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Chapter: Analysis and Design of Algorithm : Divide and Conquer, Greedy Method

Divide and Conquer Method

Divide and Conquer is one of the best-known general algorithm design technique. Divide-and-conquer algorithms work according to the following general plan:

DIVIDE AND CONQUER

 

 

Divide and Conquer is one of the best-known general algorithm design technique. Divide-and-conquer algorithms work according to the following general plan:

 

1.A problem's instance is divided into several smaller instances of the same problem, ideally of about the same size.

 

2.  The smaller instances are solved (typically recursively, though sometimes a different algorithm is employed when instances become small enough).

 

3.  If necessary, the solutions obtained for the smaller instances are combined to get a solution to the original problem.

 

The divide-and-conquer technique is diagrammed, which depicts the case of dividing a problem into two smaller sub problems,


 

Examples of divide and conquer method are Binary search, Quick sort,

 

 

 

GENERAL METHOD

 

 

e.g. Merge sort.

♦ As an example, let us consider the problem of computing the sum of n numbers a0, ... an-1.

 

  If n > 1, we can divide the problem into two instances of the same problem. They are To compute the sum of the first └ n/2numbers

 

  To compute the sum of the remaining [n/2]numbers. (Of course, if n = 1, we simply return a0 as the answer.)

 

    Once  each  of  these  two  sums  is  computed  (by  applying  the  same  method,  i.e.,

 

recursively),

 

we can add their values to get the sum in question. i.e.,

a0 + . . . + an-1 =( a0  + ....+ a n/2┘) + (a n/2 + . . . . + an-1)

 

 

  More generally, an instance of size n can be divided into several instances of size nib, with a of them needing to be solved. (Here, a and b are constants; a≥1 and b > 1.).

 

  Assuming that size n is a power of b, to simplify our analysis, we get the following recurrence for the running time T(n).

 

T(n)=aT(n/b)+f(n)

 

 

This is called the general divide and-conquer recurrence. Where f(n) is a function that accounts for the time spent on dividing the problem into smaller ones and on combining their solutions. (For the summation example, a = b = 2 and f (n) = 1.

 

Obviously, the order of growth of its solution T(n) depends on the values of the constants a and b and the order of growth of the function f (n). The efficiency analysis of many divide-and-conquer algorithms is greatly simplified by the following theorem.

 

 

1. MASTER THEOREM


 


ADVANTAGES:

 

The time spent on executing the problem using divide and conquer is smaller then other methods.

 

The divide and conquer approach provides an efficient algorithm in computer science.

 

The divide and conquer technique is ideally suited for parallel computation in which each sub problem can be solved simultaneously by its own processor.

 

  Merge sort is a perfect example of a successful application of the divide-and conquer technique.

 

  It sorts a given array Al O .. n - 1) by dividing it into two halves A [O··└ n/2 - 1] and A[

 

n/2..n - 1], sorting each of them recursively, and then merging the two smaller sorted arrays into a single sorted one.

 

 

ALGORITHM Mergesort (A[O .. n - 1])

 

//Sorts array A[O .. n - 1] by recursive mergesort //Input: An array A[O .. n - 1] of orderable elements //Output: Array A[O... n - 1] sorted in nondecreasing order if n > 1

 

copy A [O··└ n/2 - 1] to B [O··└ n/2 - 1] copy A[└ n/2..n - 1] to C[0...└ n/2 - 1] Mergesort (B [O··└ n/2 - 1])

 

Mergesort (C[0...└ n/2 - 1]) Mergesort(B,C,A)

 

STEPS TO BE FOLLOWED

 

The first step of the merge sort is to chop the list into two.

 

If the list has even length, split the list into two equal sub lists.

 

  If the list has odd length, divide the list into two by making the first sub list one entry greater than the second sub list.

 

then split both the sub list into two and go on until each of the sub lists are of size one.

 

finally, start merging the individual sub lists to obtain a sorted list.

 

The operation of the algorithm for the array of elements 8,3,2,9,7,1,5,4 is explained as follows,

 



AN EXAMPLE OF MERGE SORT OPEERATION

 

  The merging of two sorted arrays can be done as follows. Two pointers (array indices) are initialized to point to the first elements of the arrays being merged.

 

  Then the elements pointed to are compared and the smaller of them is added to a new array or list being constructed.

 

   Then the index of the smaller element is incremented to point to its immediate successor in the array.

 

   This  operation  is  continued  until  one  of  the  two  given  arrays  is  exhausted

 

♦Then         the remaining elements of the other  array  are copied to the end of the new

array.                  

          ALGORITHM : Merge Sort B(0....p-1), C(0....    q-1), A(0....p + q -1)

// Merges two sorted arrays into one sorted array .

 

// Input: Array B (0....p-1) and C (0....q-1) both sorted

 

//Output: Sorted array A (0....p + q -1) of the elements of B and C

 

i f 0; j f0 ;kf 0 while i<p and j<q do if B[i] ≤ C[j]

A[k] fB[i];i fi+1

else A[k] f C[j] jf j+1 k fk+1

if i=p

 

copy C[j....q-1] to A[k....p+q-1] else copy B[i....p-1] to A[k....p+q-1]

 

 

 

2 EFFICIENCY OF MERGE SORT

 

The recurrence relation for the number of key comparison C (n) is

C (n) = 2 C (n/2) + Cmerge (n) for n>1, C (1) = 0

 

Assume, n is a power of 2. Cmerge (n) is the number of key comparison performed during the merge sort.

 

  In the merging of two sorted array after one comparison is made the total number of elements in the two array still to be processed and sorted is reduced by one.

 

  Hence in the worst case neither of the two arrays becomes empty before the other one contains just one element . Therefore, for the worst case,

C worst(n)=2 C worst(n / 2)+n-1 for n > 1, C worst(1)=0

Cmerge(n) = n 1

 

and we have the recurrence

 

Hence, according to the Master Theorem, C worst (n)  = (n log n) . In fact, it is easy to find the exact solution to the worst-case recurrence for n = 2k

C worst(n) = n log2 n - n + 1



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