sum of the first n numbers

sum(n) = 1 + 2+ 3+ … +n-1 +n

If n = 5 sum(n) = 1+2 +3 +4 +5 =15

If n=6 sum(n) = 1 +2 +3 +4 +5 +6= 21

sum(6)=sum(5)+6

And in general

sum(n) = sum(n-1)+n

sum – non recursive function

double sumiteration( int n )

{

double sum1;

sum1 = 0 ;

for ( int i =1; i <= n ; i ++)

sum1 = sum1 +i;

return ( sum1);

}

sum-recursive function

double sum( int n )

{

if ( n ==0 ) // base case

return 0;

// recursive case

return n + sum( n - 1 );

}

Product of the first n numbers

factorial(n) = 1 x 2 x 3x … x n-1 x n

If n = 5 factorial(n) = 1x2 +3 x 4 x 5 =15

If n=6 sum(n) = 1 x2 x3 x4 x 5 x 6= 21

factorial(6)=factorial(5) x 6

And in general

factorial(n) = factorial(n-1) x n

Factorial-nonrecursive

double factorial( int n )

{

double fac1;

fac1 = 1 ;

for ( int i =1; i <= n ; i ++)

fac1 =fac1 *i;

return ( fac1);

}

Factorial recursive

double factorial( int n )

{

if ( n ==1 ) // base case

return 1;

// recursive case

return n * factorial( n - 1 );

}

Sum of the first n elements of an array

Given an array A of length n,

suma(n) =A[0]+A[1]+ …A[n-1]+A[n]

suma(n-1)= A[0]+A[1]+ …A[n-1]

suma(n)=suma(n-1)+A[n]

Sum first elements of an array -nonrecursive

double suma( double A[],int n )

{

double sum1;

sum1 = A[0] ;

for ( int i =1; i <= n ; i ++)

sum1 = sum1 +A[I];

return ( sum1);

}

Sum first n elements of an array-recursive

double suma( double A[], int n )

{

if ( n ==0 ) // base case

return A[0];

// recursive case

return (A[n] + sum(A, n - 1 ));

}

Dot product of 2 arrays A and B

Dot(n, A, B)= A[0]*B[0] + A[1]*B[1]+ .. + A[n-1]*B[n-1]+A[n]*B[n]

Dot( n, A, B)= Dot(n-1, A, B) +A[n-1]*B[n-1]

Dot product- non recursive

double dotpiteration( int n , double A[], double B[] )

{

double sum1;

sum1 = 0 ;

for ( int i =n ; i >= 0; i --)

sum1 = sum1 + A[i] * B[i];

return ( sum1);

}

Dot product- recursive

double dotp( int n , double A[], double B[])

{

if ( n ==0 ) // base case

return 0;

// recursive case

return (A[n ]* B[n ]+ dotp( n - 1 , A, B));

}

Series expansion for ln(2)

ln(2)= 1-1/2+ 1/3- 1/4 + 1/5-

Let double ln2(int n) be a function that computes the first n terms of the expansion.

We note that

ln2(1) = 1

ln2(4)= 1-1/2+1/3-1/4

ln2(5)=1-1/3+1/3-1/4+1/5=ln2(4)+1/5

ln2(n) =ln2(n-1)+1./n

Nonrecursive function ln2

double ln2(int )

{ double sum=0, sign=1.;

for (int I = 1; I <= n ; I++)

{ sum =+sign/I;

sign=-sign;}

return sum;

}

Recursive function ln2

double ln2(int n)

{

if (n==1)return 1;

if (n%2 ==0)

return ln2(n-1) - 1.0/n;

return ln2(n-1)+1.0/n;

}

Maximum element of an array A

We will first assume we have the function max which takes the maximum of 2 elements.

The function maxa(n, A) will take the maximum of the first n elements of A.

Given A={2,5,3,8,6,9}

Maxa(4, A) is 8

Maxa(5, A) is 8

Maxa(6,A) is 9

so that maxa(n,A) = max(A[n],maxa(n-1,A);

Maximum element- nonrecursive

Given the function max given by

double max( double x, double y){ if (x>y)

return x;else

return y;}

The function maxa( int n, double A) would be

double maxa( int n, double A[])

{int max1=A[0];

for (int I =0; I < n; I++)

max1=max(max1, A[I]);

return max1;}

Maximum element of an array-recursive

double maxa( int n, double A[])

{

if (n==0)

return A[0]; //base case

return max(A[n],maxa(n-1, A)); //recursive case

}

Reverse the digits of an integer

Consider the number 347.

347 % 10 = 7 so print it out.

347/10 = 34

34 % 10 = 4 so print it out.

34/10 = 3

3 % 10 = 3 so print it out.

Altogether we get 743.

Reverse digits- no recursion

void reverseiter( int number )

{

while (number >0)

{

cout << number % 10 ;

number = number /10;

}

}

Reverse digits- recursion

void reverse( int number )

{

if ( number !=0 )

{

cout << number % 10 ;

reverse (number /10);

}

}

Greatest common divisor

Consider the numbers 32 and 24.

24=3 x 2 x 2 x 2; 32 = 2 x 2 x 2 x 2 x2

gcd(32,24) = 2 x 2 x 2 = 8

Algorithm for gcd (x,y):

Let z = x%y (in our case 8)

gcd(x,y)= gcd(y,z)= gcd(24,8)

z = 24%8 = 0 so gcd(24,8)=gcd(8,0)

If y is 0 return x so solution is 8.

Greatest common divisor recursive code

int gcd( int x, int y) { int z;if ( y =0) // base case return x; { // recursive case

z= x% y; z = gcd(y, z);

return z; } }

Insertion sort

Assume you have an array A= {6,2,4,3}

Look at first 2 and sort these to 2,6

Now find place for 4 by sliding 6 over to get 2 _ 6 and putting 4 in underlined space to get 2 4 6

(Note always look at rightmost of sorted list and slide over)

Now find place for the number 3 by sliding numbers over that are bigger:

2 4 _ 6

2 _ 4 6

Put in 3 in empty slot

2 3 4 6

Insertion sort- nonrecursion:

void insertsort( int n , int A[]){ int j, k, temp; for (k=1;k<n;k++) {

temp=A[k]; j=k-1; //begin at guy directly to left while (j>=0 && temp < A[j]) {

A[j+1]=A[j]; //slide numbers over j--; }

A[j+1]=temp; //insert number in empty slot }

}

Insertion sort recursion

void insertsort( int n , int A[]){

int j, temp; if ( n>0) {

insertsort(n-1, A); //recursion temp =A[n-1]; // now find place for new entry j = n-2; while (j>=0 && temp < A[j]) {

A[j+1]=A[j]; //slide bigger numbers over

j--; } A[j+1]=temp; //insert new number in empty space}

}

Tower of hanoi

To move 2 disks from peg 1 to peg 2:

move disk 1 from peg 1 to peg 3

move disk 2 from peg 1 to peg 2

move disk 1 from peg 3 to peg 2

To move 3 disks from peg 1 to peg 2:

move the top 2 disks from peg 1 to peg 3(recursion)

move disk 3 from peg 1 to peg 2

move the top 2 disks from peg 3 to peg 2(recursion)

Count the moves: moves(2)=3

moves(3)=moves(2) +1 + moves(2) = 7

moves(4)=moves(3) + 1 + moves(3) = 15

In general

moves(n)=2*moves(n-1)+1

Using recursion to count the number of moves

int moves(int n)

{

if (n==1)

return 1; //base case

return (2*moves(n-1)+1);

}