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And one more at once, on the follow-up:
There are 13 yellow, 15 blue and 17 red chameleons living on the island. When two chameleons of different colours meet, they change to a third colour. In other cases, nothing happens. Can it happen that all the chameleons turn out to be the same colour?
Not 1 yellow, but two.
Not 1 yellow, but two.
No, that's not it.I get to
0-1-N
and that's it. No matter how you spin it, if you eliminate one colour or try to pair the same amount, nothing works
Not one yellow, but two.
About the chameleons.
There are three differences between the number of chameleons, initially equal to d1 = 2 d2 = 2 d3 = 4;
What happens when 2 chameleons meet? one difference stays the same, the other two change to 3. Since none of the differences is divisible by 3, no combination will reduce any difference to 0.
Right.
Regarding the Olympic system:
One game, one team out. That means N-1 teams have to drop out, i.e. N-1 games. That's all the proof you need.
I did the inductive proof myself, but then I saw a very simple solution - in a few words. I felt ashamed :)
Moskslomayka (this turned out to be the case for me, although the weight is only 4; I just haven't taken up cutting tasks before):
Cut a circle into several equal (overlapping) parts so that the centre of the circle does not lie on the border of at least one of them.
Explanation: the parts can be equal up to any identity transformation of the plane - including mirror symmetry. But there is a solution without the need for mirror symmetry. Any of them will do.
MD: Наши мозги привыкли замечать только победителей.
I am slowly reading Wilson. Strong man.
...
Slowly reading Wilson. Strong man.
I'm on chapter 14 right now.
Are we synchronizing, or did you run ahead?
Nah, I just started. Haven't had a chance to do it before.
Anticipating the last few chapters: who wants LSD for pennies on the pussy? :)
Anyway, you're right, the man's just tearing his brains out, rushing the wreckage of consciousness into the infinity of development.