It is based on the sparse grid method which has been shown to give good results for low- and moderate–dimensional problems. The dimension–adaptive quadrature method which is developed and presented in this paper aims to find such an expansion automatically. The problem, however, is to find a good expansion given little knowledge of the integrand itself. Due to the concentration of measure phenomenon, such functions can often be well approximated by sums of lower–dimensional terms. Here, we especially address the high–dimensional case, where in general the curse of dimension is encountered. Just not a pentagon or a triangle.We consider the numerical integration of multivariate functions defined over the unit hypercube. So a hexagon can fit inside a hypercube in this sense. (note: It is possible to do a walk of length 6 in the cube and return back where you started e.g. By restricting to tree-shaped puzzle pieces, he's sure that this is never a problem. The same's true for pentagons, heptagons, and odd length cycles in general. So if your original puzzle piece has a triangle somewhere in it, there's no way to fit it inside of a cube. One reason why his result only works for trees is that there's a natural obstruction to fitting together arbitrary puzzle pieces to make cubes: squares, cubes, and hypercubes in general don't contain triangles. So for example, a cube is not a tree because you can just travel around one of the faces.Īnother way of thinking about trees is that you start at a single vertex (the root) and branch outwards from there, with the different branches never rejoining together. One way of thinking about a tree is a graph that's connected (it's possible to get from any point to any other point), but has no cycles (You can't start from point A and get back to it unless you just backtrack the same path you took). What he has shown is that, no matter what the puzzle piece looks like, you can always combine a bunch of them to make a hypercube of some dimension. For example, the cube has 12 edges - if you want to build it out of pieces with 5 edges each, it'll never come out even. the L's), and want to put them together to get a bigger piece. Think of it as a sort of puzzle: You have a bunch of smaller pieces (e.g. So for example, you can think of the square as being made up of two L's (each centered at an opposite corner) or the cube as either being made up of two squares or of 6 L's (the latter one takes a bit of thought). Essentially it states you can take a big graph (the hypercube, in his case), and divide it up into a bunch of smaller identical graphs. What the man proved was what's termed a decomposition theorem. (☑, ☑, ☑, ☑, \dots, ☑), where two points are connected if they only differ in one place. You can think of it as all the points of the form So an n-dimensional hypercube has 2 n vertices, and each vertex is connected to n other vertices. Each time you go up a dimension, you increase the number of connections per vertex by 1. It has 8 vertices (the corners of the cube), with each vertex connected to three other vertices.Įach time you go up a dimension, you double the number of vertices. A 3 dimensional hypercube is what you normally think of as a cube.As a graph, we think of it as 4 points (the corners of the square), with each point connected to 2 other points (its neighbors) A 2 dimensional hypercube is a square.As a graph, we think of it as two points, with one edge connecting them.
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