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// If numbers remain only in list_B, move those // to the sorted list while list_B is not empty sorted_list[index] <-- list_B[1] discard list_B[1] index <-- index + 1 // Return the sorted list return sorted_list The performance of merge is related to the lengths of the lists on which it operates, the total number of items being merged. The real work of the routine is in moving the appropriate elements of the original lists into the sorted list. Since the total number of such moves is equal to the sum of the numbers in the two lists, merge has a theta of nA + nB, or (nA + nB), where nA + nB is equal to the sum of the numbers in the two lists. The merge_sort will use the merge routine, but first the merge_sort will divide the problem up into smaller and smaller sorting tasks. Then merge_sort will reassemble the small sorted lists into one fully sorted list. In fact, merge_sort divides the list of numbers until each sublist consists of a single number, which can be considered a sorted list of length 1. Then the merge_sort uses the merge procedure to join the sorted sublists. The technique used by merge_sort to divide the problem into subproblems is called recursion. The merge_sort repeatedly calls itself until the recursion bottoms out with lists whose lengths are one. Then the recursion returns, reassembling the numbers in sorted order as it does. Here is pseudocode for the merge sort. It takes the list of numbers to be sorted, and it returns a sorted list of those numbers. merge_sort(num_list) length <-- length of num_list // if there is more than 1 number in the list, if length > 1 // divide the list into two lists half as long shorter_list_A <-- first half of num_list shorter_list_B <-- second half of num_list // Perform a merge sort on each shorter list result_A <-- merge_sort(shorter_list_A) result_B <-- merge_sort(shorter_list_B) // Merge the results of the two sorted sublists sorted_list <-- merge(result_A, result_B) // Return the sorted list return sorted_list else // If there s only 1 number in the list, // just return it return num_list end Let s follow the execution of merge_sort when one calls it with this list of numbers: NUMS = { 1, 6, 4, 2 } 1 First, we call merge_sort passing the list NUMS. This is what we call the top-level of recursion, level 0.
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2 merge_sort calls merge_sort again, passing a list of the first two numbers in NUMS. This will sort the front half of the list. This is level 1 of recursion. 3 Now merge_sort calls merge_sort again, passing only the first number in NUMS. This is level 2. 4 Now merge_sort simply returns; it s down to one element in the list, merge_sort returns to level 1. 5 Now merge_sort calls merge_sort again, passing only the second of the first two numbers in NUMS. This is level 2. 6 Again, merge_sort simply returns; it s down to one element in the list, merge_sort returns to level 1. 7 At level 1 of recursion, merge_sort now has result_A and result_B. merge_sort calls merge to put those two numbers in order, and then it returns the sorted pair of numbers back to level 0. The first half of the list is sorted. 8 From level 0, merge_sort calls merge_sort again, passing a list of the last two numbers in NUMS. This will sort the back half of NUMS. It s back to level 1 of recursion. 9 merge_sort calls merge_sort again, passing only the first of the last two numbers of NUMS. This is level 2 of recursion again. 10 Since the list contains only one number, merge_sort simply returns back to level 1. 11 merge_sort calls merge_sort again, passing only the last of the numbers of NUMS. This is level 2 of recursion again. 12 Since the list contains only one number, merge_sort simply returns back to level 1. 13 At level 1 of recursion, merge_sort now has result_A and result_B. merge_sort calls merge to put the two lists in order, and then it returns the sorted set of two numbers back to level 0. 14 At level 0 of recursion, merge_sort now has result_A and result_B. merge_sort calls merge to put the two lists of numbers in order, and then it returns the entire set of four numbers in sorted order. Aside from being an interesting exercise in recursion, the merge_sort provides attractive performance. The merge sort has a theta of n(lg n), which for large problems is much better than the theta of n2 for the insertion sort. The recursion in merge_sort divides the problem into many subproblems by repeatedly halving the size of the list to be sorted. The number of times the list must be divided by two in order to create lists of length one is equal to the logarithm to the base 2 of the number of elements in the list. In the case of our 4-element example, the logarithm to the base 2 of 4 is 2, because 22 = 4. This can be written as log2 n, but in computer science, because of the ubiquity of binary math, this is usually written as lg n, meaning logarithm to the base 2 of n. The total running time T of the merge sort consists of the time to recursively solve two problems of half the size, and then to combine the results. One way of expressing the time required is this: T = 2T(n/2) + merge Since merge runs in (nA + nB), and since nA + nB = n, we will restate this: T = 2T(n/2) + (n) A recursion tree is a way to visualize the time required. At the top level, we have the time required for merge (n), plus the time required for the two subproblems: (n) T(n/2) T(n/2) At the next level, we have the time required for the two merges of the two subproblems, and for the further subdivision of the two subproblems: (n) (n/2) (n/2) T(n/4) T(n/4) T(n/4) T(n/4)
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