sum of the first n numbers

1. 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

2. sum – non recursive function double sumiteration( int n ) { double sum1; sum1 = 0 ; for ( int i =1; i <= n ; i ++) sum1 = sum1 +i; return ( sum1); }

3. sum-recursive function double sum( int n ) { if ( n ==0 ) // base case return 0; // recursive case return n + sum( n - 1 ); }

4. 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

5. Factorial-nonrecursive double factorial( int n ) { double fac1; fac1 = 1 ; for ( int i =1; i <= n ; i ++) fac1 =fac1 *i; return ( fac1); }

6. Factorial recursive double factorial( int n ) { if ( n ==1 ) // base case return 1; // recursive case return n * factorial( n - 1 ); }

7. 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]

8. 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); }

9. 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 )); }

10. 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]

11. 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); }

12. 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)); }

13. 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

14. 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; }

15. 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; }

16. 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);

17. 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;}

18. 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 }

19. 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.

20. Reverse digits- no recursion void reverseiter( int number ) { while (number >0) { cout << number % 10 ; number = number /10; } }

21. Reverse digits- recursion void reverse( int number ) { if ( number !=0 ) { cout << number % 10 ; reverse (number /10); } }

22. 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.

23. 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; } }

24. 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

25. 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 } }

26. 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 } }

27. 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

28. 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); }