[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 345
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But he knows where the hole is all the time, as the female is there. In general, I don't envy him: so many attempts - while any distance he could cover in just a couple of jumps...
Anyway, the blacksmith is at (alpha, beta), and the hole is at (x,y). Do you get it now?
Я могу посмотреть в решение, чтобы выяснить, - но не хоцца. Боюсь, она где-то в случайном месте внутри квадрата. И кузнечик тоже где-то в случайном месте.
Но он все время знает, где лунка, т.к. там самка. А вообще я ему не завидую: столько попыток - при том, что любое расстояние он мог бы одолеть всего за пару прыжков...
Короче, кузнец - в точке (альфа, бета), а лунка - в точке (x,y). Ну теперь-то тебе понятно?
There is no grail, all the believers are dead! Your statement!!!?
I believe it, just like Poincaré was searched for 100 years and there must have been those who said it was nonsense, the result is that some died without surprising the wheel.
Я могу посмотреть в решение, чтобы выяснить, - но не хоцца. Боюсь, она где-то в случайном месте внутри квадрата. И кузнечик тоже где-то в случайном месте.
Но он все время знает, где лунка, т.к. там самка. А вообще я ему не завидую: столько попыток - при том, что любое расстояние он мог бы одолеть всего за пару прыжков...
Короче, кузнец - в точке (альфа, бета), а лунка - в точке (x,y). Ну теперь-то тебе понятно?
If the grasshopper moves towards the hole, it may not hit.
If it is loose, it may hit.
There is one interesting observation: all four endpoints of jumps from the farrier's starting point in all directions he can move form a square. What is its side equal to?
2 Mer495: yes, it's mine. But I wrote it back then when I was practising on critiquing the grails posted here. Now I put a slightly different meaning in that slogan.
2 Mer495: да, моё. Но я написал его тогда, когда тренировался на критике граалей, выкладываемых здесь. Теперь я вкладываю в этот слоган несколько другой смысл.
Can you tell me what it is?
Ну да, ihor, никто и не говорил, что он всегда должен двигаться к лунке.
Есть одно интересное наблюдение: все четыре конечные точки прыжков во всех направлениях, в которые может двинуться кузнец, образуют квадрат. Чему равна его сторона?
n/2
It seems to me that it is sufficient to consider movement along one coordinate.
(it moves along the second one in exactly the same way)
2 Mer495: It's simple, the grail exists, but everyone has a different grail and probably a different one than the one he first imagined.
Believing in the grail is certainly a good incentive. But because of exaggerated expectations, the seeker is wasting a lot of time. Much better to just look for something robust and moderately profitable (by forex standards), concentrating more on how not to make money, but how not to lose it.
P.S. Seems like it's really enough to consider movement on just one coordinate.
OK, I'm off to bed. Night-night, everybody.
So, let's look at the abscissa separately. Consider that the abscissa of the left vertex is 0, the abscissa of the right vertex is 1. We solve the problem in a binary system, i.e. the coordinate of the smith is 0,x1x2x3x4.... is some fraction, where x1,x2,x3,... - 0 or 1. Similarly, the hole coordinate is 0,y1y2y3y4...
Let the coordinate of the insect at the initial time is a0. Then if it jumps to the left, its coordinate is halved a1=a0/2, which is equivalent to shifting the binary fraction one digit to the right OR that is the same, assigning zero to the left and shifting by one digit. If it jumps to the right, then the coordinate is converted by the law a1=(a0+1)/2, i.e. we add 1 to the fraction on the left and again shift it by one digit.
If the grasshopper's brain allows him to represent the hole coordinate as a binary fraction, he can, with a predetermined accuracy, approximate his coordinate to it using the following algorithm:
1. we start with the nth digit, choosing it so that approaching the coordinate of the centre of the hole to this digit does not give an exit beyond its boundary (the vicinity of the centre).
2. If the digit is 0, jump to the left. If 1, jump to the right.
3. Move to the next digit towards the higher
and so on, until we reach a point:))
At n=0 the fact is trivial. Let us make an inductive transition from n to n+1. Consider some square of size 4^(-n-1). Choose the vertex of the original square closest to it and perform homothety with centre in this vertex and coefficient 2. Then the selected cell will be one of the cells of size 4^(-n). By the induction assumption, the grasshopper can get into it. If it jumps now half the distance to the specified vertex, it will hit the desired cell.
Concerning the problem with 1999 numbers: MD, the answer is correct. But the proof is murky and not that simple.
Оказывается, кузнечику неплохо бы знать еще и что такое гомотетия...
It is supposed to be homothetic to an eighth-grader solving a problem. With fractions, it seems to me, it's more beautiful, and it's more programmatic:)