Searches and Sorts. ICT Officers. September 2018

Transcription

1 Searches and Sorts ICT Officers September Searches Searching is a powerful tool that will come up often in the USACO competition. Searching is the process of retrieving information stored within some data structure. Here we will look into two possibilities of this and discuss the advantages and disadvantages of both methods. 1.1 Linear Search Linear Search is the simplest type of search. It has complexity O(N) and searches through an array until it can find the goal element. While this might work for small amounts of numbers, when working with data this process is time consuming. Algorithm 1 Linear Search 1: for x in [0, len(array)] do 2: if array[x] = goal then return x 1.2 Binary Search and Modifications Binary Search is a more efficient search when dealing with large amounts of ordered data. It has a time complexity of O(log N). Each iteration of the loop the algorithm will split the data in half based on whether the middle number is higher or lower than the goal element. This explains it s O(log N) time complexity. One modification of the binary search is the ternary search. Instead of splitting the data in half each iteration, a ternary search splits it into thirds, making it more efficient. Modifications of the binary search may be required in some USACO problems. Note: BINARY SEARCHES AND MODIFICATIONS OF IT ONLY WORK WITH ORDERED DATA Algorithm 2 Binary Search 1: high = len(array) 2: low = 0 3: mid 4: while low < high do 5: mid = (high + low) / 2 6: if array[mid] < goal then 7: low = mid + 1 8: else 9: high = mid 10: 1

2 2 Sorting Sorting is an useful tool for it puts our data into an order that we can utilize. For example, this order can be based on ascending numerical value, descending numerical value, or another useful order. Sorting our data has many advantages, from being able to binary search it to doing calculations based on the data. Java and C++ both have methods that sort arrays automatically, but it is still important to understand the algorithms behind these sorting problems. 2.1 Bubble Sort Our first sort is Bubble Sort. In this sort, we swap adjacent numbers if they are in the wrong order. It has time complexity O(n 2 ). Algorithm 3 Bubble Sort 1: for i in [0, len(array)] do 2: for j in [0,i] do 3: if array[j] <array[i] then 4: swap(array[i],array[j]) 2.2 Selection Sort Our next sort is Selection Sort. In this sort, we find the maximum value in the unsorted part of our array and swap it with the last unsorted item. It has time complexity O(n 2 ). The findmax function is left as an exercise to the reader Algorithm 4 Selection Sort for i in [0, len(array)] do maxpos = findmax(array, len(array) - i) swap(array[maxpos], len(array) - i - 1) 2.3 Insertion Sort Our next sort is Insertion Sort. In this sort, we slide numbers into the correct position, by comparing it to adjacent numbers. In the best case scenario it has time complexity O(n), but for the worst case scenario it has time complexity O(n 2 ). What causes the difference in efficiency? Algorithm 5 Insertion Sort for i in [1, len(array)] do pos = i temp = array[i] while pos > 0 && array[pos - 1] > array[pos] do array[pos] = array[pos - 1] pos = pos - 1 array[pos] = temp 2

3 2.4 Merge Sort Our next sort is Merge Sort. This sort works using a divide and conquer method. Data is divided into smaller groups, until each groups contains one element. Then, these groups are merged together. When merged the groups are sorted to produce new, larger sorted groups. This is repeated until there is only one group remaining. It has time complexity O(n log n). As an exercise, try to write the Merge method Algorithm 6 Merge Sort function MergeSort(array,high,low) middle = (low + high) / 2 MergeSort(array, low, middle) MergeSort(array, middle + 1, high) Merge(array, low, middle, high) 2.5 Quick Sort Our next sort is Quick Sort. This sort works using a divide and conquer method, similar to the Merge Sort. A pivot point is chosen from the data and the data is reordered so that all values smaller than the pivot are moved before it and all values larger than the pivot are moved after it Thus, when this is done, the pivot is in its final position. The above step is then done for the numbers smaller than the pivot and for the numbers greater than the pivot. This is repeated until the data is sorted. It has a best case time complexity of O(n log n) but a worst case of O(N 2 ) Note: The Arrays.sort() function uses a quick sort. Algorithm 7 Quick Sort function QuickSort(array, first, last) if first < last then splitpt = split(array, first, last) QuickSort(array, first, splitpt-1) QuickSort(array, split+1, last) function split(array, first, last) splitpt = first pivot = array[first] while first <= last do if array[first] <= pivot then first++ else if array[last] >= pivot then last else swap(first, last) first++ last 2.6 Heap Sort Our final sort is a Heap Sort. Similar to the previous two sorts, the efficiency of Heap Sort is O(n log n). A heap sort utilizes a max heap, in which every node is greater than both of its children. By this logic, the first node is always the greatest element in the unsorted section of the array. First, you begin by making a max heap. Then swap the first and last elements. We now consider the last element to be sorted because it is in its final position. Next, readjust the heap by heaping down from the new root of the tree. Finally, after the structure of the heap is restored, swap the first and last unsorted 3

4 elements again. Repeat these processes till the array is sorted. The creation of the initial max heap and the heap down are left as exercises to the reader Algorithm 8 Heap Sort for i in (array.length-1, 1) do swap(1,i) heapdown(array, 1, i - 1) if array[1] > array[2] then swap(1, 2) 3 Coding Questions Question 1: 2018 BioCode Problem O Rahul is an avid lover of math and when discussing sequences, he exclaimed to the class that, Strictly increasing sequences are srue(trait of being extremely cool). Given a sequence please print the length of the longest subsequence that are strictly increasing (there can be multiple). Input The input will consist of two lines. The first line will consist of the number N which is the length of the given sequence. The second line of input will consist of the sequence (of length N) itself. Output Print out one line. The line should contain the length of the longest strictly increasing subsequence. Question 2: 2018 US Open Silver Problem 1 Bessie s favorite algorithm is bubble sort. Here is Bessie s implementation, in cow-code, for sorting an array A of length N. while not sorted do sorted = true moo for i in (i, N-2) do if A[i+1] < A[i] then swap (A[i], A[i+1]) Apparently, the moo command in cow-code does nothing more than print out moo. Strangely, Bessie seems to insist on including it at various points in her code. Given an input array, please predict how many times moo will be printed by Bessie s code. 4

5 Question 3: 2018 US Open Gold Problem 1 Similar problem to before, but now the algorithm is while not sorted do sorted = true moo for i in (0, N-2) do if A[i+1] < A[i] then swap (A[i], A[i+1]) for i in (N-2, 0) do if A[i+1] A[i]: swap A[i], A[i+1] for i = 0 to N-2: if A[i+1] A[i]: Question 4: 2017 USACO January Silver Problem 1 After several months of rehearsal, the cows are just about ready to put on their annual dance performance; this year they are performing the famous bovine ballet Cowpelia. The only aspect of the show that remains to be determined is the size of the stage. A stage of size K can support K cows dancing simultaneously. The N cows in the herd (1 N 10, 000) are conveniently numbered 1 to N in the order in which they must appear in the dance. Each cow i plans to dance for a specific duration of time d(i). Initially, cows 1 to K appear on stage and start dancing. When the first of these cows completes her part, she leaves the stage and cow K + 1 immediately starts dancing, and so on, so there are always K cows dancing (until the end of the show, when we start to run out of cows). The show ends when the last cow completes her dancing part, at time T. Clearly, the larger the value of K, the smaller the value of T. Since the show cannot last too long, you are given as input an upper bound T max specifying the largest possible value of T. Subject to this constraint, please determine the smallest possible value of K. Question 5: 2017 USACO Feburary Silver Problem 2 The long road through Farmer John s farm has N crosswalks across it, conveniently numbered 1 to N (1 N 100, 000). To allow cows to cross at these crosswalks, FJ installs electric crossing signals, which light up with a green cow icon when it is ok for the cow to cross, and red otherwise. Unfortunately, a large electrical storm has damaged some of his signals. Given a list of the damaged signals, please compute the minimum number of signals that FJ needs to repair in order for there to exist some contiguous block of at least K working signals. 4 Resources