![template<class ItemType>
bool BinarySearch(ItemType info[], ItemType item, int fromLocation,
int toLocation)
{
if (fromLocation > toLocation) // Base case 1
return false;
else
{
int midPoint;
midPoint = (fromLocation + toLocation) / 2;
if (item < info[midPoint])
return BinarySearch(info, item, fromLocation, midPoint - 1);
else if (item == info[midPoint]) // Base case 2
return true;
else
return BinarySearch(info, item, midPoint + 1, toLocation);
}
}
Binary Search: The Recursive Way
What’s its time complexity?](https://image.slidesharecdn.com/l2timecomplexityrecursivefunction-250408141819-3023537b/85/L2_Time_complexity_recursive_function-pptx-3-320.jpg)
L2_Time_complexity_recursive_function.pptx
- 1. Lecture 2 Time complexity Analysis of Recursive Algorithms CSE 373: Design & Analysis of Algorithms
- 2. Recursive Factorial Function // Computes the factorial of a nonnegative integer. // Precondition: n must be greater than or equal to 0. // Postcondition: Returns the factorial of n; n is unchanged. int Factorial(int n) { if (n == 0) return (1); else return (n * Factorial(n-1)); } void main() { int n; cin>>n; Factorial(n); } T(n) = O(1) , if n = 0 T(n) = T(n-1)+O(1) , otherwise O(1) time Let factorial(n) call takes total T(n) time So Factorial(n-1) call will take T(n-1) time O(1) time O(1) time
- 3. template<class ItemType> bool BinarySearch(ItemType info[], ItemType item, int fromLocation, int toLocation) { if (fromLocation > toLocation) // Base case 1 return false; else { int midPoint; midPoint = (fromLocation + toLocation) / 2; if (item < info[midPoint]) return BinarySearch(info, item, fromLocation, midPoint - 1); else if (item == info[midPoint]) // Base case 2 return true; else return BinarySearch(info, item, midPoint + 1, toLocation); } } Binary Search: The Recursive Way What’s its time complexity?
- 4. Tower of Hanoi Problem • There is a tower of n discs stacked in increasing order of size in the first peg • Goal: transfer the entire tower to the 3rd peg in the fewest possible moves. • Constraints: • You can only move one disk at a time and • You can never stack a larger disk onto a smaller disk.
- 5. Recursive Solution • Problem: Move N discs from source peg to destination peg using auxiliary peg as via • Base case: • Move disc from source peg to destination peg (N = 1)
- 6. Recursive Solution • Problem: Move N discs from source peg to destination peg using auxiliary peg as via • General case: • Move N-1 discs from source peg to auxiliary peg via destination peg • Move disc from source peg to destination peg • Move N-1 discs from auxiliary peg to destination peg via source peg
- 7. #include <stdio.h> void hanoi(int n, char src, char dst, char aux) { if(n == 1) printf("Move disc from peg %c to peg %cn", src, dst); else { hanoi(n-1, src, aux, dst); hanoi(1, src, dst, aux); hanoi(n-1, aux, dst, src); } } int main() { hanoi(4, 'A', 'C', 'B'); return 0; } Recursive Function for Solving Hanoi What’s its time complexity?