演算法概論 Introduction to Algorithms Department of Computer Science and Information Engineering, Chaoyang University of Technology 朝陽科技大學資工系 Speaker: Fuw-Yi Yang 楊伏夷 伏夷非征番, 道德經 察政章(Chapter 58) 伏者潛藏也 道紀章(Chapter 14) 道無形象, 視之不可見者曰夷 Fuw-Yi Yang
Reference: K. H. Rosen, Discrete Mathematics and Its Applications, 5th ed. J. A. Dossey, A. D. Otto, L. E. Spence and C. V. Eynden, Discrete Mathematics, 4th ed. R.1 Introduction R.2 Recurrence Relations R.3 Modeling with Recurrence Relations R.4 Solving Recurrence Relations R.5 First-Order Linear Difference Equation R.6 Divide-and-Conquer Algorithms and Recurrence Relations R.7 Prune-and-Search R.8 Dynamic Programming Fuw-Yi Yang
R.7 Prune-and-Search The Selection Problem The Selection Problem A prune-and-Search algorithm to find the kth smallest element Input: A set S of n elements Output: the kth smallest element of S 1. If |S| ≦ 5, solve the problem by any method 2. Divide S into n/5 subsets. 3. Sort each subset of elements. 4. Recursively find the element x which is the medians of the median of the n/5 subsets. 5. Partition S into S1, S2, and S3, which contains the elements less than, equal to, and greater than x, respectively. (next page) Fuw-Yi Yang
R.7 Prune-and-Search The Selection Problem 5. Partition S into S1, S2, and S3, which contains the elements less than, equal to, and greater than x, respectively. 6. If |S1| ≧ k, then discard S2 and S3 and solve the problem that selects the kth smallest element from S1 during the next iteration; else if |S1| + |S2| ≧ k, then x is the kth smallest element of S; otherwise, let k = k - |S1| - |S2|. Solve the problem that selects the kth smallest element from S3 during the next iteration. Fuw-Yi Yang
R.7 Prune-and-Search The Selection Problem—worst case It follows that the number of elements greater than x is at least 3(1/2 n/5 - 2) ≧ 3n/10 – 6. Similarly, at least 3n/10 – 6 elements are less than x. Thus, in the worst case, step 6 calls algorithm recursively on at most 7n/10 + 6 elements. s x b Fuw-Yi Yang
R.7 Prune-and-Search The Selection Problem—worst case 11, 10, 23, 40, 32, 25, 76, 45, 88, 28, 13, 32, 53, 29, 31, 19, 98, 54, 16, 42, 67, 73, 55, 62, 34, 88, 63, 17, 25, 74, 35, 67, 99, 48, 78, 93, 92, 97 10 13 16 25 34 17 35 92 11 29 19 28 55 48 93 23 31 42 (x) 45 62 63 67 97 32 54 76 74 78 40 53 98 88 73 99 23 63 31 67 42 (x) 97 45 62 Fuw-Yi Yang
R.7 Prune-and-Search The Selection Problem We make the assumption that any input of fewer than 140 elements requires O(1) time. We can therefore obtain the recurrence T(n) ≦ O(1) if n < 140, T(n) ≦ T(n/5 ) + T(7n/10 + 6 ) + O(n) if n > 140. T(n) ≦ c(n/5 ) + c(7n/10 + 6 ) + an ≦ cn/5 + c + 7cn/10 + 6c + an = 9cn/10 + 7c + an = cn + (-cn/10 + 7c + an) Which is at most cn if (-cn/10 + 7c + an) ≦0. To satisfy the inequality above, n > 140, c ≧20a. Fuw-Yi Yang
R.7 Prune-and-Search Example R.7.1 Binary Search. procedure binary search (x: integer, a1, a2, …, an: increasing integers) i := 1 {i is left endpoint of search interval} j := n {j is right endpoint of search interval} while i < j begin m := integer ((i + j) /2) if x > am then i := m + 1 else j : = m end if x = ai then location := i else location := 0 Fuw-Yi Yang
R.7 Prune-and-Search The binary search algorithm reduces the search for an element in a search sequence of size n to the binary search for this element in a search sequence of size n/2, when n is even. Hence, the problem of size n has been reduced to one problem of size n/2. Two comparisons are needed to implement this reduction, one to determine which half of the list to use and the other to determine whether any terms of the list remain. Hence if f(n) is the number of comparisons required to search for an element in a search sequence of size n, then f(n) = f(n/2) + 2, when n is even. Find its time complexity. Solution: Fuw-Yi Yang
R.7 Prune-and-Search Solution: The recurrence relation is f(n) = f(n/2) + 2, i.e. a = 1, c = 2. From algorithm, it can be seen f(1) = 1. If n is a power of b, n = bk, it follows that f(n) = f(1) + 2k = 1 + 2 log bn. Thus f(n) is O(log n). Fuw-Yi Yang