CS 18000 Homework 04: Repetition


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For this homework, you will be making a program that will approximate the square root of
a number from user input. Users should be able to decide how many decimal points they
want the square root approximated to. For the purposes of this lab, the number of decimal
points will be limited from 1 to 14 and all numbers will be non-negative.
First, let’s think about some properties of square roots. Think about the guess and check
method of solving a math problem. If our goal is to find the square root of real number “x”,
then for some real number “y”, y*y = x. Therefore, theoretically we should be able to
approximate the square root of x by trying various values of y and squaring them. While
y*y does not equal x to the desired precision, we can adjust and continue trying different
values of y. But how can we efficiently try values of y to calculate x?
As a starting point, let’s look at something called the binary search algorithm. You will
need to implement a modified version of this algorithm for the homework. We will provide
the binary search algorithm and it will be your task to figure out how to modify and
implement it to approximate the square root of a number.
The binary search algorithm provides an efficient way to search for certain value in an
ordered sequence. If there are n values in the sequence, the value at the n/2th
position will be our first guess (round up to the n+1th position if n is odd). Think about the
first number in the sequence as defining the minimum of the search space, and the last
number as defining the maximum position of the search space. Since we know the
sequence is ordered, if the value at the middle position is less than the value we are
searching for, then we know we can now confine our search to only the values at positions
greater than our initial guess. Now, we can redefine the minimum position of the search
Problem Solving And Object-Oriented Programming
space as being the position defined by the index of the middle position plus 1, and the
maximum position stays the same. If the initial guess was instead greater than the value
we were searching for you would instead adjust the maximum position. Now, we can
redefine the middle value and continue repeating the process until we have found the value
we are searching for (we conclude the value does not exist in the sequence if the min
position is equal to the max position). See the next image for finding 4.
In a number line, while there are an infinite number of values in between any two numbers,
there are a finite number of values between any two numbers to a defined precision. This
is what allows us to use a modified binary search algorithm to approximate the square root
below (note that ~ means “approximately equal to”, or “close enough to” according to the
defined precision).
CS 18000
Problem Solving And Object-Oriented Programming
1. Start with a range of numbers from 0 to 26.
2. Calculate the middlemost number and square it. For the first iteration, we found 13,
which squares to 169.
3. Compare the squared number to 26.00. 169 is greater than 26.00, so we will look for
another number that is less than 13, but bigger than 0. If the square were larger, we would
look at numbers between 13 and 26.
4. Repeat steps 1 through 3 until we find a number that has a square of about 26.00 when
rounded. Note that the answer is not exactly the square root of 26: this is only an estimate
precise to two decimal places. Reducing the precision of the result from 5.098755 to 5.10
may also change the accuracy of your result. For the purpose of this assignment, this is
okay. Just don’t use this algorithm for your finances.
CS 18000
Problem Solving And Object-Oriented Programming
An example run of the program is below. You should model the functionality of your
program on the one shown using the algorithm discussed above.
**NOTE: You may NOT use any pre-existing square root functions for this task (e.g.
Math.sqrt or Math.pow). You will receive a ZERO and use a submission**
Items required for submission via Blackboard:
● SqrtEstimator.java
· SqrtEstimator class — 100 points total
· Program continues running until user enters “quit” command – 20 points
(required for other criteria).
· Error checking for invalid (non-positive) input/prompting while not correct – 20
· Approximation displays correct precision – 10 points
· Approximation is mathematically correct – 50 points