N-dimensional geometric partitioning

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Discussion Overview

The discussion centers on the enumeration of polytopes defined by n+1 points in n-dimensional Euclidean space, exploring combinatorial aspects and comparing growth rates of polytopes to other mathematical constructs like partitions and the Traveling Salesman problem.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that the number of polytopes defined by n+1 points in n-dimensional space can be calculated as 2^(n+1) - n - 2.
  • Another participant suggests that any combination of 2 or more points would yield a valid polytope, leading to the same formula.
  • A different participant expresses a desire to develop a combinatorial approach that surpasses existing methods of partitioning, questioning the derivation of a heuristic for this problem.
  • One participant believes that the count of polytopes should yield a series that increases more rapidly than ordinary partitions, citing the Traveling Salesman problem as an example of rapid growth in combinatorial paths.
  • Another participant estimates that the number of polytopes could be on the order of p[p[n]], where p[n] is the partition of n, suggesting that this grows faster than the previously mentioned formula.
  • One participant contrasts the growth rates of polytopes and paths in the Traveling Salesman problem, noting that the latter grows factorially while the former grows exponentially.
  • Discussion includes references to extremely fast-growing functions, such as the Ackermann function and the Busy Beaver function, with one participant providing a brief explanation of the Busy Beaver function.

Areas of Agreement / Disagreement

Participants express differing views on the growth rates of polytopes compared to partitions and other combinatorial constructs. There is no consensus on the best approach to derive or understand these relationships, and multiple competing models are presented.

Contextual Notes

Participants mention various mathematical constructs and their growth rates without resolving the complexities involved in their comparisons. The discussion includes assumptions about the definitions and properties of polytopes and partitions that remain unexamined.

Who May Find This Useful

This discussion may be of interest to those studying combinatorial mathematics, geometric properties of polytopes, or the growth rates of mathematical functions.

Loren Booda
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Given n+1 points in n-dimensional Euclidean space, how many polytopes (generalizations of polygons of n to as few as 2 dimensions) may be defined by the representation of each point as a possible vertex?
 
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If I'm understanding you correctly, then any combination of 2 or more types would yield a valid polytope.

So 2^(n+1) - 1 - (n+1) = 2^(n+1) - n - 2
 
...n...k.....0...1...n+1
sum...C...=2^(n+1)-C...- C...- C...= 2^(n+1)-n-2;
.k=2...n+1.....n+1...n+1...n+1

(using Newton's bynom...(bad english));
Exactly as damgo said...
Take the . as space (' ')
 
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I am trying to develop a combinatorics which surpasses the progression of partitioning, thus the above definition. Even partitioning does not have an exact heuristic to determine its nth term.

Can anyone show the derivation of an approximate heuristic for this problem, or at least whether the magnitude of its nth term is significantly greater than that of partitions?
 
^^^ Can you restate that? I don't understand what you're saying. I can derive my formula if that's what you want...
 
Take n+1 points in n dimensions. Count all possible 2-dimensional polygons and their n-dimensional generalizations, formed with any or all points as vertices.

I believe that the count should yield a more rapidly increasing series than that of ordinary partitions. An example of partitioning is the simply stated Traveling Salesman problem, which asks to count the possible paths covering all "cities" on a map. The number of paths increases astronomically against the number of locations.
 
The number of polygons is 2^(n+1)-n-2, which increases astronomically against the number of vertices.
 
My guess is that the number of polytopes outlined by arbitrary connections between n points in n dimensions is on the order of p[p[n]], where p[n] is the partition of n. Even [p[n]] (recall the "Traveling Salesman" problem) eventually increases much more rapidly than 2^(n+1)-n-2, or any other exponential relation. In turn, p[p[n]] almost immediately reaches incalculable numeration.
 
The # of paths in the traveling salesman problem go as Permutations(n) = n! ~ (n/e)^n (Stirling's approx). The # of polytopes is slower, ~2^n , since order does not matter.

If you want to see *really* quickly increasing things, try Ackerman's function, or the Busy Beaver function.
 
  • #10
I've seen the Ackermann function (and boy does it grow fast!), what's the busy beaver function?

Hurkyl
 
  • #11
Consider Turing machines with 2 symbols (call them 0 and 1) and n states, starting on a blank (0's) tape. Look at just the ones that eventually halt with a consecutive series of 1's on the tape. BB(n) is the length of the longest series of 1's produced in this manner.

Roughly, it's the biggest number a computer with n states and infinite storage can create. It's known for n<5... BB(6) is crazy huge, at least 10^100. It's provably uncomputable, and IIRC you can prove that it grows faster (in a rough sense) than any computable function.
 

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