[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 356

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Mathemat >>:
Одинаковые они по прочности. Цвета имеют принципиальнейшее значение: их перекрашивать никак нельзя, т.к. это эксклюзивный каприз заказчика для кодера. Шарика только два.
P.S. Задачка действительно серьезная. Я и не подозревал, что подобные задачки дают в качестве испытательных.

For the case of "100 storeys, 2 balls", it is possible to do in the worst case with sixteen throws

// at first I thought it would take nineteen, but after your intense hints of ambush, I realised "there are options"... :)

Strategy:

Throw the red ball first (programmer's whim) :

Floors: 16, 31, 45, 58, 70, 81, 91, 100

Whichever floor the red balloon crashes on, we throw the blue one starting from the previous one "survived in red list +1", and throw it until "red one crashed -1".

Worst case total ==16.

// for the general case I'll work out the formula when I wake up, if someone doesn't beat me to it.

 
Very good for a start. But you can do even better.
 
Mathemat писал(а) >>
You have hired a man to chop wood. He will chop it for a week (7 days). You have a piece of gold of 7 grams and every day you have to pay him exactly 1 gram. But you can only chop the piece twice. How will you pay him?
The piece has to be chopped into 1 gram, 2 gram and 4 gram masses. Using these masses you can make any mass up to 7 grams with a stack of 1 gram.
But the question is how to chop it so precisely without measuring instruments. And if such devices are available, there is another way: to cut up a piece in 7 grams with 1 or 2 choppers, pre-shredding and bending a piece on a stump :)
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Mathemat, what are your impressions of the victory parade?
 
Did I understand correctly that the worst option. overkill. throw the ball on the 1st, 2nd, etc. floor until it crashes. and in principle the 2nd ball is not needed. Yes ?

Although no not the worst. we can throw a balloon on the 1st floor and it will break, then there is no point in going over everything else. A maximum of 100 (not broken).
The second balloon gives us a chance to use the division in half. Until it breaks, which reduces the maximal number, the first one we throw on the 50th floor. Broken goes from 1 to 49. If it doesn't break, we go to 25 and so on.
we get min 2 steps, max. 50.
I don't see the point in colours. if there is no condition. Like at what maximum floor the red ball will not break.
 
it's funny to see when people create difficulties for themselves - there are no limitations in the problem, so anything is possible - why think about ways to divide a piece into parts if there are no limitations?!
Although, of course, the problem condition is to blame, if you had written that there is a chain of seven links and only one can be cut, it would have been more definite...
 
Prival >>:
правильно ли я понял, что самый худший вариант. перебор. бросаем шарик на 1, 2 и т.д. этаже, пока он не разобьется. и в принципе 2-й шарик не нужен. Да ?

the worst is to throw the ball across two floors, moving from the bottom to the top until it breaks and throw a floor below the second

 
MetaDriver is moving correctly. It's just that the option found is not yet optimal.
Richie, why complicate the task when there is no hint of thinking about how to cut it? Stumps of some kind, heat. We have the ability to chop a piece into any two pieces with any accuracy twice. The problem has already been solved by you.
 
Mathemat >>:
MetaDriver правильно движется. Просто найденный вариант еще не оптимален.
You got it. Here's a variant with 14 throws.
Red: 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 98 // the last move (98 instead of 99) saves one worst roll, in case with 95 unbroken
Blue: fills the last unbroken red gap, same as last time.
 
Mathemat писал(а) >>
Richie, why complicate the task when there's no hint of thinking about how to chop it up? Stumps of some kind, heat. We have the ability to chop a piece into any two pieces with any accuracy twice. The problem has already been solved by you.
What heating? Gold is good without heat. About the stump - that's humour :)
 
MetaDriver >>:
Уговорил. Вот вариант с 14 бросками.
Красный: 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 98 // последний ход (98 вместо 99) позволяет сэкономить один худший бросок, в случае если с 95 не разбит
Синий: заполняет последний неразбитый промежуток красного, как и в прошлый раз.

Yes, interesting. In the solution given where I found, the best written option was almost the same (with 99), but still comes out 14. The problem is in the proof. Why can't we solve the problem for any case in 13 steps?

I know proofs are not liked here (especially by you, Volodya), and it seems to me that this is an optimal solution. But something is missing. Why is this algorithm is the best?

P.S. With this algorithm to prove that 14 is the minimum is not difficult. OK, we've got it. Are we going to solve for the general case or not?

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