演算法概論 Introduction to Algorithms

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.1 Introduction The number of a bacteria in a colony doubles every hour. If a colony begins with five bacteria, how many will be present in n hours? To solve this problem, let an be the number of bacteria at the end of n hours. Since the number of bacteria doubles every hours, the relationship an = 2an-1 holds whenever n is a positive integer. This relationship, together with the initial condition a0 = 5, uniquely determines an for all nonnegative integers n. Solution: 5, 10, 20, 40, … Fuw-Yi Yang

R.2 Recurrence Relations
Definition R.2.1. A recurrence relation for the sequence {an } is an equation that expresses an in terms of one or more of the previous terms of the sequence, namely, a0 , a1, …, an-1, for all integers n with n ≧n0, where n0 is a nonnegative integer. A sequence is called a solution of a recurrence relation if its terms satisfy the recurrence relation. Fuw-Yi Yang

Connection between recursion and recurrence relations Recursive algorithms provide the solution of a problem of size n in terms of the solutions of one or more instances of the same problem of a smaller size. Consequently, when we analyze the complexity of a recursive algorithm, we obtain a recurrence relation that expresses the operations required to solve a problem of size n in terms of the number of operations required to solve the problem for one or more instances of smaller size. Fuw-Yi Yang

Example R.2.1 Let {an} be a sequence that satisfies the recurrence relation an = an-1 – an-2 for n = 2, 3, 4,…, and suppose that a0 = 3 and a1 = 5. What are a2 and a3? Solution: a2 = a1 – a0 = 5 – 3 = 2, a3 = a2 – a1 = 2 – 5 = -3 Fuw-Yi Yang

Example R.2.2 Determine whether the sequence {an} is a solution of the recurrence relation an = 2an-1 – an-2 for n = 2, 3, 4,…, where an = 3n for every nonnegative integer n. Answer the same question where an = 2n and where an = 5. Solution: suppose that an = 3n for every nonnegative integer n. Then for n ≧ 2, we have an = 2an-1 – an-2 = 2(3(n - 1)) - 3(n - 2) = 3n. Therefore, the sequence {an} is a solution of the recurrence relation. Fuw-Yi Yang

Example R.2.3 Determine whether the sequence {an} is a solution of the recurrence relation an = 2an-1 – an-2 for n = 2, 3, 4,…, where an = 3n for every nonnegative integer n. Answer the same question where an = 2n and where an = 5. Solution: suppose that an =2n for every nonnegative integer n. Then for n > 2, we have an = 2*2n-1 – 2n-2 = 2n – 2n-2 ≠ 2n. Therefore, the sequence {an} is not a solution of the recurrence relation. Fuw-Yi Yang

Example R.2.4 Determine whether the sequence {an} is a solution of the recurrence relation an = 2an-1 – an-2 for n = 2, 3, 4,…, where an = 3n for every nonnegative integer n. Answer the same question where an = 2n and where an = 5. Solution: suppose that an = 5 for every nonnegative integer n. Then for n ≧ 2, we have an = 2*5 – 5 = 5. Therefore, the sequence {an} is a solution of the recurrence relation. Fuw-Yi Yang

R.3 Modeling with Recurrence Relations
We can use recurrence relations to model a wide variety of problems, such as finding compound interest, counting rabbits on an island, determining the number of moves in the Tower of Hanoi puzzle, and counting bit strings with certain properties. Fuw-Yi Yang

Example R.3.1 Compound Interest. Suppose that a person deposit $10000 in a saving account at a bank yielding 11% per year with interest compounded annually. How much will be in the account after 30 years? Solution: Let pn denote the amount in the account after n years. Then we see that the sequence {pn} satisfies the recurrence relation Fuw-Yi Yang

Solution: Let pn denote the amount in the account after n years. Then we see that the sequence {pn} satisfies the recurrence relation pn = pn pn-1 = (1.11)pn-1. The initial condition is p0 = p1 = (1.11) p0 p2 = (1.11) p1 = (1.11)2 p0 … pn = (1.11) pn-1 = (1.11)np0. Thus p30 = (1.11) p30 = Fuw-Yi Yang

Example R.3.2 Rabbits and the Fibonacci Numbers. Consider this problem, which was originally posed by Leonardo di Pisa, also known as Fibonacci, in the thirteenth century in his book Liber abaci. A young pair of rabbits (one of each sex) is placed on an island. A pair of rabbits does not breed until they are 2 months old. After they are 2 months old, each pair of rabbits produces another pair each month. Find a recurrence relation for the number of pairs of rabbits on the island after n months, assuming that no rabbits ever die. Solution: Let fn denote the number of pairs of rabbits after n months. Fuw-Yi Yang

Solution: Let fn denote the number of pairs of rabbits after n months. At the end of the first month, the number of pairs of rabbits on the island is f1 = 1. At the end of the second month, the number of pairs of rabbits on the island is f2 = 1. N.t. Fuw-Yi Yang

Month Reproducing pairs fn-2 Young pairs fn-1 Total pairs fn 1 2 3 4 5 6 8 7 The sequence {fn} satisfies the recurrence relation fn = fn-1 + fn-2 Fuw-Yi Yang

Example R.3.3 Counting bit string of length n with no two consecutive 0s. Find a recurrence relation and give initial conditions for the number of bit strings of length n that do not have two consecutive 0s. How many such bit strings are there of length five? Solution: Let an denote the number of bit strings of length n that do not have two consecutive 0s. Fuw-Yi Yang

Solution: Let an denote the number of bit strings of length n that do not have two consecutive 0s. To obtain a recurrence relation for {an}, note that by the sum rule, the number of bit strings of length n that do not have two consecutive 0s equals the number of such bit strings ending with a 0 plus the number of such bit strings ending with a 1. Fuw-Yi Yang

an: the number of bit strings of length n that do not have two consecutive 0s equals 1 the number of bit strings of length n that do not have two consecutive 0s ending with a 1 plus 2 the number of bit strings of length n that do not have two consecutive 0s ending with a 0. an-1: the number of the first term (1) is precisely the number of bit strings of length n-1 that do not have two consecutive 0s, i.e. an-1. an-2: the number of the second term (2) is precisely the number of bit strings of length n-2 that do not have two consecutive 0s, i.e. an-2. Therefore, the recurrence relation an = an-1 + an-2 satisfies the sequence {an}. Fuw-Yi Yang

R.4 Solving Recurrence Relations
Definition R.4.1 A linear homogeneous recurrence relation of degree k with constant coefficients is a recurrence relation of the form an = c1an-1+ c2an-2+…+ ckan-k, where c1, c2, …, ck are real numbers, and ck ≠0. Fuw-Yi Yang

Linear: since the right-hand side is a sum of multiples of the previous terms of the sequence. Homogeneous: since no terms occur that are not multiples of the ajs. Constants: the coefficients of the terms of the sequence are all constants, rather than functions that depend on n. Degree: the degree is k because an is expressed in terms of the previous k terms of the sequence. Fuw-Yi Yang

A sequence satisfying the recurrence relation in the definition is uniquely determined by this recurrence relation and the k initial conditions a0 = C0, a1 = C1, … ,ak-1 = Ck-1. Fuw-Yi Yang

Example R.4.1 some linear homogeneous recurrence relations: degree 1: pn = (1.11) pn-1, degree 2: fn = fn-1 + fn-2, degree 5: an = an-5, Example 2. some recurrence relations: an = an-1 + (an-2)2 , not linear (square), Hn = 2Hn , not homogeneous (1), Bn = nBn-1 , not constant coefficient (n). Fuw-Yi Yang

Linear homogeneous recurrence relations are studied for two reasons. First, they often occur in modeling of problems. Second, they can be systematically solved. Fuw-Yi Yang

R.4 Solving Recurrence Relations with Constant Coefficients
The basic approach for solving linear homogeneous recurrence relations is to look for solutions of the form an = rn, where r is a constant. Note that an = rn is a solution of the recurrence relation an = c1an-1+ c2an-2+…+ ckan-k, if and only if : rn = c1rn-1+ c2rn-2+…+ ckrn-k. Both sides dividing by rn-k, we obtain, rk - c1rk-1 - c2rk-2 -…- ckrk-k = 0. Fuw-Yi Yang

Note that an = rn is a solution of the recurrence relation an = c1an-1+ c2an-2+…+ ckan-k, if and only if : rn = c1rn-1+ c2rn-2+…+ ckrn-k. Both sides dividing by rn-k, we obtain, rk - c1rk-1 - c2rk-2 -…- ckrk-k = 0. Consequently, the sequence {an} with an = rn is a solution if and only if r is a solution of this last equation, which is called the characteristic equation of the recurrence relation. The solutions of this equation are called the characteristic roots of the recurrence relation. Fuw-Yi Yang

Example R.4.2 What is the solution of the recurrence relation: an = an-1 + 2an-2, with a0= 2 and a1 = 7 ? Solution: The characteristic equation of the recurrence relation is r2 - r1 - 2 = 0. The characteristic roots of the recurrence relation are r = 2 and r = -1. Which one is the solution ? Fuw-Yi Yang

Which one is the solution ? With the same method of solving homogeneous differential equations, we guess the solution may be the combination of both of them. The followings show that. Let the characteristic roots be r1 and r2 for a characteristic equation r2 - c1r1 - c2 = 0. The corresponding recurrence relation is an = c1an-1 + c2an-2. Also A and B denote constant coefficients. Fuw-Yi Yang

The corresponding recurrence relation is an = c1an-1 + c2an-2. Also A and B denote constant coefficients. If the combination of r1 and r2 is solution, then an = A(r1)n + B(r2)n. That is: c1an-1 + c2an-2 = c1 (A(r1)n-1 + B(r2)n-1) + c2 (A(r1)n-2 + B(r2)n-2) = A(r1)n-2(c1r1 + c2) + B(r2)n-2(c1r2 + c2) = A(r1)n-2(r1)2 + B(r2)n-2(r2)2 = an. Fuw-Yi Yang

Hence the sequence {an} is a solution to the recurrence relation iff an = A(2)n + B(-1)n, From the initial conditions, it follows that a0 = 2 = A + B, a1 = 7 = 2 – B. Thus, A = 3 and B = -1. Hence, the solution to the recurrence relation and initial conditions is the sequence {an} with an = 3*2n - (-1)n. Fuw-Yi Yang

Theorem R.4.1 Let c1 and c2 be real numbers. Suppose that r2 - c1r - c2 = 0 has two distinct roots r1 and r2. Then the sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2 iff an = A(r1)n + B(r2)n for n = 0, 1, 2, …, where A and B are constant coefficients. Fuw-Yi Yang

Theorem R.4.2 Let c1 and c2 be real numbers. Suppose that r2 - c1r - c2 = 0 has only one roots r0. Then the sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2 iff an = A(r0)n + Bn(r0)n for n = 0, 1, 2, …, where A and B are constant coefficients. Fuw-Yi Yang

Theorem R.4.3 Let c1, c2, …, ck be real numbers. Suppose that the characteristic equation rk - c1rk-1 - …- ck = 0 has k distinct roots r1, r2, …, rk. Then the sequence {an} is a solution of the recurrence relation an = c1an-1 + c2an-2 + … + ckan-k iff an = A1(r1)n + A2 (r2)n + …+Ak (rk)n for n = 0, 1, 2, …, where A1, A2, …, Ak are constant coefficients. Fuw-Yi Yang

R.5 First Order Linear Difference Equation with Constant Coefficient
The simplest type of recurrence relation give sn as a function of sn-1 for n≧ 1. We call an equation of the form sn = asn-1 + b, where a and b are constants and a ≠ 0, a first order linear difference equation with constant coefficients. It is not difficult to derive estimates of the size of functions satisfy such recurrence relations. sn = a sn-1 + b = a (a sn-2 + b) + b = a2sn-2 + ab + b = a3sn-3 + a2b + ab + b =… = ans0 + an-1b + … + ab + b Fuw-Yi Yang

sn = a sn-1 + b = ans0 + an-1b + … + ab + b IF a = 1, then sn = s0 + nb If a ≠ 1, sn = sn-1 + b = ans0 + an-1b + … + ab + b = ans0 + b(an - 1)/ (a - 1) = an(s0 + b/(a - 1)) - b/(a - 1) Fuw-Yi Yang

Example R.5.1 Sequential Search Algorithm This algorithm searches a list of n items a1, a2, …,an for a given target value t. If t = ak for some index k, then the algorithm gives the first such index k. Otherwise the algorithm gives k = 0. Step 1 j = 1 Step 2 while j ≦n and aj ≠t j = j + 1 Step 3 if j ≦n then k = j elsek k = 0 Fuw-Yi Yang

R.5 First Order Linear Difference Equation
with Constant Coefficient To search a list of n items, we compare the target value to the first item in the list and then, in the worst case, still have to search a list of n-1 items. Thus we see that sn satisfies the recurrence relation sn = sn-1 + 1 for for n≧ 1. Also, the initial condition s0 = 0. = (sn-2 + 1) + 1 = s0 + n = n + 1 Therefore the complexity of this algorithm is O(n). Fuw-Yi Yang

Example R.5.2 Find a formula for sn if sn = 3 sn-1 -1 for n ≧1. The initial condition is s0 = 2. Solution: sn = asn-1 + b = an(s0 + b/(a - 1)) - b/(a - 1), where a = 3 and b = -1. Therefore, sn = 3n(2 - 1/2) + 1/2 =1/2(3n+1 +1). The time complexity of sn is of O(3n). Fuw-Yi Yang

Example R.5.3 Bubble Sort This algorithm places the numbers in the list a1, a2, …, an in non-decreasing order. Step 1 for j = 1to n // set beginning of sublist Step for k = n-1 to j by -1 // find the smallest element Step if ak+1 < ak interchange the values of ak+1 and ak Step endfor Step endfor Step 4 Print a1, a2, …, an Fuw-Yi Yang

In performing a bubble sort, the first iteration requires comparing an-1 to an, then an-2 to an-1, and so forth until a2 is compared to a1. After each such comparison, we interchange the values of ak and ak+1, if necessary, so that ak≦ak+1. Thus when the first iteration is complete, a1 is the smallest number in the original list. A second iteration is then performed on the smaller list a2, a3,…,an in order to make a2 the second-smallest number in the original list. After n-1 such iterations, the original list will be non-decreasing order. Thus the recurrence relation is bn = bn-1 + (n-1) with initial condition b1 = 0. Find its complexity. Fuw-Yi Yang

Solution: The recurrence relation is bn = bn-1 + (n-1), for n ≧ 2 initial condition b1 = 0. b1 = 0, b2 = 0 + 1, b3 = , we conjecture that bn = …+ (n-1) = (n2 – n)/2 Thus bn is O(n2). Fuw-Yi Yang

R.6 Divide-and-Conquer Algorithms and Recurrence Relations
Recurrence relations of the form f(n) = af(n/b) + g(n) This form of recurrence relation arise in many differential situations. It implies the concepts of divide and conquer. It is possible to derive estimates of the size of functions satisfy such recurrence relations. Suppose that f satisfies this recurrence relation whenever n is divisible by b. Let n = bk, where k is a positive integer. Then f(n) = a f(n/b) + g(n) = a (af((n/b)/b) + g(n/b)) + g(n) = a2f(n/b2) + ag(n/b) + g(n) = a3f(n/b3) + a2g(n/b2) + ag(n/b) + g(n) Fuw-Yi Yang

f(n) = a f(n/b) + g(n) = a (af((n/b)/b) + g(n/b)) + g(n) = a2f(n/b2) + ag(n/b) + g(n) = a3f(n/b3) + a2g(n/b2) + ag(n/b) + g(n) =… = akf(n/bk) + ak-1g(n/bk-1) + … + ag(n/b) + g(n) = akf(1) + ak-1g(n/bk-1) + … + ag(n/b) + g(n) Fuw-Yi Yang

Recurrence relations of the form f(n) = af(n/b) + g(n) f(n) = a f(n/b) + g(n) = akf(n/bk) + ak-1g(n/bk-1) + … + ag(n/b) + g(n) = akf(1) + ak-1g(n/bk-1) + … + ag(n/b) + g(n) Since n = bk, thus k = logbn. We can use this equation for f(n) to estimate the size of functions that satisfy divide and conquer relation. Fuw-Yi Yang

Theorem R.6.1 Let f be an increasing function that satisfies the recurrence relation f(n) = af(n/b) + c. Whenever n is divisible by b, where a ≧1, b is an integer greater than 1, and c is a positive real number. Then f(n) is O(nlog ba) if a > 1, O(log n) if a = 1. Furthermore, when n = bk, where k > 0, f(n) = Anlogba + B, where A = f(1) + c/(a - 1) and B = -c/(a - 1). Proof: Fuw-Yi Yang

f(n) = af(n/b) + c = akf(n/bk) + ak-1c + … + ac + c = akf(1) + ak-1c + … + ac + c ∵n = bk = akf(1) + (ak – 1)c / (a - 1) If a = 1, then f(n) = f(1) + kc if n is a power of b, n = bk, f(n) = f(1) + c logbn if n is not a power of b, bk < n < bk+1, f(n) = f(1) + c logbn < f(1) + c(k + 1) = (f(1) + c) + ck < (f(1) + c) + c logbn for both cases, f(n) is O(log n). Fuw-Yi Yang

If a > 1, and n is a power of b, n = bk, it follows that f(n) = akf(1) + c(ak – 1)/(a - 1) = ak(f(1) + c/(a – 1)) - c/(a - 1) = A n logba + B, ∵ak = alogbn = nlogba where A = f(1) + c/(a – 1) , B = - c/(a - 1). If a > 1, and n is not a power of b, by similar processing we obtain the same result. Thus f(n) is O(n logba). End of proof Fuw-Yi Yang

R.6 Divide-and-Conquer Algorithms and Recurrence Relations ---Merge Sort
Example R.6.2 Merge Sort Procedure mergesort (L = a1, a2, …, an) If n > 1 then m := int(n/2) L1 := a1, a2, …, am L2 := am+1, am+2, …, an L = merge(mergesort(L1), mergesort(L2)) {L is sorted into elements in nondecreasing order} Fuw-Yi Yang

R.6 Divide-and-Conquer Algorithms and Recurrence Relations --- Merge Sort
Procedure merge (L1, L2: lists) L := empty list While L1 and L2 are both nonempty remove smaller of first element of L1 and L2 from the list it is in and put it at the left end of L if removal of this element makes one list empty then remove all elements form the other list and append them to L Endwhile {L is the merged list with elements in nondecreasing order} Two sorted lists with m elements and n elements can be merged into a sorted list using no more than m+n-1 comparisons. Fuw-Yi Yang

R.6 Divide-and-Conquer Algorithms and Recurrence Relations – Merge Sort
Thus, the recurrence relation for merge sort algorithm is : T(n) = 2T(n/2) + n. Let n = 2k, T(n) = 2T(n/2) + n = 2kT(n/2k) + 2k-12 + … + 2(n/2) + n = 2kT(1) + 2k-12 + … + 2(n/2) + n = nT(1) + n log2n ∵n = 2k Therefore T(n) is O(n log n). Fuw-Yi Yang

R.6 Divide-and-Conquer Algorithms and Recurrence Relations --- Quick Sort
Example R.6.3 Quick Sort Quick sort applies the divide-and-conquer paradigm. Here is the three-step divide-and-conquer process for sorting a typical subarray A[p...r], i.e. Divide, Conquer, and Combine. Divide: partition the array A[p..r] into two subarrays (possibly empty) A[p...q-1] and A[q+1...r] such that for each element of A[p...q-1] is less than or equal to A[q], which is, in turn, less than or equal to each element of A[q+1...r]. Compute the index q as part of this partitioning procedure. Fuw-Yi Yang

Divide: partition the array A[p..r] into two subarrays (possibly empty) A[p...q-1] and A[q+1...r] such that for each element of A[p...q-1] is less than or equal to A[q], which is, in turn, less than or equal to each element of A[q+1...r]. Compute the index q as part of this partitioning procedure. Conquer: Sort the two subarrays A[p...q-1] and A[q+1...r] by recursive calls to quicksort. Combine: Because the subarrays are already sorted, no work is needed to combine them: the entire array A[p...r] is now sorted. Fuw-Yi Yang

Procedure Quicksort(A, p, r) If p < r then q = Partition(A, p, r) Quicksort(A, p, q-1) Quicksort(A, q+1, r) Fuw-Yi Yang

Procedure Partition(A, p, r) x = A[r] i = p - 1 for j = p to r – 1 if A[j]  x i = i + 1 swap(A[i], A[j]) swap(A[i+1], A[r]) return i + 1 Fuw-Yi Yang

Fuw-Yi Yang

Worst-case partitioning The worst-case behavior for quick sort occurs when the partitioning routing produces one subproblem with n-1 elements and one with 0 elements. Let us assume that this unbalanced partitioning arises in each recursive call. The partitioning cost (n) time. Since the recursive call on an array of size 0 just returns, T(0) = (1), and the recurrence for the running time is: T(n) = T(n-1) + T(0) + (n) = T(n-1) + (n)  O(n2). Fuw-Yi Yang

Best-case partitioning The best-case behavior for quick sort occurs when the partitioning routing produces two subproblems, each of size no more than n/2 , since one is n/2 and one of size n/2 -1. The partitioning cost (n) time and the recurrence for the running time is: T(n) = 2T(n/2) + (n)  O(n log n). Fuw-Yi Yang

Procedure Randomized-Partition(A, p, r) i = Random(p, r) swap(A[i], A[r]) return Partition(A, p, r) Procedure Randomized-Quicksort(A, p, r) If p < r then q = Randomized-Partition(A, p, r) Randomized-Quicksort(A, p, q-1) Randomized-Quicksort(A, q+1, r) Fuw-Yi Yang

Average-case partitioning Fuw-Yi Yang

Average-case partitioning Prove that Tavg(n) ≦kn log n Basis: n = 2 and k = 2(b+c), Note: Tavg(0) ≦b, Tavg(1) ≦b Assumption: Tavg(n) ≦kn log n for 1 ≦n < m Inductive step: for m we want to prove that Tavg(m) ≦km log m Fuw-Yi Yang

Average-case partitioning (logb(x) = logk(x)/logk(b)) Inductive step: for m we want to prove that Tavg(m) ≦km log m Fuw-Yi Yang

Average-case partitioning 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 kth 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/ 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/ ) + O(n) if n > 140. T(n) ≦ c(n/5 ) + c(7n/ ) + 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 Binary 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 = log bn. Thus f(n) is O(log n). Fuw-Yi Yang

R.8 Dynamic Programming 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 = log bn. Thus f(n) is O(log n). Fuw-Yi Yang 72

演算法概論 Introduction to Algorithms

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