Search results
Results From The WOW.Com Content Network
Conversely, it is possible to 2-colour a K 5 without creating any monochromatic K 3, showing that R(3, 3) > 5. The unique [b] colouring is shown to the right. Thus R(3, 3) = 6. The task of proving that R(3, 3) ≤ 6 was one of the problems of William Lowell Putnam Mathematical Competition in 1953, as well as in the Hungarian Math Olympiad in 1947.
When x i = 0, one has RP n−1. Therefore the n−1 skeleton of RP n is RP n−1, and the attaching map f : S n−1 → RP n−1 is the 2-to-1 covering map. One can put =. Induction shows that RP n is a CW complex with 1 cell in every dimension up to n.
R 4 can be imagined using the fact that 16 points (x 1, x 2, x 3, x 4), where each x k is either 0 or 1, are vertices of a tesseract (pictured), the 4-hypercube (see above). The first major use of R 4 is a spacetime model: three spatial coordinates plus one temporal .
The animation that is looped above right provides an example of a homotopy between two embeddings, f and g, of the torus into R 3. X is the torus, Y is R 3, f is some continuous function from the torus to R 3 that takes the torus to the embedded surface-of-a-doughnut shape with which the animation starts; g is some continuous function that ...
Two metrics and on X are strongly or bilipschitz equivalent or uniformly equivalent if and only if there exist positive constants and such that, for every ,, (,) (,) (,).In contrast to the sufficient condition for topological equivalence listed above, strong equivalence requires that there is a single set of constants that holds for every pair of points in , rather than potentially different ...
The hydraulic diameter is the equivalent circular configuration with the same circumference as the wetted perimeter. The area of a circle of radius R is . Given the area of a non-circular object A, one can calculate its area-equivalent radius by setting = or, alternatively:
In order to prove that 01-Permanent is #P-hard, it is therefore sufficient to show that the number of satisfying assignments for a 3-CNF formula can be expressed succinctly as a function of the permanent of a matrix that contains only the values 0 and 1. This is usually accomplished in two steps:
A space is T 1 if and only if it is both R 0 and T 0. A finite T 1 space is necessarily discrete (since every set is closed). A space that is locally T 1, in the sense that each point has a T 1 neighbourhood (when given the subspace topology), is also T 1. [4] Similarly, a space that is locally R 0 is also R 0.