[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 535
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No. This is the only figure without a distinguishing feature.
10*(х^2)+5(y^2)-2xy-38x-6y+41=0
3*(x^2)-2(y^2)+5xy-17x-6y+20=0
the roots are only valid, I forgot to say.
Well, it's not so interesting anymore. The previous one was better.
This system can be solved through the discriminant of the system, without much trouble (but I don't remember how).
Neutron:
По сюжету утверждается, что для установления контакта с любым жителем планеты, в среднем, достаточно шести человек первый из которых является твоим знакомым, второй - знакомым первого и т.д. Это так называемая теория шести рукопожатий.
MetaDriver:
Well, let's have some fun. It's Friday, after all. :)
What shall we analytically decide? We shall check and estimate reasonableness of the theory (which is easier) or we shall search concrete "friends in the sixth degree" (which is more difficult, because it is necessary to make something like a database).
And I turned off the Internet on Friday and went to the sauna (which I own) to get it ready for Saturday. On Saturday it started... Anyway, it's only now that I'm slowly becoming aware of myself in the workplace.
As for the proposed task, I haven't progressed an inch in solving it. It seems more reasonable to get a theoretical proof, the rule of six handshakes.
I see the following scheme: Let there be a two-dimensional rectangular coordinate system in the plane. In the nodes of the coordinate grid there are identical people with the same Gaussian profile of probability density distribution of having an acquaintance depending on the distance from the node, i.e. it is an analogue of "circle of acquaintances". The integral of the two-dimensional Gaussian should give the total number of acquaintances for a given node. Let this number for all node-persons is the same and equal to N.
Then you have to come up with a condition of detecting the familiar at a distance from node R. Somehow...
An ordinary tree, with six branches from the beginning, then five each.
The graphic representation is a honeycomb.
An attempt at a rough estimate.
About 25 years ago, as one part of a brain exercise, I sat down to write a list of my acquaintances. I wrote about two hundred people, and then stopped when I discovered that the criterion was not clear.
If I had started writing about people I don't know very well, I would have written the same number of people. But I'm not a very sociable person. Rather the opposite.
I was very surprised then, when I started to write the list it seemed to me that it would come to 40 people at the most... :)
But let's assume that the "standard citizen" knows fewer people. Let it be for example 150. (Let the figure be "slightly" underestimated).
Further suppose that my "circle of acquaintances" with each of my acquaintances overlaps by 50%. (The estimate of overlap is exaggerated. I think the real figure is 30 per cent, at the most).
That leaves 75 "fresh acquaintances" per iteration step from each acquaintance from the previous step.
So with each handshake we have an expansion of the circle as a power function of 75. The calculator says 75^6 = 177,978,515,625. There are about 7,000,000,000 people living on Earth.
Even taking into account the uneven distribution of my (and not only) acquaintances over the Earth, one has to admit that the "theory of six" is quite reasonable, and possibly over-insured. :)
--
A little more reasoning. The habitat of my familiars is definitely not distributed according to Gauss. Looking around I observe similar structures in others. Something like forex, with obviously thick tails.
I can write a list of twenty or thirty people I know from out of town quite easily. These are only people I've met/crossed paths with in person. Absentee Internet acquaintances do not count.
Plus to the Russian from other parts of Russia, there are eight or ten foreigners.
With such a structure of distribution of acquaintances by territory, it seems to me that distances in iterations are overcome quite easily.
--
Logically.
You can write an identity: N^6=7*10^9, where N is the average number of acquaintances in a large sample. Therefore N=exp{10/6*ln(10)}=46 people. Each of us can give out up to fifty new friends. Sounds about right. It wasn't a difficult task. Thanks, MetaDriver.
Integer:
The graphic is a honeycomb.
Can you explain the solution in more detail?
Here is another problem that I managed to solve and if anyone has a ready solution, let's compare:
We need to find formulas for uniquely determining the coefficients a,b and c of an equation with two unknowns by the MNC Gaussian method, if the necessary and unbounded array of raw data on the values of Y is known with corresponding values of X and Z :
Y = a + bX + cZ