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Questions - divide-and-conquer

divide rectangle in random sized smaller rectangles

Im doing an art project where i really need to be able to divide a rectangle in a random way. Ive have been searching alot for this but have not found anything i really can implement in proccesing. Th...
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Soma Juice

processing

divide-and-conquer

kdtree

Votes: 0

Answers: 1

Latest Answer

This (p5js) sketch divides a rectangular plane in random sized smaller rectangles and seems suitable for your need. The sketch slices (either horizontally or vertically) the plane recursively until a ...
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micycle

Closest pair algorithm - Optimizing my code for huge sets of points

I'm trying to create an algorithm that finds the closest pair of points (2D) within a set of points. I'm using a divide and conquer method which is explained here Closest Pair of Points Algorithm. How...
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Lamett

python

optimization

divide-and-conquer

Votes: 0

Answers: 0

Checking whether an Array equals another Array Backwards using divide and conquer

I've been trying to make a simple function which checks if an array is the same as another array reversed. For instance, comparing [0, 1, 2] and [2, 1, 0] would return true, but comparing [0, 1, 2] an...
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AWDn0n

java

arrays

algorithm

recursion

divide-and-conquer

Votes: 0

Answers: 1

Latest Answer

There's a couple of issues in your solution: The base case has a logical flaw in it. The return value represents the result of comparison a[start] == b[end] when indices are equal. Which isn't the co...
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Alexander Ivanchenko

Solving recurrence of divide and conquer using master theorem

Can T(n) = 3T(n/2) + c T(1)=0, solved using master theorem? If yes, I am struggling on understand master theorem now, could someone tell me which case it falls to and why.
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MangHatDa

time-complexity

recurrence

divide-and-conquer

master-theorem

Votes: 0

Answers: 1

Latest Answer

There are different possible approaches for the master theorem. I like the one proposed by Cormen et al. in their book Introduction to Algorithms. If f(n) = O(n^(logb(a-e))) for some constant e>0,...
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Joao Paulo

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