[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 475
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At Alexey's request and my personal interest in understanding the process of speculative trading ;) I will duplicate my post https://www.mql5.com/ru/forum/101846/page15:
In order to define the concept of volumes and the most trivial notion of how the market works, we can try to simulate the market with a primitive model:
- let's have 10 people, 5 of them have 100 EUR and the other 5 have 100 USD.
- in the initial state, the price is 1EUR=1USD.
- All 10 people want to exchange their money at a certain profit, i.e. nobody is willing to do it at the rate of 1:1.
_______________________________________________________________________________________________________
How would the exchange rate look like if
1. one of the participants leaves with its money in USD, and comes back a few hours later?
2. one of the exchangers left with their USD money, and a few hours later came back, but somewhere along the way managed to buy another 100USD?
Igor,
Such a model, and this can be said beforehand, will knowingly have nothing in common with the real market, because we lose its most important feature - fractality. I.e. in reality it is a big amount of traders that creates the picture we see: for example (roughly), if we take a group of 10000 traders and see how its behavior is influenced by subgroups of, say, 1000 people, then we will get the same picture as if we take 1000 people and break them into subgroups of 100. All scales together give a self-similarity in both the price chart and the statistical characteristics. Without this effect, what we see on the graph would be very different.
People are solving the problem on the forum of the Faculty of Mechanics:
дана матрица 5х5, состоящая из нулей и единиц, причем в каждой строке и каждом столбце ровно по 3 единицы. Найти количество способов составить такую матрицу.
(The correct answer has already been found by brute force, but there is no analytical solution yet)
P.S. no peeking:)))
Give me the numbers, we'll come up with some
People are solving the problem on the forum of Mechmatas:
(the right answer has already been found by brute force, but no analytical solution yet)
P.S. no peeking:)))
5! * 5!
?
Петя заметил, что у всех его 25 одноклассников различное число друзей в этом классе. Сколько друзей может быть у Пети?
Comment:
1. Petya is also in this class, so there are 26 people in the class.
2. If A is friends with B, then B is friends with A.
Find all the solutions.
How many friends can Peter have?
Answer: as many as he wants...
w.s. Whatever the condition is, so is the solution.
lol)))
Mathematics is a corrupt science wench that is ready to derive any formula under any condition and give the scientist what he wants from it...5! * 5!
?
lol101:
lol)))
A fun problem about arranging the units in a matrix. Well, we have to start somewhere. Trying to match at least one such matrix leads to this result:
1 0 0 1 1
1 1 0 0 1
1 1 1 0 0
0 1 1 1 0
0 0 1 1 1
Comparison of the first upper horizontal row with the second leads us to the conclusion that the second row is nothing else but the first one shifted by one position to the right. The rightmost character (the last one in the row) goes out of the matrix and we just put it in the first position, in the vacant place of the first character. Comparing all subsequent lines with previous ones leads to the same conclusion: each subsequent line is the previous one shifted by one position to the right. It is the same for the columns, only shifted vertically. So every line is a looped ribbon and every column is a looped ribbon. It turns out that this is not just a matrix - it is a Karno map. So the problem is not how many ways you can build such a matrix, but how many ways you can build such Karno maps.
Frankly, it seems to me that the ribbon has a single sequence of symbols, namely 00111, where the first zero and the last one are two adjacent symbols of the looped ribbon. If this assumption is correct (about the uniqueness of the sequence), the number of combinations is not difficult to calculate.
It is clear that if the upper ribbon is shifted horizontally, then all other horizontal ribbons should be shifted in the same direction and by the same number of positions. So we have 5 vertical and 5 horizontal shifts of the whole map field. For every vertical shift, there are 5 horizontal ones. The total is 5*5. But we can rotate the box. Let's paint the top line blue. How many positions will the square have? Blue top, blue right, blue bottom, blue left. In total there are 4 positions. Therefore we have 5*5*4 = 100 ways to build the given Karno map.
It remains to prove that the arrangement of symbols in the looped tape 00111 is the only one. For example, at no shifts and no turns do we encounter the sequence - 01011