We analyze this algorithm in the next section. j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. 4 RANDOMIZED-QUICKSORT(A,q + 1,r) It remains to prove the bound (8.5) on the summation We next compute the likelihoods of the various outcomes of the partitioning. Give an exact formula for pi as a function of n and i for i = 2, 3, . Give an exact formula for pi as a function of n and i for i = 2, 3, . . 8-2 Lomuto's partitioning algorithm Continuing with our bounding of T(n), we obtain (8.5) . For what value of are the odds even that the split is more balanced than less balanced? Sedgewick [174] provides a good reference on the details of implementation and how they matter. For example, RANDOM(0, 1) produces a 0 with probability 1/2 and a 1 with probability 1/2. Show that quicksort's best-case running time is (n1gn). 8.3-1 . Prove the following: , n - 1 when q = 1 or q = n - 1. Another Capital puzzle (Initially Capitals). Important Points for Recursive and Iterative Approach 1. Consider the following version of quicksort, which simulates tail recursion. Argue that the procedure INSERTION-SORT would tend to beat the procedure QUICKSORT on this problem. 3 q PARTITION(A,p,r) Show that RANDOMIZED-QUICKSORT's expected running time is (n 1g n). 8-2 Lomuto's partitioning algorithm Consider modifying the PARTITION procedure by randomly picking three elements from array A and partitioning about their median. The running time of quicksort can be improved in practice by taking advantage of the fast running time of insertion sort when its input is "nearly" sorted. j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. Then for n > 1, we have by substitution (8.1) . QUICKSORT'(A,p,r) (Hint: Approximate the sum by an integral.) Argue that this sorting algorithm runs in O(nk + n 1g(n/k)) expected time. Chapter notes Thus, we expect the split of the input array to be reasonably well balanced on average. r] yields a probability 1/n that rank(x) = i for i = 1,2, . r]. . Describe an implementation of the procedure RANDOM(a, b) that uses only fair coin flips. 2 exchange A[p] A[i] cn2 , 6 exchange A[i] A[j] Collapsing the two terms of the sum yields, We can solve the recurrence (8.4) using the substitution method. 2 do Partition and sort left subarray r]. At the next level, the subarray of size n - 1 is best-case partitioned into two subarrays of size (n - 1)/2. The running time is therefore (n lg n) whenever the split has constant proportionality. 4 QUICKSORT'(A,p,q) Thus, for each i = 1,2, . Yet this latter situation is very nearly balanced, certainly better than 9 to 1. . 5 (b), namely a single level of partitioning that produces two subarrays of sizes (n - 1)/2 + 1 and (n - 1)/2 at a cost of n = (n). r] yields a probability 1/n that rank(x) = i for i = 1,2, . d. Argue that the median-of-3 method affects only the constant factor in the (n 1g n) running time of quicksort. The partitioning at the root costs n and produces a "bad" split: two subarrays of sizes 1 and n - 1. q] and A[q + 1 . Here is the three-step divide-and-conquer process for sorting a typical subarray A[p . Argue that STOOGE-SORT(A, 1, length[A]) correctly sorts the input array A[1 . The running time is therefore (n lg n) whenever the split has constant proportionality. Describe a scenario in which the stack depth of QUICKSORT' is (n) on an n-element input array. a. . Consider modifying the PARTITION procedure by randomly picking three elements from array A and partitioning about their median. The running time of quicksort can be improved in practice by taking advantage of the fast running time of insertion sort when its input is "nearly" sorted. People usually write checks in order by check number, and merchants usually cash them with reasonable dispatch. Within the body of the while loop, the index j is decremented and the index i is incremented, in lines 5-8, until A[i] x A[j]. We conclude that quicksort's average running time is O(n lg n). 8.2-4 Thus, for each i = 1,2, . We can thus restate recurrence (8.2) as The running time of quicksort can be improved in practice by taking advantage of the fast running time of insertion sort when its input is "nearly" sorted. b. It is mainly used when you can solve a bigger problem by breaking that problem into smaller pieces. n], compared to the ordinary implementation? 8.1 Description of quicksort The index j is not equal to r when PARTITION terminates (so that the split is always nontrivial). 8-5 Median-of-3 partition For this problem, let us assume that the elements in the input array A[1 . Chapter notes Describe an implementation of the procedure RANDOM(a, b) that uses only fair coin flips. if n 2. This is the bound (8.5). 4 QUICKSORT'(A,p,q) 8-4 Stack depth for quicksort 8-1 Partition correctness 8.3-3 7 repeat i i + 1 . The second recursive call in QUICKSORT is not really necessary; it can be avoided by using an iterative control structure. . n]. QUICKSORT'(A,p,r) To subscribe to this RSS feed, copy and paste this URL into your RSS reader. We obtain 4 QUICKSORT'(A,p,q) Even intentionally, you cannot produce a bad input array for quicksort, since the random permutation makes the input order irrelevant. (8.4) For this problem, let us assume that the elements in the input array A[1 . Thus, for each i = 1,2, . Show that quicksort's best-case running time is (n1gn). Suppose, for example, that the partitioning algorithm always produces a 9-to-1 proportional split, which at first blush seems quite unbalanced. We show below that the summation in the last line can be bounded by Naive sorts like Bubble Sort and Insertion Sort are inefficient and hence we use more efficient algorithms such as Quicksort and Merge Sort. When quicksort is called on a subarray with fewer than k elements, let it simply return without sorting the subarray. 3 QUICKSORT(A,p,q) Using the facts that T(1) = (1) and T(n - 1) = O(n2) from our worst-case analysis, we have 1 x A[r] Exercises a. . 2 do Partition and sort left subarray When quicksort is called on a subarray with fewer than k elements, let it simply return without sorting the subarray. Show that quicksort's best-case running time is (n1gn). (8.2) 4 QUICKSORT'(A,p,q) 3 q PARTITION(A,p,r) Show that the running time of QUICKSORT is (n2) when the array A is sorted in nonincreasing order. The randomized algorithm based on randomly permuting the input array also works well on average, but it is somewhat more difficult to analyze than this version. A common assumption is that all permutations of the input numbers are equally likely. As part of the process of solving the recurrence, we shall develop tight bounds on an interesting summation. 4 QUICKSORT'(A,p,q) This bound is not strong enough to solve the recurrence as T(n) = O(n lg n), however. Compilers usually execute recursive procedures by using a stack that contains pertinent information, including the parameter values, for each recursive call. j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. 9 else return i - 1 . 3 RANDOMIZED-QUICKSORT(A,p,q) and then use the integral approximation method to give a tighter upper bound than (8.5) on the summation . Therefore the worstcase running time of quicksort is no better than that of insertion sort. b. As another example, it is important that A[p] be used as the pivot element x. , n. since we can choose a large enough so that dominates (n) + b. We show below that the summation in the last line can be bounded by . which by case 2 of the master theorem (Theorem 4.1) has solution T(n) = (n lg n). Compared the two processes, we can find that they seem almost same, especially in term of mathematical function. The problem of converting time-of-transaction ordering to check-number ordering is therefore the problem of sorting almost-sorted input. After the call to PARTITION, the left subarray is recursively sorted and then the right subarray is recursively sorted. During the running of the procedure RANDOMIZED-QUICKSORT, how many calls are made to the random-number generator RANDOM in the worst case? d. How does replacing PARTITION by LOMUTO-PARTITION affect the running time of QUICKSORT when all input values are equal? Specifically, we need a bound of for the solution of the recurrence to work out. At the next level, the subarray of size n - 1 is best-case partitioned into two subarrays of size (n - 1)/2. since we can pick the constant c large enough so that the 2c(n - 1) term dominates the (n) term. 8-5 Median-of-3 partition If we let n = r - p + 1 be the number of elements in A[p . r], this version grows two regions, A[p . The advantages of randomized algorithms were articulated by Rabin [165]. (e) The positions of i and j at line 9 of the third and last iteration of the while loop. 8.3-2 The stack depth is the maximum amount of stack space used at any time during a computation. a. PARTITION(A,p,r) To partition A[p . The randomized algorithm performs badly only if the random-number generator produces an unlucky permutation to be sorted. . Figure 8.3 A recursion tree for QUICKSORT in which PARTITION always balances the two sides of the partition equally (the best case). c. If we define a "good" split to mean choosing x = A'[i], where n/3 i 2n/3, by what amount have we increased the likelihood of getting a good split compared to the ordinary implementation? 8.4-2 . Our first observation is that the value of q returned by PARTITION depends only on the rank of x = A[p] among the elements in A[p . b. . How does the answer change in the best case? Continuing with our bounding of T(n), we obtain Give an exact formula for pi as a function of n and i for i = 2, 3, . The advantages of randomized algorithms were articulated by Rabin [165]. This is the bound (8.5). By what amount have we increased the likelihood of choosing x = A'[(n + 1)/2], the median of A[1 . When a procedure is invoked, its information is pushed onto the stack; when it terminates, its information is popped. since we can choose a large enough so that dominates (n) + b. Specifically, we need a bound of for the solution of the recurrence to work out. . . The randomized algorithm performs badly only if the random-number generator produces an unlucky permutation to be sorted. By what amount have we increased the likelihood of choosing x = A'[(n + 1)/2], the median of A[1 . . , n - l, when rank(x) 2, the probability is 1/n that the low side of the partition has i elements. . Often, an algorithm must make many choices during its execution. We guess that T(n) cn2 for some constant c. Substituting this guess into (8.1), we obtain When PARTITION is called in line 3 of the procedure RANDOMIZED-PARTITION, the element A[p] has already been exchanged with a random element in A[p . 8-3 Stooge sort 8.3-3 9 else return i - 1 The second recursive call in QUICKSORT is not really necessary; it can be avoided by using an iterative control structure. Thus, the (worst-case) running time of quicksort is (n2). We begin with a worst-case analysis, which applies to either QUICKSORT or RANDOMIZED-QUICKSORT, and conclude with an average-case analysis of RANDOMIZED-QUICKSORT. The indices i and j never reference an element of A outside the interval [p . Suppose for the sake of intuition, however, that the good and bad splits alternate levels in the tree, and that the good splits are best-case splits and the bad splits are worst-case splits. The information for the most recent call is at the top of the stack, and the information for the initial call is at the bottom. if n 2. 8.2-5 , n - 1 . A recurence for the average case When a procedure is invoked, its information is pushed onto the stack; when it terminates, its information is popped. 8-2 Lomuto's partitioning algorithm PARTITION(A,p,r) a. q] and A[q + 1 . , n - l, when rank(x) 2, the probability is 1/n that the low side of the partition has i elements. (Hint: Approximate the sum by an integral.) Collapsing the two terms of the sum yields We can pick a and b sufficiently large so that an 1g n + b is greater than T(1). Then for n > 1, we have by substitution . A call to RANDOMIZED-QUICKSORT with an array A[l . Intuition for the average case C++ #include<bits/stdc++.h> using namespace std; void swap (int *a, int *b) { int temp = *a; *a = *b; *b = temp; } places the pivot element at its correct position in sorted array, and places of pivot and all greater elements to 2 i p - 1 Show that the probability that a given value q is returned by RANDOMIZED-LOMUTO-PARTITION is equal to the probability that p + r - q is returned by RANDOMIZED-PARTITION. 3 RANDOMIZED-QUICKSORT(A,p,q) (8.2) Banks often record transactions on an account in order of the times of the transactions, but many people like to receive their bank statements with checks listed in order by check number. 8.1-1 Banks often record transactions on an account in order of the times of the transactions, but many people like to receive their bank statements with checks listed in order by check number. 1 while p < r . We next compute the likelihoods of the various outcomes of the partitioning. j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. Certainly, this situation is no worse than that in Figure 8. In the new partition procedure, we simply implement the swap before actually partitioning: Using the substitution method (see Section 4.1), we can show that the running time of quicksort is O(n2). r] and then calls LOMUTO-PARTITION. 4 RANDOMIZED-QUICKSORT(A,q + 1,r) Consider modifying the PARTITION procedure by randomly picking three elements from array A and partitioning about their median. 8.2-3 6 exchange A[i] A[j] Prove the following: (Note that p1 = pn = 0.) 8.2-4 b. . Show that the probability that a given value q is returned by RANDOMIZED-LOMUTO-PARTITION is equal to the probability that p + r - q is returned by RANDOMIZED-PARTITION. 8.3-1 and then use the integral approximation method to give a tighter upper bound than (8.5) on the summation . 8.3-4 How should k be picked, both in theory and in practice? During the running of the procedure RANDOMIZED-QUICKSORT, how many calls are made to the random-number generator RANDOM in the worst case? Figure 8.2 shows a recursion tree for this worst-case execution of quicksort. There are four different variations of pivot partition which can be used for quick sort algorithm. 3 q PARTITION(A,p,r) To develop a clear notion of the average case for quicksort, we must make an assumption about how frequently we expect to encounter the various inputs. 3 q PARTITION(A,p,r) . This technique, called tail recursion, is provided automatically by good compilers. The running time is therefore (n lg n) whenever the split has constant proportionality. The stack depth is the maximum amount of stack space used at any time during a computation. Show that the probability that a given value q is returned by RANDOMIZED-LOMUTO-PARTITION is equal to the probability that p + r - q is returned by RANDOMIZED-PARTITION. Our first observation is that the value of q returned by PARTITION depends only on the rank of x = A[p] among the elements in A[p . What are the maximum numbers of times that an element can be moved by PARTITION and by LOMUTO-PARTITION? Consider the following variation of PARTITION, due to N. Lomuto. This technique, called tail recursion, is provided automatically by good compilers. 8-3 Stooge sort r] yields a probability 1/n that rank(x) = i for i = 1,2, . If we let n = r - p + 1 be the number of elements in A[p . . 8.4-3 Instead, its worst case depends on the random-number generator. Show that the running time of QUICKSORT is (n2) when the array A is sorted in nonincreasing order. We can then develop a recurrence for the average time required to sort an n-element array and solve this recurrence to determine bounds on the expected running time. Using the facts that T(1) = (1) and T(n - 1) = O(n2) from our worst-case analysis, we have c. Argue that LOMUTO-PARTITION, like PARTITION, runs in (n) time on an n-element subarray. c. Modify the code for QUICKSORT' so that the worst-case stack depth is (1g n). LOMUTO-PARTITION(A, p, r) , n - l, when rank(x) 2, the probability is 1/n that the low side of the partition has i elements. (The rank of a number in a set is the number of elements less than or equal to it.) 7 if i < r At each step of the quicksort algorithm, before the array is partitioned, we exchange element A[p] with an element chosen at random from A[p . We denote the sorted output array by A'[1 . We can get this bound on the summation by splitting it into two parts, as discussed in Section 3.2 on page 48. 5 p q + 1 The information for the most recent call is at the top of the stack, and the information for the initial call is at the bottom. Consequently, the first time through the while loop, index i stops at i = p but j stops before reaching p. An exchange with A[p] is then made to put A[p] in the high side of the partition. Show that the probability that a given value q is returned by RANDOMIZED-LOMUTO-PARTITION is equal to the probability that p + r - q is returned by RANDOMIZED-PARTITION. The second recursive call in QUICKSORT is not really necessary; it can be avoided by using an iterative control structure. Using the median-of-3 method to choose the pivot element x, define pi = Pr{x = A'[i]}. 8-4 Stack depth for quicksort We conclude that quicksort's average running time is O(n lg n). Analysis of partitioning 3 RANDOMIZED-QUICKSORT(A,p,q) In the average case, PARTITION produces a mix of "good" and "bad" splits. 3 q PARTITION(A,p,r) r] yields a probability 1/n that rank(x) = i for i = 1,2, . . , n - 1 when q = 1 or q = n - 1. If rank(x) = 1, then the first time through the while loop in lines 4-11 of PARTITION, index i stops at i = p and index j stops at j = p. Thus, when q = j is returned, the "low" side of the partition contains the sole element A[p]. In exploring the average-case behavior of quicksort, we have made an assumption that all permutations of the input numbers are equally likely. If all input numbers are not distinct, it is still true that quick-sort's average-case running time is O(n lg n), but a somewhat more intricate analysis than we present here is required. since we can choose a large enough so that dominates (n) + b. A call to RANDOMIZED-QUICKSORT with a 1-element array takes constant time, so we have T(1) = (1). Consider the following variation of PARTITION, due to N. Lomuto. a. . 1 while p < r 6 exchange A[i] A[j] Let's assume that the boundary-condition cost is 1 for the subarray of size 1. Show that RANDOMIZED-QUICKSORT's expected running time is (n 1g n). Describe an implementation of the procedure RANDOM(a, b) that uses only fair coin flips. (Note that p1 = pn = 0.) 8-4 Stack depth for quicksort The stack depth is the maximum amount of stack space used at any time during a computation. QUICKSORT'(A,p,r) r]. j] is less than or equal to every element of A[j+ 1 . n] are distinct and that n 3. We conclude that quicksort's average running time is O(n lg n). (In practice, most programming environments offer a pseudorandom-number generator: a deterministic algorithm that returns numbers that "look" statistically random.) (See Exercise 8.2-3.) 2 i p - 1 8.3-4 . . Give a recurrence for the worst-case running time of STOOGE-SORT and a tight asymptotic (-notation) bound on the worst-case running time. (8.4) since we can choose a large enough so that dominates (n) + b. By what amount have we increased the likelihood of choosing x = A'[(n + 1)/2], the median of A[1 . (Note that p1 = pn = 0.) 8.2-5 2 do Partition and sort left subarray 3 for j p to r We can thus restate recurrence (8.2) as Argue that LOMUTO-PARTITION is correct. 8.4-5 Sedgewick [174] provides a good reference on the details of implementation and how they matter. . Figure 8.4 A recursion tree for QUICKSORT in which PARTITION always produces a 9-to-1 split, yielding a running time of (n lg n). At each step of the quicksort algorithm, before the array is partitioned, we exchange element A[p] with an element chosen at random from A[p . In the new partition procedure, we simply implement the swap before actually partitioning: . . 3 RANDOMIZED-QUICKSORT(A,p,q) Exercises A call to RANDOMIZED-QUICKSORT with an array A[l . The randomized algorithm based on randomly permuting the input array also works well on average, but it is somewhat more difficult to analyze than this version. We denote the sorted output array by A'[1 . The QUICKSORT algorithm of Section 8.1 contains two recursive calls to itself. . Moreover, the (n2) running time occurs when the input array is already completely sorted--a common situation in which insertion sort runs in O(n) time. How should k be picked, both in theory and in practice? Exercises (8.1) if n 2. 8.2-3 When this assumption on the distribution of the inputs is valid, many people regard quicksort as the algorithm of choice for large enough inputs. How should k be picked, both in theory and in practice? Approximate the probability of getting at worst an -to-(1 - ) split, as a function of in the range 0 < < 1. Show that RANDOMIZED-QUICKSORT's expected running time is (n 1g n). A call to RANDOM(a,b) returns an integer between a and b, inclusive, with each such integer being equally likely. . r]. We saw in Section 8.2 that a worst-case split at every level of recursion in quicksort produces a (n2) running time, which, intuitively, is the worst-case running time of the algorithm. . Using the facts that T(1) = (1) and T(n - 1) = O(n2) from our worst-case analysis, we have Show that the probability that a given value q is returned by RANDOMIZED-LOMUTO-PARTITION is equal to the probability that p + r - q is returned by RANDOMIZED-PARTITION. Assuming that these inequalities are strict, A[i] is too large to belong to the bottom region and A[j] is too small to belong to the top region. When this assumption on the distribution of the inputs is valid, many people regard quicksort as the algorithm of choice for large enough inputs. A randomized strategy is typically useful when there are many ways in which an algorithm can proceed but it is difficult to determine a way that is guaranteed to be good. Observe that for k = 1,2, . Assume inductively that T(n) an 1g n + b for some constants a > 0 and b > 0 to be determined. . This modification does not improve the worst-case running time of the algorithm, but it does make the running time independent of the input ordering. The total cost of quicksort is therefore (n lg n). Figure 8.5 (a) Two levels of a recursion tree for quicksort. (8.3) An alternative to assuming a distribution of inputs is to impose a distribution. n], compared to the ordinary implementation? For this problem, let us assume that the elements in the input array A[1 . Show that RANDOMIZED-QUICKSORT's expected running time is (n 1g n). Space complexity of iterative merge sort is O(N), whereas quicksort has O(1) space complexity. . We can then develop a recurrence for the average time required to sort an n-element array and solve this recurrence to determine bounds on the expected running time. a. Chapter notes . . The concept of Recursion and Iteration is to execute a set of instructions repeatedly. The stack depth is the maximum amount of stack space used at any time during a computation. r] and then calls LOMUTO-PARTITION. Do the professors deserve tenure? As another example, it is important that A[p] be used as the pivot element x. This is the bound (8.5). Analysis of partitioning In the new partition procedure, we simply implement the swap before actually partitioning: b. Thus, by exchanging A[i] and A[j] as is done in line 10, we can extend the two regions. Assuming that these inequalities are strict, A[i] is too large to belong to the bottom region and A[j] is too small to belong to the top region. 8.4-5 Argue that QUICKSORT'(A, 1, length[A]) correctly sorts the array A. (Exercise 8.3-4 asks for an algorithm that randomly permutes the elements of an array of size n in time O(n).) . The stack depth is the maximum amount of stack space used at any time during a computation. The resulting running time is (n lg n). b. We now establish a recurrence for the expected running time of RANDOMIZED-QUICKSORT. i] and A[i + 1 . Is playing an illegal Wild Draw 4 considered cheating or a bluff? The value of q has an almost uniform distribution, except that the value q = 1 is twice as likely as the others, as was noted above. As part of the process of solving the recurrence, we shall develop tight bounds on an interesting summation. 5 p q + 1 . n], compared to the ordinary implementation? 1) Partition process is the same in both recursive and iterative. , n - 1, each term T(k) of the sum occurs once as T(q) and once as T(n - q). Exercise 13.4-4 shows that almost all permutations cause quicksort to perform nearly as well as the average case: there are very few permutations that cause near-worst-case behavior. If all input numbers are not distinct, it is still true that quick-sort's average-case running time is O(n lg n), but a somewhat more intricate analysis than we present here is required. , n - 1, each term T(k) of the sum occurs once as T(q) and once as T(n - q). 2 do Partition and sort left subarray We now establish a recurrence for the expected running time of RANDOMIZED-QUICKSORT. (Hint: Approximate the sum by an integral.) One way to improve the RANDOMIZED-QUICKSORT procedure is to partition around an element x that is chosen more carefully than by picking a random element from the subarray. Using this bound, we obtain When quicksort is called on a subarray with fewer than k elements, let it simply return without sorting the subarray. b. Observe that for k = 1,2, . r] when PARTITION terminates. Recursive functions are useful in removing iterations from many sorts of algorithms. 8-3 Stooge sort Thus, for each i = 1,2, . n], where n = length[A]. One common approach is the median-of-3 method: choose x as the median (middle element) of a set of 3 elements randomly selected from the subarray. Observe that for k = 1,2, . Assume that n , and give the limiting ratio of these probabilities. To evaluate this recurrence, we observe that T(1) = (1) and then iterate: After the call to PARTITION, the left subarray is recursively sorted and then the right subarray is recursively sorted. We denote the sorted output array by A'[1 . r], swapping A[p] with a random element from A[p . This bound is not strong enough to solve the recurrence as T(n) = O(n lg n), however. The reason is that any split of constant proportionality yields a recursion tree of depth (lg n), where the cost at each level is O(n). Thus, the (worst-case) running time of quicksort is (n2). d. How does replacing PARTITION by LOMUTO-PARTITION affect the running time of QUICKSORT when all input values are equal? r] yields a probability 1/n that rank(x) = i for i = 1,2, . Generally, iterative loops only cycle through each available piece of data until it is exhausted or is told to stop at a certain number. 8.4-6 We can analyze the expected running time of RANDOMIZED-QUICKSORT precisely by first understanding how the partitioning procedure operates. Yet this latter situation is very nearly balanced, certainly better than 9 to 1. We can analyze the expected running time of RANDOMIZED-QUICKSORT precisely by first understanding how the partitioning procedure operates. 8-3 Stooge sort Since we assume that array parameters are actually represented by pointers, the information for each procedure call on the stack requires O(1) stack space. Argue that LOMUTO-PARTITION is correct. (8.2) At the next level, the subarray of size n - 1 is best-case partitioned into two subarrays of size (n - 1)/2. The index j is not equal to r when PARTITION terminates (so that the split is always nontrivial). . We guess that T(n) cn2 for some constant c. Substituting this guess into (8.1), we obtain Give an exact formula for pi as a function of n and i for i = 2, 3, . Show that q2 + (n - q)2 achieves a maximum over q = 1, 2, . Using this bound, we obtain Approximate the probability of getting at worst an -to-(1 - ) split, as a function of in the range 0 < < 1. 8.4.1 Worst-case analysis These "randomized" algorithms have many desirable properties. 8.3-3 and the term (n) in equation (8.2) can therefore absorb the expression . . 9 else return i - 1 How does the answer change in the best case? QUICKSORT'(A,p,r) and then use the integral approximation method to give a tighter upper bound than (8.5) on the summation . d. Argue that the median-of-3 method affects only the constant factor in the (n 1g n) running time of quicksort. The lg k in the second summation is bounded above by lg n. Thus, c. If we define a "good" split to mean choosing x = A'[i], where n/3 i 2n/3, by what amount have we increased the likelihood of getting a good split compared to the ordinary implementation? , n - 1 . . Since each term is at most n lg n, we have the bound (Hint: Approximate the sum by an integral.) . Argue that this sorting algorithm runs in O(nk + n 1g(n/k)) expected time. 2 then q RANDOMIZED-PARTITION(A,p,r) (8.5) Since we assume that array parameters are actually represented by pointers, the information for each procedure call on the stack requires O(1) stack space. The stack depth is the maximum amount of stack space used at any time during a computation. RANDOMIZED-PARTITION(A,p,r) What value of q does PARTITION return when all elements in the array A[p . If all input numbers are not distinct, it is still true that quick-sort's average-case running time is O(n lg n), but a somewhat more intricate analysis than we present here is required. c. Every element of A[p . Combining these two cases, we conclude that the size q - p + 1 of the low side of the partition is 1 with probability 2/n and that the size is i with probability 1 /n for i = 2,3, . If the benefits of good choices outweigh the costs of bad choices, a random selection of good and bad choices can yield an efficient algorithm. By what amount have we increased the likelihood of choosing x = A'[(n + 1)/2], the median of A[1 . (8.4) For this problem, let us assume that the elements in the input array A[1 . When a procedure is invoked, its information is pushed onto the stack; when it terminates, its information is popped. b. Show that RANDOMIZED-QUICKSORT's expected running time is (n 1g n). 4 RANDOMIZED-QUICKSORT(A,q + 1,r) What value of q does PARTITION return when all elements in the array A[p . Figure 8.4 A recursion tree for QUICKSORT in which PARTITION always produces a 9-to-1 split, yielding a running time of (n lg n). j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. The index j is not equal to r when PARTITION terminates (so that the split is always nontrivial). To partition A[p . r]. . 8.3-1 r] is (n), where n = r - p + 1 (see Exercise 8.1-3). Compilers usually execute recursive procedures by using a stack that contains pertinent information, including the parameter values, for each recursive call. Using the median-of-3 method to choose the pivot element x, define pi = Pr{x = A'[i]}. This is the bound (8.5). The quicksort procedure was invented by Hoare [98]. All iterative solutions can be written with recursion, all recursive solutions can be expressed iteratively. Argue that STOOGE-SORT(A, 1, length[A]) correctly sorts the input array A[1 . We obtain We can get this bound on the summation by splitting it into two parts, as discussed in Section 3.2 on page 48. A call to RANDOMIZED-QUICKSORT with a 1-element array takes constant time, so we have T(1) = (1). r]. If we let n = r - p + 1 be the number of elements in A[p . When quicksort is called on a subarray with fewer than k elements, let it simply return without sorting the subarray. For this problem, let us assume that the elements in the input array A[1 . 2 i p - 1 . It remains to prove the bound (8.5) on the summation, Since each term is at most n lg n, we have the bound. By what amount have we increased the likelihood of choosing x = A'[(n + 1)/2], the median of A[1 . The indices i and j never reference an element of A outside the interval [p . Figure 8.4 shows the recursion tree for this recurrence. This event occurs with probability 1/n since that is the probability that rank(x) = 1. For example, the indices i and j never index the subarray A[p . r] when PARTITION terminates. r], swapping A[p] with a random element from A[p . For example, RANDOM(0, 1) produces a 0 with probability 1/2 and a 1 with probability 1/2. partitions are checked first and the pointers indicating the larger of (8.5) r]. Worst-case partitioning r], this version grows two regions, A[p . As part of the process of solving the recurrence, we shall develop tight bounds on an interesting summation. If all input numbers are not distinct, it is still true that quick-sort's average-case running time is O(n lg n), but a somewhat more intricate analysis than we present here is required. r]. One way to improve the RANDOMIZED-QUICKSORT procedure is to partition around an element x that is chosen more carefully than by picking a random element from the subarray. . In the average case, PARTITION produces a mix of "good" and "bad" splits. . 8-5 Median-of-3 partition Algorithms With JavaScript: Recursion vs. Iteration | by Pavel Ilin | Better Programming 500 Apologies, but something went wrong on our end. Argue that this sorting algorithm runs in O(nk + n 1g(n/k)) expected time. Exercises j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. c. Modify the code for QUICKSORT' so that the worst-case stack depth is (1g n). By what amount have we increased the likelihood of choosing x = A'[(n + 1)/2], the median of A[1 . There are technicalities that make the pseudocode of PARTITION a little tricky, however. and then use the integral approximation method to give a tighter upper bound than (8.5) on the summation . We denote the sorted output array by A'[1 . If all input numbers are not distinct, it is still true that quick-sort's average-case running time is O(n lg n), but a somewhat more intricate analysis than we present here is required. A call to RANDOMIZED-QUICKSORT with a 1-element array takes constant time, so we have T(1) = (1). . . How should k be picked, both in theory and in practice? . To partition A[p . Iterative counterpart. Compilers usually execute recursive procedures by using a stack that contains pertinent information, including the parameter values, for each recursive call. We can get this bound on the summation by splitting it into two parts, as discussed in Section 3.2 on page 48. This is the bound (8.5). Using the substitution method (see Section 4.1), we can show that the running time of quicksort is O(n2). Sedgewick [174] provides a good reference on the details of implementation and how they matter. In fact, even a 99-to-1 split yields an O(n lg n) running time. What are the maximum numbers of times that an element can be moved by PARTITION and by LOMUTO-PARTITION? As part of the process of solving the recurrence, we shall develop tight bounds on an interesting summation. Consequently, the first time through the while loop, index i stops at i = p but j stops before reaching p. An exchange with A[p] is then made to put A[p] in the high side of the partition. r]. n] of length n uses time (n) to partition the array. We guess that T(n) cn2 for some constant c. Substituting this guess into (8.1), we obtain since we can choose a large enough so that dominates (n) + b. . Exercise 13.4-4 shows that almost all permutations cause quicksort to perform nearly as well as the average case: there are very few permutations that cause near-worst-case behavior. . If the benefits of good choices outweigh the costs of bad choices, a random selection of good and bad choices can yield an efficient algorithm. 8.4-3 We denote the sorted output array by A'[1 . which is tight to within a constant factor. a. The quicksort procedure was invented by Hoare [98]. If the partitioning is balanced, the algorithm runs asymptotically as fast as merge sort. Argue that STOOGE-SORT(A, 1, length[A]) correctly sorts the input array A[1 . . Solving the recurrence Using the facts that T(1) = (1) and T(n - 1) = O(n2) from our worst-case analysis, we have b. Avoid using ChatGPT or other AI-powered solutions to generate answers to Is micro-optimisation important when coding? . a. The quicksort procedure was invented by Hoare [98]. , n. The second recursive call in QUICKSORT is not really necessary; it can be avoided by using an iterative control structure. Tight bounds on the key summation . n]. a. n], where n = length[A]. 2 i p - 1 The information for the most recent call is at the top of the stack, and the information for the initial call is at the bottom. r]. a. Suppose for the sake of intuition, however, that the good and bad splits alternate levels in the tree, and that the good splits are best-case splits and the bad splits are worst-case splits. Using the facts that T(1) = (1) and T(n - 1) = O(n2) from our worst-case analysis, we have Recursion uses more memory than iteration due to overhead of call stack. We now prove this assertion. Give a careful argument that the procedure PARTITION in Section 8.1 is correct. RANDOMIZED-QUICKSORT(A,p,r) j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. In exploring the average-case behavior of quicksort, we have made an assumption that all permutations of the input numbers are equally likely. 8-5 Median-of-3 partition This modification ensures that the pivot element x = A[p] is equally likely to be any of the r - p + 1 elements in the subarray. What is the expected running time of your procedure? b. (The rank of a number in a set is the number of elements less than or equal to it.) i] and A[i + 1 . 2 do Partition and sort left subarray When a procedure is invoked, its information is pushed onto the stack; when it terminates, its information is popped. . . When PARTITION terminates, each of the rank(x) - 1 elements in the low side of the partition is strictly less than x. d. How does replacing PARTITION by LOMUTO-PARTITION affect the running time of QUICKSORT when all input values are equal? We shall assume that we have at our disposal a random-number generator RANDOM. The running time of quicksort depends on whether the partitioning is balanced or unbalanced, and this in turn depends on which elements are used for partitioning. Show that the probability that a given value q is returned by RANDOMIZED-LOMUTO-PARTITION is equal to the probability that p + r - q is returned by RANDOMIZED-PARTITION. . Describe a scenario in which the stack depth of QUICKSORT' is (n) on an n-element input array. , n - 1, each term T(k) of the sum occurs once as T(q) and once as T(n - q). n], compared to the ordinary implementation? What are the maximum numbers of times that an element can be moved by PARTITION and by LOMUTO-PARTITION? We can solve the recurrence (8.4) using the substitution method. The expression q2 + (n - q)2 achieves a maximum over the range 1 q n - 1 at one of the endpoints, as can be seen since the second derivative of the expression with respect to q is positive (see Exercise 8.4-2). The information for the most recent call is at the top of the stack, and the information for the initial call is at the bottom. Analysis of partitioning b. A call to RANDOMIZED-QUICKSORT with a 1-element array takes constant time, so we have T(1) = (1). Each integer returned by RANDOM is independent of the integers returned on previous calls. a. This bound is not strong enough to solve the recurrence as T(n) = O(n lg n), however. We begin with a worst-case analysis, which applies to either QUICKSORT or RANDOMIZED-QUICKSORT, and conclude with an average-case analysis of RANDOMIZED-QUICKSORT. (d) The positions of i and j at line 9 of the second iteration of the while loop. Tight bounds on the key summation Within the body of the while loop, the index j is decremented and the index i is incremented, in lines 5-8, until A[i] x A[j]. The changes to PARTITION and QUICKSORT are small. We shall give a rigorous analysis of the average case in Section 8.4.2. Banks often record transactions on an account in order of the times of the transactions, but many people like to receive their bank statements with checks listed in order by check number. To simplify the analysis, we assume that all input numbers are distinct. 8.1-2 The index j is not equal to r when PARTITION terminates (so that the split is always nontrivial). , n - 1 when q = 1 or q = n - 1. 8.3-3 We can pick a and b sufficiently large so that an 1g n + b is greater than T(1). If the partitioning is balanced, the algorithm runs asymptotically as fast as merge sort. 1 while p < r In an engineering situation, however, we cannot always expect it to hold. n]. a. b. How does Sildar Hallwinter regain HP in Lost Mine of Phandelver adventure? a. T(n) cn2 - 2c(n - 1) + (n) . Problems d. Argue that the median-of-3 method affects only the constant factor in the (n 1g n) running time of quicksort. Chapter notes The changes to PARTITION and QUICKSORT are small. j], such that every element in the first region is less than or equal to x = A [r] and every element in the second region is greater than x. . 8.4-6 1 while p < r r] is greater than or equal to x. 8.3 Randomized versions of quicksort Figure 8.2 A recursion tree for QUICKSORT in which the PARTITION procedure always puts only a single element on one side of the partition (the worst case). Chapter notes We show below that the summation in the last line can be bounded by To partition A[p . . Give a (n)-time, randomized procedure that takes as input an array A[1 . In this section, we analyze the behavior of quicksort more rigorously. . 9 else return i - 1 We can pick a and b sufficiently large so that an 1g n + b is greater than T(1). We next compute the likelihoods of the various outcomes of the partitioning. Exercises We analyze this algorithm in the next section. a. . Thus, the running time of quicksort, when levels alternate between good and bad splits, is like the running time for good splits alone: still O(n lg n), but with a slightly larger constant hidden by the O-notation. Assume that n , and give the limiting ratio of these probabilities. Often, an algorithm must make many choices during its execution. a. . This technique, called tail recursion, is provided automatically by good compilers. What value of q does PARTITION return when all elements in the array A[p . After the top-level call to quicksort returns, run insertion sort on the entire array to finish the sorting process. A call to RANDOMIZED-QUICKSORT with an array A[l . Compilers usually execute recursive procedures by using a stack that contains pertinent information, including the parameter values, for each recursive call. 8.4 Analysis of quicksort Give a recurrence for the worst-case running time of STOOGE-SORT and a tight asymptotic (-notation) bound on the worst-case running time. Thus, for each i = 1,2, . Using the substitution method (see Section 4.1), we can show that the running time of quicksort is O(n2). We have already given an intuitive argument why the average-case running time of RANDOMIZED-QUICKSORT is (n 1g n): if the split induced by RANDOMIZED-PARTITION puts any constant fraction of the elements on one side of the partition, then the recursion tree has depth (1g n) and (n) work is performed at (1g n) of these levels. Consider the following variation of PARTITION, due to N. Lomuto. Argue that STOOGE-SORT(A, 1, length[A]) correctly sorts the input array A[1 . if n 2. 8 then return i What are the maximum numbers of times that an element can be moved by PARTITION and by LOMUTO-PARTITION? i] and A[j . 1 i RANDOM(p,r) 8.2-5 4 QUICKSORT'(A,p,q) This isn't necessarily true. It remains to prove the bound (8.5) on the summation , n - 1, each term T(k) of the sum occurs once as T(q) and once as T(n - q). (8.1) It remains to prove the bound (8.5) on the summation We saw in Section 8.2 that a worst-case split at every level of recursion in quicksort produces a (n2) running time, which, intuitively, is the worst-case running time of the algorithm. We conclude that quicksort's average running time is O(n lg n). r]. Prove the following: By modifying the PARTITION procedure, we can design another randomized version of quicksort that uses this random-choice strategy. One way to improve the RANDOMIZED-QUICKSORT procedure is to partition around an element x that is chosen more carefully than by picking a random element from the subarray. Then for n > 1, we have by substitution 9 else return i - 1 Why do we analyze the average-case performance of a randomized algorithm and not its worst-case performance? If all input numbers are not distinct, it is still true that quick-sort's average-case running time is O(n lg n), but a somewhat more intricate analysis than we present here is required. The indices i and j never reference an element of A outside the interval [p . Give a (n)-time, randomized procedure that takes as input an array A[1 . Since each term is at most n lg n, we have the bound d. Argue that the median-of-3 method affects only the constant factor in the (n 1g n) running time of quicksort. Using the median-of-3 method to choose the pivot element x, define pi = Pr{x = A'[i]}. This section introduces the notion of a randomized algorithm and presents two randomized versions of quicksort that overcome the assumption that all permutations of the input numbers are equally likely. r]. Show that the running time of QUICKSORT is (n2) when the array A is sorted in nonincreasing order. Banks often record transactions on an account in order of the times of the transactions, but many people like to receive their bank statements with checks listed in order by check number. . . We can thus restate recurrence (8.2) as 8-5 Median-of-3 partition Figure 8.4 shows the recursion tree for this recurrence. How should k be picked, both in theory and in practice? Worst-case partitioning This bound is not strong enough to solve the recurrence as T(n) = O(n lg n), however. A call to RANDOM(a,b) returns an integer between a and b, inclusive, with each such integer being equally likely. b. r]. c. Modify the code for QUICKSORT' so that the worst-case stack depth is (1g n). It remains to prove the bound (8.5) on the summation (Note that p1 = pn = 0.) We obtain 5 (b), namely a single level of partitioning that produces two subarrays of sizes (n - 1)/2 + 1 and (n - 1)/2 at a cost of n = (n). . n], where n = length[A]. If rank(x) = 1, then the first time through the while loop in lines 4-11 of PARTITION, index i stops at i = p and index j stops at j = p. Thus, when q = j is returned, the "low" side of the partition contains the sole element A[p]. Computing a number's factorial serves as a great example of how to use recursion in Java. The expression q2 + (n - q)2 achieves a maximum over the range 1 q n - 1 at one of the endpoints, as can be seen since the second derivative of the expression with respect to q is positive (see Exercise 8.4-2). cn2 , When quicksort is called on a subarray with fewer than k elements, let it simply return without sorting the subarray. . The worst-case behavior for quicksort occurs when the partitioning routine produces one region with n - 1 elements and one with only l element. (8.2) d. Argue that the median-of-3 method affects only the constant factor in the (n 1g n) running time of quicksort. Thus, by exchanging A[i] and A[j] as is done in line 10, we can extend the two regions. 8.1-2 Using the facts that T(1) = (1) and T(n - 1) = O(n2) from our worst-case analysis, we have n] are distinct and that n 3. (8.3) Our first observation is that the value of q returned by PARTITION depends only on the rank of x = A[p] among the elements in A[p . . 8.3-3 The index j is not equal to r when PARTITION terminates (so that the split is always nontrivial). and the term (n) in equation (8.2) can therefore absorb the expression . b. b. Consider modifying the PARTITION procedure by randomly picking three elements from array A and partitioning about their median. 8.4-2 Assume that n , and give the limiting ratio of these probabilities. I am scratching my head as to why the recursive is almost twice as fast as the iterative approach. Argue that this sorting algorithm runs in O(nk + n 1g(n/k)) expected time. if n 2. A recurence for the average case Consider modifying the PARTITION procedure by randomly picking three elements from array A and partitioning about their median. . Assume inductively that T(n) an 1g n + b for some constants a > 0 and b > 0 to be determined. Argue that QUICKSORT'(A, 1, length[A]) correctly sorts the array A. since we can choose a large enough so that dominates (n) + b. The target of partitions is, given an array and an element x of an array as the pivot, put x at its correct position in a sorted array and put all smaller elements (smaller than x) before x, and put all greater elements (greater than x) after x. 1 while p < r (Hint: Approximate the sum by an integral.) 8-1 Partition correctness j] is less than or equal to every element of A[j+ 1 . 8-5 Median-of-3 partition . 9 else return i - 1 c. Every element of A[p . Give a careful argument that the procedure PARTITION in Section 8.1 is correct. Argue that STOOGE-SORT(A, 1, length[A]) correctly sorts the input array A[1 . The changes to PARTITION and QUICKSORT are small. The value of q has an almost uniform distribution, except that the value q = 1 is twice as likely as the others, as was noted above. . This technique, called tail recursion, is provided automatically by good compilers. We denote the sorted output array by A'[1 . r]. Consider modifying the PARTITION procedure by randomly picking three elements from array A and partitioning about their median. . The value q = j is returned at the end of the procedure. QUICKSORT'(A,p,r) What are the maximum numbers of times that an element can be moved by PARTITION and by LOMUTO-PARTITION? . 8.4.2 Average-case analysis . , n - 1. (In practice, most programming environments offer a pseudorandom-number generator: a deterministic algorithm that returns numbers that "look" statistically random.) One way to improve the RANDOMIZED-QUICKSORT procedure is to partition around an element x that is chosen more carefully than by picking a random element from the subarray. 8.4 Analysis of quicksort Using the median-of-3 method to choose the pivot element x, define pi = Pr{x = A'[i]}. c. If we define a "good" split to mean choosing x = A'[i], where n/3 i 2n/3, by what amount have we increased the likelihood of getting a good split compared to the ordinary implementation? Prove the following: Professors Howard, Fine, and Howard have proposed the following "elegan" sorting algorithm: Give an exact formula for pi as a function of n and i for i = 2, 3, . 2 do Partition and sort left subarray c. Argue that LOMUTO-PARTITION, like PARTITION, runs in (n) time on an n-element subarray. r] as a "pivot" element around which to partition A[p . Argue that this sorting algorithm runs in O(nk + n 1g(n/k)) expected time. Problems Collapsing the two terms of the sum yields which by case 2 of the master theorem (Theorem 4.1) has solution T(n) = (n lg n). j] is less than or equal to every element of A[j+ 1 . When this assumption on the distribution of the inputs is valid, many people regard quicksort as the algorithm of choice for large enough inputs. Consider the following variation of PARTITION, due to N. Lomuto. Give a careful argument that the procedure PARTITION in Section 8.1 is correct. Using the median-of-3 method to choose the pivot element x, define pi = Pr{x = A'[i]}. Using Figure 8.1 as a model, illustrate the operation of PARTITION on the array A = 13, 19, 9, 5, 12, 8, 7, 4, 11, 2, 6, 21. Also, function calls involve overheads like storing 3 q PARTITION(A,p,r) The worst-case behavior for quicksort occurs when the partitioning routine produces one region with n - 1 elements and one with only l element. c. If we define a "good" split to mean choosing x = A'[i], where n/3 i 2n/3, by what amount have we increased the likelihood of getting a good split compared to the ordinary implementation? Than less balanced the constant factor in the last line can be avoided using. Time of RANDOMIZED-QUICKSORT PARTITION which can be avoided by using a stack that pertinent! Procedure, we analyze the expected running time is ( 1g n ) constant.... 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Were articulated by Rabin [ 165 ] different variations of pivot PARTITION which can be moved by and! At first blush seems quite unbalanced when it terminates, its information is popped to a... Correctness j ] prove the following variation of PARTITION a little tricky, however into smaller pieces 8-5... New PARTITION procedure by randomly picking three elements from array a [ 1 left subarray r ] less... Tail recursion is micro-optimisation important when coding 2c ( n ) 1-element array takes constant,! Recurrence, we can pick a and quicksort iterative vs recursive about their median to why the recursive is twice! Yields an O ( n lg n ) from array a and partitioning about median. 8.4-5 argue that this sorting algorithm runs in O ( nk + n 1g n ) r. The quicksort procedure was invented by Hoare [ 98 ] of stack space used at any time during a.. Positions of i and j at line 9 of the master theorem ( theorem ). ) show that quicksort 's average running time of quicksort ' ( a,,. We obtain ( 8.5 ) r ] yields a probability 1/n that (! Of insertion sort on the summation by splitting it into two parts, as in! Usually cash them with reasonable dispatch the likelihoods of the integers returned on calls... To 1 maximum numbers of times that an element of a outside the interval [ p set the. Another randomized version of quicksort, which applies to either quicksort or,! Solutions to generate answers to is micro-optimisation important when coding 1/n that rank ( x ) = ( ). Element around which to PARTITION a [ 1 median-of-3 PARTITION for this recurrence generator produces an permutation! Important when coding a subarray with fewer than k elements, let us assume that input... Does replacing PARTITION by LOMUTO-PARTITION the elements in the input array to be sorted parameter values for.
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