Recursion Trees1 Recursion is a concept of defining a method that makes a call to itself.

Recursion Trees2 Recursion is a concept of defining a method that makes a call to itself.

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); } Trees3

L16 4 Fibonacci sequence Fibonacci sequence: {f n } = 0,1,1,2,3,5,8,13,21,34,55,… Initialization:f 0 = 0, f 1 = 1 Recursive Call: f n = f n-1 +f n-2 for n > 1. Java code: public static int recursiveFibonacci(int n) { if (n==0) return 0; if (n==1) return 1; else return recursiveFibonacci(n-1)+recursiveFibonacci (n-2); }

LinearSum Trees5 LinearSum(A,5) LinearSum(A,4) LinearSum(A,3) LinearSum(A,2) LinearSum(A,1) Algorithm LinearSum(A, n) Input: an integer array A of n elements Output: The sum of the n elements if n=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] ” return A[0]=4 return 4+A[1]=7 return 7+A[2]=13 return 13+A[3]=15 return 15+A[4]=20 A={4,3,6,2,5} 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]

Factorial Trees6 recursiveFactorial(4) recursiveFactorial(3) recursiveFactorial(2) recursiveFactorial(1) recursiveFactorial (0) 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 ” return f(0)=1 return f(1)=1*1=1 return f(2)=2*f(1)=2 return f(3)=3*f(2)=6 return f(4)=4*f(3)=24 f(n)=n*f(n-1) for n>0 f(0)=1. n=4

L16 7 Fibonacci sequence public static int recursiveFibonacci(int n) { if (n==0) return 0; if (n==1) return 1; return recursiveFibonacci(n-1) +recursiveFibonacci (n-2); }

ReverseArray Algorithm ReverseArray(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)} } Trees8 ReverseArray(A, 0, 3) ReverseArray(A, 1 2) A=(4,3,2,1} A={4,2,3,1} A={1, 2, 3, 4}. What is the base case?

FindMax Algorithm FindMax(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 Trees9 Running time: T(n)=2T(n/2)+c 1 T(1)=c 2 where c 1 and c 2 are some constants. T(n)=2T(n/2)+c 1 =2[2T(n/4)+c 1 ]+c 1 =4T(n/4)+3c 1 =… =2 k T(1)+(1+2+4+…2 k )c 1 =nT(1)+2 k+1 c 1 =O(n)

Binary Search Algorithm BinarySearch(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 if A[mid] < key return BinarySearch(A, mid+1, j, key) else return BinarySearch(A, i, mid-1, key) return -1 Trees10 Running time: T(n)=T(n/2)+c 1 T(1)=c 2 where c 1 and c 2 are some constants. T(n)=T(n/2)+c 1 =[T(n/4)+c 1 ]+c 1 =T(n/4)+2c 1 =… =T(1) + kc 1 =?