Recursive Methods in Java for Factorial and Fibonacci with Examples

1. Recursion • Recursionis a concept of defining a method that makes a call to itself. Trees

2. Recursionis a concept of defining a method that makes a call to itself. Recursion Trees

3. Factorial Example: f(n)=n!=n×(n-1)×(n-2)×…×2×1 Initialization: f(0)=1 Recursive Call: f(n)=n×f(n-1) and. Java code: public static int recursiveFactorial(int n) { if (n==0) return 1; else return n*recursiveFactorial(n-1); } Trees

4. Fibonacci sequence Fibonacci sequence: {fn } = 0,1,1,2,3,5,8,13,21,34,55,… Initialization: f0 = 0, f1 = 1 Recursive Call: fn = fn-1+fn-2 for n > 1. Java code: public static int recursiveFibonacci(int n) { if (n==0) return 0; if (n==1) return 1; else returnrecursiveFibonacci(n-1)+recursiveFibonacci (n-2); }

5. A={4,3,6,2,5} return 15+A[4]=20 Algorithm LinearSum(A, n) Input: an integer array A of n elements Output: The sum of the n elements ifn=1 then return A[0] return LinearSum(A, n-1)+A[n-1] • The recursive method should always possess—the method terminates. • We did it by setting : • ” if n=1 then return A[0] ” LinearSum(A,5) return 13+A[3]=15 LinearSum LinearSum(A,4) return 7+A[2]=13 LinearSum(A,3) return 4+A[1]=7 LinearSum(A,2) return A[0]=4 LinearSum(A,1) The compiler of any high level computer language uses a stack to handle recursive calls. f(n)=A[n-1]+f(n-1) for n>0 and f(1)=A[0] Trees

6. n=4 return f(4)=4*f(3)=24 public static int recursiveFactorial(int n) if (n==0) return 1; return n*recursiveFactorial(n-1);} The recursive method should always possess—the method terminates. • We did it by setting: • ” if n=0 then return 1 ” recursiveFactorial(4) return f(3)=3*f(2)=6 Factorial recursiveFactorial(3) return f(2)=2*f(1)=2 recursiveFactorial(2) return f(1)=1*1=1 recursiveFactorial(1) return f(0)=1 f(n)=n*f(n-1) for n>0 f(0)=1. recursiveFactorial (0) Trees

7. Fibonacci sequence public static int recursiveFibonacci(int n) { if (n==0) return 0; if (n==1) return 1; returnrecursiveFibonacci(n-1) +recursiveFibonacci (n-2); }

8. ReverseArray AlgorithmReverseArray(A, i, j): input: An array A and nonnegative integer indices i and j output: The reversal of the elements in A starting at index i and ending at j if i<j then { swap A[i] and A[j] ReverseArray(A, i+1, j-1)} } A={1, 2, 3, 4}. ReverseArray(A, 0, 3) A={4,2,3,1} What is the base case? ReverseArray(A, 1 2) A=(4,3,2,1} Trees

9. FindMax Running time: T(n)=2T(n/2)+c1 T(1)=c2 where c1 and c2 are some constants. T(n)=2T(n/2)+c1 =2[2T(n/4)+c1]+c1 =4T(n/4)+3c1 =… =2kT(1)+(1+2+4+…2k)c1 =nT(1)+2k+1 c1 =O(n) AlgorithmFindMax(A, i, j): input: Array A , indices i and j, i≤j output: The maximum element starting i and ending at j if i<j then 1 { a←FindMax(A, i, (i+j)/2) T(n/2)+1 b←FindMax(A, (i+j)/2+1, j) T(n/2)+1 return max(a, b) 1 } return A[i] 1 Trees

10. Binary Search Running time: T(n)=T(n/2)+c1 T(1)=c2 where c1 and c2 are some constants. T(n)=T(n/2)+c1 =[T(n/4)+c1]+c1 =T(n/4)+2c1 =… =T(1) + kc1 =? AlgorithmBinarySearch(A, i, j, key): input: Sorted Array A , indices i and j, i≤j, and key output: If key appears between elements from i to j, inclusively if i≤j mid  (i + j) / 2 if A[mid] = key return mid ifA[mid] < key return BinarySearch(A, mid+1, j, key) else return BinarySearch(A, i, mid-1, key) return -1 Trees