We strongly recommend to refer below articles as a prerequisite of this.

Randomized Algorithms | Set 1 (Introduction and Analysis)

Randomized Algorithms | Set 2 (Classification and Applications)

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 Â½ Approximate Median Algorithm &epsilom; is 1/2 => rank should be in the range [n/4, 3n/4]

We can find k’th smallest element in O(n) expected time and O(n) worst case time.

**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/n^{2}.

- 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

/* C++ program to find Approximate Median using 1/2 Approximate Algorithm */ #include<bits/stdc++.h> using namespace std; // This function returns the Approximate Median int randApproxMedian(int arr[],int n) { // Declaration for the random number generator random_device rand_dev; mt19937 generator(rand_dev()); // Random number generated will be in the range [0,n-1] uniform_int_distribution<int> distribution(0, n-1); if (n==0) return 0; int k = 10*log2(n); // Taking c as 10 // A set stores unique elements in sorted order set<int> s; for (int i=0; i<k; i++) { // Generating a random index int index = distribution(generator); //Inserting into the set s.insert(arr[index]); } set<int> ::iterator itr = s.begin(); // Report the median of the set at k/2 position // Move the itr to k/2th position advance(itr, (s.size()/2) - 1); // Return the median return *itr; } // Driver method to test above method int main() { int arr[] = {1, 3, 2, 4, 5, 6, 8, 7}; int n = sizeof(arr)/sizeof(int); printf("Approximate Median is %d\n",randApproxMedian(arr,n)); return 0 }

Output:

Approximate Median is 4

**Time Complexity:**

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/n ^{2}?**

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/n^{2}

Probability of selecting at least k/2 elements from the left or right quarter) <= 2/n^{2}

Therefore algorithm produces incorrect result with probability less that or equal to 2/n^{2}.

**
References:**www.cse.iitk.ac.in/users/sbaswana/CS648/Lecture-2-CS648.pptx

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