We strongly recommend to refer below articles as a prerequisite of this.
In this post, a Monte Carlo algorithm is discussed.
Problem Statement : Given an unsorted array A of n numbers and ε > 0, compute an element whose rank (position in sorted A) is in the range [(1 – ε)n/2, (1 + ε)n/2].
For ½ Approximate Median Algorithm &epsilom; is 1/2 => rank should be in the range [n/4, 3n/4]
What if we want in less than O(n) time with low probable error allowed?
Following steps represent an algorithm that is O((Log n) x (Log Log n)) time and produces incorrect result with probability less than or equal to 2/n2.
- Randomly choose k elements from the array where k=c log n (c is some constant)
- Insert then into a set.
- Sort elements of the set.
- Return median of the set i.e. (k/2)th element from the set
- Randomized Algorithms | Set 2 (Classification and Applications)
- Randomized Algorithms | Set 1 (Introduction and Analysis)
- Randomized Algorithms | Set 0 (Mathematical Background)
- Randomized Binary Search Algorithm
- Load Balancing on Servers (Randomized Algorithm)
- Mean and Median of a matrix
- Geometric Median
- Maximize the median of an array
- Median after K additional integers
- Median of Stream of Running Integers using STL
- Program for Mean and median of an unsorted array
- Find median of BST in O(n) time and O(1) space
- Median of sliding window in an array
- Median and Mode using Counting Sort
- Median of two sorted arrays of different sizes
- Median of two sorted arrays with different sizes in O(log(min(n, m)))
- Median of sliding window in an array | Set 2
- Median of two sorted arrays of same size
- Find median in row wise sorted matrix
Approximate Median is 4
We use a set provided by the STL in C++. In STL Set, insertion for each element takes O(log k). So for k insertions, time taken is O (k log k).
Now replacing k with c log n
=>O(c log n (log (clog n))) =>O (log n (log log n))
How is probability of error less than 2/n2?
Algorithm makes an error if the set S has at least k/2 elements are from the Left Quarter or Right Quarter.
It is quite easy to visualize this statement since the median which we report will be (k/2)th element and if we take k/2 elements from the left quarter(or right quarter) the median will be from the left quarter (or the right quarter).
An array can be divided into 4 quarters each of size n/4. So P(selecting left quarter) is 1/4. So what is the probability that at least k/2 elements are from the Left Quarter or Right Quarter? This probability problem is same as below :
Given a coin which gives HEADS with probability 1/4 and TAILS with 3/4. The coin is tossed k times. What is the probability that we get at least k/2 HEADS is less than or equal to?
If we put k = c log n for c = 10, we get P <= (1/2)2log n P <= (1/2)log n2 P <= n-2
Probability of selecting at least k/2 elements from the left quarter) <= 1/n2
Probability of selecting at least k/2 elements from the left or right quarter) <= 2/n2
Therefore algorithm produces incorrect result with probability less that or equal to 2/n2.
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