Reference no: EM131028668
1. Submission Instructions
Create a document using your favorite word processor and type your exercise solutions. At the top of the document be sure to include your name and the homework assignment number, e.g. HW3. Convert this document into Adobe PDF format and name the PDF file <asuriteid>.pdf where <asuriteid> is your ASURITE user id (for example, my ASURITE user id is kburger2 so my file would be named kburger2.pdf). To convert your document into PDF format, Microsoft Office versions 2008 and newer will export the document into PDF format by selecting the proper menu item from the File menu. The same is true of Open Office and Libre Office. Otherwise, you may use a freeware PDF converter program, e.g., CutePDF is one such program.
Next, create a folder named <asuriteid> and copy <asuriteid>.pdf to that folder. Copy any requested Java source code files to this folder (note: Java source code files are the files with a .java file name extension; do not copy the .class files as we do not need those).
Next, compress the <asuriteid> folder creating a zip archive file named <asuriteid>.zip. Upload <asuriteid>.zip to the Homework Assignment link: by the assignment deadline. The deadline is in course schedule info. Consult the online syllabus for the late and academic integrity policies.
2. Learning Objectives
1. To apply recursion solutions to solve problems.
2. To use the linear search and binary search algorithms to search a list for a key.
3. To analyze an algorithm to determine the asymptotic time complexity.
4. To determine the order of growth of a function and express it in Big O notation.
5. To implement sorting algorithms to sort a list of elements.
6. To analyze the time complexity of sorting algorithms.
3. Recursion
3.1. The sum of the first n positive integers is:
sum (n ) = i=1∑n i
and can easily be computed using a for loop:
public int sum1toN(int n) {
int sum = 0;
for (int i = 1; i <= n; ++i) {
sum += i;
}
return sum;
}
It is also possible to recursively compute the sum by recognizing that sum(n) = n + sum(n - 1). For this exercise, write a recursive version of sum1toN().
3.2. Name your source code file Hw3_1.java. Write a recursive method double power(double x, int n) that computes and returns xn where n ≥ 0. Hint: remember that xn = x · xn-1. Write a driver routine that computes (and prints) 2p for p = 0, 2, 3, ..., 10 by calling power(). This driver routine will test your method to ensure it is working properly.
3.3. Modify Hw3_1.java to implement a recursive method double powerFaster(double x, int n) that computes and returns xn. However, write the code so that when n is even the method will return (xn/2)2. Modify your driver routine to compute and print 2p for p = 0, 2, 3, ..., 10 by calling powerFaster().
3.4. Modify Hw1_1.java and declare two int counter variables (as private instance variables) named calls and callsFaster. In the loop of the test driver, set both instance variables to 0 before calling power() and powerFaster(). Increment the corresponding counter variable as the first statement in each of the power methods. After each method returns the computed power 2p back to the driver routine, print the values of calls and callsFaster (if you have not figured it out yet, calls and callsFaster are counting the number of times each method is called during the computation of 2p). Explain why powerFaster() is faster in some cases than power().
3.5. Write a recursive method String reverse(String s) that returns the reverse of s. For example, if s is "Hello world" then the method would return "dlrow olleH". Hint: remove the first character c at index 0 of s forming a substring t which consists of the characters at indices 1, 2, 3, ..., s.length()-1 of s; then concatenate the reverse of t and c and return this new string.
4. Linear and Binary Search
4.1 Write a recursive method int recLinearSearch(ArrayList<String> pList, String pKey, int pBeginIdx, int pEndIdx) that searches pList elements pBeginIdx up to and including pEndIdx for pKey and returns the index of pKey in pList if found and -1 if not found. The method will be called in this manner to search the entire list:
ArrayList<String> list = new ArrayList<>();
// populate list with some Strings...
int idx = recLinearSearch(list, "some key", 0, list.size() - 1);
Hint: the base case is reached when pBeginIdx is greater than pEndIdx (what does this mean?). Otherwise, check to see if the element at pBeginIdx is pKey. If it is, then return pBeginIdx. If it is not, then make a recursive method call.
4.2. Suppose list is an ArrayList of Integers and contains these elements:
list = { 2, 3, 5, 10, 16, 24, 32, 48, 96, 120, 240, 360, 800, 1600 }
and we call the recursive binary search method from the lecture notes:
int idx = recursiveBinarySearch(list, 10, 0, list.size() - 1);
Trace the method and show the values of pLow and pHigh on entry to each method call. Show the value of pMiddle that is computed and state which clause of the if-elseif-elseif statement will be executed, i.e., specify if return middle; will be executed, or if return recursiveBinarySearch(pList, pKey, pLow, middle - 1); will be executed, or if return recursiveBinarySearch(pList, pKey, middle + 1, pHigh); will be executed. Finally, at the end, specify the value assigned to index and the total number of times that recursiveBinary Search() was called (including the original call).
4.3. Repeat Exercise 4.2 but this time let pKey be 150.
5. Analysis of Algorithms and Big O Notation
5.1. Using the formal definition of big O, prove mathematically that f(n) = 2.5n + 4 is O(n).
5.2. Using the formal definition of big O, prove mathematically that f(n) = -4 × 106,00,000 is O(1).
5.3. An important concept to know regarding big O notation is that for logarithmic complexity the base is irrelevant. In other words, if f(n) is a function that counts the number of times the key operation is performed as a function of n and f(n) is O(loga n) then it it also true that f(n) is O(logb n). For example, binary search-which is usually stated as being O(lg n)-is also O(ln n), O(log10 n), and O(log3.14159265 n). Using the formal definition of big O, prove mathematically that if f(n) is O(loga n) then f(n) is also O(logb n).
5.4. Consider this split() method where: pList is an ArrayList of Integers containing zero or more elements; pEvenList is an empty ArrayList of Integers; and pOddList is an empty ArrayList of Integers. On return, pEvenList will contain the even Integers of pList and pOddList will contain the odd Integers.
void split(ArrayList<Integer> pList, ArrayList<Integer> pEvenList, ArrayList<Integer> pOddList) {
for (int n : pList) {
if (n 
2 == 0) pEvenList.add(n); else pOddList.add(n);
}
}
To analyze the worst case time complexity of an algorithm we first identify the "key operation" and then derive a function which counts how many times the key operation is performed as a function of the size of the input. What is the key operation in this algorithm? Explain.
5.5. Continuing with the previous exercise, derive a function f(n) which equates to the number of times the key operation is performed as a function of n, where n is the size of pList. State the worst case time complexity of split() in big O notation.
5.6. Would the time complexity of split() change if the elements of pList were sorted into ascending order? Explain.
5.7. Binary search is such an efficient searching algorithm because during each pass of the loop (in the iterative version) or in each method call (in the recursive version) the size of the list is essentially halved. If reducing the size of the list to be searched by one half is so effective, it may seem that reducing it by two thirds each time would be even more effective. To that end, consider this iterative ternary search method. Rewrite it is as a recursive ternary search method named int recTernarySearch(ArrayList<Integer> pList, Integer pKey, int pLow, int pHigh).
int ternarySearch(ArrayList<Integer> pList, Integer pKey) {
int low = 0, high = pList.size() - 1;
while (low <= high) {
int range = high - low;
int oneThirdIdx = (int)Math.round(low + range / 3.0);
int twoThirdIdx = (int)Math.round(low + range / 1.33);
if (pKey.equals(pList.get(oneThirdIdx))) {
return oneThirdIdx;
} else if (pKey.equals(pList.get(twoThirdIdx))) { return twoThirdIdx;
} else if (pKey < pList.get(oneThirdIdx)) { high = oneThirdIdx - 1;
} else if (pKey > pList.get(twoThirdIdx)) { low = twoThirdIdx + 1;
} else {
low = oneThirdIdx + 1;
high = twoThirdIdx - 1;
}
}
return -1;
}
5.8. For recTernarySearch() the key operations are the four comparisons of pKey to the the elements of pList. Treat these four comparisons as one comparison and provide an informal proof of the worst case time complexity of ternary search. To simplify the analysis, assume that the size of the list on entry to recTernarySearch() is always a power of 3, e.g., on the first call assume the size of the list is 3p, on the second call the size of the list is 3p-1, on the third call 3p-2, and so on.
6. Sorting
6.1. Consider the Point class discussed in Objects and Classes : Section 1 (in burger-cse205-note-objs-classes-01.pdf in the Week 1 Notes zip archive; the Point.java source code file can be found in the Week 1 Source Code zip archive). Modify this class so it implements the Comparable<Point> interface. We define Point p1 to be less than Point p2 if the distance from the origin to p1 is less than the distance from the origin to p2; p1 is greater than p2 if the distance from the origin to p1 is greater than the distance from the origin to p2; otherwise, if they distances are equal then p1 is equal to p2. Submit your modified source code file in your homework submission zip archive.
6.2. Consider this ArrayList of Integers, which is to be sorted into ascending order:
list = { 13, 75, 12, 4, 18, 6, 9, 10, 7, 14, 15 }.
Trace the insertionSort() method and show the contents of list: (1) on entry to insertionSort(); (2) after the for j loop terminates each time but before the for i loop is repeated; and (3) after the for i loop terminates. The objective of this exercise is to help you understand how the insertion sort algorithm sorts a list.
6.3. Your after-hours research finally pays off: you discover a way to build a computer system that is one billion times faster than the fastest system currently in existence. Therefore, on your new computer system each comparison of selection sort requires only 25 × 10-18 seconds, which is 25 attoseconds. Create a table similar to the one in Sorting Algorithms : Section 6 showing how long selection sort will require to sort various sized lists.
6.4. If we would like to require insertionSort() to sort a list of 10-billion elements in no more than one minute, how many times faster would your new computer system need to be than the fastest system currently in existence?