[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 559

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alsu:
exactly the same as the probability of it hitting the "right" plane, i.e. zero ))
and we don't care which one it hits, as long as it's not on the "right" one. Everyone else is the "right" one. :))
 
MetaDriver:
and we don't care which one he ends up in as long as it's not the "unnecessary" one. Everyone else is the right one. :))
There's only one right one, there's an endless number of unnecessary ones. The task is to calculate the right one.
 
Put an arbitrary vector into my example, and you will see that the result is different from the desired one, and each time in a different way.
 
alsu:
the necessary one, the unnecessary ones an infinite number. The challenge is to calculate the right one
It is exactly the opposite - there is only one unnecessary one (i.e. according to the full algorithm several == CountInput), while the right ones are a dime a dozen.
 
alsu:

Checked))

The transformation is, of course, strictly planar, and the result is generally accurate to a sign regardless of the choice of the original arbitrary vector - but! only in this plane. Who told us that out of an infinite number of options to draw a plane through a given vector, we have chosen the right one?

Here is an example. Suppose you have two vectors in 3-dimensional space: (1,0,0) and (0,sqrt(2),sqrt(2)). They are orthogonal, as you can see. You started by taking an arbitrary x1 in the plane z=0 and using it to construct an orthogonal vector (0,1,0) to the first vector. We obtain that the algorithm is complete, but the result is not obtained - the third vector is not orthogonal to the remaining second vector. And in order to get the right answer, you need to take care beforehand to choose the right plane during the first construction - and then you will come to the variant (0,-sqrt(2),sqrt(2)) or the second possible solution.

That's not the end of the algorithm at all !!!

Read my pseudocode. There the algorithm does not stop here, but skips to the next iteration, until the input vectors are exhausted.

And I argue that orthogonality with previous processed input vectors is not destroyed by the described iterations. This follows from the condition of orthogonality and normality of input vectors.

 
MetaDriver:

That's not the end of the algorithm at all!

Read my pseudocode script. There the algorithm does not end there, but just moves on to the next iteration - until the input vectors are exhausted.

And I claim that orthogonality with previous processed input vectors is not broken during the described iterations. This follows from the condition of orthogonality and normalization of input vectors.

OK, maybe I'm stupid. Spell out the next step - there aren't many vectors left.
 
alsu:
OK, maybe I'm stupid. Spell out the next step - there aren't many vectors left.
That's it, no need, three-dimensional case I've got it.
 

The pseudocode already has all the steps. look again.

there's a pass through all the inputs.

 
alsu:
That's it, I've got the three-dimensional case.

Can you confirm?

;)

 

In the case of N=M+1 you really get the result immediately in the desired plane and can rotate your vector to complete orthogonality.

But if N>M+1 it is possible that after the next iteration you find yourself in the region of space where there are simply no planes containing vectors from the initial set. What to do in this case?

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