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External sorting algorithms generally fall into two types, distribution sorting, which resembles quicksort, and external merge sort, which resembles merge sort. External merge sort typically uses a hybrid sort-merge strategy. In the sorting phase, chunks of data small enough to fit in main memory are read, sorted, and written out to a temporary ...
An example of such is the classic merge that appears frequently in merge sort examples. The classic merge outputs the data item with the lowest key at each step; given some sorted lists, it produces a sorted list containing all the elements in any of the input lists, and it does so in time proportional to the sum of the lengths of the input lists.
Example C-like code using indices for top-down merge sort algorithm that recursively splits the list (called runs in this example) into sublists until sublist size is 1, then merges those sublists to produce a sorted list. The copy back step is avoided with alternating the direction of the merge with each level of recursion (except for an ...
A list containing a single element is, by definition, sorted. Repeatedly merge sublists to create a new sorted sublist until the single list contains all elements. The single list is the sorted list. The merge algorithm is used repeatedly in the merge sort algorithm. An example merge sort is given in the illustration.
In computer science, divide and conquer is an algorithm design paradigm. A divide-and-conquer algorithm recursively breaks down a problem into two or more sub-problems of the same or related type, until these become simple enough to be solved directly. The solutions to the sub-problems are then combined to give a solution to the original problem.
For instance, the array might be subdivided into chunks of a size that will fit in RAM, the contents of each chunk sorted using an efficient algorithm (such as quicksort), and the results merged using a k-way merge similar to that used in merge sort. This is faster than performing either merge sort or quicksort over the entire list. [40] [41]
In the most balanced case, each time we perform a partition we divide the list into two nearly equal pieces. This means each recursive call processes a list of half the size. Consequently, we can make only log 2 n nested calls before we reach a list of size 1. This means that the depth of the call tree is log 2 n.
The partition problem is NP hard. This can be proved by reduction from the subset sum problem. [6] An instance of SubsetSum consists of a set S of positive integers and a target sum T; the goal is to decide if there is a subset of S with sum exactly T.