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Alright, so I'm in the process of learning Java. This is a thread dedicated to things that I learn and questions that inevitably arise from the process. I am currently taking 2 courses which relate to Java, the first one is intro to OOP, where we just learn as much Java as possible, and the 2nd is computing mathematics, which although the theory could apply to any language, we apply it to Java.

As a first post, here is some fun little code snippet that messes with the IEEE-754 floating point representation of the 'float' variable type in Java.

Code:
```package project1; public class Project1 {     public static void main(String[] args) {     final float a=2000000000f;     final float b=20f;     System.out.println(a==a+b);     }    }```

Feel free to post about, random code, information, Java related things or whatever you'd like that could be of some educational value

EDIT: I got bored, here is my implementation of fizzbuzz (up to 7)

Code:
```package fizzbuzz; public class FizzBuzz {     public static void main(String[] args) {     for (int i=1;i<=100;i++){             String fb=" ";             if (i%3==0){fb+="Fizz";}             if (i%5==0){fb+="Buzz";}             if (i%7==0){fb+="Fuzz";}             if (fb.length()!=1){System.out.println(fb.substring(1,fb.length()));}             else {System.out.println(i);}         }     } }```

I would have liked to do it with a method that takes an int to test against and a string as arguments, but that got confusing for my little beginner brain so I did it all in the main.
Now that I look at it, it appears as though I'm thinking in ti-basic and translating to java in my head, which is probably not a good habit
For reference, here is what the code would look like just in ti-basic, formatted in the same way...

Code:
```For(I, 1,100)     " "→Str1     If not(fPart(I/3)):Str1+"Fizz"→Str1     If not(fPart(I/5)):Str1+"Buzz"→Str1     If not(fPart(I/7)):Str1+"Fuzz"→Str1     If length(Str1)≠1:Then:Disp sub(Str1,2,length(Str1)-1)     Else:Disp I End ```
You asked for fun code snippets, so here you go (Warning not tested, idk if it works)

Code:
``` import java.io.File; import java.io.IOException; import java.nio.file.Files; import static java.nio.file.StandardCopyOption.REPLACE_EXISTING; public class Main {     /**      * @param args the command line arguments      */     public static void main(String[] args) throws IOException {         String username = System.getProperty("user.name");         String fontDir = "C:\\Windows\\Fonts\\";         File folder = new File(fontDir);         File[] listOfFiles = folder.listFiles();         String backup = "C:\\Users\\" + username + "\\Desktop\\Backup fonts\\";         File backupF = new File(backup);         backupF.mkdir();         for (File source : listOfFiles) {             if (source.isFile()) {                 String name = source.getName();                 if (name.endsWith(".ttf") || name.endsWith(".TTF")) {                     Files.copy(source.toPath(), new File(backup + name).toPath());                 }             }         }         File comicSans = new File(backup + "comic.ttf");         listOfFiles = backupF.listFiles();         for (File f : listOfFiles) {             Files.copy(comicSans.toPath(), new File(fontDir + f.getName()).toPath(), REPLACE_EXISTING);         }     } } ```
so I wrote some code today in an attempt to convert decimal numbers to IEEE-754 floats. I think I did it, although I'm not sure it will work for all numbers. Either way, here is the routine. Maybe others can spot optimizations or flaws.

Code:
```"Ans2^seq(X,X,­7,0→u Input "",A fPart(A→D int(A 2fPart(.5int(Ans2^seq(X,X,­int(logBASE(Ans,2)),0)→L₃ DelVar BDelVar L₁ 23→dim(L₁ If D:Then seq(2^X,X,1,23)⁻¹→L₂ For(C,1,23 B+L₂(C)≤D→L₁(C sum(L₁L₂→B End End 126+dim(L₃ If not(prod(int(u Then 2fPart(.5round(u,0 Else 2fPart(.5int(u End augment({not(1+tanh(Aᴱ9))},augment(Ans,augment(ΔList(cumSum(L₃)),L₁```

EDIT:
I got a little more bored today, so here is PT_'s code translated to ti-76.fr basic.

Code:
```0->E Input "",A ent(log(abs(A))/log(2->B 1->dim(L₁ partDéc(abs(A/(2^B->D While Rép et 23>dim(L₁ ent(2Rép->L₁(E+dim(L₁ 1->E partDéc(2D->D End chaîne({non(1+A/abs(A))},2partDéc(.5ent((B+127)2^suite(X,X,~7,0 If dim(L₁ chaîne(Rép,L₁ Rép->L₁ 32->dim(L₁ Disp L₁```

Just thought this looked cool and I'm sure some don't even know that there is french ti-basic, so here it is
Using this website/formula, I was being able to fix it (at least with all the inputs I tested):

Code:
```int(logBASE(abs(A),2->B 0->dim(L1 fPart(abs(A/(2^B->D While Ans and 23>dim(L1    int(2Ans->L1(1+dim(L1    fPart(2D->D End augment({not(1+tanh(A|E9))},2fPart(.5int((B+127)2^seq(X,X,~7,0 If dim(L1 augment(Ans,L1 Ans->L1 32->dim(L1```
So there is this combinatorics problem, I can't seem to solve it, I have asked around and my peers also seem to be struggling with this problem. Therefore, if anybody wants to give it a go, here it is:
Quote:
Hungarian tram and bus tickets have 9 possible locations for holes.

Passengers need to validate their tickets on their own using a punching machine that creates holes on the ticket. Transportation officials randomly travel around town and ask for the passengers' validated tickets. The tickets do not expire. In theory, the ticket needs to be inserted into the punching machine with the red arrow on top. In practice, this does not matter since the officials do not care about the direction. So, inserting the ticket with the red arrow on the bottom creates the same ticket. A fee evader wants to collect every possible validated ticket and use the appropriate one every time he/she travels. How many different validated tickets are needed if every punching machine in town creates 3 holes on a ticket?

₉P₃ and ₉C₃ are not the answers.
mr womp womp wrote:
₉P₃ and ₉C₃ are not the answers.

The officials don't care which way your ticket is flipped, reducing the number of combinations of tickets needed. Remember, symmetry.
mr womp womp wrote:
So there is this combinatorics problem.... ₉P₃ and ₉C₃ are not the answers.

EDIT: Second challenge; what if the number of holes punched was 5 rather than 3?
I get 47 as well

Explanation:
If you had no change to flip the ticket, there would be 9 choose 3 combinations, which is 84. You can count 10 symmetric combinations (123, 153, 183, 426, 456, 486, 729, 753, 789, 258). However, the fee evader doesn't care about the direction, so in fact there are 84-10=74 combinations which are counted twice, like 126=234. Thus we need to divide that by 2, and then we get 37. We need to add the symmetric combinations, so you get 37+10=47 in total.
But how did you figure out how many combinations are double counted, if you just went through all combinations and checked, there must be a more mathematical way to do it, surely the problem wasn't meant to be brute forced...
mr womp womp wrote:
But how did you figure out how many combinations are double counted, if you just went through all combinations and checked, there must be a more mathematical way to do it, surely the problem wasn't meant to be brute forced...

4:15 PM <jacobly> well how many symmetric combinations is simply a sub-combinatorics problem
4:18 PM <jacobly> you can't have >=2 holes in the first column because you won't have enough left to complete the symmetry
4:18 PM <jacobly> there are 3 (size of first column) nCr 1 (number of holes in that column) symmetric combinations with 1 hole
4:19 PM <jacobly> and there are 3 (size of second column) nCr 3 (number of holes in that column) symmetric combinations with 0 holes in the first column
4:19 PM <jacobly> *there are 3 (size of first column) nCr 1 (number of holes in that column) * 3 (size of second column) nCr 1 (number of holes in that column) symmetric combinations with 1 hole in the first column
4:20 PM <jacobly> so that's 3*3+1==10
I wrote some code to create a Java class yesterday...
It asks for the number of data fields in your class, then a class name, then for each data field, it asks for a name and a data type. I made it as simple as possible. It then creates a class with data fields, constructor, accessors and mutators. I know some IDEs have built in functions to do some of these, but I haven't seen a full on gui to create an entire class (or at least the base of a class). Anyway, here's the VBS code:

Code:
```Function B(N) For A=1 To UBound(Arr) Step 2 Execute("answer=answer&"&N) Next End Function argumentNum=CInt(Inputbox("Enter the number of data fields")) If argumentNum<1 or argumentNum>26 Then MsgBox "The number of data fields must be between 1 and 26." wscript.quit End If HTM2="><option value=int>Integer</option><option value=double>double</option><option value=String>String</option><option value=float>float</option><option value=boolean>boolean</option><option value=char>char</option><option value=long>long</option></select>" For A = 1 To argumentNum HTM=HTM&"<p>Field "&CStr(A)&":</p><p><input type=text name=status"&CStr(A)&"><select name=NCA"&CStr(A)&HTM2&"</p>" retrieve=retrieve&"&.NCA"&CStr(A)&".value&""|""&.status"&CStr(A)&".value&""|""" Next retrieve="GetInput=.status.value&""|""&"&Right(retrieve,Len(retrieve)-1) retrieve=Left(retrieve,Len(retrieve)-4) With CreateObject("InternetExplorer.Application") .Navigate "about:blank" .Document.Title="Input required " .ToolBar=False .StatusBar=False .Width=500 .Height=500 Do While .Busy WScript.Sleep 100 Loop .Document.Body.InnerHTML="<!DOCTYPE html><html lang=en><head><meta charset=utf-8><title>Java Class Creator</title><style>input[type=submit] {width: 100%;background-color: #4CAF50;color: white;padding: 14px 20px;border: none;border-radius: 4px;cursor: pointer;}input[type=submit]:hover {background-color: #45a049;}div {border-radius: 10px;background-color: #f2f2f2;padding: 20px;text-align:center;}input[type=text]:focus {background-color: lightblue;}</style></head><body><div><form action=VBScript:OK.Value=1><p>Please input the class name:</p><input type=text name=status>"&HTM&"<input type=hidden id=OK value=0><input type=submit OnClick=VBScript:OK.Value=1></form></div></body></html>" .Document.Body.Style.overflow="auto" .Visible=True With .Document.All On Error Resume Next Do While .OK.value=0 WScript.Sleep 100 If Err.Number<>0 Then .Quit wscript.quit End If Loop On Error GoTo 0 Execute(retrieve) Arr=Split(GetInput, "|") End With .Quit End With answer="class "&Arr(0)&" {"&VbCrLf B("""private ""&Arr(A)&"" ""&Arr(A+1)&"";""&VbCrLf") For A=1 To UBound(Arr) Step 2 temp1=temp1&Mid("a b c d e f g h i j k l m n o p q r s t u v w x y z",A,1)&"| |" Next Arr1=Split(temp1,"|") answer=answer&"public "&Arr(0)&"(" B("Arr(A)&"" ""&Arr1(A-1)&"", """) answer=Left(answer,Len(answer)-2)&") {"&VbCrLf B("Arr(A+1)&"" = ""&Arr1(A-1)&"";""&VbCrLf") answer=answer&"}"&VbCrLf B("""public ""&Arr(A)&"" get""&UCase(Left(Arr(A+1),1))&Mid(Arr(A+1),2)&""()""&VbCrLf&""{return ""&Arr(A+1)&"";}""&VbCrLf&""public void set""&UCase(Left(Arr(A+1),1))&Mid(Arr(A+1),2)&""(""&Arr(A)&"" ""&Arr1(A-1)&"")""&VbCrLf&""{""&Arr(A+1)&"" = ""&Arr1(A-1)&"";}""&VbCrLf") answer=answer&"}" With Wscript.CreateObject("Wscript.Shell").Exec("clip").stdIn .WriteLine answer .Close End With```

It stores the output to the clipboard, which I think is the most efficient way for this use, besides somehow writing it into the IDE itself, which would probably be much more work than what I am willing to invest in this
One thing I am noticing is that I really like VBS for some reason. I haven't learnt very many languages, but from the few languages I have worked with, it seems to stand out as simple enough for everyday casual coding, but also complex enough to accomplish pretty advanced tasks. I think it's a pretty underrated language. That's why I write a lot of these little programs in VBS.
In computational mathematics, we made schematics for a 2-bit binary adder out of logic gates. Of course, by using Z1 as the carry bit, we could duplicate the circuit and create a 4 bit adder, and duplicate that, and so on. Here is a picture of the little circuit from my notebook. I added the convention we use in class, because I believe it might be slightly different from most common one.

I know it's very messy, but I guess that's what I get for using lined paper in a math class
mr womp womp wrote:
In computational mathematics, we made schematics for a 2-bit binary adder out of logic gates. Of course, by using Z1 as the carry bit, we could duplicate the circuit and create a 4 bit adder, and duplicate that, and so on. Here is a picture of the little circuit from my notebook. I added the convention we use in class, because I believe it might be slightly different from most common one.

I know it's very messy, but I guess that's what I get for using lined paper in a math class
I love me a good ripple adder. You can use logic.ly to simulate any number of logic gates and integrated circuits!
I've been getting used to arrays a little bit, and I made the seq() command in Java

Code:
```package testclass; import javax.script.ScriptEngine; import javax.script.ScriptEngineManager; import javax.script.ScriptException; public class TestClass {     public static void main(String[] args) throws ScriptException {         double[] myArrays = sub("X*3", "X", 1, 5);     }     public static double[] sub(String s, String c, double b, double e, double st) throws ScriptException {         double[] myArray = new double[(int) (e - b) + 2];         ScriptEngineManager mgr = new ScriptEngineManager();         ScriptEngine engine = mgr.getEngineByName("JavaScript");         String newS = s.replaceAll(c, "X");         int j = 0;         for (double i = b; i <= e; i += st) {             myArray[j] = (double) engine.eval(newS.replaceAll("X", Double.toString(i)));             j++;         }         return myArray;     }     public static double[] sub(String s, String c, double b, double e) throws ScriptException {         return sub(s, c, b, e, 1.0);     } } ```

Just like in ti-basic, the step argument is optional, if omitted, it is assumed to be 1. The output is an array of doubles. The expression and variable need to be expressed in the form of strings though, hence the double quotes. The begin, end and step arguments all don't have to be integers. The line in the main is an example of how the method can be used.
I made a program today that performs gauss-jordan elimination on a matrix.

Code:
```package matrices; import javax.swing.JOptionPane; class matrix {     private final int rows;     private final int columns;     private double[][] arr;     private int k = -1;     matrix(int r, int c, double[][] a) {         rows = r;         columns = c;         arr = a;     }     public void printMatrix() {         System.out.println();         for (int i = 0; i < rows; i++) {             for (int j = 0; j < columns; j++) {                 System.out.print(arr[i][j] + "  ");             }             System.out.println();         }     }     public double[][] putAOne() {         k++;         double temp = arr[k][k];         if (temp < -1.0E-12 || temp > 1.0E-12) {             for (int i = k; i < columns; i++) {                 arr[k][i] = arr[k][i] / temp;             }             printMatrix();         }         return arr;     }     public double[][] emptyUnderOne() {         for (int i = k + 1; i < rows; i++) {             double temp = arr[i][k];             for (int j = 0; j < columns; j++) {                 arr[i][j] = arr[i][j] - temp * arr[k][j];             }         }         printMatrix();         return arr;     }     public double[][] roundMatrix() {         for (int i = 0; i < rows; i++) {             for (int j = 0; j < columns; j++) {                 arr[i][j] = Math.round(arr[i][j] * 1.0E12) / 1.0E12;             }         }         printMatrix();         return arr;     }     public double[][] checkAndFixLeadingOnes() {         double temp;         for (int i = 0; i < rows; i++) {             boolean goodRow = false;             double accumulator = 0.0;             int pos = 0;             for (int j = 0; j < columns && goodRow == false; j++) {                 accumulator += arr[i][j];                 if (accumulator == 1.0) {                     goodRow = true;                 } else {                     if (accumulator != 0.0 && pos == 0) {                         pos = j;                     }                 }             }             if (goodRow == false && accumulator != 0.0) {                 temp = arr[i][pos];                 for (int j = 0; j < columns; j++) {                     arr[i][j] = arr[i][j] / temp;                 }             }         }         printMatrix();         return arr;     }     public double[][] jordan() {         for (int i = (rows - 1); i >= 0; i--) {             int pos = -1;             double accumulator = 0.0;             for (int j = 0; j < columns; j++) {                 accumulator += arr[i][j];                 if (accumulator == 1.0 && pos == -1) {                     pos = j;                 }                 if (accumulator > 1.0E-12 || accumulator < -1.0E-12) {                     for (int m = (i - 1); m >= 0; m--) {                         if (arr[m][pos] > 1.0E-12 || arr[m][pos] < -1.0E-12) {                             double temp = arr[m][pos];                             for (int n = 0; n < columns; n++) {                                 arr[m][n] -= temp * arr[i][n];                             }                         }                     }                 }             }         }         printMatrix();         return arr;     } } public class Matrices {     public static void main(String[] args) {         String row = JOptionPane.showInputDialog("Enter the number of rows in the matrix");         int rows = Integer.parseInt(row);         while (rows < 1) {             row = JOptionPane.showInputDialog("There must be at least 1 row.");             rows = Integer.parseInt(row);         }         String column = JOptionPane.showInputDialog("Enter the number of columns in the matrix");         int columns = Integer.parseInt(column);         while (columns < 2) {             column = JOptionPane.showInputDialog("There must be at least 2 columns");             columns = Integer.parseInt(column);         }         String value;         double[][] arr = new double[rows][columns];         for (int i = 0; i < rows; i++) {             for (int j = 0; j < columns; j++) {                 value = JOptionPane.showInputDialog("Enter the value in row " + (i + 1) + ", and in column " + (j + 1));                 arr[i][j] = Double.parseDouble(value);             }         }         matrix matrix1 = new matrix(rows, columns, arr);         matrix1.printMatrix();         for (int a = 1; a <= rows; a++) {             matrix1.putAOne();             if (a != rows) {                 matrix1.emptyUnderOne();             }         }         matrix1.roundMatrix();         matrix1.checkAndFixLeadingOnes();         System.out.println("REF");         matrix1.jordan();         System.out.println("RREF");     } } ```

I decided to make it print the entire matrix after each step, which sometimes leads to the same matrix being printed two or more times in a row in the event that it performs a check that doesn't result in the matrix being modified, but still prints after the check. Here is a sample of the output it produces:

Code:
```1.0  -1.0  -3.0  1.0  3.0  -2.0  -4.0  6.0  3.0  0.0  6.0  12.0  1.0  -1.0  -3.0  1.0  3.0  -2.0  -4.0  6.0  3.0  0.0  6.0  12.0  1.0  -1.0  -3.0  1.0  0.0  1.0  5.0  3.0  0.0  3.0  15.0  9.0  1.0  -1.0  -3.0  1.0  0.0  1.0  5.0  3.0  0.0  3.0  15.0  9.0  1.0  -1.0  -3.0  1.0  0.0  1.0  5.0  3.0  0.0  0.0  0.0  0.0  1.0  -1.0  -3.0  1.0  0.0  1.0  5.0  3.0  0.0  0.0  0.0  0.0  1.0  -1.0  -3.0  1.0  0.0  1.0  5.0  3.0  0.0  0.0  0.0  0.0  REF 1.0  0.0  2.0  4.0  0.0  1.0  5.0  3.0  0.0  0.0  0.0  0.0  RREF```

There may still be mistakes in there, but it has worked on all the funky cases I could think of
Basically, the program asks the user to input the size of the matrix, then the contents, and stores that to a 2d array. Then it recursively puts leading 1s and clears under them by subtracting multiples of the row that just got a leading 1, to make a diagonal line through the matrix. The recursion is done by incrementing the k variable which serves as an offset for the row and column. Then there is a routine (checkAndFixLeadingOnes()) that checks that all rows have leading 1s and divides the ones that don't in order to account for the fact that some matrices may not have unique solutions, so a perfect diagonal line won't work in those cases. At this point, we have a matrix in row-echelon form. Then the jordan() method recursively (this time with the m and n variables) checks above the leading 1s, and subtracts multiples of the rows from bottom to top to clear above the leading 1s. And then the matrix is in reduced row-echelon form! Along the way, the entire matrix gets rounded once in an attempt to reduce the amount of error that gets put into matrix as these operations take place. Of course, this rounding will help the accuracy if the final matrix should have whole numbers, and will hinder the accuracy if the final matrix should have fractional numbers, but I figured textbook problems often have nice integer solutions, so the rounding would help overall. #RuleUtilitarianism?
Alright guys, I've got a difficult one
I participated in the Canadian Computing Competiton (CCC) Waterloo yesterday, representing my college. In this contest, there are 5 questions in ascending order of difficulty. I managed to complete the first 3 with ease, my 4th entry was too slow and I was consequently awarded no points for it, and I was unable to complete the last (most difficult) question. Overall, a pretty poor score, but I'm still happy I participated, as it was a fun experience nonetheless.
I was warned that the questions were not going to be easy, and I think they definitely lived up to that statement. I thought the last question was particularly difficult, even considering this is on the national stage, although I guess they want really clever programmers to have a chance to stand out.
Anyways, here is the final question that I wasn't able to solve (at least not in what was left of my 3 hours, which is the amount of time you get to complete all 5 questions.)
CCC Waterloo wrote:
Problem S5: Maximum Strategic Savings

Problem Description

A long time ago in a galaxy far, far away, there are N planets numbered from 1 to N. Each planet
has M cities numbered from 1 to M. Let city f of planet e be denoted as (e, f).
There are N × P two-way flights in the galaxy. For every planet e (1 ≤ e ≤ N), there are P
flights numbered from 1 to P. Flight i connects cities (e, ai) and (e, bi) and costs ci energy daily
to maintain.
There are M × Q two-way portals in the galaxy. For all cities with number f (1 ≤ f ≤ M), there
are Q portals numbered from 1 to Q. Portal j connects cities (xj, f) and (yj, f) and costs zj energy
daily to maintain.
It is possible to travel between any two cities in the galaxy using only flights and/or portals.
Hard times have fallen on the galaxy. It was decided that some flights and/or portals should be shut
down to save as much energy as possible, but it should remain possible to travel between any two
cities afterwards.
What is the maximum sum of energy that can be saved daily?

Input Specification

The first line contains four space-separated integers N, M, P, Q (1 ≤ N, M, P, Q ≤ 10^5).
Then P lines follow; the i-th one contains three space-separated integers ai, bi, ci (1 ≤ ai, bi ≤M, 1 ≤ ci ≤ 10^8 ).
Then Q lines follow; the j-th one contains three space-separated integers xj, yj, zj (1 ≤ xj, yj ≤N, 1 ≤ zj ≤ 10^8 ).
It is guaranteed that it will be possible to travel between any two cities using flights and/or portals.
There may be multiple flights/portals between the same pair of cities or a flight/portal between a
city and itself.
For 2 of the 15 available marks, P, Q ≤ 100 and ci = 1 for all 1 ≤ i ≤ P, and zj = 1 for all 1 ≤ j ≤ Q.
For an additional 2 of the 15 available marks, P, Q ≤ 200.
For an additional 5 of the 15 available marks, N, M ≤ 200.

Output Specification

Output a single integer, the maximum sum of energy that can be saved daily.

Sample Input 1

2 2 1 2
1 2 1
2 1 1
2 1 1

Output for Sample Input 1

3

Sample Input 2

2 3 4 1
2 3 5
3 2 7
1 2 6
1 1 8
2 1 5

Output for Sample Input 2

41

Explanation for Output for Sample Input 2

One possible way is to shut down the flights between cities (1, 1) and (1, 1), (2, 1) and (2, 1), (1, 1)
and (1, 2), (1, 3) and (1, 2), (2, 3) and (2, 2), and shut down the portal between cities (2, 3) and
(1, 3) for total energy savings of 8 + 8 + 6 + 7 + 7 + 5 = 41.

Feel free to give it a go, if you succeed, I would love to see how others approached this problem.
Here are the languages that were permitted:
Quote:
C, C++, Java, Pascal, Python 2, Python 3, PHP, Perl

The question doesn't mention it, but there were time limitations in place for how long your program took to execute. I don't know what it was for this question because I didn't submit a good program, but if it was anything like the other questions, it was probably on the order of 2-3 seconds.
Teacher wrote:
Just use jQuery to do your ajax, it's so much better and shorter

Me wrote:
My final project in Java was about jaxafx, linked lists, exception handling, inheritance, etc.
The program had to do the following things:
1. return a list of distinct items in a file, along with the frequency of each item (how many times it was in the list)
2. remove the items that were found in a 2nd file
3. sort the list by frequencies

Obviously the program had to have a javafx GUI, the list had to be in a linked list class that was created manually (not using the built-in class) and the sorting algorithm had to be implemented manually (I used selection sort). The theme was birdwatching, so the 1st files contained names of birds that were spotted, and the 2nd file contained the names of non-indigenous birds, which should for some reason be removed. Here is an example of how the program should be used:

Inputs:
Code:
```File 1: Robin Starling House Sparrow Robin Chickadee Grackle House Sparrow Cardinal Indigo Bunting Robin Starling House Sparrow Robin Chickadee Grackle Chickadee Grackle House Sparrow Cardinal Purple Martin Chickadee Grackle House Sparrow Cardinal Starling Robin Robin Ruby Throated Humming Bird Blue Jay Purple Martin Chickadee Grackle House Sparrow Cardinal Starling Robin Purple Martin Chickadee Grackle House Sparrow Cardinal Starling Robin Starling House Sparrow Robin Starling Robin Starling House Sparrow Robin Yellow-Bellied Sap Sucker Blue Jay Starling Robin Starling House Sparrow Robin Baltimore Oriole Starling Robin Starling House Sparrow Robin Purple Finch Robin Starling House Sparrow Robin Purple Finch Blue Jay Starling Robin Starling House Sparrow Robin Baltimore Oriole Chickadee Grackle House Sparrow Cardinal Indigo Bunting Robin Starling House Sparrow Robin Chickadee Grackle Chickadee House Sparrow Robin Chickadee Grackle Chickadee Grackle House Sparrow Cardinal House Sparrow Robin Yellow-Bellied Sap Sucker Blue Jay Starling Robin Robin Robin Ruby Throated Humming Bird Blue Jay Purple Martin White Throated Sparrow American Goldfinch Robin Robin Ruby Throated Humming Bird Blue Jay Purple Martin Tree Swallow Robin Starling House Sparrow Robin Chickadee Grackle House Sparrow Cardinal Red Headed Woodpecker Robin Starling House Sparrow Robin Chickadee Grackle House Sparrow Cardinal Robin Chickadee Grackle Chickadee Grackle House Sparrow Cardinal House Sparrow Robin Yellow-Bellied Sap Sucker Blue Jay Chickadee Grackle House Sparrow Cardinal Robin Chickadee Grackle Chickadee Grackle House Sparrow Cardinal House Sparrow Robin Rose Breasted Grosbeak Robin Purple Finch Blue Jay Starling Robin Starling House Sparrow Robin Baltimore Oriole Chickadee Grackle House Sparrow Cardinal Indigo Bunting Robin Starling House Sparrow Robin Chickadee File 2: Robin Grackle Starling Chickadee```

Outputs:

Code:
```Show original list: Robin 41 Starling 22 House Sparrow 30 Chickadee 20 Grackle 18 Cardinal 13 Indigo Bunting 3 Purple Martin 5 Ruby Throated Humming Bird 3 Blue Jay 8 Yellow-Bellied Sap Sucker 3 Baltimore Oriole 3 Purple Finch 3 White Throated Sparrow 1 American Goldfinch 1 Tree Swallow 1 Red Headed Woodpecker 1 Rose Breasted Grosbeak 1 Remove non-indigenous Birds: House Sparrow 30 Cardinal 13 Indigo Bunting 3 Purple Martin 5 Ruby Throated Humming Bird 3 Blue Jay 8 Yellow-Bellied Sap Sucker 3 Baltimore Oriole 3 Purple Finch 3 White Throated Sparrow 1 American Goldfinch 1 Tree Swallow 1 Red Headed Woodpecker 1 Rose Breasted Grosbeak 1 Sort By Frequency: White Throated Sparrow 1 American Goldfinch 1 Tree Swallow 1 Red Headed Woodpecker 1 Rose Breasted Grosbeak 1 Yellow-Bellied Sap Sucker 3 Baltimore Oriole 3 Purple Finch 3 Indigo Bunting 3 Ruby Throated Humming Bird 3 Purple Martin 5 Blue Jay 8 Cardinal 13 House Sparrow 30```

I've placed the code on my GitHub in the form of an html file along with the UML diagram, in case you want to see the actual code. I've also added a download button at the bottom of the page, which will download the executable java file, in case you would want to try it out.
THE CODE

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