Computer Science II Assignment 2 Solution




  • Recursion (10 points)


Create a  le called This     le should have the following method:


public static long calculate(long n)


Factorial.calculate should recursively calculate n!, where 0! = 1 and n! = n(n 1)!. The method should also print out an error and exit if n < 0 or n > 20, since factorial is not de ned for negative numbers and will over ow Java’s long variable with larger numbers (if we used an int, it would over ow even sooner!).


Inside, also include a main method which runs a couple of tests on Facto-rial.calculate:


java Factorial


Factorial.calculate(0) returned 1.                Test passed!


Factorial.calculate(5) returned 120.                Test passed!


(Obviously, if these tests do not return 1 and 120 respectively, the tests should fail and print out an appropriate message.)


  • Fibonacci revisited (10 points)


Create a le called Refactor your from assignment 1 to be a static method in FibTest. Odds are that your assignment 1 answer is coded in iterative style, where you are looping from 1 to n and adding up numbers as you go. If not, you should write an iterative Fibonacci calculator that does so (and refactor the version that you wrote for assignment 1 for the next part of the problem). This method should be called


public static int fibIter(int n)


The recurrence relation for the Fibonacci sequence is the following:

f ib(1) = 1
f ib(2) = 1


f ib(n)     =      f ib(n      1) + f ib(n     2)


Write another static method in the same FibTest class, this time in recursive style, using this relationship.


public static int fibRecur(int n)

In FibTest’s main method, write a series of tests that establishes that both of these functions work correctly. Finally, using Java’s System.currentTimeMillis() function, print out the time it takes to execute FibTest. bIter(40) and FibTest. bRecur(40).


  • revisited (10 points)


In Assignment 1, you used the Gregory formula to compute an approximation of . Here, you will use Ramanujan’s series, discovered in 1910. It converges much faster than Gregory’s, and has been used to calculate to billions of digits. Implement exactly like, taking a number n speci ed by the user on the command line and calculating using the rst n terms of the Ramanujan series. The program should print this approximate value of , as well as the percentage error between this value and the one provided by Java in the constant Math.PI.

1 2p   1 (4k)!(1103+26390k)
Ramanujan’s series: = kP  
9801 4 396 4k
=0 (k!)

Notice that this formula makes use of the factorial function. Call your Factorial.calculate method from Problem 1.


Extra credit: Look at the Java API de nitions of BigDecimal and BigInteger. Rework your Factorial and Ramanujan code (call the classes something else, like RamanujanBig, so that you don’t overwrite your regular-credit work) to use these instead, and calculate to 100 digits or more. Note


that, in order for the terms to be perfectly accurate, you also need a BigDecimal representation p


of 2. There is unfortunately no native library support for calculating this; however, you can implement it yourself using a BigDecimal version of the evaluate method you will write in Problem 6.


You will get a little extra credit simply by explaining (in comments in your Ramanujan code) how BigDecimal and BigInteger di er from ordinary doubles and ints, and why the latter are insu cient for calculating very precise values of .


  • Root nding (20 points)


The roots of a continuous function are points at which the value of the function is equal to zero. If we happen to know a value for which the function is positive and another value for which it is negative, then we know for certain that somewhere between the two values lies a zero. In other words, if f(a) > 0 and f(b) < 0, then there exists a value x between a and b for which f(x) = 0.


This fact suggests that we can zero in on the value of x using the following algorithm, where is the amount of error we are willing to tolerate:


  1. Compute x, where x is halfway between a and b, ie x = (a+2b) .


  1. If ja xj <  return x.


  1. If f(x) has the same sign as f(a), the root lies between x and b. Recursively perform the algorithm with x as the new a.


  1. Otherwise, the root lies between a and x. Recursively perform the algorithm with x as the new b.


Write a class and implement the following method:



2public static double findRoot(double a, double b, double epsilon)



Within ndRoot(), use Math.sin() as the function to be evaluated. In your main function, print out the root of sin(x) that falls between 3 and 4, to within 0.00000001. Does the number look familiar?


  • Polynomials (30 points)


Write a class to represent polynomials (ie, functions of the form axn + bxn 1 + + cx2 + dx + e). Implement the following methods:


Poly(int[] coe cients): Constructor for an array of coe cients where c[n] is the coe cient of xn. In other words, the polynomial 2x5 + 3x4 8x2 + 4 would be represented by the array [4; 0; 8; 0; 3; 2]. This way, if I want to know the coe cient of x2, I look at c[2], and get 8. This convenience is worth the confusion that arises because we usually write arrays left-to-right starting from the smallest index, whereas we usually write polynomials left-to-right starting from the largest coe cient.


int degree(): Return the power of the highest non-zero term


String toString(): Overriding the Object class’s toString method, this should return a String representation of the polynomial using x as the variable, arranged in decreasing order of exponent and printing nothing for terms with a coe cient of zero. For example, the Poly represented by the array [4; 0; 8; 0; 3; 2] should return the string:




Poly add(Poly a): To add two polynomials, simply add together each scale degree in turn. Thus, (2x5 + 3x4 8x2 + 4) + (x3 + 4x2 2x) = (2x5 + 3x4 + x3 4x2 2x + 4). Create a method that constructs and returns a new Poly object with a new coe cient array created by adding the coe cients of the current object and the Poly object a passed as an argument to the add() function.


double evaluate(double x): Return the value of the function at the point x. In other words, if the Poly object represents 2x5 + 3x4 8x2 + 4 and evaluate(2.0) is called, the method should calculate 2(2)5 + 3(2)4 8(2)2 + 4 and return 84.


Within the main() method of the Poly class, create a series of tests which exercise each of these methods and report their success or failure.


  • Inheritance (20 points)


The ndRoot() function in works great, but only speci cally for the sine function. That’s nice and all, but hardly general. We would like to be able to use the same code for any function at all, and we will use inheritance to make that happen. Create a new class called that supports an abstract method:

public abstract double evaluate(double x)


Refactor your FunctionTest. ndRoot() method so that it belongs to the Function class, and instead of calling Math.sin(), it calls this evaluate() method. Note that ndRoot() will no longer be static.


Write a class that extends Function and implements evaluate() using Math.sin(). Write a similar class that does the same for Math.cos(). Copy your code from into a new class which also extends Function (you shouldn’t have to make any substantive changes, but you will have to rename occurrences of the \Poly” type inside your code to \PolyFunc”). Look at that { you’ve already implemented evaluate() for polynomials!


In Function.main(), create some tests. Instantiate a few functions and nd some roots. Verify that the root of sin(x) between 3 and 4 is the same as it was before. Find the root of cos(x) between 1 and 3. Find the positive root (x > 0) of x2 3 and x2 x 2.


Finally, for all of the classes in this problem (,, and, write javadoc comments explaining the purpose of each class and the uses of each method.


Turning in


You should have created nine Java les for this assignment:,,,,,,, and Ensure that all of them can be compiled and run from the command line. Create a zip le called assignment 2 your and upload it to the Dropbox at This assignment is due Wednesday, September 14, at noon.



































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