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Recursive Methods in Java for Factorial and Fibonacci with Examples

Explore the concepts of recursion in Java with examples of calculating factorial and Fibonacci sequences. Understand recursive calls, initialization, and termination conditions. Java code snippets provided.

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Recursive Methods in Java for Factorial and Fibonacci with Examples

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  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

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