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Given $N$ points where the coordinates of the $i$th point are $(x_i,y_i)$ find $(x,y)$ that minimizes the following sum. $$\sum_{i=1}^ N\max(|x-x_i|,|y-y_i|)$$ I've done a similar problem minimize $$\sum_{i=1}^ N(|x-x_i|+|y-y_i|)$$ Which I found that the minimum is achieved when we take $x$ as the median of $x_1,x_2,\ldots,x_N$ and $y$ as median of $y_1,y_2,\ldots,y_N$.

I'm stuck on the first one because I don't know how to manipulate the $\max$.

Edit 1:

After tinkering with the suggestion below I am still struggling to make progress here: The $\frac{|x-x_i|+|y-y_i|}{2}$ can be minimized by taking the median but I am not sure how to minimize the other part: $\frac{||x|-|x_i||-||y|-|y_i||}{2}$. I also don't know what to do if I could minimize the other part. How can I minimize the sum of these two?

Mason
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kingW3
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    Not a complete answer which finds an expression for the $(x,y)$. However, here is an identity that I think will probably get you where you want to go. $\max{{a,b}} = \frac{a+b}{2} + \frac{|a-b|}{2}$ – Mason Dec 23 '18 at 15:55
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    I hope I didn't rob you of more complete answers by putting something more worthy as a comment as an answer. Anyway: I put a suggested edit and deleted my answer to get this back on the list of unanswered questions+ an edit will bump this back on to active questions. – Mason Dec 23 '18 at 16:06
  • @Mason Nice of you to do that, I don't think you robbed me of an answer. I feel that the question isn't interesting to people and your observation is helpful. – kingW3 Dec 23 '18 at 16:36
  • I gave an answer! Hope it's helpful – user103341 May 06 '24 at 14:44

1 Answers1

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Here is my attempt based on the hint given in the comment: We can take the gradient: $$ \nabla f(x, y) = \nabla \sum_{i=1}^n max(|x-x_i|, |y-y_i|) $$

$$= \frac12 \sum_{i=1}^n \nabla \left( |x-x_i| + |y-y_i| + ||x-x_i| + |y - y_i|| \right) $$

The first element of the gradient is:

$$ \frac{d}{d x} \left(|x-x_i| + |y-y_i| + ||x-x_i| + |y - y_i|| \right) \\ = sign(x - x_i) + \frac{d}{du} |u| * \frac{d u} {dx} \\ = sign(x-x_i) + sign(u) * \frac{d}{dx} \left(|x-x_i| + |y - y_i|\right) \\ = sign(x-x_i) + sign(|x-x_i| + |y-y_i|) * sign(x-x_i) $$

Leaving aside the case $sign(|x-x_i| + |y-y_i|) = 0$ , we can assume $sign(|x-x_i| + |y-y_i|) = 1$ in which case the above simplifies to $$ \frac{d}{d x} \left(|x-x_i| + |y-y_i| + ||x-x_i| + |y - y_i|| \right) = 2 sign(x-x_i) .$$

If you sum over i, it seems the median of $x$ and $y$ will set the gradient to zero, or any point $(x,y)$ so that there are the same number of $x_i$ to the left and right of $x$ and likewise for $y$.

Now we still need to think about what happens when $x=x_i$ and $y=y_i$ for some $i$. Then the gradient looks the same but with the one $i$ term missing. In this case the gradient is zero if and only if $x_i$ is in the middle of the remaining $x_j$ and likewise for $y_i$. But this point falls under the description above.

user103341
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