The dog died on the fourth day, but on the third day, the answer was three.