Unit 5

1.     Write an Algorithms for finding roots of a quadratic equations

The standard form of a quadratic equation is:

Ax2 + bx + c = 0, where

a, b and c are real numbers and

a != 0

The term b2-4ac is known as the discriminant of a quadratic equation. It tells the nature of the roots.

• If the discriminant is greater than 0, the roots are real and different.
• If the discriminant is equal to 0, the roots are real and equal.
• If the discriminant is less than 0, the roots are complex and different.

Program to Find Roots of a Quadratic Equation

#include <math.h>

#include <stdio.h>

Int main() {

    double a, b, c, discriminant, root1, root2, realPart, imagPart;

    printf("Enter coefficients a, b and c: ");

    scanf("%lf %lf %lf", &a, &b, &c);

    discriminant = b * b - 4 * a * c;

    // condition for real and different roots

    if (discriminant > 0) {

        root1 = (-b + sqrt(discriminant)) / (2 * a);

        root2 = (-b - sqrt(discriminant)) / (2 * a);

        printf("root1 = %.2lf and root2 = %.2lf", root1, root2);

    }

    // condition for real and equal roots

    else if (discriminant == 0) {

        root1 = root2 = -b / (2 * a);

        printf("root1 = root2 = %.2lf;", root1);

    }

    // if roots are not real

    else {

        realPart = -b / (2 * a);

        imagPart = sqrt(-discriminant) / (2 * a);

        printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi", realPart, imagPart, realPart, imagPart);

    }

    return 0;

}

Output

Enter coefficients a, b and c: 2.3

4

5.6

Root1 = -0.87+1.30i and root2 = -0.87-1.30i

In this program, the sqrt() library function is used to find the square root of a number.

2.     Write a program for Finding minimum and maximum numbers of a given set explain with different methods

First of all, how do we return multiple values from a C function? We can do it either using structures or pointers.

We have created a structure named pair (which contains min and max) to return multiple values. 

And the function declaration becomes: struct pair getMinMax(int arr[], int n) where arr[] is the array of size n whose minimum and maximum are needed.

METHOD 1 (Simple Linear Search)
Initialize values of min and max as minimum and maximum of the first two elements respectively. Starting from 3rd, compare each element with max and min, and change max and min accordingly (i.e., if the element is smaller than min then change min, else if the element is greater than max then change max, else ignore the element)

Output:

Minimum element is 1

Maximum element is 3000

Time Complexity: O(n)

In this method, total number of comparisons is 1 + 2(n-2) in worst case and 1 + n – 2 in best case.
In the above implementation, worst case occurs when elements are sorted in descending order and best case occurs when elements are sorted in ascending order.

METHOD 2 (Tournament Method)
Divide the array into two parts and compare the maximums and minimums of the two parts to get the maximum and the minimum of the whole array.

Pair MaxMin(array, array_size)

   if array_size = 1

      return element as both max and min

   else if arry_size = 2

      one comparison to determine max and min

      return that pair

   else    /* array_size  > 2 */

      recur for max and min of left half

      recur for max and min of right half

      one comparison determines true max of the two candidates

      one comparison determines true min of the two candidates

      return the pair of max and min

Implementation

Output:

Minimum element is 1

Maximum element is 3000

Time Complexity: O(n)
Total number of comparisons: let number of comparisons be T(n). T(n) can be written as follows:
Algorithmic Paradigm: Divide and Conquer

  T(n) = T(floor(n/2)) + T(ceil(n/2)) + 2 

  T(2) = 1

  T(1) = 0

If n is a power of 2, then we can write T(n) as:

   T(n) = 2T(n/2) + 2

After solving above recursion, we get

  T(n)  = 3n/2 -2

Thus, the approach does 3n/2 -2 comparisons if n is a power of 2. And it does more than 3n/2 -2 comparisons if n is not a power of 2.

METHOD 3 (Compare in Pairs)
If n is odd then initialize min and max as first element.
If n is even then initialize min and max as minimum and maximum of the first two elements respectively.
For rest of the elements, pick them in pairs and compare their
maximum and minimum with max and min respectively.

Output:

Minimum element is 1

Maximum element is 3000

Time Complexity: O(n)

Total number of comparisons: Different for even and odd n, see below:

       If n is odd:    3*(n-1)/2 

       If n is even:   1 Initial comparison for initializing min and max,

                           and 3(n-2)/2 comparisons for rest of the elements 

                      =  1 + 3*(n-2)/2 = 3n/2 -2

Second and third approaches make equal number of comparisons when n is a power of 2.

In general, method 3 seems to be the best.

3.     Write a program for Finding if a number is prime number

A prime number is a whole number greater than 1, which is only divisible by 1 and itself. First few prime numbers are : 2 3 5 7 11 13 17 19 23 …..

Some interesting fact about Prime numbers

1. Two is the only even Prime number.
2. Every prime number can represented in form of 6n+1 or 6n-1 except 2 and 3, where n is natural number.
3. Two and Three are only two consecutive natural numbers which are prime too.
4. Goldbach Conjecture  : Every even integer greater than 2 can be expressed as the sum of two primes.
5. Wilson Theorem : Wilson’s theorem states that a natural number p > 1 is a prime number if and only if
6.     (p - 1) ! ≡  -1   mod p

OR  (p - 1) ! ≡  (p-1) mod p

7.     Fermat’s Little Theorem : If n is a prime number, then for every a, 1 <= a < n,

8.     an-1 ≡ 1 (mod n)

9.     OR

an-1 % n = 1

10. Prime Number Theorem : The probability that a given, randomly chosen number n is prime is inversely proportional to its number of digits, or to the logarithm of n.

11. Lemoine’s Conjecture : Any odd integer greater than 5 can be expressed as a sum of an odd prime (all primes other than 2 are odd) and an even semiprime. A semiprime number is a product of two prime numbers. This is called Lemoine’s conjecture.

How we check whether a number is Prime or not?

1. Naive solution.
   A naive solution is to iterate through all numbers from 2 to n-1 and for every number check if it divides n. If we find any number that divides, we return false.

Output :
Time complexity :O(n)

True

4.     Explain linear search

Searching

Searching is the process of finding some particular element in the list. If the element is present in the list, then the process is called successful and the process returns the location of that element, otherwise the search is called unsuccessful.

There are two popular search methods that are widely used in order to search some item into the list. However, choice of the algorithm depends upon the arrangement of the list.

• Linear Search
• Binary Search

Linear Search

Linear search is the simplest search algorithm and often called sequential search. In this type of searching, we simply traverse the list completely and match each element of the list with the item whose location is to be found. If the match found then location of the item is returned otherwise the algorithm return NULL.

Linear search is mostly used to search an unordered list in which the items are not sorted. The algorithm of linear search is given as follows.

Algorithm

• LINEAR_SEARCH(A, N, VAL)
• Step 1: [INITIALIZE] SET POS = -1
• Step 2: [INITIALIZE] SET I = 1
• Step 3: Repeat Step 4 while I<=N
• Step 4: IF A[I] = VAL
  SET POS = I
  PRINT POS
  Go to Step 6
  [END OF IF]
  SET I = I + 1
  [END OF LOOP]
• Step 5: IF POS = -1
  PRINT " VALUE IS NOT PRESENTIN THE ARRAY "
  [END OF IF]
• Step 6: EXIT

Complexity of algorithm

C Program

1. #include<stdio.h>   
2. Void main ()  
3. {  
4.    Int a[10] = {10, 23, 40, 1, 2, 0, 14, 13, 50, 9};  
5.    Int item, i,flag;  
6.    Printf("\nEnter Item which is to be searched\n");  
7.    Scanf("%d",&item);  
8.    For (i = 0; i< 10; i++)  
9.    {  
10.        If(a[i] == item)   
11.        {  
12.            Flag = i+1;  
13.            Break;  
14.        }   
15.        Else   
16.        Flag = 0;  
17.    }   
18.    If(flag != 0)  
19.    {  
20.        Printf("\nItem found at location %d\n",flag);  
21.    }  
22.    Else  
23.    {  
24.        Printf("\nItem not found\n");   
25.    }  
26. }   

Output:

Enter Item which is to be searched

20

Item not found

Enter Item which is to be searched

23

Item found at location 2

5.     Explain binary search

Binary Search

Binary search is the search technique which works efficiently on the sorted lists. Hence, in order to search an element into some list by using binary search technique, we must ensure that the list is sorted.

Binary search follows divide and conquer approach in which, the list is divided into two halves and the item is compared with the middle element of the list. If the match is found then, the location of middle element is returned otherwise, we search into either of the halves depending upon the result produced through the match.

Binary search algorithm is given below.

BINARY_SEARCH(A, lower_bound, upper_bound, VAL)

• Step 1: [INITIALIZE] SET BEG = lower_bound
  END = upper_bound, POS = - 1
• Step 2: Repeat Steps 3 and 4 while BEG <=END
• Step 3: SET MID = (BEG + END)/2
• Step 4: IF A[MID] = VAL
  SET POS = MID
  PRINT POS
  Go to Step 6
  ELSE IF A[MID] > VAL
  SET END = MID - 1
  ELSE
  SET BEG = MID + 1
  [END OF IF]
  [END OF LOOP]
• Step 5: IF POS = -1
  PRINT "VALUE IS NOT PRESENT IN THE ARRAY"
  [END OF IF]
• Step 6: EXIT

Complexity

Example

Let us consider an array arr = {1, 5, 7, 8, 13, 19, 20, 23, 29}. Find the location of the item 23 in the array.

In 1st step :

1. BEG = 0   
2. END = 8ron  
3. MID = 4   
4. a[mid] = a[4] = 13 < 23, therefore  

In Second step:

1. Beg = mid +1 = 5   
2. End = 8  
3. Mid = 13/2 = 6    
4. a[mid] = a[6] = 20 < 23, therefore;   

In third step:

1. Beg = mid + 1 = 7   
2. End = 8   
3. Mid = 15/2 = 7  
4. a[mid] = a[7]   
5. a[7] = 23 = item;   
6. Therefore, set location = mid;   
7. The location of the item will be 7.   

Binary Search Program using Recursion

C program

1. #include<stdio.h>  
2. Int binarySearch(int[], int, int, int);  
3. Void main ()  
4. {  
5.    Int arr[10] = {16, 19, 20, 23, 45, 56, 78, 90, 96, 100};  
6.    Int item, location=-1;   
7.    Printf("Enter the item which you want to search ");  
8.    Scanf("%d",&item);  
9.    Location = binarySearch(arr, 0, 9, item);  
10.    If(location != -1)   
11.    {  
12.        Printf("Item found at location %d",location);  
13.    }  
14.    Else  
15.    {  
16.        Printf("Item not found");  
17.    }  
18. }   
19. Int binarySearch(int a[], int beg, int end, int item)  
20. {  
21.    Int mid;  
22.    If(end >= beg)   
23.    {     
24.        Mid = (beg + end)/2;  
25.        If(a[mid] == item)  
26.        {  
27.            Return mid+1;  
28.        }  
29.        Else if(a[mid] < item)   
30.        {  
31.            Return binarySearch(a,mid+1,end,item);  
32.        }  
33.        Else   
34.        {  
35.            Return binarySearch(a,beg,mid-1,item);  
36.        }  
37.      
38.    }  
39.    Return -1;   
40. }  

Output:

Enter the item which you want to search

19

Item found at location 2

6. Explain bubble sort

Bubble Sort

In Bubble sort, Each element of the array is compared with its adjacent element. The algorithm processes the list in passes. A list with n elements requires n-1 passes for sorting. Consider an array A of n elements whose elements are to be sorted by using Bubble sort. The algorithm processes like following.

1. In Pass 1, A[0] is compared with A[1], A[1] is compared with A[2], A[2] is compared with A[3] and so on. At the end of pass 1, the largest element of the list is placed at the highest index of the list.
2. In Pass 2, A[0] is compared with A[1], A[1] is compared with A[2] and so on. At the end of Pass 2 the second largest element of the list is placed at the second highest index of the list.
3. In pass n-1, A[0] is compared with A[1], A[1] is compared with A[2] and so on. At the end of this pass. The smallest element of the list is placed at the first index of the list.

Algorithm :

• Step 1: Repeat Step 2 For i = 0 to N-1
• Step 2: Repeat For J = i + 1 to N - I
• Step 3: IF A[J] > A[i]
  SWAP A[J] and A[i]
  [END OF INNER LOOP]
  [END OF OUTER LOOP
• Step 4: EXIT

Complexity

C Program

1. #include<stdio.h>  
2. Void main ()  
3. {  
4.    Int i, j,temp;   
5.    Int a[10] = { 10, 9, 7, 101, 23, 44, 12, 78, 34, 23};   
6.    For(i = 0; i<10; i++)  
7.    {  
8.        For(j = i+1; j<10; j++)  
9.        {  
10.            If(a[j] > a[i])  
11.            {  
12.                Temp = a[i];  
13.                a[i] = a[j];  
14.                a[j] = temp;   
15.            }   
16.        }   
17.    }   
18.    Printf("Printing Sorted Element List ...\n");  
19.    For(i = 0; i<10; i++)  
20.    {  
21.        Printf("%d\n",a[i]);  
22.    }  
23. }  

Output:

Printing Sorted Element List . . .

7

9

10

12

23

34

34

44

78

101

7. Explain selection sort

Selection Sort

Selection sort is a simple sorting algorithm. This sorting algorithm is an in-place comparison-based algorithm in which the list is divided into two parts, the sorted part at the left end and the unsorted part at the right end. Initially, the sorted part is empty and the unsorted part is the entire list.

The smallest element is selected from the unsorted array and swapped with the leftmost element, and that element becomes a part of the sorted array. This process continues moving unsorted array boundary by one element to the right.

This algorithm is not suitable for large data sets as its average and worst case complexities are of Ο(n2), where n is the number of items.

How Selection Sort Works?

Consider the following depicted array as an example.

For the first position in the sorted list, the whole list is scanned sequentially. The first position where 14 is stored presently, we search the whole list and find that 10 is the lowest value.

So we replace 14 with 10. After one iteration 10, which happens to be the minimum value in the list, appears in the first position of the sorted list.

For the second position, where 33 is residing, we start scanning the rest of the list in a linear manner.

We find that 14 is the second lowest value in the list and it should appear at the second place. We swap these values.

After two iterations, two least values are positioned at the beginning in a sorted manner.

The same process is applied to the rest of the items in the array.

Following is a pictorial depiction of the entire sorting process −

Now, let us learn some programming aspects of selection sort.

Algorithm

Step 1 − Set MIN to location 0

Step 2 − Search the minimum element in the list

Step 3 − Swap with value at location MIN

Step 4 − Increment MIN to point to next element

Step 5 − Repeat until list is sorted

Pseudocode

Procedure selection sort

   list  : array of items

   n     : size of list

   for i = 1 to n - 1

   /* set current element as minimum*/

      min = i   

      /* check the element to be minimum */

      for j = i+1 to n

         if list[j] < list[min] then

            min = j;

         end if

      end for

      /* swap the minimum element with the current element*/

      if indexMin != i  then

         swap list[min] and list[i]

      end if

   end for

End procedure

8.     Explain Basic concept of order of complexity through the example programs

Analysis of efficiency of an algorithm can be performed at two different stages, before implementation and after implementation, as

A priori analysis − This is defined as theoretical analysis of an algorithm. Efficiency of algorithm is measured by assuming that all other factors e.g. Speed of processor, are constant and have no effect on implementation.

A posterior analysis − This is defined as empirical analysis of an algorithm. The chosen algorithm is implemented using programming language. Next the chosen algorithm is executed on target computer machine. In this analysis, actual statistics like running time and space needed are collected.

Algorithm analysis is dealt with the execution or running time of various operations involved. Running time of an operation can be defined as number of computer instructions executed per operation.

Algorithm Complexity

Suppose X is treated as an algorithm and N is treated as the size of input data, the time and space implemented by the Algorithm X are the two main factors which determine the efficiency of X.

Time Factor − The time is calculated or measured by counting the number of key operations such as comparisons in sorting algorithm.

Space Factor − The space is calculated or measured by counting the maximum memory space required by the algorithm.

The complexity of an algorithm f(N) provides the running time and / or storage space needed by the algorithm with respect of N as the size of input data.

Space Complexity

Space complexity of an algorithm represents the amount of memory space needed the algorithm in its life cycle.

Space needed by an algorithm is equal to the sum of the following two components

A fixed part that is a space required to store certain data and variables (i.e. simple variables and constants, program size etc.), that are not dependent of the size of the problem.

A variable part is a space required by variables, whose size is totally dependent on the size of the problem. For example, recursion stack space, dynamic memory allocation etc.

Space complexity S(p) of any algorithm p is S(p) = A + Sp(I) Where A is treated as the fixed part and S(I) is treated as the variable part of the algorithm which depends on instance characteristic I. Following is a simple example that tries to explain the concept

Algorithm

SUM(P, Q)

Step 1 - START

Step 2 - R ← P + Q + 10

Step 3 - Stop

Here we have three variables P, Q and R and one constant. Hence S(p) = 1+3. Now space is dependent on data types of given constant types and variables and it will be multiplied accordingly.

Time Complexity

Time Complexity of an algorithm is the representation of the amount of time required by the algorithm to execute to completion. Time requirements can be denoted or defined as a numerical function t(N), where t(N) can be measured as the number of steps, provided each step takes constant time.

For example, in case of addition of two n-bit integers, N steps are taken. Consequently, the total computational time is t(N) = c*n, where c is the time consumed for addition of two bits. Here, we observe that t(N) grows linearly as input size increases.

9.     Explain built in user define function

User defined functions are the functions defined by the programmer or user.

C++ provides the facility to user to define their own functions.

When the function gets call anywhere from program the function body gets execute.

Structure

#include<iostream>

Void function_name(){

}

Int main(){

Function_name();

}

Example

#include<iostream>

Using namespace std;

Int add(int, int);

Int main(){

int a, b, sum;

 cout<<"Enters two numbers  ";

              cin >> a >> b;

              sum = add(a, b);

              cout << "Sum = " << sum;

               return 0;

}

Int add(int a, int b)

{

   int add;

  add = a + b;

  return add;   }

}

10. What is function definition and declaration?

Function is declared globally to inform the compiler about function name, function parameters and return type.

Return_type function_name([arguments type)];

Example

Void create_function(){

}

Function call

Function call has parameter list with function name. Using function call     function can be called anywhere in program.

Function_name([actual arguments]);

Example

Void createfunction() {
  cout << "Function created";
}

int main() {
  createfunction();
  return 0;
}

Function definition.

Function definition consists of block of statements which contains actual execution code. When function gets called in program control comes to the function definition.

Return_type function_name([arguments])

{

Statements;

}

Example

Int main() {
  createfunction();
  return 0;
}
void createfunction() {
  cout <<”Functoin created.”;
}

Void mycreateunction();
int main() {
  createfunction(); 
  return 0;
}
void createfunction() {
  cout << “Function created”;
}