Re:A fast 3D collision detection algorithm
by OlympicMarmoto on Wednesday July 09, 2025 @03:45PM
(#44514056)
This is the epigraph form of the problem. You try to find the point with the lowest height in the epigraph.
Re:A fast 3D collision detection algorithm
by msteffen on Thursday July 10, 2025 @12:32AM
(#44517198)
Ah, got it, thanks!!
But why would d be much greater. The problem asks to minimise d, and so it cannot be greater than the smallest |x-y|^2.
Re:A fast 3D collision detection algorithm
by mathgradthrow on Wednesday July 09, 2025 @01:53PM
(#44512964)
I can't read substack on my phone, so I can't see the article, but the correct statement that is closest to what you have written is just that d is any real number satisfying this inequality. We define a subset U of AxBxR by
U={(a,b,x):x>|a-b|^2}
and then were looking for the infimum of (the image of) U under the third coordinate function
d(a,b,x)=x
Re:A fast 3D collision detection algorithm
by markisus on Wednesday July 09, 2025 @05:43PM
(#44515013)
I think you are missing that d, x, and y are variables that get optimized over. Any choice of d lower than the the solution to 1) is infeasible. Any d higher than the solution to 1) is suboptimal.
edit: I see now that the problem 2) is missing d in the subscript of optimization variables. I think this is a typo.
1) min_{x,y} |x-y|^2
2) = min_{x,y} d What is 'd'? If d is much greater than |x-y|^2 at the actual (x, y) with minimal distance, and equal to |x-y|^2 at some other (x', y'), couldn't (2) yield a different, wrong solution? Is it implied that 'd' is a measure or something, such that it's somehow constrained or bounded to prevent this?