Searching. The process used to find the location of a target among a list of objects Searching an...

Searching

The process used to find the location of a target among a list of objectsSearching an array finds the index of first element in an array

containing that value

Searching

Unordered Linear Search

Search an unordered array of integers for a value and return its index if the value is found. Otherwise, return -1. A[0] A[1] A[2] A[3] A[4] A[5] A[6] A[7]

Algorithm:Start with the first array element (index 0) while(more elements in array){

if value found at current index, return index;

Try next element (increment index);}Value not found, return -1;

14 2 10 5 1 3 17 2

Unordered Linear Search

// Searches an unordered array of integersint search(int data[], //input: array int size, //input: array size int value){ //input: search value // output: if found, return index; // otherwise, return –1. for(int index = 0; index < size; index++){ if(data[index] == value) return index;

} return -1;}

821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

Binary Search

lo

Binary search. Given value and sorted array a[], find index isuch that a[i] = value, or report that no such index exists.

Invariant. Algorithm maintains a[lo] value a[hi].

Ex. Binary search for 33.

hi

Binary Search




821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

lo himid

Binary Search




821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

lo hi

Binary Search




821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

lo mid hi

Binary Search




821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

lo hi

Binary Search




821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

lo himid

Binary Search




821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

lohi

Binary Search




821 3 4 65 7 109 11 12 14130

641413 25 33 5143 53 8472 93 95 97966

lohimid

17

Example: Binary SearchSearching for an element k in a sorted array A with n elementsIdea:

Choose the middle element A[n/2] If k == A[n/2], we are done If k < A[n/2], search for k between A[0] and

A[n/2 -1] If k > A[n/2], search for k between A[n/2 + 1]

and A[n-1] Repeat until either k is found, or no more

elements to searchRequires less number of comparisons than linear search in the worst case (log2n instead of n)

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int main() { int A[100], n, k, i, mid, low, high; scanf(“%d %d”, &n, &k); for (i=0; i<n; ++i) scanf(“%d”, &A[i]);

low = 0; high = n – 1; mid = (high + low)/2;

while (high >= low) {

if (A[mid] == k) { printf(“%d is found\n”, k);

break; } if (k < A[mid]) high = mid – 1; else low = mid + 1; mid = (high + low)/2; }

If (high < low) printf(“%d is not found\n”, k);}

Sorting

20

Bubble Sort

Stage 1: Compare each element (except the last one) with its neighbor to the right

If they are out of order, swap them This puts the largest element at the very end The last element is now in the correct and final place

Stage 2: Compare each element (except the last two) with its neighbor to the right

If they are out of order, swap them This puts the second largest element next to last The last two elements are now in their correct and final

placesStage 3: Compare each element (except the last three) with its neighbor to the right

Continue as above until you have no unsorted elements on the left

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Example of Bubble Sort

7 2 8 5 4

2 7 8 5 4

2 7 8 5 4

2 7 5 8 4

2 7 5 4 8

2 7 5 4 8

2 5 7 4 8

2 5 4 7 8

2 7 5 4 8

2 5 4 7 8

2 4 5 7 8

2 5 4 7 8

2 4 5 7 8

2 4 5 7 8

(done)

Stage 1 Stage 2 Stage 3 Stage 4

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Code for Bubble Sort

void bubbleSort(int[] a) { int stage, inner, temp; for (stage = a.length - 1; stage > 0; stage--) { // counting down for (inner = 0; inner < stage; inner++) { // bubbling up if (a[inner] > a[inner + 1]) { // if out of order... temp = a[inner]; // ...then swap a[inner] = a[inner + 1]; a[inner + 1] = temp; } } }}

23

Insertion Sort

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Insertion Sort

Suppose we know how to insert a new element x in its proper place in an already sorted array A of size k, to get a new sorted array of size k+1

Use this to sort the given array A of size n as follows: Insert A[1] in the sorted array A[0]. So now A[0],A[1]

are sorted Insert A[2] in the sorted array A[0],A[1]. So now

A[0],A[1],A[2] are sorted Insert A[3] in the sorted array A[0],A[1],A[2]. So now

A[0],A[1],A[2],A[3] are sorted ….. Insert A[i] in the sorted array A[0],A[1],…,A[i-1]. So

now A[0],A[1],…A[i] are sorted Continue until i = n-1 (outer loop)

25

How to do the first step

Compare x with A[k-1] (the last element) If x ≥ A[k-1], we can make A[k] = x (as x is the

max of all the elements) If x < A[k-1], put A[k] = A[k-1] to create a hole

in the k-th position, put x thereNow repeat by comparing x with A[k-2] (inserting

x in its proper place in the sorted subarray A[0],A[1],…A[k-1] of k-2 elements)

The value x bubbles to the left until it finds an element A[i] such that x ≥ A[i]

No need to compare any more as all elements A[0], A[1], A[i] are less than x

26

Example of first step

5 7 11 13 20 22A Insert x = 15

27


5 7 11 13 20 22

5 7 11 13 20 15 22

A Insert x = 15

Compare with 22. x < 22, so move 22 right

28


5 7 11 13 20 22

5 7 11 13 20 15 22

5 7 11 13 15 20 22

A Insert x = 15



29


5 7 11 13 20 22

5 7 11 13 20 15 22

5 7 11 13 15 20 22

5 7 11 13 15 20 22

A Insert x = 15



Compare with 13. x > 13, so stop

A

30

Sort using the insertion

7 5 13 11 22 20

5 7 13 11 22 20

5 7 13 11 22 20

5 7 11 13 22 20

A

Insert 5 in 7

Insert 13 in 5, 7

Insert 11 in 5, 7, 13

Insert 22 in 5, 7, 11, 13

Insert 20 in 5, 7, 11, 13, 22

5 7 11 13 20 22

5 7 11 13 22 20

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Insertion Sort Code

void InsertionSort (int A[ ], int size)

{

int i, j, item;

for (i=1; i<size; i++)

{ /* Insert the element in A[i] */

item = A[i] ;

for (j = i-1; j >= 0; j--)

if (item < A[j])

{ /* push elements down*/

A[j+1] = A[j];

A[j] = item ; /* can do this once finally also */

}

else break; /*inserted, exit loop */

}

}

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Look at the sorting!

8 2 9 4 7 6 2 1 5 i = 1:: 2, 9, 4, 7, 6, 2, 1, 5, i = 2:: 9, 2, 4, 7, 6, 2, 1, 5, i = 3:: 9, 4, 2, 7, 6, 2, 1, 5, i = 4:: 9, 7, 4, 2, 6, 2, 1, 5, i = 5:: 9, 7, 6, 4, 2, 2, 1, 5, i = 6:: 9, 7, 6, 4, 2, 2, 1, 5, i = 7:: 9, 7, 6, 4, 2, 2, 1, 5, Result = 9, 7, 6, 5, 4, 2, 2, 1,

void InsertionSort (int A[ ], int size) {

int i,j, item;

for (i=1; i<size; i++) {

printf("i = %d:: ",i);

for (j=0;j<size;j++) printf("%d, ",A[j]);

printf("\n"); item = A[i] ;

for (j=i-1; j>=0; j--)

if (item > A[j])

{ A[j+1] = A[j]; A[j] = item ; }

else break;

} }

int main() { int X[100], i, size;

scanf("%d",&size);

for (i=0;i<size;i++) scanf("%d",&X[i]);

InsertionSort(X,size);

printf("Result = ");

for (i=0;i<size;i++) printf("%d, ",X[i]);

printf("\n"); return 0;

}

33

Mergesort

34

Basic Idea

Divide the array into two halvesSort the two sub-arraysMerge the two sorted sub-arrays into a

single sorted arrayStep 2 (sorting the sub-arrays) is done

recursively (divide in two, sort, merge) until the array has a single element (base condition of recursion)

35

Merging Two Sorted Arrays

3 4 7 8 9

2 5 7

2

3 4 7 8 9

2 5 7

2 3

3 4 7 8 9

2 5 7

2 3 4

3 4 7 8 9

2 5 7

Problem: Two sorted arrays A and B are given. We are required to produce a final sorted array C which contains all elements of A and B.

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2 3 4 5

3 4 7 8 9

2 5 7

2 3 4 5 7

3 4 7 8 9

2 5 7

2 3 4 5 7 7

3 4 7 8 9

2 5 7

2 3 4 5 7 7 8

3 4 7 8 9

2 5 7

37

Merge Code

2 3 4 5 7 7 8 9

3 4 7 8 9

2 5 7

void

merge (int A[], int B[], int C[], int m,int n)

{

int i=0,j=0,k=0;

while (i<m && j<n)

{

if (A[i] < B[j]) C[k++] = A[i++];

else C[k++] = B[j++];

}

while (i<m) C[k++] = A[i++];

while (j<n) C[k++] = B[j++];

}

38

Merge Sort: Sorting an array recursively

void mergesort (int A[], int n)

{

int i, j, B[max];

if (n <= 1) return;

i = n/2;

mergesort(A, i);

mergesort(A+i, n-i);

merge(A, A+i, B, i, n-i);

for (j=0; j<n; j++) A[j] = B[j];

free(B);

}

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