- #1
jacobrhcp
- 169
- 0
Hi!
I stumbled upon something I need to implement my problem in a computer. Because I think its actually fun to puzzle on, I thought it might be nice to post it here. The problem is this:
If you have a 4D hypercube, [itex](a_1,b_1)\times(a_2,b_2)\times(a_3,b_3)\times(a_4,b_4)[/itex], what is an easy algorithm to find a set of max 4+1 points that encompasses the hypercube?
For 2D, ie [itex](a_1,b_1)\times(a_2,b_2)[/itex], its easier, as you can find 2+1 points by taking the points [itex](a_1-|a_1-b_1|,a_2), (b_1+|a_1-b_1|,a_2)[/itex], and the point that is the intersection of the lines defined by the points [itex](a_1-|a_1-b_1|,a_2), (a_1, b_2)[/itex] and [itex](b_1+|a_1-b_1|,c), (b_1,b_2)[/itex], respectively (if you draw this it becomes clear immediately =P). If you expand this procedure to 4D, things get complicated fast. Is there a simple way to find a set of max 5 points that encompasses my hypercube?
Edit: I noticed I say 'cube', but as [itex]a_i, b_i \in ℝ[/itex], it doesn't need to be a cube at all of course.
I stumbled upon something I need to implement my problem in a computer. Because I think its actually fun to puzzle on, I thought it might be nice to post it here. The problem is this:
If you have a 4D hypercube, [itex](a_1,b_1)\times(a_2,b_2)\times(a_3,b_3)\times(a_4,b_4)[/itex], what is an easy algorithm to find a set of max 4+1 points that encompasses the hypercube?
For 2D, ie [itex](a_1,b_1)\times(a_2,b_2)[/itex], its easier, as you can find 2+1 points by taking the points [itex](a_1-|a_1-b_1|,a_2), (b_1+|a_1-b_1|,a_2)[/itex], and the point that is the intersection of the lines defined by the points [itex](a_1-|a_1-b_1|,a_2), (a_1, b_2)[/itex] and [itex](b_1+|a_1-b_1|,c), (b_1,b_2)[/itex], respectively (if you draw this it becomes clear immediately =P). If you expand this procedure to 4D, things get complicated fast. Is there a simple way to find a set of max 5 points that encompasses my hypercube?
Edit: I noticed I say 'cube', but as [itex]a_i, b_i \in ℝ[/itex], it doesn't need to be a cube at all of course.
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