The explanations have been kept elementary without sacrificing depth of coverage or mathematical rigor. The first edition became a widely used text in universities worldwide as well as the standard reference for professionals. The second edition featured new chapters on the role of algorithms, probabilistic analysis and randomized algorithms, and linear programming. The third edition has been revised and updated throughout.
It features improved treatment of dynamic programming and greedy algorithms and a new notion of edge-based flow in the material on flow networks. Many exercises and problems have been added for this edition. We will prove a lower bound, then beat it by playing a different game. All sorts seen so far are comparison sorts: insertion sort, selection sort, merge sort, quicksort, heapsort, treesort.
Abstracts away everything else: control and data movement. Each leaf is labeled by the permutation of orders that the algorithm determines.
View the tree as if the algorithm splits in two at each node, based on the information it has determined up to that point.
The tree models all possible execution traces. What is the length of the longest path from root to leaf? Why is this useful? Basis: h D 0. Tree is just one node, which is a leaf. Inductive step: Assume true for height D h 1.
Extend tree of height h 1 by making as many new leaves as possible. Each leaf becomes parent to two new leaves. Sorting in linear time Non-comparison sorts. Counting sort Depends on a key assumption: numbers to be sorted are integers in f0; 1; : : : ; kg. Array A and values n and k are given as parameters. B is assumed to be already allocated and is given as a parameter. How big a k is practical? Probably not. Maybe, depending on n. Probably unless n is really small.
Counting sort will be used in radix sort. Radix sort How IBM made its money. Card sorters, worked on one column at a time. The human operator was part of the algorithm! Key idea: Sort least significant digits first. Assume digits 1; 2; : : : ; i 1 are sorted. If 2 digits in position i are equal, numbers are already in the right order by inductive hypothesis. The stable sort on digit i leaves them in the right order. Analysis Assume that we use counting sort as the intermediate sort.
How to break each key into digits? Break into r-bit digits. Use counting sort, k D 2r 1. Example: bit words, 8-bit digits. So, to sort bit numbers, use r D lg D 16 bits. How does radix sort violate the ground rules for a comparison sort?
Used keys as array indices. Distribute the n input values into the buckets. Sort each bucket. Then go through buckets in order, listing elements in each one. If earlier bucket, concatenation of lists fixes up. Intuitively, if each bucket gets a constant number of elements, it takes O.
But we need to do a careful analysis. Because insertion sort runs in quadratic time, bucket sort time is T. Used a function of key values to index into an array. Different from a randomized algorithm, where we use randomization to impose a distribution.
Use the same argument as in the proof of Theorem 8. Proof First notice that, as pointed out in the hint, we cannot prove the lower bound by multiplying together the lower bounds for sorting each subsequence.
That would only prove that there is no faster algorithm that sorts the subsequences Solutions for Chapter 8: Sorting in Linear Time independently. This was not what we are asked to prove; we cannot introduce any extra assumptions. Now, consider the decision tree of height h for any comparison sort for S. Thus, any decision tree for sorting S must have at least. We implicitly assume here that k is even. We could adjust with floors and ceilings if k were odd.
Since there exists at least one path in any decision tree for sorting S that has length at least. Solution to Exercise 8. The algorithm is correct no matter what order is used!
But the modified algorithm is not stable. As before, in the final for loop an element equal to one taken from A earlier is placed before the earlier one i. The original algorithm was stable because an element taken from A later started out with a lower index than one taken earlier. But in the modified algorithm, an element taken from A later started out with a higher index than one taken earlier.
Radix sort sorts separately on each digit, starting from digit 1. Thus, radix sort of d digits, which sorts on digits 1; : : : ; d is equivalent to radix sort of the low-order d 1 digits followed by a sort on digit d.
By our induction hypothesis, the sort of the low-order d 1 digits works, so just before the sort on digit d , the elements are in order according to their low-order d 1 digits.
The sort on digit d will order the elements by their d th digit. Consider two elements, a and b, with d th digits ad and bd respectively. If ad D bd , the sort will leave a and b in the same order they were in, because it is stable. But that order is already correct, since the correct order of a and b is determined by the low-order d 1 digits when their d th digits are equal, and the elements are already sorted by their low-order d 1 digits. If the intermediate sort were not stable, it might rearrange elements whose d th digits were equal—elements that were in the right order after the sort on their lower-order digits.
Each digit ranges from 0 to n Sort these 3-digit numbers with radix sort. If, for example, all the input ends up in the first bucket, then in the insertion sort phase it needs to sort all the input, which takes O. A simple change that will preserve the linear expected running time and make the worst-case running time O. Any remaining leaves will have probability 0, since they are not reached for any input.
To prove this last assertion, let dT. To show that d. Then d. Take the tree T with k leaves such that D. Then k i is the number of leaves in LT and d. Let fk. Now we use substitution to prove d.
The base case of the induction is satisfied because d. For the inductive step we assume that d. We will show how to modify a randomized decision tree algorithm to define a deterministic decision tree algorithm that is at least as good as the randomized one in terms of the average number of comparisons. At each randomized node, pick the child with the smallest subtree the subtree with the smallest average number of comparisons on a path to a leaf. Delete all Solutions for Chapter 8: Sorting in Linear Time the other children of the randomized node and splice out the randomized node itself.
The deterministic algorithm corresponding to this modified tree still works, because the randomized algorithm worked no matter which path was taken from each randomized node. The average number of comparisons for the modified algorithm is no larger than the average number for the original randomized tree, since we discarded the higher-average subtrees in each case.
The randomized algorithm thus takes at least as much time on average as the corresponding deterministic one. The usual, unadorned radix sort algorithm will not solve this problem in the required time bound. The number of passes, d , would have to be the number of digits in the largest integer.
We assume that the range of a single digit is constant. Let us assume without loss of generality that all the integers are positive and have no leading zeros. If there are negative integers or 0, deal with the positive numbers, negative numbers, and 0 separately. Under this assumption, we can observe that integers with more digits are always greater than integers with fewer digits. Thus, we can first sort the integers by number of digits using counting sort , and then use radix sort to sort each group of integers with the same length.
Noting that each integer has between 1 and n digits, let mi be the number of integers with Pn i digits, for i D 1; 2; : : : ; n. It takes O.
The time to sort all groups, therefore, is! One way to solve this problem is by a radix sort from right to left. Since the strings have varying lengths, however, we have to pad out all strings that are shorter than the longest string.
Unfortunately, this scheme does not always run in the required time bound. Suppose that there are m strings and that the longest string has d characters. To solve the problem in O.
We take advantage of this property by sorting the strings on the first letter, using counting sort. We take an empty string as a special case and put it first.
We gather together all strings with the same first letter as a group. Then we recurse, within each group, based on each string with the first letter removed. The correctness of this algorithm is straightforward. Analyzing the running time is a bit trickier. Let us count the number of times that each string is sorted by a call of counting sort. Suppose that the ith string, si , has length li. Then si is sorted by at most li C 1 counting sorts.
The string a is sorted its length, 1, time plus one more time. Thus, the total time for all calls of counting sort is! Compare each red jug with each blue jug. To solve the problem, an algorithm has to perform a series of comparisons until it has enough information to determine the matching.
We can view the computation of the algorithm in terms of a decision tree. Every internal node is labeled with two jugs one red, one blue which we compare, and has three outgoing edges red jug smaller, same size, or larger than the blue jug. The leaves are labeled with a unique matching of jugs.
The height of the decision tree is equal to the worst-case number of comparisons the algorithm has to make to determine the matching. To bound that size, let us first compute the number of possible matchings for n red and n blue jugs. If we label the red jugs from 1 to n and we also label the blue jugs from 1 to n before starting the comparisons, every outcome of the algorithm can be represented as a set f.
Now we can bound the height h of our decision tree. Every tree with a branching factor of 3 every inner node has at most three children has at most 3h leaves.
Assume that the red jugs are labeled with numbers 1; 2; : : : ; n and so are the blue jugs. The numbers are arbitrary and do not correspond to the volumes of jugs, but are just used to refer to the jugs in the algorithm description. Moreover, the output of the algorithm will consist of n distinct pairs.
We will call the procedure only with inputs that can be matched; one necessary condition is that jRj D jBj. Termination is also easy to see: since jR j. When we compare bi to every jug in R fri g, jug bi is not put into either B. Then jugs ri and bj will be compared if and only if the first jug in Rij to be chosen is either ri or rj.
Still following the quicksort analysis, until a jug from Rij is chosen, the entire set Rij is together. Any jug in Rij is equally likely to be first one chosen. The remainder of the analysis is the same as the quicksort analysis, and we arrive at the solution of O. Just like in quicksort, in the worst case we always choose the largest or smallest jug to partition the sets, which reduces the set sizes by only 1. The running time then obeys the recurrence T.
Therefore, algorithm X performs the same sequence of exchanges on array B as it does on array A. Hence algorithm X fails to sort array B correctly. The even steps perform fixed permutations. The odd steps sort each column by some sorting algorithm, which might not be an oblivious compare-exchange algorithm. But the result of sorting each column would be the same as if we did use an oblivious compare-exchange algorithm. After step 1, each column has 0s on top and 1s on the bottom, with at most one transition between 0s and 1s, and it is a 0!
As we read the array in column-major order, all 1! All 1! Step 3 moves the 0s to the top rows and the 1s to the bottom rows. The s dirty rows are somewhere in the middle. The dirty area after step 3 is at most s rows high and s columns wide, and so its area is at most s 2. Step 4 turns the clean 0s in the top rows into a clean area on the left, the clean 1s in the bottom rows into a clean area on the right, and the dirty area of size s 2 is between the two clean areas.
In the former case, step 5 sorts the column containing the dirty area, and steps 6—8 maintain that the array is sorted. In the latter case, step 5 cannot increase the size of the dirty area, step 6 moves the entire dirty area into the same column, step 7 sorts it, and step 8 moves it back. If s does not divide r, then after step 2, we can see up to s 0!
After step 3, we would have up to 2s 1 dirty rows, for a dirty area size of at most 2s 2 s. We can reduce the number of transitions in the rows after step 2 back down to at most s by sorting every other column in reverse order in step 1. Now if we have a transition either 1!
The minimum is the first order statistic i D 1. The maximum is the nth order statistic i D n. When n is odd, the median is unique, at i D. Output: The element x 2 A that is larger than exactly i 1 other elements in A. In other words, the ith smallest element of A. We can easily solve the selection problem in O. Then return the ith element in the sorted array. There are faster algorithms, however. This is the best we can do, because each element, except the minimum, must be compared to a smaller element at least once.
To do so, the program must first find the minimum and maximum of each coordinate. A simple algorithm to find the minimum and maximum is to find each one independently. There will be n 1 comparisons for the minimum and n 1 comparisons for the maximum, for a total of 2n 2 comparisons.
Process elements in pairs. Compare the elements of a pair to each other. Then compare the larger element to the maximum so far, and compare the smaller element to the minimum so far. This leads to only 3 comparisons for every 2 elements. Setting up the initial values for the min and max depends on whether n is odd or even. Then process the rest of the elements in pairs.
If n is odd, set both min and max to the first element. If the pivot element is the ith smallest element i. Otherwise, recurse on the subarray containing the ith smallest element. Because it is randomized, no particular input brings out the worst-case behavior consistently. To obtain an upper bound, we assume that T. When Xk D 1, the two subarrays have sizes k 1 and n k. If n is odd, these terms appear twice and T. Looking at the expression max. Assume that T. Selection in worst-case linear time We can find the ith smallest element in O.
It executes the following steps: 1. Divide the n elements into groups of 5. Takes O. Then just pick the median from each group, in O. Let x be the kth element of the array after partitioning, so that there are k 1 elements on the low side of the partition and n k elements on the high side. If i k, return the. Analysis Start by getting a lower bound on the number of elements that are greater than the partitioning element x: x [Each group is a column.
Each white circle is the median of a group, as found in step 2. Arrows go from larger elements to smaller elements, based on what we know after step 4. Elements in the region on the lower right are known to be greater than x. Symmetrically, the number of elements that are 6. Step 4: partitioning the n-element array around x takes O.
Step 3 takes time T. Assume that c is large enough that T. Substitute the inductive hypothesis in the right-hand side of the recurrence: T. We conclude that T. Why ? We could have used any integer strictly greater than Sorting algorithms that run in linear time need to make assumptions about their input. Linear-time selection algorithms do not require any assumptions about their input. To see that this algorithm does exactly n 1 comparisons, notice that each number except the smallest loses exactly once.
To show this more formally, draw a binary tree of the comparisons the algorithm does. The n numbers are the leaves, and each number that came out smaller in a comparison is the parent of the two numbers that were compared.
Each non-leaf node of the tree represents a comparison, and there are n 1 internal nodes in an n-leaf full binary tree see Exercise B. In the search for the smallest number, the second smallest number must have come out smallest in every comparison made with it until it was eventually compared with the smallest.
So the second smallest is among the elements that were compared with the smallest during the tournament. To find it, conduct another tournament as above to find the smallest of these numbers. At most dlg ne the height of the tree of comparisons elements were compared with the smallest, so finding the smallest of these takes dlg ne 1 comparisons in the worst case.
The total number of comparisons made in the two tournaments was n 1 C dlg ne in the worst case. For groups of 3, however, the algorithm no longer works in linear time.
Observe also that the O. Thus, we get the recurrence T. You can also see that T. Then k elements of X are less than or equal to m and n k elements of X are greater than or equal to m. We know that in the two arrays combined, there must be n elements less than or equal to m and n elements greater than or equal to m, and so there must be n k elements of Y that are less than or equal to m and n.
A boundary case occurs for k D n. If n is odd, then on the oil well whose y-coordinate is the median. Proof We examine various cases. In each case, we will start out with the pipeline at a particular y-coordinate and see what happens when we move it. We start with the case in which n is even. Let us start with the pipeline somewhere on or between the two oil wells whose y-coordinates are the lower and upper medians. Now suppose that the pipeline goes through the oil well whose y-coordinate is the upper median.
We conclude that moving the pipeline up from the oil well at the upper median increases the total spur length. A symmetric argument shows that if we start with the pipeline going through the oil well whose y-coordinate is the lower median and move it down, then the total spur length increases.
We see, therefore, that when n is even, an optimal placement of the pipeline is anywhere on or between the two medians. Now we consider the case when n is odd. All oil wells at or below the median become d units farther from the pipeline, and there are at least.
There are at most. A symmetric argument shows that moving the pipeline down from the median also increases the total spur length, and so the claim optimal placement of the pipeline is on the median.
Since we know we are looking for the median, we can use the linear-time medianfinding algorithm. Solution to Problem This solution is also posted publicly We assume that the numbers start out in an array. Implement the priority queue as a heap. Note that method c is always asymptotically at least as good as the other two methods, and that method b is asymptotically at least as good as a.
Comparing c to b is easy, but it is less obvious how to compare c and b to a. The sum of two things that are O. We first sort the n elements into increasing order by xi values. The sorting phase can be done in O. The total running time in the worst case, therefore, is O. Although the first paragraph of the section only claims an O. The weighted-median algorithm works as follows.
Otherwise, we proceed as follows. We find the actual median xk of the n elements and then partition around it.
We then compute the total weights of the two halves. The solution of the recurrence is T. Let the n points be denoted by their coordinates x1 ; x2 ; : : : ; xn , let the corresponding weights be w1 ; w2 ; P : : : ; wn , and let x D xk be the weighted median. Let y be any point real number other than x. We show the optimality of the weighted median x by showing that f. Solutions for Chapter 9: Medians and Order Statistics e. I ran into the alley and aimed my Colt at the garage door.
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Last week, but now he turned his attention to her. Solutions manual available upon adoptions. Introduction to Applied Optimization is intended for advanced undergraduate and graduate students and will benefit scientists from diverse areas, including engineers. You could not only going in imitation of book collection or library or borrowing from your associates to way in them.
This is an enormously easy means to specifically get guide by on-line. And lines perform the action to move the smallest element of the subarray A[i..
So incrementing i reestablishes the loop invariant for the next iteration. By the loop invariant, the subarray A[ Also, this subarray is sorted. So the element A[A. So loop invariant holds. The worst-case running time is the same as Merge Sort, i.
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