**4-1.** The Grinch is given the job of partitioning 2n players into two teams of n players each. Each player has a numerical rating that measures how good he/she is at the game. He seeks to divide the players as unfairly as possible, so as to create the biggest possible talent imbalance between team A and team B. Show how the Grinch can do the job in time.

** Solution** First sort the players with an algorithm which runs in time and afterwards form two teams with the first n player in the first team and the last n players in the second team.

**4-3.** Take a sequence of 2n real numbers as input.Design an algorithm that partitions the numbers into n pairs, with the property that the partition minimizes the maximum sum of a pair. For example, say we are given the numbers (1,3,5,9). The possible partitions are ((1,3),(5,9)), ((1,5),(3,9)), and ((1,9),(3,5)). The pair sums for these partitions are (4,14), (6,12), and (10,8). Thus the third partition has 10 as its maximum sum, which is the minimum over the three partitions.

**Solution** We can minimize the maximum sum if we pair up the lowest and the highest element. But first we have to sort our numbers which takes and create the pairs which takes .

1 2 3 4 | Let S be the sequence with length 2n sort S for i := 1 to n pair[i] = (S[i], S[2n - i - 1]) |

**4-6.** Given two sets and (each of size n), and a number x, describe an algorithm for finding whether there exists a pair of elements, one from and one from , that add up to x. (For partial credit, give a algorithm for this problem.)

**Solution** We could sort and then iterate through and calculate the second number Now we just have to search for in which takes time. Therefore we take time.

**4-7.** Outline a reasonable method of solving each of the following problems. Give the order of the worst-case complexity of your methods.

(a) You are given a pile of thousands of telephone bills and thousands of checks sent in to pay the bills. Find out who did not pay.

(b) You are given a list containing the title, author, call number and publisher of all the books in a school library and another list of 30 publishers. Find out how many of the books in the library were published by each company.

(c) You are given all the book checkout cards used in the campus library during the past year, each of which contains the name of the person who took out the book. Determine how many distinct people checked out at least one book.

**Solution** (a) We can use quick sort in this scenario which works in expected time and match the ith bill with the ith check which takes time.

(b) We can use a dictionary quite easily which counts the books by publisher. This takes where time.

(c) We can sort the list by name and then iterate through it and count each different name. This takes about time.

**4-9.** Give an efficient algorithm to compute the union of sets A and B, where The output should be an array of distinct elements that form the union of the sets, such that they appear more than once in the union.

(a) Assume that A and B are unsorted. Give an algorithm for the problem.

(b) Assume that A and B are sorted. Give an O(n) algorithm for the problem.

** Solution** (a) We can create one array with every element of A and B. Afterwards we sort this array and iterate through it. In this iteration we just have to check if ith item differs from the (i+1)th item which takes therefore we need

(b)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | Let A got n elements and B got m elements with n >= m. i := 1 # counter for A, A[1] is first element j := 1 # counter for B while i <= n if j >= m # no more elements in B add A[i] to union set i := i + 1 else if A[i] == B[j] # same element add A[i] to union set i := i + 1 j := j + 1 else if A[i] > B[j] # increase j until B[j] is big enough add B[j] to union set j := j + 1 else if A[i] < B[j] add A[j] to union set i := i + 1 |

**4-12.** Devise an algorithm for finding the k smallest elements of an unsorted set of n integers in

**Solution** We can build an “unsorted” heap in , i.e. only bring the smallest item at the top. Now we can extract k times the smallest item which takes $O(k \log n)$ time.

**4-13.** You wish to store a set of n numbers in either a max-heap or a sorted array. For each application below, state which data structure is better, or if it does not matter. Explain your answers.

(a) Want to find the maximum element quickly.

(b) Want to be able to delete an element quickly.

(c) Want to be able to form the structure quickly.

(d) Want to find the minimum element quickly.

**Solution** (a) They are both equally fast. In the max-heap it’s the first element in the sorted array the last one.

(b) A max-heap is here better because it takes only instead of for a sorted array.

(c) Both the heap and the sorted array take time to be formed.

(d) The minimum element in a sorted array is the first. In a max-heap every leaf could be the minimum element, therefore it takes

**4-14.** Give an -time algorithm that merges k sorted lists with a total of n elements into one sorted list. (Hint: use a heap to speed up the elementary – time algorithm).

**Solution** We can basically do an heap sort on these lists. We iterate from i := 1 to n and select the smallest item from the head of each list which takes time. Therefore the algorithm takes time.

**4-15.** (a) Give an efficient algorithm to find the second-largest key among n keys. You can do better than comparisons.

(b) Then, give an efficient algorithm to find the third-largest key among n keys. How many key comparisons does your algorithm do in the worst case? Must your algorithm determine which key is largest and second-largest in the process?

**Solution** (a) There’s a faster method for construction heaps which runs in . Afterwards we just have to call find-min two times which takes

(b) We can use the same method here as well. The largest second-largest key is implicitly found by constructing the heap.

**4-16.** Use the partitioning idea of quicksort to give an algorithm that finds the median element of an array of n integers in expected time.

**Solution** The partition part of quicksort basically can help us. It determines partitions which are bigger respectively smaller than the pivot element. We just have to find n/2 elements which are smaller or equal to our element m which is then the median. This takes expected time.

**4-19.** An inversion of a permutation is a pair of elements that are out of order.

(a) Show that a permutation of n items has at most inversions. Which permutation(s) have exactly inversions?

(b) Let P be a permutation and Pr be the reversal of this permutation. Show that and have a total of exactly inversions.

(c) Use the previous result to argue that the expected number of inversions in a random permutation is

**Solution** (a) Let’s take for example the items 1..5. We can get the maximum inversions if we reverse this list, i.e. [5, 4, 3, 2, 1] or more general [n, n-1, …, 1]. Now we can start at the right and count the inversions.

1 2 3 4 5 6 | Sublist Inversions #Inversions 1 0 2, 1 (2,1) 1 3, 2, 1 (3, 2), (3, 1) 2 4, 3, 2, 1 (4, 3), (4, 2), (4, 1) 3 5, 4, 3, 2, 1 (5, 4), (5, 3), (5, 2), (5,1) 4 |

Therefore we got inversions at most.

(b) Let’s prove this relation. Assume we got a permutation with n items. Let’s add an other item. The reversal for the old permutation got inversions. The (n+1)th item adds n inversions which are:

Therefore we get and we’re done.

(c) A rough approach is to take the smallest and the highest value and assume that inversions are uniformly distribute. Therefore we get

**4-24.** Let A[1..n] be an array such that the first elements are already sorted (though we know nothing about the remaining elements). Give an algorithm that sorts A in substantially better than steps.

**Solution** Merge sort can used quite nicely. We need to sort the remaining which takes time. Afterwards we have to use merge on our old and new sorted lists which takes time.

** 4-29.** Mr. B. C. Dull claims to have developed a new data structure for priority queues that supports the operations Insert, Maximum, and Extract-Max—all in worst-case time. Prove that he is mistaken. (Hint: the argument does not involve a lot of gory details—just think about what this would imply about the lower bound for sorting.)

**Solution** If insert, maximum and extract-max would be possible in we could use the following algorithm to sort data.

1 2 3 4 5 6 7 | A[1...n] for i := 1 to n insert(A[i]) for i := 1 to n A[i] = maximum() Extract-max() |

This would sort data in which is smaller than lower bound

**4-30.** A company database consists of 10,000 sorted names, 40% of whom are known as good customers and who together account for 60% of the accesses to the database. There are two data structure options to consider for representing the database:

• Put all the names in a single array and use binary search.

• Put the good customers in one array and the rest of them in a second array. Only if we do not find the query name on a binary search of the first array do we do a binary search of the second array.

Demonstrate which option gives better expected performance. Does this change if linear search on an unsorted array is used instead of binary search for both options?

**Solution** Single array, binary search:

Good and bad, binary search:

The first variant is a bit faster here.

Single array, unsorted:

Good and bad, unsorted:

However in this case the second variant is far superior.

**4-32.** Consider the numerical 20 Questions game. In this game, Player 1 thinks of a number in the range 1 to n. Player 2 has to figure out this number by asking the fewest number of true/false questions. Assume that nobody cheats.

(a) What is an optimal strategy if n in known?

(b) What is a good strategy is n is not known?

**Solution** (a) Binary search.

(b) We can start asking if the number is between 1 and two. If not we can double our range from two to four, four and eight, etc.

**4-35.** Let M be an integer matrix in which the entries of each row are sorted in increasing order (from left to right) and the entries in each column are in increasing order (from top to bottom). Give an efficient algorithm to find the position of an integer x in M, or to determine that x is not there. How many comparisons of x with matrix entries does your algorithm use in worst case?

**Solution**

1 2 3 4 5 6 | for j:= 1 to m if m[1][j] <= x <= m[n][j] do binary search on this row if found return value return Not found |

This algorithm runs in time.

**4-39.** Design and implement a parallel sorting algorithm that distributes data across several processors. An appropriate variation of mergesort is a likely candidate. Mea- sure the speedup of this algorithm as the number of processors increases. Later, compare the execution time to that of a purely sequential mergesort implementation. What are your experiences?

**Solution** Non-threaded version:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 | from random import randint def mergesort(lst): n = len(lst) if n < 2: return lst else: middle = n / 2 left = mergesort(lst[:middle]) right = mergesort(lst[middle:]) return merge(left, right) def merge(left, right): result = [] i = 0 j = 0 while i < len(left) and j < len(right): if left[i] <= right[j]: result.append(left[i]) i += 1 else: result.append(right[j]) j += 1 result += left[i:] # append the rest result += right[j:] return result unsorted = [randint(0, 150000) for i in xrange(0, 10 ** 6)] mergesort(unsorted) |

Threaded version:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 | from random import randint from multiprocessing import Process, Queue def mergesort(lst): n = len(lst) if n < 2: return lst else: middle = n / 2 left = mergesort(lst[:middle]) right = mergesort(lst[middle:]) return merge(left, right) def merge(left, right): result = [] i = 0 j = 0 while i < len(left) and j < len(right): if left[i] <= right[j]: result.append(left[i]) i += 1 else: result.append(right[j]) j += 1 result += left[i:] # append the rest result += right[j:] return result def callMergesort(lst, q): q.put(mergesort(lst)) def main(): unsorted = [randint(0, 150000) for i in xrange(0, 10 ** 6)] middle = 5 * 10 ** 5 q = Queue() p1 = Process(target = callMergesort, args = (unsorted[:middle], q)) p2 = Process(target = callMergesort, args = (unsorted[middle:], q)) p1.start() p2.start() merge(q.get(), q.get()) if __name__ == '__main__': main() |

For small to medium sizes the threaded version is a bit slower than the non-threaded one. However, for very big sizes the threaded version works faster.

**4-45.** Given a search string of three words, find the smallest snippet of the document that contains all three of the search words—i.e. , the snippet with smallest number of words in it. You are given the index positions where these words in occur search strings, such as word1: (1, 4, 5), word2: (4, 9, 10), and word3: (5, 6, 15). Each of the lists are in sorted order, as above.

**Solution** We can sort these positions with its identifier. Afterwards we iterative through the list and put identifiers on a stack. If we got all of them, we found the snippet. We replace each identifier if we find a nearer one. At the end, we just have to search for the smallest snippet in all snippets.