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Explain the analysis of quick sort and apply the same to sort following data. [1 0 7 5 9 12 3].
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Analysis of Quick Sort:

T(N) = T(i) + T(N - i -1) + cN

The time to sort the file is equal to

o the time to sort the left partition with i elements, plus

o the time to sort the right partition with N-i-1 elements, plus

o the time to build the partitions

  1. Worst case analysis

The pivot is the smallest element

T(N) = T(N-1) + cN, N > 1

Telescoping:

T(N-1) = T(N-2) + c(N-1)

T(N-2) = T(N-3) + c(N-2)

T(N-3) = T(N-4) + c(N-3)

T(2) = T(1) + c.2

Add all equations:

T(N) + T(N-1) + T(N-2) + … + T(2) =

= T(N-1) + T(N-2) + … + T(2) + T(1) + c(N) + c(N-1) + c(N-2) + … + c.2

T(N) = T(1) + c(2 + 3 + … + N)

T(N) = 1 + c(N(N+1)/2 -1)

Therefore T(N) = O(N2)

  1. Best-case analysis:

The pivot is in the middle

T(N) = 2T(N/2) + cN

Divide by N:

T(N) / N = T(N/2) / (N/2) + c

Telescoping:

T(N/2) / (N/2) = T(N/4) / (N/4) + c

T(N/4) / (N/4) = T(N/8) / (N/8) + c

……

T(2) / 2 = T(1) / (1) + c

Add all equations:

T(N) / N + T(N/2) / (N/2) + T(N/4) / (N/4) + …. + T(2) / 2 =

= (N/2) / (N/2) + T(N/4) / (N/4) + … + T(1) / (1) + c.logN

After crossing the equal terms:

T(N)/N = T(1) + cLogN = 1 + cLogN

T(N) = N + NcLogN

Therefore T(N) = O(NlogN)

  1. Average case analysis

Similar computations, resulting in T(N) = O(NlogN)

The average value of T(i) is 1/N times the sum of T(0) through T(N-1)

1/N  T(j), j = 0 thru N-1

T(N) = 2/N ( T(j)) + cN

Multiply by N

NT(N) = 2( T(j)) + cN*N

To remove the summation, we rewrite the equation for N-1:

(N-1)T(N-1) = 2( T(j)) + c(N-1)2, j = 0 thru N-2

and subtract:

NT(N) - (N-1)T(N-1) = 2T(N-1) + 2cN -c

Prepare for telescoping. Rearrange terms, drop the insignificant c:

NT(N) = (N+1)T(N-1) + 2cN

Divide by N(N+1):

T(N)/(N+1) = T(N-1)/N + 2c/(N+1)

Telescope:

T(N)/(N+1) = T(N-1)/N + 2c/(N+1)

T(N-1)/(N) = T(N-2)/(N-1)+ 2c/(N)

T(N-2)/(N-1) = T(N-3)/(N-2) + 2c/(N-1)

….

T(2)/3 = T(1)/2 + 2c /3

Add the equations and cross equal terms:

T(N)/(N+1) = T(1)/2 +2c  (1/j), j = 3 to N+1

T(N) = (N+1)(1/2 + 2c (1/j))

The sum  (1/j), j =3 to N-1, is about LogN

Thus T(N) = O(NlogN)

We have set a pivot for the given numbers by which we can apply quick sort algorithm.

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Hence after applying the quick sort technique the sorted numbers are

0, 1, 3, 5, 9, 7, 12

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