[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 225
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OK, people, no inaccuracies. People with a mathematical mind can understand that. No "to the gram" or "to two atoms". Milk is infinitely divisible and has no atomic nature.
So there are 100 grams, 100 and 130 in three glasses. Prove that for a finite number of steps - you can't equalise (for an infinite number you probably can). Or construct a finite algorithm that disproves my statement (I admit it, as I'm not 100 grams sure I'm right).
Seriously, the problem cannot be solved in a finite number of steps in the general case.
The only question is how to construct a proof in the simplest way and specify the boundary conditions of solvability.
Доказать, что за конечное число шагов - нельзя уравнять (за бесконечное, вероятно, можно). Или построить конечный алгоритм, опровергающий мое заявление (я это допускаю, т.к. не на все 100 граммов уверен в своей правоте).
Not so -- you can't equate x, x, x + a grams, a and x can be any non-zero numbers.
Не не так -- нельзя уравнять х, х, х + а граммов, а и х могут быть любыми ненулевыми числами.
Yes, this is one particular case. Here the intractability is obvious. And in the general case, how do you describe it? Or is a counter-example (like this one) enough?
Да, это один из частных случаев. Здесь неразрешимость очевидна. А в общем случае как расписать? Или достаточно контрпримера (типа этого)?
This is not a special case, it is the state of the system after any overflow. I.e. the problem for 3 cups can only be solved in one transfusion.
Да, это один из частных случаев. Здесь неразрешимость очевидна. А в общем случае как расписать? Или достаточно контрпримера (типа этого)?
If it's obvious to you, don't rush to say so, let them guess. A counterexample for 30 glasses is enough. The answer to the problem simply gives a counterexample without proof. But here you will have to prove it.
It is interesting that in the problem 3, 4, 5 (solvable) just equalize the first two glasses and it becomes insoluble. I.e. steps are irreversible: solvable problem can be spoiled by wrong step.
Here's another hint: take 4 glasses, each has a, b, c, d milk. In this case the problem is always solvable (in 4 correct steps), there are no counterexamples in principle.
Mathemat писал(а) >>
Interestingly, in the problem 3, 4, 5 (solvable) it is enough to equalize the first two glasses and it becomes unsolvable. I.e. steps are irreversible: a solvable problem can be "spoiled" by wrong step.
Problem 4 (8, 16, 32 ...) cannot be spoiled.
I like the direction of your thought :) I'm really not sure if it's impossible.
Направление твоей мысли мне нравится :) Я, правда, не уверен, что невозможно.
Easily proved by induction starting at 2.
Induction makes it easy to construct the correct algorithm by reducing it to a base (2 glasses). But does it prove the impossibility of spoilage? I'll think about it.