Timeline for Is the direction of the longest line of a polytope unique?
Current License: CC BY-SA 2.5
11 events
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| Jan 9, 2010 at 5:22 | comment | added | j.c. | You are right that the plane retains its orientation, however the shape of the rectangle that is the intersection with the cube changes. For my example, p=3, m=1, hence X u = y can be written as N.u=a where N is a normal vector to the plane L and a is the distance of L to the origin along N. Let N = (1,1,0) and a=1. Then the intersection of L with A_v (for v>1) is the rectangle with vertices (1,0,0), (0,1,0), (0,1,v), (1,0,v). One longest segment is the one joining (1,0,0) and (0,1,v) which is parallel to (1,-1,-v) which clearly changes as v increases. | |
| Jan 9, 2010 at 2:41 | comment | added | some_random_guy | Never mind my comment reg $P_v$. The argument does not work. However, I am confused about why you think a rectangle that is perpendicular to the top and the bottom faces of a cube becomes skinnier as $v$ increases. Such a plane will always have the same shape irrespective of the value of $v$. As $v$ increases, the plane shifts towards the origin but retains its orientation. (Note: The longest line is not unique as either diagonal is a longest line but that is a different issue.) | |
| Jan 8, 2010 at 21:40 | comment | added | j.c. | I've updated my answer to show that the answer to your question is no. For planes L which approach only one vertex, I still think the answer is yes. @Srikant, I'm not sure what you mean exactly. P_v is defined to be the intersection of the hypercube A_v (unit hypercube in the rescaled coordinates) with L, so it always lies in a hypercube... Considering the boundaries of P_v as you suggested might be a good approach though. | |
| Jan 8, 2010 at 21:34 | history | edited | j.c. | CC BY-SA 2.5 |
the answer is no.
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| Jan 8, 2010 at 21:29 | comment | added | some_random_guy | How about this: Define $S$ as the set of points where $L$ intersects the half-planes defined by $u'_i = 0$ and u'_j >= 0 for all $j$ not equal to $i$. I guess we are done if we can show that for some $v > v^*$ all the points in the set $S$ are less than $1$. The above would suggest that $P_v$ lies entirely in the hypercube and hence 'retains' its shape. Does that make sense? ps; I am unable to edit my comments after posting. I hope the above is clear. | |
| Jan 8, 2010 at 21:23 | history | edited | j.c. | CC BY-SA 2.5 |
mistake
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| Jan 8, 2010 at 20:48 | history | edited | j.c. | CC BY-SA 2.5 |
oops typo
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| Jan 8, 2010 at 20:48 | comment | added | j.c. | Yes, sorry. I've updated my answer. | |
| Jan 8, 2010 at 20:33 | comment | added | some_random_guy | A clarification: Do you mean the transformation $u'\rightarrow u/v$ instead of $u \rightarrow vu$? | |
| Jan 8, 2010 at 20:29 | history | edited | j.c. | CC BY-SA 2.5 |
more details
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| Jan 8, 2010 at 20:16 | history | answered | j.c. | CC BY-SA 2.5 |