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I'm trying to understand the closest pair of points problem. I am beginning to understand the two-dimensional case from a question a user posted some years ago. I'll link it in case someone wants to look at it: For 2-D case (plane) - "Closest pair of points" algorithm.

What I'm trying to do is: Given two points $p_1$ and $p_2$ in the plane, the $L_\infty$-distance between them is given by max$(|x_1, x_2|, |y_1,y_2|)$. Modify the closest-pair algorithm to use the $L_\infty$ distance.

From what I understand (thinking about two points in an xy plane) the Euclidean distance would be the line that directly connects the two points. To see what the L-infinity distance looks like, draw a rectangle with the two points at two opposite corners. The L-infinity distance would then be the length of the longest side of the rectangle.

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The only difference is in the "merging" part of the recursion. Let the two sets created by the dividing line be called as $L$ and $R$, respectively (for left and right sets). Via recursion, we have our temporary current closest pair -- let's say its at distance $M$. Similar to the original algorithm, we filter only the points that is within $M$ distance from the dividing line, sort the points by their $y$ coordinates, and sweep the points from top to bottom. For each point $(x', y')$ in $L$, consider the square of size $2M \times 2M$ centered at this point. How many points in $R$ within $M$ distance from the dividing line can be inside this square at most? $6$ points (I'm being ultra conservative here), and those corresponds to the points in $R$ whose $y$ coordinate is between $y' - M'$ and $y' + M$. Thus, a similar reasoning holds as in the original algorithm here.

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