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Summary

The purpose of this lab is to practice transforming recurrence relationships into working Java code. Start by reviewing the example code to gain an intuition and understanding for recursive behavior, and then build your own custom recursive methods. You’ll be modifying RecursionLab.java and turning that in, along with FractalFlake.java.

Summation: An Example Recursive Function

RecursionLab.java

In the main function, experiment with the two variations of summation functions (iterative versus recursive) by uncommenting the desired code paths. Read the comments throughout the code before moving on to modify this algorithm, and consider how you might rewrite the recursive function as an iterative function and vice versa.

From Summation to Factorial

The definition of the factorial function is defined below. Using the summation code as a guide, write a very similar recursive method that calculates the factorial of some input n. This method should be like all recursive functions in that it knows how to solve a trivial case (case1), and the function knows how to decompose larger cases into smaller cases (case2). Just as for and while loops use a boolean expression to govern loop termination, so too will we use a boolean condition in combination with an if to control repetition, as demonstrated here:

if (case1) {

return base case solution; // for summation, factorials, exponent, and Fibonacci, this is 1

} else (case 2) {

result = recursive call // typically the recursive call decrements by 1 or n/2

return result;

}

Factorial(n) is defined as 1 * 2 * 3 *…. * n. So Factorial(5) = 1 * 2 * 3 * 4 * 5 = 120

Build this new recursive method and test your code with a short main driver. (Do this in RecursionLab.java.)

The Exponent Function

In this section,

we will build two recursive functions (call them Power1(x,n) and Power2(x,n) ) and roughly compare their respective performance.

(I hope this would be obvious, but don’t use Math.pow() in your functions!) Write a method that produces the result of xn by observing the following recurrence relationships:

Power1(x,n):

(1) x0 = 1

(2) xn = x * x(n-1) when n > 0

Verify your method produces the correct value by writing a main that tests it.

In the same class, write another function (Power2()) that will also recursively solve for exponential values, but will do so by cutting our problem in half at each step (rather than moving the solution along incrementally by subtracting one from n at each step). This new relationship is defined by the piecewise function:

Power2(x,n):

(1) x0 = 1

(2) xn = ( x(n/2) )2 when n is even

(3) xn = x * ( x(( n-1 )/2 ) )2 when n is odd

To obtain the desired speedup, make sure your new function recursively calls itself no more than once in the body of the method. Test both the functions in main to make sure they work right. Then, in order to check which of these methods is faster, add a System.out.println statement as the first line of each of these methods, so that they prints out something like

“Calling Power1(x==5, n== 4)”

Now when you test them you can count how many recursive calls get made to each one. Test them with a couple of different values of n. Also compare the execution times for large n with the old implementation â€“ which is faster, and why?

ASK YOUR LAB INSTRUCTOR IF YOU NEED TO DO THIS (If you’re in Nixon’s Lab, you don’t):

Build a polynomial f1(x) for the first exponent function

Next, select a reference function g(x) that will dominate the f1(x) function in question

Finally, for g(x) find witnesses c, k such that |f1(x) < |c * g(x)| for all x > k?

Build a polynomial f2(x) for the second exponent function

Select a g(x) reference function

Find witnesses c, k to prove the class of Big O for the second

The Fibonacci Numbers

This series is described by the following recursive relationship:

f(0) = 0

f(1) = 1

f(n) = f(n-1) + f(n-2) for n > 1

With these formulae, you can write a recursive function that tells you the nth Fibonacci number. (Call this function FibNum(n).) Then you can write a recursive function (call it FibSequence(n)) to print out the Fibonacci sequence up to n, which is all the Fibonacci numbers starting at 0. So for example FibNum(6) == 8 and FibSequence(6) prints out

0 1 1 2 3 5 8

After you write FibNum and FibSequence, write tests in main for them that invoke these with various sample values of n. (Note that it would be easy to write an iterative version of FibSequence: for(int i=0;i= 3) { //base case is depth <3 for ( int i = 0; i < numBranches; i++ ) { int x2 = startX + ( int ) (size * Math. cos ( (2 * Math. PI / numBranches) * i )); int y2 = startY - ( int ) (size * Math. sin ( (2 * Math. PI / numBranches) * i )); g.drawLine( startX, startY, x2, y2 ); //do a branch draw(g, x2, y2, limit/3); //recursive call } } } Finally, modify the PolyDemo driver so that in the getRandomShape() function, you can produce FractalFlakes as well as Sprays. This type of recursion is limited by the value â€œlimitâ€ â€“ when weâ€™ve divided size enough times, eventually we will arrive at our base case (limit<3). In this way the limit variable provides a limit on the number of times the function will call itself; hence the name â€œdepth-limited recursionâ€ Execute your driver and verify your recursive fractal patterns visually; this should look something like the snowflake patterns below. Try altering the values for limit, numBranches, and the magic constant â€œ3â€ above to obtain different snowflakes. optional: In my version, I also added a color variable in my FractalFlake, and then I made a randomColor() function to set that color in the constructor, and then in my draw function, I set the color like we did in the homework: Graphics2D g2d = (Graphics2D) g; g2d.setColor(this.color) if(limit>=3) {….}

By messing around with limit and size and numBranches, I got things like this:

Recursive File Searching

(IF YOU ARE IN NIXON’S LAB, YOU MAY SKIP #7, and 7.1, and 7.2 and 7.3)

In general, searching can take multiple forms depending on the structure and order of the set to search. If we can make promises about the data (this data is sorted, or deltas vary by no more than 10, etc.), then we can leverage those constraints to perform a more efficient search. Files in a file system are exposed to clients of the operating system and can be organized by filename, file creation date, size, and a number of other properties. Weâ€™ll just be interested in the file names here, and we want perform a brute force (i.e, sequential) search of these files looking for a specific file. The way in which weâ€™ll get file information from the operating system, there will be no imposed ordering; as a result, a linear search is the best we can do. Weâ€™d like to search for a target file given a specified path and return the location of the file, if found. Letâ€™s sketch out this logic linearly before we attempt to tackle it recursively. Before attempting any of this section, be sure you have ready your recursive homework assignment in its entirety, as bits of that assignment will be referenced here. Weâ€™ll start by imagining the steps required to find a file in a given directory or subdirectory on a very high level in pseudocode.

7.1 Pseudocode & Stepwise Refinement

Pseudocode is a useful tool for mapping out complex algorithms in a language-independent way before even writing one line of code. Google

Stepwise Refinement

and

Divide-and-Conquer.

Weâ€™re going to practice this approach in detail here. Weâ€™ll start with a very high level description of the program, and flesh out bits of detail in each stepwise refinement. Given the small amount of pseudocode below, can you take a small, incremental step to describe in more detail the steps required to accomplish this task?

Pseudocode Refinement Step 1 â€“ The Problem Statement

This should be as short as possible, yet convey the full requirement of the program (however vague). Consider the following starting point for your stepwise refinement, which was pulled from the homework description.

â€œGiven a target file to find and a starting directory, determine if and where the target file exists.â€

Pseudocode Refinement Step 2

Using the above sentence as an incomplete guide split the embedded concepts into 4 distinct steps, such as:

(1) Setup & Initialization (for the program, or a specific function, similar to preconditions)

______________________________________________________

(2) Input

______________________________________________________

(3) Processing

______________________________________________________

(4) Output

______________________________________________________

Pseudocode Refinement Step 3 to (N-1)

Using Step 2 above as a slightly more complete guide, split up the processing phase into more discrete steps required to search a directory. For example, a start on the next step might beâ€¦

//just for the (3) step above (Processing))

while(more directories to look at) {

look at one file or directory and check for a match and return it if found

if the current item is a directory, we must repeat these steps for this directory as well

}

(Spoiler Alert!) Stepwise Refinement, Step N

Compare your final stepwise refinement to the refinement below of the processing step we examined above. Does the logic appear similar?

Add the given directory to some structure to manage directories

while(more directories to examine) {

Get a directory

If a file, check for match

If a directory,

for(each item (file or directory) in the directory) {

if item is a file, check for a match and return if found

if item is a directory, save this in a structure

}

}

7.2 From Pseudocode to Java – File Search, Version 1.0

Build a linear search following the pseudocode approach above

. You can put this code in simply a main, or you could design a static helper function in your HP static class that searches for files. Did you notice we used a stack to accomplish this directory searching? Using the main below and the outline above, complete the iterative file searching algorithm in the method

searchFiles()

. Use fileObject.listFiles(), fileObject.isDirectory(), and fileObject.getName() to accomplish this below.

public static void main(String[] args) {

System.out.println( searchFiles(new File(â€œc:\â€), â€œhw3.zipâ€) );

}

public static String searchFiles(File path, String target) {

// to do

}

7.3 File Search, Version 2.0

We made use of an explicit stack to track new directories as they appear in the above iterative code. Do you think we could use the method call stack in place of our Java stack to accomplish this recursively? If we remove the stack definition and code above and produce a recursive function, might that be shorter? Change the main function above to call recursiveSearch() instead, which is a new method that you will create. This method will take the same input as searchFiles, but instead of searching through files using explicit for loops and a Stack from java.util, weâ€™ll be using recursive function calls to facilitate some of the looping and use the method call stack to store our directory list instead.

Build a recursive linear search following the pseudocode you have refined to this point, or using the starter pseudocode approach below.

Note that multiple distinct correct solutions exist to this problem, so donâ€™t worry if your pseudocode isnâ€™t exactly like the code below.

public static String searchFiles(File path, String target) {

check if path is actually a dir, abort if not

loop over all files & dirs in the current directory

if a file, check for a match and return if found

if a directory, repeat these steps

if found in the directory, return found

}