[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 12
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Задачка с мехматовского форума, тут.
В той же ветке приведено решение - 12 или 13.
Такой категорический ответ вызывает изумление. Я начал размышлять на досуге и пришел к некоторым заключениям. Но до решения задачи далековато. Кому интересно, присоединяйтесь.
Только прошу не гуглить и не рэмблить, а то станет неинтересно. Наверняка задачка решается элементарно.
5 friends can have Pete +1 person no one is friends with
sanyooooook, ну ладно в верху всё перепутал, но ты обкурился что-ли, какие 5 ?
Why can't Pete have a maximum of five friends?
Prove me wrong and I'll admit I'm wrong! ;)
hasavatara decided to apply derivative functions?
So, let's say there are N students in the class. If there is one among them who is not friends with anyone, this simply leads to the case of N-1 students. So from now on, we will assume that everyone among these N students is friends with someone.
Let us put all the students in one row. It is not too convenient to draw circles here, so I will denote each student as follows: (М). The brackets instead of a circle, and the letter M stands for the number of its friendships. In total we get N notation of the form (M).
Now we draw the friendly relations. Suppose the last (rightmost) student is a friend of all the previous students. This means he has N-1 friendships. He is also friends with the first (leftmost in the row) student. That is, for the first one there is already one friend. Therefore for him there will be no more friends. We obtain the row: (1), (...), ... (...), (N-1)
The second and penultimate have no relations yet, that is why there are dots in brackets.
Now repeat the procedure for the penultimate one. We connect it to all previous ones, but without the first ! We have N-2 connections: N-3 with the previous ones and 1 with the last one.
For the penultimate one we connect it with the previous ones, except for the first and the second. You will have N-3 connections: N-5 with the previous and 2 with the last and penultimate. So the picture is as follows:
(1), (2), (3), ... (N-3), (N-2), (N-1)
This operation can be continued until the numbering from the end and from the beginning meet.
What happens at the meeting point can be figured out by hand, but it's not very obvious. There is a simpler method.
We have N elements in a string. The procedure provides consecutive filling in ascending order from 1 from the beginning, and in descending order from N-1 from the end. Is it possible to number N elements, starting from 1, so that N-1 is at the end and all elements have different numbers? Obviously not. Two elements must have the same values.
It is easy to check that when N=26 (i.e. there is no student in the class with zero connections), this repeating number = 13.
If N=25 (i.e. one renegade is present), the number = 12.
Petya can only have this repeating number of friends. Only in this case (as already said here) all others will have different number of friends.
Робяты :) Вам похоже совсем уже думать типа не о чем... Что такое фигней маятесь :)
Well offer not bullshitYurixx писал(а) >>
Suppose that the last (rightmost) student is friends with all the previous students.
With N=25 (i.e. one renegade is still present) this number = 12.
Petya can only have this repeating number of friends. Only in this case (as already said here) all others will have different number of friends.
If the rightmost one is friends with everyone, then the maximal number is 25 (why can't Petya be the rightmost?)
and your answer is 12
Ну предложи не фигнюI suggested it, remember the 'architecture' and there was a circus going on... :) I was just asking :)
1. if the rightmost one is friends with all, the maximum equals 25
2. (why can't Petya be the rightmost?)
3. your answer is 12.
1. ) Correct. That is if there is no one who is friends with anyone.
2. I don't recommend Petya to be touched for the time being. He's a tough guy, he can punch you in the eye.
3. I get 12 if I'm the only one who isn't friends with anyone. In this case, the maximum for the rightmost one is 24.
Сорри, сегодня нет времени, ужЕ не смогу подсчитать.
Take your time and don't get frustrated, you'll finish the wikipendia later :o)