This is a simple-to-state problem my friend spent too long explaining to me today about which I know nothing. I thought it might be a good candidate for MMM (Massively Multiplayer Mathematics) of the smart people here. So please, throw out ideas, however small.

Let's say you have a sequence of vertices [imath]v_k, k \in [h], n \le h \le 2^n[/imath] from the n-dimensional {0,1} hypercube. From this we create an [imath]n+1[/imath]-dimensional set [imath]S = conv(<v_k,k>)[/imath], the convex hull of the sequence of vectors with their index appended. Now, we choose a constant r, such that each level set in S, [imath]L_k=\{x \in S:x_{n+1} = k\}[/imath] is contained in an n-dimensional ball, centered appropriately, of radius r.

The problem: For any h, choose the vector sequence [imath]v_k[/imath] so that r is minimized.

Questions: Is there an LP formulation of this problem? (I would guess that such a formulation would be unwieldy or unhelpful, or else this problem would not still be a problem.)

Is there a minimum h below which all solutions are trivial? (It seems to me that if h<2^i, then there is an easy upper bound on r depending on i.)

Are there any greedy-ish heuristic worth trying? Like using consecutive vertices in any hamilton cycle (i.e. a gray code ordering)? (I doubt things are this simple, but stupid ideas are worth taking a stab at first.)

## Stacked Hypercube Vertex Convex Hull Boundary Minimization

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### Re: Stacked Hypercube Vertex Convex Hull Boundary Minimization

Can you describe this better? Here is my best translation:

For some n, take an h such the n<=h<=2^n. Pick a sequense of points v_k in a n-dimensional {0,1} hypercube. From this we create an n+1-dimensional set S as the convex closure of all of the <v_k,k>. Now, we choose the minimal constant r, such that each L_k, defined as the set of points x in S such that *something*=k, is contained in an n-dimensional ball of radius r.

The problem: For any n and h, choose the sequence such that r is minimized.

Is there any misunderstanding? and what exactly is x

For some n, take an h such the n<=h<=2^n. Pick a sequense of points v_k in a n-dimensional {0,1} hypercube. From this we create an n+1-dimensional set S as the convex closure of all of the <v_k,k>. Now, we choose the minimal constant r, such that each L_k, defined as the set of points x in S such that *something*=k, is contained in an n-dimensional ball of radius r.

The problem: For any n and h, choose the sequence such that r is minimized.

Is there any misunderstanding? and what exactly is x

_{n+1}supposed to mean?I NEVER use all-caps.

### Re: Stacked Hypercube Vertex Convex Hull Boundary Minimization

You repeated it the same. [imath]x_{n+1}[/imath] is simply the last component of the vector x. What would be better notation for that? Anyway, the point is that it is a level set: The set of points whose height in dimension n+1 is k.

- parallax
**Posts:**157**Joined:**Wed Jan 31, 2007 5:06 pm UTC**Location:**The Emergency Intelligence Incinerator

### Re: Stacked Hypercube Vertex Convex Hull Boundary Minimization

Choose h vertices from a n-dimensional hypercube. Assign each point a "Temperature" T from 1 to h, without replacement. The temperature varies linearly between the points. Minimize the size of all the isothermal surfaces.

Cake and grief counseling will be available at the conclusion of the test.

- Yakk
- Poster with most posts but no title.
**Posts:**11129**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

### Re: Stacked Hypercube Vertex Convex Hull Boundary Minimization

quintopia wrote:This is a simple-to-state problem my friend spent too long explaining to me today about which I know nothing. I thought it might be a good candidate for MMM (Massively Multiplayer Mathematics) of the smart people here. So please, throw out ideas, however small.

Let's say you have a sequence of vertices [imath]v_k, k \in [h],[/imath]

What is [h]? The set of integers from 1 to h inclusive? 0 to h-1 inclusive? The operator that takes polynomials in h and extracts the linear term?

[imath]n \le h \le 2^n[/imath] from the n-dimensional {0,1} hypercube. From this we create an [imath]n+1[/imath]-dimensional set [imath]S = conv(<v_k,k>)[/imath], the convex hull of the sequence of vectors with their index appended.

<,> usually refers to inner products, so your use is very confusing. I think you are using it to mean <a,b>: (F^n x F) -> F^(n+1) := (a_0, ..., a_n, b)?

Now, we choose a constant r, such that each level set in S, [imath]L_k=\{x \in S:x_{n+1} = k\}[/imath] is contained in an n-dimensional ball, centered appropriately, of radius r.

So L_k is your level set in S. So r measures how 'fat' the level set is at a point is being measured by the r, I suppose, in a round about sense.

The problem: For any h, choose the vector sequence [imath]v_k[/imath] so that r is minimized.

So we want the 'fattest' part of our convex hull to be thin, in the above round sense.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

### Re: Stacked Hypercube Vertex Convex Hull Boundary Minimization

Yakk wrote:What is [h]? The set of integers from 1 to h inclusive?

Yes. It's a fairly common shorthand.

Yakk wrote:I think you are using it to mean <a,b>: (F^n x F) -> F^(n+1) := (a_0, ..., a_n, b)?

yes, hence "the sequence of vectors with their index appended." I just made up the notation on the spot, because everything else I thought of would have been more confusing.

- parallax
**Posts:**157**Joined:**Wed Jan 31, 2007 5:06 pm UTC**Location:**The Emergency Intelligence Incinerator

### Re: Stacked Hypercube Vertex Convex Hull Boundary Minimization

For example, if n=2 and h=4 you assign:

(0,0) -> 1

(1,0) -> 2

(0,1) -> 3

(1,1) -> 4

The level set for 2 goes from (0,0.5) to (1,0). so r is [imath]\sqrt 5 / 2[/imath].

This is the minimal arrangement.

(0,0) -> 1

(1,0) -> 2

(0,1) -> 3

(1,1) -> 4

The level set for 2 goes from (0,0.5) to (1,0). so r is [imath]\sqrt 5 / 2[/imath].

This is the minimal arrangement.

Cake and grief counseling will be available at the conclusion of the test.

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