![Binary Search
2. Calculate
Middle = int (low + high)
/2
If 37 == array[middle]
return middle
Else if 37 < array[middle]
high = middle -1
Else if 37 > array[middle]
low = middle +1
Here Mid=int(0+8)/2
=int(8/2)
Mid =4
Array[Mid]=45
37<45
Hence,high=4-1=3
Now,
Low=0 and high=3](https://image.slidesharecdn.com/demopresentation-220625071533-23a2efcc/85/Binary-Search-Algorithm-pptx-5-320.jpg)
![Repeat 2. Calculate Mid =int (low + high) / 2.
= (0 + 3) / 2 = 1.
Array[mid]=35
37>35
Hence, low=mid+1=1+1=2
Now, Low=2 & high=3](https://image.slidesharecdn.com/demopresentation-220625071533-23a2efcc/85/Binary-Search-Algorithm-pptx-6-320.jpg)
![Repeat 2. Calculate Mid = int(low + high) / 2.
= (2 + 3) / 2 = 2.
Array[Mid]=37=Item](https://image.slidesharecdn.com/demopresentation-220625071533-23a2efcc/85/Binary-Search-Algorithm-pptx-7-320.jpg)
Binary Search Algorithm.pptx
- 1. Binary Search Algorithm By : Bhagyashree Madan
- 2. Introduction A Binary search algorithm finds the position of a specified input value (the search "key") within a sorted array . For binary search, the array should be arranged in ascending or descending order.
- 3. Binary search algorithm Algorithm is quite simple. It can be done either recursively or iteratively: Get the middle element; If the middle element equals to the searched value, the algorithm stops. Otherwise, two cases are possible: Searched value is less, than the middle element. In this case, go to the step 1 for the part of the array, before middle element and value high=mid-1 Searched value is greater, than the middle element. In this case, go to the step 1 for the part of the array, after middle element. And value of low=mid+1 Now, The iterations should stop when searched element is found & when sub array has no elements. In this case, we can conclude, that searched value is not present in the array.
- 4. Binary Search Item to be search=37 1. Sort Array.
- 5. Binary Search 2. Calculate Middle = int (low + high) /2 If 37 == array[middle] return middle Else if 37 < array[middle] high = middle -1 Else if 37 > array[middle] low = middle +1 Here Mid=int(0+8)/2 =int(8/2) Mid =4 Array[Mid]=45 37<45 Hence,high=4-1=3 Now, Low=0 and high=3
- 6. Repeat 2. Calculate Mid =int (low + high) / 2. = (0 + 3) / 2 = 1. Array[mid]=35 37>35 Hence, low=mid+1=1+1=2 Now, Low=2 & high=3
- 7. Repeat 2. Calculate Mid = int(low + high) / 2. = (2 + 3) / 2 = 2. Array[Mid]=37=Item
- 8. Item Found at position 2
- 9. If not found stop when low > high.
- 10. Analysis of binary search To analyze the binary search algorithm, we need to recall that each comparison eliminates about half of the remaining items from consideration. If we start with n items, about n/2 items will be left after the first comparison. After the second comparison, there will be about n/4. Then n/8, n/16, and so on. When we split the list enough times, we end up with a list that has just one item. Either that is the item we are looking for or it is not. Either way, we are done. The number of comparisons necessary to get to this point is i where n/2^i=1. Solving for i gives us i=(log n). The maximum number of comparisons is log n. Therefore, the complexity of binary search is O(log n).
- 11. THANK YOU
- 12. QUESTIONS