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We can continue this sort of expansion until the tree is deep enough for the size of the overall problem: (n) (n/2) (n/2) (n/4) (n/4) (n/4) (n/4) ... ... Adding across each row, we find: Sum (n) (n) (n/2) (n/2) (n) (n/4) (n/4) (n/4) (n/4) (n) ... ... For any particular problem, because we repetitively divide the problem in two, we will have as many levels as (lg n). For instance, our example with four numbers had only two levels of recursion. A problem with eight numbers will have three levels, and a problem with 16 numbers will have four. Summing over the whole problem, then, we find the merge sort has a theta of n(lg n). There are (lg n) levels, each with a theta of n. So the merge sort has an order of growth of (n(lg n)). This is a very big deal, because for large sets of numbers, n(lg n) is very much smaller than n2. Suppose that one million numbers must be sorted. The insertion sort will require on the order of (10 6)2, or 1,000,000,000,000 units of time, while the merge sort will require on the order of 10 6 (lg 10 6), or 106 (20), or 20,000,000 units of time. The merge sort will be almost five orders of magnitude faster. If a unit of time is one millionth of a second, the merge sort will complete in 20 seconds, and the insertion sort will require a week and a half! Binary search An example of order of growth of (lg n) Q(lg n) Earlier we discussed the sequential search algorithm and found its performance to be (n). One can search much more efficiently if one knows the list is in order to start with. The improvement in efficiency is akin to the improved usefulness of a telephone book when the entries are sorted by alphabetical order. In fact, for most communities, a telephone book where the entries were not sorted alphabetically would be unthinkably inefficient! If the list to be searched is already ordered from smallest to largest, the binary search algorithm can find any entry in (lg n) time. If the list contains 1,000,000 entries, that means the binary search will locate the item after reading fewer than 20 entries. The sequential search, on average, will have to read 500,000 entries. What a difference! The binary search works by repetitively dividing the list in half. It starts by comparing the element in the middle of the list with the item sought. If the search item is smaller than the element in the middle of the list, the binary search reads the element at the middle of the first half of the list. Then, if the search item is larger than that element, the binary search next reads the element at the middle of the second half of the front half of the list. Eventually, the search finds the element sought, or concludes that the element is not present in the list. Here is pseudocode for a binary search: BinarySearch(list, search_item) begin <-- 1 end <-- length of list match_found <-- false // Repeat search as long as no match has been found // and we have not searched the entire list. while match_found = false AND begin <= end // Find the item at the midpoint of the list midpoint <-- (begin + end) / 2
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// If it s the one we re looking for, we re done if list[midpoint] = search_item match_found = true // If the search item is smaller, the next // list item to check is in the first half else if search_item < list[midpoint] end <-- midpoint - 1 // Otherwise, the next list item to check // is in the back half of the list else begin <-- midpoint + 1 // Return true or false, depending on whether we // found the search_item return match_found With each iteration, the binary search reduces the size of the list to be searched by a factor of 2. So, the binary search generally will find the search item, or conclude that the search item is not in the list, when the algorithm has executed (lg n) iterations or fewer. If there are seven items in the list, the algorithm will complete in three iterations or fewer. If there are 1,000,000 items in the list, the algorithm will complete in 20 iterations or fewer. If the original list happens to be a perfect power of 2, the maximum number of iterations of the binary search can be 1 larger than (lg n). When the size of the list is a perfect power of 2, there are two items at the (lg n) level, so one more iteration may be necessary in that circumstance. For instance, if there are eight items in the list, the algorithm will complete in (3 + 1) iterations or fewer. In any case, the running time of the binary search is (lg n). This efficiency recommends it as a search algorithm, and also, therefore, often justifies the work of keeping frequently searched lists in order. Intractable problems The algorithms discussed so far all have an order of growth that can be described by some polynomial equation in n. A polynomial in n means the sum of some number of terms, where each term consists of n raised to some power and multiplied by a coefficient. For instance, the insertion sort order of growth is (n2/2 - n/2). When an algorithm has an order of growth that is greater than can be expressed by some polynomial equation in n, then computer scientists refer to the algorithm as intractable. If no better algorithm can be discovered to solve the problem, computer scientists refer to the problem as an intractable problem. As an example of an intractable problem, consider a bioinformatics problem. The Department of Genetics at Yale School of Medicine maintains a database of genetic information obtained from different human populations. ALFRED (ALlele FREquency Database) is a repository of genetic data on 494 anthropologically defined human populations, for over 1600 polymorphisms (differences in DNA sequences between individuals). However, researchers have collected data for only about 6 percent of the possible population polymorphism combinations, so most of the possible entries in the database are absent. When population geneticists seek to find the largest possible subset of populations and polymorphisms for which complete data exist (that is, measures exist for all polymorphisms for all populations), the researchers are confronted by a computationally intractable problem. This problem requires that every subset of the elements in the matrix be examined, and the number of subsets is very large! The number of subsets among n elements is 2n, since each element can either be in a particular subset or not. For our problem, the number of elements of our set is the number of possible entries in the database. That is, the ALFRED database presents us with 2 (494 1600) subsets to investigate! To exhaustively test for the largest subset with complete data, we would have to enumerate all the subsets, and test each one to see if all entries in the subset contained measurements! Clearly, the order of growth of such an algorithm is 2n; (2n). This is an exponential function of n, not a polynomial, and it makes a very important difference. An exponential algorithm becomes intractable quickly.
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