[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 9
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Yes, the existence of the solution, by the way, is not at all obvious. But we can do it this way: firstly build explicit examples (with relation matrices), and then, knowing that the solution exists, prove that there are no other options.
2 Richie: for 5 people there are only two possible configurations {Other} - {"0", "1", "2", "3"} and {"1", "2", "3", "4"}.
Reasoning.
Can Petya be "0"? No, because then only configure {Other}|Petya = {"0", "1", "2", "3"}|"0" is possible. Contradiction, as "3" must have three friends, not a maximum of two as here.
Can Petya be "1"? If {Other}|Petya = {"0", "1", "2", "3"}|"1", then the sum of the relations is 7 - a contradiction, since it must be even. The same applies to {"1", "2", "3", "4"}|"1" (the sum is 11).
Can Petya be "3"? No - for the same reasons as with "1".
Can Petya be "4"? Only configure {Other}|Petya = {"1", "2", "3", "4"}|"4" is possible. Contradiction, as "1" must be friends with both "4".
That leaves Petya = "2". Well there remains to build an explicit example for both config cases.
Avals, can you comment on what you have written.
on page 6 commented
It all starts with 3 people in a class and solving the problem set for that case. As the number of students increases, the same pattern is observed.
Drawing for 5 people in class please.
1-2 (1 friend)
2-1,3,4,5,П (5)
3-2,4,5,П (4)
4-2,3,П (3)
5-2,3 (2)
Total: Petya has 3. No doodle board handy, only a scribbler.
Draw a picture for the 5 people in the class please.
Рисунок для 5 человек в классе пожалуйста нарисуйте.
Richie, well go ahead and draw it yourself. I've proved to you that Petya can only be '2'. It's not 26 people, after all :)
The number of friends can be from zero to 25,
zero and 25 are mutually exclusive,
There are only two options, zero to 24 or 1 to 25.
intuitively it is clear to me that half of the class will have to be friends with Petya to meet the conditions of the problem))
but how will it be in the form of a formula...
Good thing you drew it, in the second version the zero is already gone.
-
Maximum number of "connections" in the system:
C=(n^2)/2;
where n is the number of students in the class.
Figaro, you have an error in the second line.
Petya can only be a '2', I proved that (for 5 people in the class).
I'll show you the possible matrices.
Both matrices are symmetric. I only filled the green cells, as all the white ones depend on them. As you can see, there are explicit solutions, in both cases Petya = "2". Under the matrices are the numbers of friends (also calculated by Excel). Swetten we have the friendliest.
Figaro, you have an error in the second line.
Petya can only be "2", I proved that.
I don't see the mistake, but I believe my eyes. I think the picture fits the problem. Is it easier to check the drawing than to multiply matrices?)
Excuse me, I'm not an artist.)