[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 292
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И вот еще одна, парадоксальная:
To do this, we minimize the number of men.
For this purpose let's suppose that the number of men who were in the first campaign 100% coincides with the number of men in the second one.
I.e. X1*0.60 = X2*0.75 // X1 and X2 - number of men in the first and the second campaigns correspondingly
Concerning women let's suppose the opposite, that those who were in the first campaign were not in the second, and vice versa. // This way we will potentially maximize them.
I.e. the number of women = X1*0.4+X2*0.25, or the same thing X1*0.4 + (X1*0.6 / 0.75)*0.25 = X1*0.6.6, which is exactly equal to the minimum number of men
Since this is the minimum case for men and the maximum for women, there can only be fewer women and more men.
Proved.
--
Example of the considered distribution: X1 = 3M +2G; X2 = 3M + 1G
// Vapchet problem for the third grade, like. :)
Давай определение сложного обмена, MetaDriver.
Пусть даны семьи F = {f1, f2, f3, ... fn}. Каждой из них в том же порядке соответствуют квартиры K = {k1, k2, ..., kn}. Сложный обмен - это такая перестановка квартир К1 = T(K), при которой ни одна из них не находится на прежнем месте. Так пойдет?
Если да, то тут, наверно, можно индукцией справиться.
Nah. That doesn't seem to work to me. It's a weak condition.
We need to prove that whatever the predetermined initial and final family/flat pairs in the set of changelings, an exchange is always possible in two moves.
I mean, it's not enough to put them anywhere. They have to go exactly where they're aiming. And in all variations of targeting.
множество мужчин бывших в первом походе 100%-но совпадает с множеством мужчин во втором
// Вапче задачка для третьего класса вроде. :)
i.e. all the men went on both hikes and the women were different each time... God, this is so familiar. It's definitely a third-grade challenge, the little ones won't get it right away:))))))))
Zadacha with a root is, I hope, not above the fourth, i.e. and not worth solving?
In any case the final numbers of flats after the exchanges will be a transposition with respect to the ordered set K = (1, 2, ..., n). Denote the elementary exchange between i and j as i<->j. Any complex one will be represented as a product of elementary ones.
Then, since the complex exchange is completely reversible, we obtain that any transposition T(K) can be transformed to K by the product of a finite number of elementary ones such that any particular number i occurs at most 2 times in the product.
The very number of elementary exchanges can be any, since the square of the elementary transposition is still equal to the identical element.
Ну тогда - еще одна попытка формализации задачи.
В любом случае окончательные номера квартир после разменов будут транспозицией относительно упорядоченного множества К = (1, 2, ..., n). Обозначим элементарный размен межу i и j как i<->j. Любой сложный представим в виде произведения элементарных.
Тогда, т.к. этот сложный размен полностью обратим, получается так: любую транспозицию Т(К) можно превратить в К с помощью произведения конечного числа элементарных так, что любой конкретный номер i встречается в произведении не более чем 2 раза.
Само количество элементарных обменов может быть каким угодно, т.к. квадрат элементарной транспозиции все равно равен тождественному элементу.
I've decided.
To begin with, let us note that any complex exchange consisting only of pairs is by necessity either a cyclic chain or decomposes into several cyclic chains.
Therefore it is sufficient, though necessary, to solve the problem for a cyclic chain of arbitrary length.
I solve it by explicitly specifying the strategy that leads to the desired result.
Let's write the initial chain as a chain of digits, where the digit represents the family and the position number in the entry represents the flat. In the final chain all families should be shifted by positions to the right, with the last digit going to the beginning of the chain. I.e. for a chain of 4 families, the entry would look like this: (1234)->(4123). Then, if the chain is of arbitrary length, exchange algorithm can be: // I will describe an example of chains of eight (even) and nine (odd) families.
1) Change between them inhabitants equidistant from the chain ends (12345678)->(87654321), [123456789]->[987654321].
2) Separate the first element of the resulting chain, and repeat the chip with the remainder (87654321)->(81234567), [987654321]->[912345678].
That's it.
You haven't specified how you will do the partitioning of arbitrary transposition into cyclic ones.
Secondly, the algorithm for handling a cyclic is only specified for a special case. Let's say there is one: (78123456). You haven't shown with it.
And in general - show me, say, using (12345678) -> (63814257) as an example, how you allocate cycles.
Наблюдение насчет цикличности верное, так оно и есть. Осталось аккуратно завершить доказательство.
Ты не указал, как ты будешь делать расчлененку произвольной транспозиции на циклические.
Во-вторых, алгоритм обработки циклической указан только для частного случая. [1] Скажем, есть и такой: (78123456). Ты с ним не показал.
Ну и вообще - покажи, скажем, на примере (12345678) -> (63814257), как ты циклы выделяешь.
[1] There is no such thing. What you wrote breaks down into two chains (one for even and one for odd).
And in fact, the numbering and recording of the positions takes place after the chains are drawn up. That is, we make up the chains first, then we number them. This removes all the complications.
Algorithm of chain-building: Take a map of this totalitarian city (you can use GoogleMap). Circle the flats with oppressed tenant-exchangers.
Starting with an arbitrary circle, connect the source flats with arrows to the target flats. If the starting point is reached and there are uncovered flats, repeat the procedure starting with any uncovered flats. And so on, until full coverage is achieved.
You have created allocated sub-chains or a long one.
All that remains is to number each flat in the chain in the direction of the move and go to the procedure from the previous post.
Хитер, черт. ОК, уговорил, а в фирме-риэлторе работают математики, знающие теорию транспозиций.
And they are crooks, too. They take bribes from those who want to move at one time (there are two of them in each chain). However, I won't repeat myself - it was talked about a lot at the rally.