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December 26

Need the definitions

Hi friends, I need to know the difference between these two phrases 1 Asymptotically perfect 2 Asymptotically ideal

If this is not the right link or way to ask a query, then please do let me know whom to contact regarding any queries?

Any help would be much obliged. Thanks Manjula R

I've not heard those terms, but can guess that one is a curve that approaches an asymptote from one side, while another approaches from both sides, like sine(x)/x. StuRat (talk) 01:04, 30 December 2013 (UTC)[reply]

Finding your way out of a catacomb

A girl died after getting lost in a catacomb. Here is a partial map of the catacombs. Is there a solution that you could follow methodically to find your way out? — Preceding unsigned comment added by 78.148.110.243 (talk) 04:26, 26 December 2013 (UTC)[reply]

Wikipedia has an article on everything: see Maze solving algorithm, but note that the Trémaux's algorithm requires some way to mark positions and recognise the marks. This would be very difficult in the dark. Dbfirs 08:25, 26 December 2013 (UTC)[reply]

For a certain type of maze, where there are no loops, either putting your right hand or left hand on a wall and following that wall through every turn might find your way out. If one hand doesn't work, you can try the other, for a second try, but if there is a loop on both sides you are out of luck, and, unfortunately, those catacombs are full of loops. StuRat (talk) 14:36, 26 December 2013 (UTC)[reply]

I suppose in catacombs, there are lots of bones to be had, so you could line each corridor with bones to show you have already explored that one. Still, based on the size of those catacombs, I'm not sure you could find your way out in time. StuRat (talk) 14:43, 26 December 2013 (UTC)[reply]

Find some maps here http://www.atlasobscura.com/articles/mind-boggling-catacomb-maps
The system of tunnels is 2,500 km long in total. --CiaPan (talk) 09:16, 27 December 2013 (UTC)[reply]

Yea, with a maze that long, an exhaustive search, where you try every path, just isn't likely to work in time. You need to have some indication as to which path is correct, before you try it out. StuRat (talk) 01:02, 30 December 2013 (UTC)[reply]

Find the coefficients of a convex combination for an arbitrary point of the unit cube

For an arbitrary point in the unit cube, how do I find the coefficients of a convex combination?

It blows my mind. Didn't learn anything in school, can't make it out myself and don't find anything on google.

As far as I can think, in the general case I really need 8 coefficients (where 4 can be chosen as zero, but different ones depending on the point).

And another question arises in this context: as the choice of coefficients is not unique and mathematicians tend to make a theory out of everything, what is the proper name for the theory of this coefficient space?

77.3.148.237 (talk) 17:02, 26 December 2013 (UTC)[reply]

I gather you want to write an arbitrary point in the interior of the unit cube as a convex combination of the eight vertices. Sort the coordinates to get three numbers a≤b≤c with 0≤a and c≤1. Write a=x, b=a+y=x+y, c=b+z=x+y+z, 1=c+w=x+y+z+w. Then (a, b, c) = x(1, 1, 1) + y(0, 1, 1) + z(0, 0, 1) + w(0, 0, 0). It only remains to rearrange the coordinates into their original order. For example (.5, .2, .3) = (.2+.1+.2, .2, .1+.2) = .2(1, 1, 1)+.1(1, 0, 1)+.2(1, 0, 0)+.5(0, 0, 0). The space of coordinates would be 8-tuples of numbers (x1, x2, ... x8) with xi ≥ 0 and sum(xi)=1; this is called the 7-simplex. It maps to the cube by mapping each of its eight vertices to a vertex of the cube and extending affinely. --RDBury (talk) 03:11, 27 December 2013 (UTC)[reply]

I don't think the general case is as easy as that. If (a, b, c) lies within the tetrahedron given by your base points it will work, otherwise your coefficients will lie outside [0,1] or will not sum up to 1. 95.112.154.39 (talk) 09:17, 27 December 2013 (UTC)[reply]

Now I see what you meant by "sort the coordinates", you choose the appropriate tetrahedron, and (0,1,1) can be any of (1,1,0), (1,0,1). (0,1,1), same with (0,0,1). 95.112.154.39 (talk) 10:00, 27 December 2013 (UTC)[reply]

There is a solution using polynomial expressions as well, namely:

(u, v, w) = uvw(1, 1, 1) + uv(1-w)(1, 1, 0) + u(1-v)w(1, 0, 1) + u(1-v)(1-w)(1, 0, 0) + (1-u)vw(0, 1, 1) + (1-u)v(1-w)(0, 1, 0) + (1-u)(1-v)w(0, 0, 1) + (1-u)(1-v)(1-w)(0, 0, 0)

--RDBury (talk) 03:55, 28 December 2013 (UTC)[reply]