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I'm trying to work through the math here, and I don't understand why these two propositions are equivalent: 1) min_{x,y} |x-y|^2 x ∈ A y ∈ B 2)
by msteffen 1y ago
I'm trying to work through the math here, and I don't understand why these two propositions are equivalent:
1)
min_{x,y} |x-y|^2
x ∈ A
y ∈ B
2)
= min_{x,y} d
d ≥ |x-y|^2
x ∈ A
y ∈ B
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?
- mathgradthrow 1y agoI 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
- Jaxan 1y agoBut why would d be much greater. The problem asks to minimise d, and so it cannot be greater than the smallest |x-y|^2.
- OlympicMarmoto 1y agoThis is the epigraph form of the problem. You try to find the point with the lowest height in the epigraph. https://en.wikipedia.org/wiki/Epigraph_(mathematics) https://en.wikipedia.org/wiki/Epigraph_(mathematics)
- msteffen 1y agoAh, got it, thanks!!
- markisus 1y agoI 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.