This solving fractional part of number pdf a good article. Follow the link for more information. This article is about searching a finite sorted array. Visualization of the binary search algorithm where 7 is the target value.
If the search ends with the remaining half being empty, the target is not in the array. Although the idea is simple, implementing binary search correctly requires attention to some subtleties about its exit conditions and midpoint calculation. There are numerous variations of binary search. Binary search works on sorted arrays. Binary search begins by comparing the middle element of the array with the target value. If the target value matches the middle element, its position in the array is returned.
If the target value is less than or greater than the middle element, the search continues in the lower or upper half of the array, respectively, eliminating the other half from consideration. 1 and go to step 2. This iterative procedure keeps track of the search boundaries with the two variables. Some implementations may check whether the middle element is equal to the target at the end of the procedure. This results in a faster comparison loop, but requires one more iteration on average.
Uniform binary search stores, linear search has lower initial complexity because it requires minimal computation, dynamic Perfect Hashing: Upper and Lower Bounds”. The records of the tree are arranged in sorted order, which may improve the algorithm’s performance on some systems. Interpolation search estimates the position of the target value, uniform binary search stores the difference between the current and the two next possible middle elements instead of specific bounds. The loop must be exited when the target element is found, there are numerous variations of binary search.
However, due to the ordered nature of sorted arrays, it is trivial to extend binary search to perform approximate matches. Rank queries can be performed using a modified version of binary search. Predecessor and successor queries can be performed with rank queries. The nearest neighbor of the target value is either its predecessor or successor, whichever is closer. Range queries are also straightforward.
Once the ranks of the two values are known, the number of elements greater than or equal to the first value and less than the second is the difference of the two ranks. This count can be adjusted up or down by one according to whether the endpoints of the range should be considered to be part of the range and whether the array contains keys matching those endpoints. The performance of binary search can be analyzed by reducing the procedure to a binary comparison tree, where the root node is the middle element of the array. The middle element of the lower half is the left child node of the root and the middle element of the upper half is the right child node of the root. The rest of the tree is built in a similar fashion.
The worst case is reached when the search reaches the deepest level of the tree, equivalent to a binary search that has reduced to one element and, in each iteration, always eliminates the smaller subarray out of the two if they are not of equal size. On average, assuming that each element is equally likely to be searched, the procedure will most likely find the target value the second-deepest level of the tree. However, the tree may be unbalanced, with the deepest level partially filled. Equivalently, the array may not be divided exactly in half by the search in some iterations, half of the time resulting in the smaller subarray being eliminated. In the best case, where the target value is the middle element of the array, its position is returned after one iteration. In terms of iterations, no search algorithm that works only by comparing elements can exhibit better average and worst-case performance than binary search. This is because the comparison tree representing binary search has the fewest levels possible as each level is filled completely with nodes if there are enough.
Otherwise, the search algorithm can eliminate few elements in an iteration, increasing the number of iterations required in the average and worst case. This problem is solved by binary search, as dividing the array in half ensures that the size of both subarrays are as similar as possible. Each iteration of the binary search procedure defined above makes one or two comparisons, checking if the middle element is equal to the target in each iteration. Again assuming that each element is equally likely to be searched, each iteration makes 1. A variation of the algorithm checks whether the middle element is equal to the target at the end of the search, eliminating on average half a comparison from each iteration.
If the target value is less than or greater than the middle element, but requires one more iteration on average. This slightly cuts the time taken per iteration on most computers, a posts is at fact wonderful : Chemical. Although its time complexity grows more slowly than binary search, noisy binary search can find the correct position of the target with a given probability that controls the reliability of the yielded position. Any algorithm that does lookup — textbook errors in binary searching”. Or in the case of an implementation where this check is moved to the end, your answer was very helpful for the prepartion of my exam. So we’ll have to select it partially i.