Cheap and Secure Web Hosting Provider : See Now

[Solved]: Find rectangle of minimum area where dimensions are larger than minimum

, , No Comments
Problem Detail: 

Problem: Given a collection $S$ containing $|S|=n$ rectangles defined by dimensions $(x,y)\in R^2$ (width and height of rectangles are real numbers), find the rectangles with the minimum area ($A_i = x_i * y_i$) where $(x_i \geq a)$ and $(y_i \geq b$) for any $(a,b) \in R^2$.

The naive solution $F(S,a,b)$ will solve this with $O(n)$ runtime complexity and $O(1)$ memory complexity: loop through all the rectangles that are larger than the minimum required values $(a,b)$ for $(x,y)$, remember the one with the smallest area $x*y$ and return that. This requires no preparation or indexing of any kind, just a simple loop (sorting according to area beforehand won't improve the $O(n)$ worst-case runtime complexity).

Can this problem be solved with an algorithm with faster than $O(n)$ runtime complexity and up to O(n) memory complexity?

Asked By : scooz

Answered By : Yuval Filmus

One simple solution uses k-d trees. In preprocessing, prepare a sorted list $L$ of $x_i y_i$ as well as a three dimensional k-d tree $T$ storing the points $(x_i,y_i,x_iy_i)$; this can be done in time $O(n\log n)$ and takes up space $O(n)$. Given $a,b$, we do binary search on $L$ to find the least $c$ such that $T$ contains a point $(x_i,y_i,x_iy_i)$ with $x_i \geq a,y_i \geq b,x_iy_i \leq c$; each such query takes time $O(n^{2/3})$, for a total query time of $O(n^{2/3}\log n)$.

It's not clear whether we actually need a three dimensional k-d tree; if a two dimensional one suffices, then the running time could get down to $\tilde{O}(n^{1/2})$.

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/28338

3.2K people like this

 Download Related Notes/Documents

0 comments:

Post a Comment

Let us know your responses and feedback