concepts of algorithms csc-244 unit 15 & 16 divide-and-conquer algorithms ( binary search and...

Concepts of AlgorithmsCSC-244

Unit 15 & 16Divide-and-conquer Algorithms

( Binary Search and Merge Sort )

Shahid Iqbal LoneComputer College

Qassim University K.S.A.

2

Divide and Conquer

CSC 244 Concepts of Algorithms

The most well known algorithm design strategy:1. Divide a large problem into two or more smaller

problems

2. Solve smaller problems recursively

3. Obtain solution to original (larger) problem by combining/merging these solutions of small problems.

Divide-and-conquer technique

subproblem 2 of size n/2

subproblem 1 of size n/2

a solution to subproblem 1

a solution tothe original problem

a solution to subproblem 2

a problem of size n

CSC 244 Concepts of Algorithms 3

Divide and Conquer Examples

• Searching: binary search• Sorting: mergesort • Sorting: quicksort• Tree traversals• Matrix multiplication-Strassen’s algorithm


Binary SearchRecall your mind, we have studied something about Binary

Search in Unit-1 & Unit-2. Binary search is an efficient algorithm for searching in a

sorted array. Search requires the following steps:

1. Inspect the middle item of an array of size N.2. Inspect the middle of an array of size N/2.3. Inspect the middle item of an array of size N/4 and

so on until Lower Bound becomes > Upper bound.– This implies k = log2N– k is the number of partitions.


Binary Search• Requires that the array be sorted. • Rather than start at either end, binary

searches split the array in half and works only with the half that may contain the value.

• This action of dividing continues until the desired value is found or the remaining values are either smaller or larger than the search value.


Pseudo code for Binary Search BinarySearch(A, START, END, key)

{ If (START <= END) { MID = (START + END) / 2 [ compute mid point] if (key equal to A[MID] return MID [ found it, Successful Search.] else if ( key < A[MID] ) [ Recursive Call itself for the lower part of the array ] return BinarySearch( START, MID-1, key) else[Recursive Call itself for the upper part of the array ] return BinarySearch(MID+1, END, key) } return -1 [ failed to find key, Un-Successful Search. ]}


Binary Search• Suppose that the data set is n = 2k – 1 sorted items, where k

is the number of partitions.• Each time examine middle item. If larger, look left, if

smaller look right.• Second ‘chunk’ to consider is n/2 in size, third is n/4, fourth

is n/8, etc.• Worst case, examine k chunks. n = 2k – 1 so, k = log2(n ).

(Decision binary Search: one comparison on each level)• Best case, O(1). (Found at n/2).• Worst Case: O(log2 (n))


5 7 9 13 32 33 42 54 56 880 1 2 3 4 5 6 7 8 9Indices

Contents

Target/Skey is 33The array a looks like this:

Binary Search Example-1 (Successful Search)

start + End

mid = (0 + 9) / 2 (which is 4)33 > A[mid] (that is, 33 > A[4]) Start = Mid + 1So, if 33 is in the array, then 33 is one of:

33 42 54 56 88 5 6 7 8 9

Eliminated half of the remaining elements from consideration because array elements are sorted.


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Contents


Binary Search Example

mid = (5 + 6) / 2 (which is 5)33 == A[mid]So we found 33 at index 5:

33 5

mid = (5 + 9) / 2 (which is 7)33 < A[mid] (that is, 33 < A[7]) End= Mid - 1So, if 33 is in the array, then 33 is one of:

33 42 5 6

Eliminate half of the remaining elements


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Binary Search Example-2(Un-Successful Search)

start + End

mid = (0 + 9) / 2 (which is 4)8 < A[mid] (that is, 8 < A[4]) End = Mid - 1So, if 8 is in the array, then 8 is one of:

Eliminated half of the remaining elements from consideration because array elements are sorted.


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Contents

5 7 9 13 32 33 42 54 56 880 1 2 3 4 5 6 7 8 9Indices

Contents


Binary Search Example

mid = (2 + 3) / 2 (which is 2)8 < A[mid] (that is, 8 < A[2]) change End = Mid - 1So, now Start = 2 and End = 1, ( that is, Start > End ) which indicates that 8 is not in the array ( Un-successful Search )

mid = (0 + 3) / 2 (which is 1)8 > A[mid] (that is, 8 > A[1]) change Start = Mid + 1So, if 8 is in the array, then 8 is one of:

9 13 2 3 Eliminate half

of the remaining elements


14CSC 244 Concepts of Algorithms

C-Language Code for Recursive Binary Search

#include <iostream.h>#include <process.h> int size; // Globle Variable int *A; // integer pointer to store the Base // Address of an integer arrayvoid get_Size( ) { cout<<" Put the size of array: ? ";

cin>>size; A = new int[size]; // Creation of a

// Dynamic Array}

void get_numbers( ){ cout<<"\n\n Put "<<size<<“ Numbers : \n";

for( int i = 0 ; i < size ; i++) { cout<<" Put "<< i+1<<" value: "; cin>> A[i];}

}


void BubbleSort( ){

int i, pass, hold, sw=1;for (pass=1; pass<= size-1; pass++) { sw=0; for (i=0; i< size-pass; i++) { if(A[i] > A[i+1])

{ hold =A[i]; A[i]=A[i+1]; A[i+1]=hold; sw=1;}

} if(sw==0) break;}

}

void print_sorted_numbers(){ cout<<"\n\n\n Sorted numbers are:\n\n"; for ( int i=0; i< size; i++) { cout<<A[i]<<"\t"; }}

C-Language Code for Recursive Binary Searchint main(){ int SKEY, LOC; get_Size(); get_numbers(); BubbleSort(); print_sorted_numbers( ); while(1) { cout<<"\n\n Put Search Key or 0 to stop: ";

cin>>SKEY;if (SKEY == 0 ) break;LOC = BinarySearch ( 0, size-1, SKEY );if(LOC == -1) cout<<" \n "<<SKEY<<" not found"<<endl;elsecout<<SKEY<< “ found at Index/location: "<<LOC<<endl;

} return 0;} // end of main() function


int BinarySearch(int START, int END, int key) { if (START <= END) { int MID = (START + END) / 2; // compute // mid point. if (key == A[MID]) return MID; // found it, Successful Search. else if (key < A[MID] ) // Call itself for // the lower part of the array return BinarySearch( START, MID-1, key); else // Call itself for the upper part of the array return BinarySearch(MID+1, END, key); } return -1; // Un-Sucessfull Search.}

Merge Sort


Divide and Conquer1. Base case/criteria: the problem is small

enough, solve directly

2. Divide the problem into two or more similar and smaller subproblems

3. Recursively solve the subproblems

4. Combine solutions to the subproblems


Divide and Conquer - Sort

Problem: • Input: A[n] – unsorted array of n ≥1 integers. • Output: A[n] – sorted in non-decreasing order


Divide and Conquer - Sort• Base case

single element (n=1), return

• Divide A into two subarrays: FirstPart, SecondPart Two Subproblems:

• sort the FirstPart • sort the SecondPart

• Recursively• sort FirstPart• sort SecondPart

• Combine sorted FirstPart and sorted second part


Merging

Suppose A is sorted list with r elements and B is a sorted list with s elements. The operation that combine the elements of A and B into a single sorted list C with n= r+s elements is called merging. One simple way to merge is to place the elements of B after the elements of A and then use some sorting algorithm on the entire lsit. This method does not take advantage of the fact that A and B are individually sorted. A much more efficient algorithm is merge sort algorithm.

Suppose one is given two sorted decks of cards. The decks are merged as:


Merging (cont.)

3 10 23 54 1 5 25 75X: Y:

Result:


Merging (cont.)

3 10 23 54 5 25 75

1

X: Y:

Result:


Merging (cont.)

10 23 54 5 25 75

1 3

X: Y:

Result:


Merging (cont.)

10 23 54 25 75

1 3 5

X: Y:

Result:


Merging (cont.)

23 54 25 75

1 3 5 10

X: Y:

Result:


Merging (cont.)

54 25 75

1 3 5 10 23

X: Y:

Result:


Merging (cont.)

54 75

1 3 5 10 23 25

X: Y:

Result:


Merging (cont.)

75

1 3 5 10 23 25 54

X: Y:

Result:


Merging (cont.)

1 3 5 10 23 25 54 75

X: Y:

Result:


Merge Sort Example

99 6 86 15 58 35 86 4 0


Merge Sort Example

99 6 86 15 58 35 86 4 0

99 6 86 15 58 35 86 4 0


Merge Sort Example

99 6 86 15 58 35 86 4 0

99 6 86 15 58 35 86 4 0

86 1599 6 58 35 86 4 0


Merge Sort Example

99 6 86 15 58 35 86 4 0

99 6 86 15 58 35 86 4 0

86 1599 6 58 35 86 4 0

99 6 86 15 58 35 86 4 0


Merge Sort Example

99 6 86 15 58 35 86 4 0

99 6 86 15 58 35 86 4 0

86 1599 6 58 35 86 4 0

99 6 86 15 58 35 86 4 0

4 0


Merge Sort Example

99 6 86 15 58 35 86 0 4

4 0


Merge Sort Example

15 866 99 35 58 0 4 86

99 6 86 15 58 35 86 0 4


Merge Sort Example

6 15 86 99 0 4 35 58 86

15 866 99 35 58 0 4 86


Merge Sort Example

0 4 6 15 35 58 86 86 99

6 15 86 99 0 4 35 58 86


Merge Sort Example

0 4 6 15 35 58 86 86 99


Another Example MergeSort

Original 24 13 26 1 12 27 38 15Divide in 2 24 13 26 1 12 27 38 15Divide in 4 24 13 26 1 12 27 38 15Divide in 8 24 13 26 1 12 27 38 15Merge 2 13 24 1 26 12 27 15 38Merge 4 1 13 24 26 12 15 27 38Merge 8 1 12 13 15 24 26 27 38


Merge Sort: Algorithm

Algorithm: MERGESORT (A, N)1- If N=1, Return.

2- Set N1=N/2, N2=N-N1. 3- Repeat for i=0,1,2 . . . . . (N1-1)

Set L [i]=A [i].4- Repeat for j=0,1,2 . . . . . (N2-1)

Set R [j]=A [N1+j].5- Call MERGESORT (L, N1).

6 - Call MERGESORT (R, N2). 7- Call MERGE (A, L, N1, R,N2). 8- Return.


Merge : Algorithm

Algorithm: MERGE (A, L, N1, R, N2)1- Set i=0, j:=0.

2- Repeat for k=0,1,2 . . . . . (N1+N2-1)If i<N1, then:

If j=N2 or L [i] ≤ R [j], then:Set A [k] =L [i].Set i=i+1;

Else:If j < N2, then:

Set A [k]=R [j].Set j=j+1.

3- Return.



Merge-Sort (A, n) if n=1 return else

n1 = n/2 and n2 = n – n1

create array L[n1], R[n2]for i = 0 to n1-1 do L[i] ← A[i]for j = 0 to n2-1 do R[j] ← A[n1+j]

Merge-Sort(L, n1) Merge-Sort(R, n2) Merge(A, L, n1, R, n2 )

Space: n

Recursive Call

Time: n


merge(A,L,n1,R,n2)i = j = 0for k = 0 to n1+n2-1

if i < n1 if j = n2 or L[i] ≤ R[j]

A[k] = L[i]i = i + 1

else if j < n2

A[k] = R[j]j = j + 1

Number of iterations: (n1+n2)

Total time: c(n1+n2) for some c



Home WorkUsing the algorithm described in above

slides, write a c-language code for MERGE-SORT


END


concepts of algorithms csc-244 unit 15 & 16 divide-and-conquer algorithms ( binary search and...

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