Median Finding Algorithm: A Linear Time Selection Algorithm, Schemes and Mind Maps of Algorithms and Programming

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Median Finding Algorithm

Submitted By:

Arjun Saraswat

Nishant Kapoor

Problem Definition

 Given a set of "n" unordered numbers we

want to find the "k

th

" smallest number. (k is

a n i n t e g e r b e t w e e n 1 a n d n ).

Linear Time selection

algorithm

 Also called Median Finding Algorithm.

 Find k

th

smallest element in O (n) time

in worst case.

 Uses Divide and Conquer strategy.

 Uses elimination in order to cut down

the running time substantially.

Steps to solve the problem

 Step 1: If n is small, for example n<6,

just sort and return the k

th

smallest

number in constant time i.e; O(1) time.

 Step 2: Group the given number in

s u b s e t s o f 5 i n O ( n ) t i m e.

Arrange the numbers in groups of five

Find median of N/5 groups

Median of each group

Find the sets L and R

 Compare each n-1 elements with the median m and find two

sets L and R such that every element in L is smaller than M and

every element in R is greater than m.

m

L R

3n/10<L<7n/10 3n/10<R<7n/

Description of the Algorithm step

 If n is small, for example n<6, just sort and return the k the smallest

number.( Bound time- 7)

 If n>5, then partition the numbers into groups of 5.(Bound time n/5)

 Sort the numbers within each group. Select the middle elements (the

medians). (Bound time- 7n/5)

 Call your "Selection" routine recursively to find the median of n/

medians and call it m. (Bound time-T

n/

)

 Compare all n-1 elements with the median of medians m and

determine the sets L and R, where L contains all elements <m, and R

contains all elements >m. Clearly, the rank of m is r=|L|+1 (|L| is the

size or cardinality of L). (Bound time- n)

Recursive formula

 T (n)=O (n) + T (n/5) +T (7n/10)

We will solve this equation in order to get the complexity.

We assume that T (n)< C*n

T (n) = a*n + T (n/5) + T (7n/10)

Cn >= T(n/5) +T(7n/10) + an

Cn >= Cn/5+ C7n/10 + a*n

C >= 9*C/10 +a

C/10 >= a

C >= 10*a

There is such a constant that exists….so T (n) = O (n)

Why group of 5 why not some other term??

 If we divide elements into groups of 3 then we will have

T (n) = O (n) + T (n/3) + T (2n/3) so T (n) > O (n)…..

 If we divide elements into groups of more than 5, the value of

constant 5 will be more, so grouping elements in to 5 is the

optimal situation.