Rename year directories to allow natural ordering
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import javax.swing.JOptionPane;
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/**
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* Array implementation of Queue ADT
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*/
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public class ArrayQueue implements Queue
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{
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protected Object Q[]; // array used to implement the queue
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protected int rear = -1; // index for the rear of the queue
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protected int capacity; // The actual capacity of the queue array
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public static final int CAPACITY = 1000; // default array capacity
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public ArrayQueue() {
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// default constructor: creates queue with default capacity
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this(CAPACITY);
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}
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public ArrayQueue(int cap) {
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// this constructor allows you to specify capacity
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capacity = (cap > 0) ? cap : CAPACITY;
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Q = new Object[capacity];
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}
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public void enqueue(Object n)
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{
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if (isFull()) {
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JOptionPane.showMessageDialog(null, "Cannot enqueue object; queue is full.");
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return;
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}
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rear++;
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Q[rear] = n;
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}
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public Object dequeue()
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{
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// Can't do anything if it's empty
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if (isEmpty())
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return null;
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Object toReturn = Q[0];
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// shuffle all other objects towards 0
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int i = 1;
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while (i <= rear) {
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Q[i-1] = Q[i];
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i++;
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}
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rear--;
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return toReturn;
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}
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public boolean isEmpty() {
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return (rear < 0);
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}
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public boolean isFull() {
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return (rear == capacity-1);
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}
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public Object front()
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{
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if (isEmpty())
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return null;
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return Q[0];
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}
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}
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@ -0,0 +1,63 @@
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/** Array implementation of Stack ADT */
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import javax.swing.JOptionPane;
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public class ArrayStack implements Stack
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{
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protected int capacity; // The actual capacity of the stack array
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protected static final int CAPACITY = 1000; // default array capacity
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protected Object S[]; // array used to implement the stack
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protected int top = -1; // index for the top of the stack
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public ArrayStack() {
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// default constructor: creates stack with default capacity
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this(CAPACITY);
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}
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public ArrayStack(int cap) {
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// this constructor allows you to specify capacity of stack
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capacity = (cap > 0) ? cap : CAPACITY;
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S = new Object[capacity];
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}
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public void push(Object element) {
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if (isFull()) {
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JOptionPane.showMessageDialog(null, "ERROR: Stack is full.");
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return;
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}
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top++;
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S[top] = element;
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}
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public Object pop() {
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Object element;
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if (isEmpty()) {
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JOptionPane.showMessageDialog(null, "ERROR: Stack is empty.");
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return null;
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}
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element = S[top];
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S[top] = null;
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top--;
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return element;
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}
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public Object top() {
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if (isEmpty()) {
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JOptionPane.showMessageDialog(null, "ERROR: Stack is empty.");
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return null;
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}
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return S[top];
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}
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public boolean isEmpty() {
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return (top < 0);
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}
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public boolean isFull() {
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return (top == capacity-1);
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}
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public int size() {
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return (top + 1);
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}
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}
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@ -0,0 +1,21 @@
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// class to implement Method 2
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public class IVersusNMinusI implements PalindromeChecker {
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// method 2 - comparing each element at index i to the element at n - i where n is the last index
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@Override
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public boolean checkPalindrome(String str) {
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// looping through the first half of the String
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NewPalindrome.operations[1]++;
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for (int i = 0; i < Math.floor(str.length() / 2); i++) {
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NewPalindrome.operations[1] += 1 + 1 + 1 + 1; // 1 for the getting str.length(), 1 for Math,floor, 1 for checking condition, 1 for incrementing
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// returning false if the digits don't match
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NewPalindrome.operations[1] += 1 + 1 + 1 + 1; // 1 for str.charAt(i), 1 for ((str.lenght() -1) - 1), 1 for the other str.charAt(), 1 for checking the condition
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if (str.charAt(i) != str.charAt((str.length()-1) - i)) {
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return false;
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}
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}
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// returning true as default
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return true;
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}
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}
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@ -0,0 +1,93 @@
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import java.io.*;
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public class NewPalindrome {
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public static long[] operations = new long[4]; // array to contain the global operations count for each method
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public static int[] decCount = new int[4]; // array to hold the count of decimal palindromes found using each method
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public static int[] binCount = new int[4]; // array to hold the count of binary palindromes found using each method
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public static int[] bothCount = new int[4]; // array to hold the count of numbers that are palindromes in both decimal & binary found using each method
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public static long[] startTime = new long[4]; // array to hold the start time of each method's test loop
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public static long[] totalTime = new long[4]; // array to hold the total time of each method's test loop
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// array to hold all the String versions of the numbers so that they don't have to be generated for each method
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// 0th column will be decimal, 1st column will be binary
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public static String[][] strings = new String[1_000_001][2];
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// array of StringBuilder objects used to hold the csv data (size of problem, number of operations) for each method
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public static StringBuilder[] data = new StringBuilder[4];
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// array of the four classes that will be tested
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public static PalindromeChecker[] palindromeCheckers = {new ReverseVSOriginal(), new IVersusNMinusI(), new StackVSQueue(), new RecursiveReverse()};
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public static void main(String args[]) {
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// initialising the data array to StringBuilder objects
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for (int i = 0; i < 4; i++) {
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data[i] = new StringBuilder("operations,size\n");
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}
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// filling up the strings array
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for (int i = 0; i <= 1_000_000; i++) {
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strings[i][0] = Integer.toString(i, 10); // converting i to a String base 10
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strings[i][1] = binary2string(strings[i][0]); // converting the decimal String to a binary String
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}
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// looping through each PalindromeChecker object in the palindromeCheckers array
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for (int j = 0; j < 4; j++) {
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// getting start time
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startTime[j] = System.currentTimeMillis(); operations[j]++;
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// looping through the numbers 0 to 1,000,000 and checking if their binary & decimal representations are palindromic
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operations[j]++;
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for (int i = 0; i <= 1_000_000; i++) {
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// incrementing the operations count by 2, 1 for the loop condition check and 1 for incrementing i
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operations[j] += 2;
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// converting the number to a decimal or binary String and checking if is a palindrome
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boolean isDecPalindrome = palindromeCheckers[j].checkPalindrome(strings[i][0]); operations[j]++;
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boolean isBinPalindrome = palindromeCheckers[j].checkPalindrome(strings[i][1]); operations[j]++;
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// incrementing the appropriate counter if the number is a palindrome in that base
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decCount[j] = isDecPalindrome ? decCount[j] + 1 : decCount[j]; operations[j] += 1 + 1; // incremnting by 2, 1 for assignment, 1 for condition check
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binCount[j] = isBinPalindrome ? binCount[j] + 1 : binCount[j]; operations[j] += 1 + 1;
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bothCount[j] = isDecPalindrome && isBinPalindrome ? bothCount[j] + 1 : bothCount[j]; operations[j] += 1 + 1 +1; // 2 condition checks and one assignment, so incrementing by 3
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// appending to the data StringBuilder at intervals of 50,000
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if (i % 50_000 == 0) {
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data[j].append(operations[j] + "," + i + "\n");
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}
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}
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// calculating total time taken for method 1 and printing out the results
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totalTime[j] = System.currentTimeMillis() - startTime[j]; operations[j] += 1 + 1; // incrementing by 2, 1 for getting current time and subtracting start time, 1 for assignment
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System.out.println("Number of decimal palindromes found using Method " + j + ": " + decCount[j]);
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System.out.println("Number of binary palindromes found using Method " + j + ": " + binCount[j]);
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System.out.println("Number of palindromes in both decimal & binary found using Method " + j + ": " + bothCount[j]);
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System.out.println("Number of primitive operations taken in Method " + j + ": " + operations[j]);
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System.out.println("Time taken for Method " + j + ": " + totalTime[j] + " milliseconds");
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System.out.println();
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// outputting the data to separate csv files
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try {
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String filename = "method" + j + ".csv";
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File csv = new File(filename);
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// creating file if it doesn't already exist
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csv.createNewFile();
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FileWriter writer = new FileWriter(filename);
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writer.write(data[j].toString());
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writer.close();
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} catch (IOException e) {
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System.out.println("IO Error occurred");
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e.printStackTrace();
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System.exit(1);
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}
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}
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}
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// utility method to convert a decimal String to its equivalent binary String
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public static String binary2string(String decimalStr) {
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return Integer.toString(Integer.parseInt(decimalStr), 2); // parsing the String to an int and then parsing that int to a binary String
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}
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}
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@ -0,0 +1,320 @@
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import java.io.*;
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public class Palindrome {
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// global operations count for each method
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public static long operations1 = 0;
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public static long operations2 = 0;
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public static long operations3 = 0;
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public static long operations4 = 0;
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// String objects used to hold the csv data (size of problem, number of operations) for each method
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public static StringBuilder data1 = new StringBuilder("operations,size\n");
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public static StringBuilder data2 = new StringBuilder("operations,size\n");
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public static StringBuilder data3 = new StringBuilder("operations,size\n");
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public static StringBuilder data4 = new StringBuilder("operations,size\n");
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public static void main(String args[]) {
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// generating all the String versions of the numbers so that they don't have to be generated for each method
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// 0th column will be decimal, 1st column will be binary
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String[][] strings = new String[1_000_001][2];
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for (int i = 0; i <= 1_000_000; i++) {
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strings[i][0] = Integer.toString(i, 10); // converting i to a String base 10
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strings[i][1] = binary2string(strings[i][0]); // converting the decimal String to a binary String
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}
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// variables for method 1 - reversed order String vs original String
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int decCount1 = 0; operations1++; // count of decimal palindromes for use by method 1
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int binCount1 = 0; operations1++; // count of binary palindromes for use by method 1
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int bothCount1 = 0; operations1++; // count of numbers that are palindromic in both binary & decimal for use by method 1
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long startTime1 = System.currentTimeMillis(); operations1 += 1 + 1;
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// testing method 1 - reversed order String vs original String
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// incrementing the operations counter for the initialisation of i below
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operations1++;
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for (int i = 0; i <= 1_000_000; i++) {
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// incrementing the operations count by 2, 1 for the loop condition check and 1 for incrementing i
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operations1 += 2;
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// converting the number to a decimal or binary String and checking if is a palindrome
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boolean isDecPalindrome = reverseVSoriginal(strings[i][0]); operations1++;
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boolean isBinPalindrome = reverseVSoriginal(strings[i][1]); operations1++;
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// incrementing the appropriate counter if the number is a palindrome in that base
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decCount1 = isDecPalindrome ? decCount1 + 1 : decCount1; operations1 += 1 + 1; // incremnting by 2, 1 for assignment, 1 for condition check
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binCount1 = isBinPalindrome ? binCount1 + 1 : binCount1; operations1 += 1 + 1;
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bothCount1 = isDecPalindrome && isBinPalindrome ? bothCount1 + 1 : bothCount1; operations1 += 1 + 1 +1; // 2 condition checks and one assignment, so incrementing by 3
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// appending to the data StringBuilder at intervals of 50,000
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if (i % 50_000 == 0) {
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data1.append(operations1 + "," + i + "\n");
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}
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}
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// calculating total time taken for method 1 and printing out the results
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long totalTime1 = System.currentTimeMillis() - startTime1; operations1 += 1 + 1; // incrementing by 2, 1 for getting current time and subtracting start time, 1 for assignment
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System.out.println("Number of decimal palindromes found using Method 1: " + decCount1);
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System.out.println("Number of binary palindromes found using Method 1: " + binCount1);
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System.out.println("Number of palindromes in both decimal & binary found using Method 1: " + bothCount1);
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System.out.println("Number of primitive operations taken in Method 1: " + operations1);
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System.out.println("Time taken for Method 1: " + totalTime1 + " milliseconds");
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// variables for method 2 - comparing each element at index i to the element at n - i where n is the last index
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int decCount2 = 0; operations2++; // count of decimal palindromes for use by method 2
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int binCount2 = 0; operations2++; // count of binary palindromes for use by method 2
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int bothCount2 = 0; operations2++; // count of numbers that are palindromic in both binary & decimal for use by method 2
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long startTime2 = System.currentTimeMillis(); operations2 += 1 + 1;
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// testingmethod 2 - comparing each element at index i to the element at n - i where n is the last index
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operations2++;
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for (int i = 0; i <= 1_000_000; i++) {
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operations2 += 1 + 1;
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// converting the number to a decimal or binary String and checking if is a palindrome
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boolean isDecPalindrome = iVSiMinusn(strings[i][0]); operations2++;
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boolean isBinPalindrome = iVSiMinusn(strings[i][1]); operations2++;
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// incrementing the appropriate counter if the number is a palindrome in that base
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decCount2 = isDecPalindrome ? decCount2 + 1 : decCount2; operations2 += 1 + 1;
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binCount2 = isBinPalindrome ? binCount2 + 1 : binCount2; operations2 += 1 + 1;
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bothCount2 = isDecPalindrome && isBinPalindrome ? bothCount2 + 1 : bothCount2; operations2 += 1 + 1 +1;
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// appending to the data StringBuilder at intervals of 50,000
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if (i % 50_000 == 0) {
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data2.append(operations2 + "," + i + "\n");
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}
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}
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// calculating total time taken for method 2 and printing out the results
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long totalTime2 = System.currentTimeMillis() - startTime2; operations2 += 1 + 1;
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System.out.println();
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System.out.println("Number of decimal palindromes found using Method 2: " + decCount2);
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System.out.println("Number of binary palindromes found using Method 2: " + binCount2);
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System.out.println("Number of palindromes in both decimal & binary found using Method 2: " + bothCount2);
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System.out.println("Number of primitive operations taken in Method 2: " + operations2);
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System.out.println("Time taken for Method 2: " + totalTime2 + " milliseconds");
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// variables for method 3 - comparing each element at index i to the element at n - i where n is the last index
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int decCount3 = 0; operations3++; // count of decimal palindromes for use by method 3
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int binCount3 = 0; operations3++; // count of binary palindromes for use by method 3
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int bothCount3 = 0; operations3++; // count of numbers that are palindromic in both binary & decimal for use by method 3
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long startTime3 = System.currentTimeMillis(); operations3 += 1 + 1;
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// testingmethod 3 - comparing each element at index i to the element at n - i where n is the last index
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operations3++;
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for (int i = 0; i <= 1_000_000; i++) {
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operations3 += 1 + 1;
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// converting the number to a decimal or binary String and checking if is a palindrome
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boolean isDecPalindrome = stackVsqueue(strings[i][0]); operations3++;
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boolean isBinPalindrome = stackVsqueue(strings[i][1]); operations3++;
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// incrementing the appropriate counter if the number is a palindrome in that base
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decCount3 = isDecPalindrome ? decCount3 + 1 : decCount3; operations3 += 1 + 1;
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binCount3 = isBinPalindrome ? binCount3 + 1 : binCount3; operations3 += 1 + 1;
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bothCount3 = isDecPalindrome && isBinPalindrome ? bothCount3 + 1 : bothCount3; operations3 += 1 + 1 + 1;
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// appending to the data StringBuilder at intervals of 50,000
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if (i % 50_000 == 0) {
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data3.append(operations3 + "," + i + "\n");
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}
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}
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// calculating total time taken for method 3 and printing out the results
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long totalTime3 = System.currentTimeMillis() - startTime3; operations3 += 1 + 1;
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System.out.println();
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System.out.println("Number of decimal palindromes found using Method 3: " + decCount3);
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System.out.println("Number of binary palindromes found using Method 3: " + binCount3);
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System.out.println("Number of palindromes in both decimal & binary found using Method 3: " + bothCount3);
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System.out.println("Number of primitive operations taken in Method 3: " + operations3);
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System.out.println("Time taken for Method 3: " + totalTime3 + " milliseconds");
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// variables for method 4 - comparing each element at index i to the element at n - i where n is the last index
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int decCount4 = 0; operations4++; // count of decimal palindromes for use by method 4
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int binCount4 = 0; operations4++; // count of binary palindromes for use by method 4
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int bothCount4 = 0; operations4++; // count of numbers that are palindromic in both binary & decimal for use by method 4
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long startTime4 = System.currentTimeMillis(); operations4 += 1 + 1;
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// testingmethod 4 - comparing each element at index i to the element at n - i where n is the last index
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operations4++;
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for (int i = 0; i <= 1_000_000; i++) {
|
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operations4 += 2;
|
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|
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// converting the number to a decimal or binary String and checking if is a palindrome
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boolean isDecPalindrome = recursiveReverseVSoriginal(strings[i][0]); operations4++;
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boolean isBinPalindrome = recursiveReverseVSoriginal(strings[i][1]); operations4++;
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// incrementing the appropriate counter if the number is a palindrome in that base
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decCount4 = isDecPalindrome ? decCount4 + 1 : decCount4; operations4 += 1 + 1;
|
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binCount4 = isBinPalindrome ? binCount4 + 1 : binCount4; operations4 += 1 + 1;
|
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bothCount4 = isDecPalindrome && isBinPalindrome ? bothCount4 + 1 : bothCount4; operations4 += 1 + 1 + 1;
|
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|
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// appending to the data StringBuilder at intervals of 50,000
|
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if (i % 50_000 == 0) {
|
||||
data4.append(operations4 + "," + i + "\n");
|
||||
}
|
||||
}
|
||||
|
||||
// calculating total time taken for method 4 and printing out the results
|
||||
long totalTime4 = System.currentTimeMillis() - startTime4; operations4 += 1 + 1;
|
||||
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System.out.println();
|
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System.out.println("Number of decimal palindromes found using Method 4: " + decCount4);
|
||||
System.out.println("Number of binary palindromes found using Method 4: " + binCount4);
|
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System.out.println("Number of palindromes in both decimal & binary found using Method 4: " + bothCount4);
|
||||
System.out.println("Number of primitive operations taken in Method 4: " + operations4);
|
||||
System.out.println("Time taken for Method 4: " + totalTime4 + " milliseconds");
|
||||
|
||||
// outputting the data to separate csv files
|
||||
try {
|
||||
File csv1 = new File("method1.csv");
|
||||
File csv2 = new File("method2.csv");
|
||||
File csv3 = new File("method3.csv");
|
||||
File csv4 = new File("method4.csv");
|
||||
|
||||
// creating files if they don't already exist
|
||||
csv1.createNewFile();
|
||||
csv2.createNewFile();
|
||||
csv3.createNewFile();
|
||||
csv4.createNewFile();
|
||||
|
||||
FileWriter writer1 = new FileWriter("method1.csv");
|
||||
writer1.write(data1.toString());
|
||||
writer1.close();
|
||||
|
||||
FileWriter writer2 = new FileWriter("method2.csv");
|
||||
writer2.write(data2.toString());
|
||||
writer2.close();
|
||||
|
||||
FileWriter writer3 = new FileWriter("method3.csv");
|
||||
writer3.write(data3.toString());
|
||||
writer3.close();
|
||||
|
||||
FileWriter writer4 = new FileWriter("method4.csv");
|
||||
writer4.write(data4.toString());
|
||||
writer4.close();
|
||||
} catch (IOException e) {
|
||||
System.out.println("IO Error occurred");
|
||||
e.printStackTrace();
|
||||
System.exit(1);
|
||||
}
|
||||
}
|
||||
|
||||
// method 1 - reversed order String vs original String
|
||||
public static boolean reverseVSoriginal(String str) {
|
||||
String reversedStr = ""; operations1++;
|
||||
|
||||
// looping through each character in the String, backwards
|
||||
// incrementing operations counter by 2, 1 for initialisating i, 1 for getting str.length()
|
||||
operations1 += 1 + 1;
|
||||
for (int i = str.length(); i > 0; i--) {
|
||||
operations1 += 1 + 1; // for loop condition check & incrementing i
|
||||
|
||||
reversedStr += str.charAt(i-1); operations1 += 1 + 1;
|
||||
}
|
||||
|
||||
// returning true if the Strings are equal, false if not
|
||||
operations1 += str.length(); // the equals method must loop through each character of the String to check that they are equal so it is O(n)
|
||||
return str.equals(reversedStr);
|
||||
}
|
||||
|
||||
// method 2 - comparing each element at index i to the element at n - i where n is the last index
|
||||
public static boolean iVSiMinusn(String str) {
|
||||
// looping through the first half of the String
|
||||
operations2++;
|
||||
for (int i = 0; i < Math.floor(str.length() / 2); i++) {
|
||||
operations2 += 1 + 1 + 1 + 1; // 1 for the getting str.length(), 1 for Math,floor, 1 for checking condition, 1 for incrementing
|
||||
|
||||
// returning false if the digits don't match
|
||||
operations2 += 1 + 1 + 1 + 1; // 1 for str.charAt(i), 1 for ((str.lenght() -1) - 1), 1 for the other str.charAt(), 1 for checking the condition
|
||||
if (str.charAt(i) != str.charAt((str.length()-1) - i)) {
|
||||
return false;
|
||||
}
|
||||
}
|
||||
|
||||
// returning true as default
|
||||
return true;
|
||||
}
|
||||
|
||||
// method 3 - using a stack and a queue to do, essentially, what method 2 does
|
||||
public static boolean stackVsqueue(String str) {
|
||||
ArrayStack stack = new ArrayStack(); operations3 += 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1;
|
||||
ArrayQueue queue = new ArrayQueue(); operations3 += 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1;
|
||||
|
||||
// looping through each character in the String and adding the character to the stack & queue
|
||||
operations3++;
|
||||
for (int i = 0; i < str.length(); i++) {
|
||||
operations3 += 1 + 1 + 1;
|
||||
|
||||
stack.push(str.charAt(i)); operations3 += 1 + 1 + 1 + 1;
|
||||
queue.enqueue(str.charAt(i)); operations3 += 1 + 1 + 1 + 1;
|
||||
}
|
||||
|
||||
// looping through each character on the stack & queue and comparing them, returning false if they're different
|
||||
operations3++;
|
||||
for (int i = 0; i < str.length(); i++) {
|
||||
operations3 += 1 + 1 + 1;
|
||||
|
||||
operations3 += 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1;
|
||||
if (!stack.pop().equals(queue.front())) {
|
||||
return false;
|
||||
}
|
||||
|
||||
// the complexity of ArrayQueue.dequeue() is 3n+2, where n is the number of items in the queue when dequeue() is called.
|
||||
// we need to determine the number of items in the queue so that we can determine the number of primitive operations performed when queue.dequeue() is called.
|
||||
// to do this, we'll loop through the queue, dequeuing each object and enqueueing it in another ArrayQueue. once complete, we'll reassign the variable queue to point to the new ArrayQueue containing all the objects
|
||||
ArrayQueue newQueue = new ArrayQueue(); // not counting the operations for this as it's not part of the algorithm, it's part of the operations counting
|
||||
int n = 0; // n is the number of items in the ArrayQueue when dequeue() is called
|
||||
|
||||
while (!queue.isEmpty()) {
|
||||
newQueue.enqueue(queue.dequeue());
|
||||
n++;
|
||||
}
|
||||
|
||||
queue = newQueue; // setting queue to point to the newQueue, which is just the state that queue would have been in if we didn't do this to calculate the primitive operations
|
||||
newQueue = null; // don't need the newQueue object reference anymore
|
||||
|
||||
operations3 += 3*n + 2; // complexity of dequeue is 3n+2
|
||||
queue.dequeue();
|
||||
}
|
||||
|
||||
return true;
|
||||
}
|
||||
|
||||
// method 4 - comparing the String reversed using recursion to the original String (essentially method 1 but with recursion)
|
||||
public static boolean recursiveReverseVSoriginal(String str) {
|
||||
// returning true if the original String is equal to the reversed String, false if not
|
||||
operations4++;
|
||||
return str.equals(reverse(str));
|
||||
}
|
||||
|
||||
// method to reverse the characters in a String using recursion
|
||||
public static String reverse(String str) {
|
||||
// base case - returning an empty String if there is no character left in the String
|
||||
operations4++;
|
||||
if (str.length() == 0) {
|
||||
return "";
|
||||
}
|
||||
else {
|
||||
char firstChar = str.charAt(0); operations4 += 1 + 1;
|
||||
String remainder = str.substring(1); operations4 += 1 + 1; // selecting the rest of the String, excluding the 0th character
|
||||
|
||||
// recursing with what's left of the String
|
||||
String reversedRemainder = reverse(remainder); operations4++;
|
||||
|
||||
// returning the reversed rest of String with the first character of the String appended
|
||||
return reversedRemainder + firstChar;
|
||||
}
|
||||
}
|
||||
|
||||
// utility method to convert a decimal String to its equivalent binary String
|
||||
public static String binary2string(String decimalStr) {
|
||||
return Integer.toString(Integer.parseInt(decimalStr), 2); // parsing the String to an int and then parsing that int to a binary String
|
||||
}
|
||||
}
|
||||
@ -0,0 +1,3 @@
|
||||
public interface PalindromeChecker {
|
||||
public boolean checkPalindrome(String str);
|
||||
}
|
||||
@ -0,0 +1,14 @@
|
||||
/*
|
||||
* Abstract Queue interface
|
||||
*/
|
||||
|
||||
public interface Queue {
|
||||
// most important methods
|
||||
public void enqueue(Object n); // add an object at the rear of the queue
|
||||
public Object dequeue(); // remove an object from the front of the queue
|
||||
|
||||
// others
|
||||
public boolean isEmpty(); // true if queue is empty
|
||||
public boolean isFull(); // true if queue is full (if it has limited storage)
|
||||
public Object front(); // examine front object on queue without removing it
|
||||
}
|
||||
@ -0,0 +1,29 @@
|
||||
// class to implement method 4
|
||||
public class RecursiveReverse implements PalindromeChecker {
|
||||
// comparing the String reversed using recursion to the original String (essentially method 1 but with recursion)
|
||||
@Override
|
||||
public boolean checkPalindrome(String str) {
|
||||
// returning true if the original String is equal to the reversed String, false if not
|
||||
NewPalindrome.operations[3]++;
|
||||
return str.equals(reverse(str));
|
||||
}
|
||||
|
||||
// method to reverse the characters in a String using recursion
|
||||
public static String reverse(String str) {
|
||||
// base case - returning an empty String if there is no character left in the String
|
||||
NewPalindrome.operations[3]++;
|
||||
if (str.length() == 0) {
|
||||
return "";
|
||||
}
|
||||
else {
|
||||
char firstChar = str.charAt(0); NewPalindrome.operations[3] += 1 + 1;
|
||||
String remainder = str.substring(1); NewPalindrome.operations[3] += 1 + 1; // selecting the rest of the String, excluding the 0th character
|
||||
|
||||
// recursing with what's left of the String
|
||||
String reversedRemainder = reverse(remainder); NewPalindrome.operations[3]++;
|
||||
|
||||
// returning the reversed rest of String with the first character of the String appended
|
||||
return reversedRemainder + firstChar;
|
||||
}
|
||||
}
|
||||
}
|
||||
@ -0,0 +1,21 @@
|
||||
// class to implement Method 3
|
||||
public class ReverseVSOriginal implements PalindromeChecker {
|
||||
// method 1 - reversed order String vs original String
|
||||
@Override
|
||||
public boolean checkPalindrome(String str) {
|
||||
String reversedStr = ""; NewPalindrome.operations[0]++;
|
||||
|
||||
// looping through each character in the String, backwards
|
||||
// incrementing operations counter by 2, 1 for initialisating i, 1 for getting str.length()
|
||||
NewPalindrome.operations[0] += 1 + 1;
|
||||
for (int i = str.length(); i > 0; i--) {
|
||||
NewPalindrome.operations[0] += 1 + 1; // for loop condition check & incrementing i
|
||||
|
||||
reversedStr += str.charAt(i-1); NewPalindrome.operations[0] += 1 + 1;
|
||||
}
|
||||
|
||||
// returning true if the Strings are equal, false if not
|
||||
NewPalindrome.operations[0] += str.length(); // the equals method must loop through each character of the String to check that they are equal so it is O(n)
|
||||
return str.equals(reversedStr);
|
||||
}
|
||||
}
|
||||
@ -0,0 +1,13 @@
|
||||
/** Abstract Stack interface */
|
||||
|
||||
public interface Stack
|
||||
{
|
||||
// most important methods
|
||||
public void push(Object n); // push an object onto top of the stack
|
||||
public Object pop(); // pop an object from top of the stack
|
||||
|
||||
// others
|
||||
public Object top(); // examine top object on stack without removing it
|
||||
public boolean isEmpty(); // true if stack is empty
|
||||
public boolean isFull(); // true if stack is full (if it has limited storage)
|
||||
}
|
||||
@ -0,0 +1,48 @@
|
||||
// class to implement method 3
|
||||
public class StackVSQueue implements PalindromeChecker {
|
||||
// method 3 - using a stack and a queue to do, essentially, what method 2 does (compare the first index to the last index, etc.)
|
||||
@Override
|
||||
public boolean checkPalindrome(String str) {
|
||||
ArrayStack stack = new ArrayStack(); NewPalindrome.operations[2] += 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1;
|
||||
ArrayQueue queue = new ArrayQueue(); NewPalindrome.operations[2] += 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1;
|
||||
|
||||
// looping through each character in the String and adding the character to the stack & queue
|
||||
NewPalindrome.operations[2]++;
|
||||
for (int i = 0; i < str.length(); i++) {
|
||||
NewPalindrome.operations[2] += 1 + 1 + 1;
|
||||
|
||||
stack.push(str.charAt(i)); NewPalindrome.operations[2] += 1 + 1 + 1 + 1;
|
||||
queue.enqueue(str.charAt(i)); NewPalindrome.operations[2] += 1 + 1 + 1 + 1;
|
||||
}
|
||||
|
||||
// looping through each character on the stack & queue and comparing them, returning false if they're different
|
||||
NewPalindrome.operations[2]++;
|
||||
for (int i = 0; i < str.length(); i++) {
|
||||
NewPalindrome.operations[2] += 1 + 1 + 1;
|
||||
|
||||
NewPalindrome.operations[2] += 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1;
|
||||
if (!stack.pop().equals(queue.front())) {
|
||||
return false;
|
||||
}
|
||||
|
||||
// the complexity of ArrayQueue.dequeue() is 3n+2, where n is the number of items in the queue when dequeue() is called.
|
||||
// we need to determine the number of items in the queue so that we can determine the number of primitive operations performed when queue.dequeue() is called.
|
||||
// to do this, we'll loop through the queue, dequeuing each object and enqueueing it in another ArrayQueue. once complete, we'll reassign the variable queue to point to the new ArrayQueue containing all the objects
|
||||
ArrayQueue newQueue = new ArrayQueue(); // not counting the operations for this as it's not part of the algorithm, it's part of the operations counting
|
||||
int n = 0; // n is the number of items in the ArrayQueue when dequeue() is called
|
||||
|
||||
while (!queue.isEmpty()) {
|
||||
newQueue.enqueue(queue.dequeue());
|
||||
n++;
|
||||
}
|
||||
|
||||
queue = newQueue; // setting queue to point to the newQueue, which is just the state that queue would have been in if we didn't do this to calculate the primitive operations
|
||||
newQueue = null; // don't need the newQueue object reference anymore
|
||||
|
||||
NewPalindrome.operations[2] += 3*n + 2; // complexity of dequeue is 3n+2
|
||||
queue.dequeue();
|
||||
}
|
||||
|
||||
return true;
|
||||
}
|
||||
}
|
||||
@ -0,0 +1,14 @@
|
||||
public class Test {
|
||||
public static void main(String[] args) {
|
||||
String str = "1234567890";
|
||||
ArrayQueue queue = new ArrayQueue();
|
||||
|
||||
for (int i = 0; i < str.length(); i++) {
|
||||
queue.enqueue(str.charAt(i));
|
||||
}
|
||||
|
||||
for (int i = 0; i < str.length(); i++) {
|
||||
queue.dequeue();
|
||||
}
|
||||
}
|
||||
}
|
||||
@ -0,0 +1,22 @@
|
||||
operations,size
|
||||
29,0
|
||||
5716904,50000
|
||||
11989234,100000
|
||||
18683879,150000
|
||||
25533879,200000
|
||||
32383879,250000
|
||||
39423164,300000
|
||||
46523164,350000
|
||||
53623164,400000
|
||||
60723164,450000
|
||||
67823164,500000
|
||||
75051729,550000
|
||||
82401729,600000
|
||||
89751729,650000
|
||||
97101729,700000
|
||||
104451729,750000
|
||||
111801729,800000
|
||||
119151729,850000
|
||||
126501729,900000
|
||||
133851729,950000
|
||||
141201734,1000000
|
||||
|
@ -0,0 +1,22 @@
|
||||
operations,size
|
||||
15,0
|
||||
1881959,50000
|
||||
3768823,100000
|
||||
5660295,150000
|
||||
7552719,200000
|
||||
9445167,250000
|
||||
11337607,300000
|
||||
13230031,350000
|
||||
15122471,400000
|
||||
17014895,450000
|
||||
18907343,500000
|
||||
20800207,550000
|
||||
22693407,600000
|
||||
24586615,650000
|
||||
26479855,700000
|
||||
28373063,750000
|
||||
30266287,800000
|
||||
32159487,850000
|
||||
34052711,900000
|
||||
35945943,950000
|
||||
37839175,1000000
|
||||
|
@ -0,0 +1,22 @@
|
||||
operations,size
|
||||
105,0
|
||||
17250278,50000
|
||||
36003532,100000
|
||||
55780375,150000
|
||||
75979194,200000
|
||||
96178166,250000
|
||||
116909827,300000
|
||||
137812321,350000
|
||||
158715267,400000
|
||||
179617761,450000
|
||||
200521097,500000
|
||||
221777541,550000
|
||||
243367974,600000
|
||||
264958463,650000
|
||||
286548836,700000
|
||||
308139668,750000
|
||||
329729920,800000
|
||||
351320353,850000
|
||||
372910963,900000
|
||||
394501277,950000
|
||||
416092270,1000000
|
||||
|
@ -0,0 +1,22 @@
|
||||
operations,size
|
||||
29,0
|
||||
6590279,50000
|
||||
13847075,100000
|
||||
21610649,150000
|
||||
29560649,200000
|
||||
37510649,250000
|
||||
45687791,300000
|
||||
53937791,350000
|
||||
62187791,400000
|
||||
70437791,450000
|
||||
78687791,500000
|
||||
87092069,550000
|
||||
95642069,600000
|
||||
104192069,650000
|
||||
112742069,700000
|
||||
121292069,750000
|
||||
129842069,800000
|
||||
138392069,850000
|
||||
146942069,900000
|
||||
155492069,950000
|
||||
164042075,1000000
|
||||
|
Binary file not shown.
@ -0,0 +1,214 @@
|
||||
\documentclass[a4paper]{article}
|
||||
|
||||
\usepackage{listings}
|
||||
\usepackage{xcolor}
|
||||
\definecolor{codegreen}{rgb}{0,0.6,0}
|
||||
\definecolor{codegray}{rgb}{0.5,0.5,0.5}
|
||||
\definecolor{codeorange}{rgb}{1,0.49,0}
|
||||
\definecolor{backcolour}{rgb}{0.95,0.95,0.96}
|
||||
|
||||
\usepackage{pgfplots}
|
||||
\pgfplotsset{width=\textwidth,compat=1.9}
|
||||
|
||||
\lstdefinestyle{mystyle}{
|
||||
backgroundcolor=\color{backcolour},
|
||||
commentstyle=\color{codegray},
|
||||
keywordstyle=\color{codeorange},
|
||||
numberstyle=\tiny\color{codegray},
|
||||
stringstyle=\color{codegreen},
|
||||
basicstyle=\ttfamily\footnotesize,
|
||||
breakatwhitespace=false,
|
||||
breaklines=true,
|
||||
captionpos=b,
|
||||
keepspaces=true,
|
||||
numbers=left,
|
||||
numbersep=5pt,
|
||||
showspaces=false,
|
||||
showstringspaces=false,
|
||||
showtabs=false,
|
||||
tabsize=2,
|
||||
xleftmargin=10pt,
|
||||
}
|
||||
|
||||
\lstset{style=mystyle}
|
||||
|
||||
\input{head}
|
||||
\begin{document}
|
||||
|
||||
%-------------------------------
|
||||
% TITLE SECTION
|
||||
%-------------------------------
|
||||
|
||||
\fancyhead[C]{}
|
||||
\hrule \medskip % Upper rule
|
||||
\begin{minipage}{0.295\textwidth}
|
||||
\begin{raggedright}
|
||||
\footnotesize
|
||||
Andrew Hayes \hfill\\
|
||||
21321503 \hfill\\
|
||||
a.hayes18@nuigalway.ie
|
||||
\end{raggedright}
|
||||
\end{minipage}
|
||||
\begin{minipage}{0.4\textwidth}
|
||||
\begin{centering}
|
||||
\large
|
||||
CT2109 Assignment 3\\
|
||||
\normalsize
|
||||
\end{centering}
|
||||
\end{minipage}
|
||||
\begin{minipage}{0.295\textwidth}
|
||||
\begin{raggedleft}
|
||||
\footnotesize
|
||||
\today \hfill\\
|
||||
\end{raggedleft}
|
||||
\end{minipage}
|
||||
|
||||
\medskip\hrule
|
||||
\medskip
|
||||
\begin{centering}
|
||||
Double-Base Palindromes \& Complexity Analysis\\
|
||||
\end{centering}
|
||||
\medskip\hrule
|
||||
\bigskip
|
||||
|
||||
\section{Problem Statement}
|
||||
The overall purpose of this assignment is to perform some basic complexity analysis on four distinct algorithms, each of which aim to achieve the same result: determining whether a String is palindromic.
|
||||
More specifically, each algorithm will be fed the binary \& decimal representations of each integer between $0_{10}$ and $1,000,000_{10}$, in String form.
|
||||
Each algorithm will then determine whether the String is palindromic, returning either \verb|true| or \verb|false|.
|
||||
Each method will have three counter variables to record the number of palindromes found: one for the numbers whose decimal form is palindromic, one for the numbers whose binary form is palindromic, and one
|
||||
for the numbers which are palindromic in both decimal \& binary form.
|
||||
\\\\
|
||||
The number of primitive operations will be recorded using a separate global variable for each method.
|
||||
Additionally, the real time taken by each method (in milliseconds) will be recorded.
|
||||
These measurements will form the basis of the complexity analysis performed.
|
||||
The number of primitive operations will be recorded for each method and plotted against the size of the problem (the number of values being checked for palindromicity, $n$).
|
||||
This graph should reflect the ``order'' of the problem, i.e. it should match the big-O representation derived from a time-complexity analysis of the algorithm.
|
||||
For example, an exponential curve on the graph would indicate that the algorithm is of $O(n^2)$ complexity, while a straight line through the origin would indicate that the algorithm is of $O(n)$ complexity.
|
||||
|
||||
\section{Analysis \& Design Notes}
|
||||
The first thing that we'll want to do in this program is declare the variables that we'll use to record the data for each method being tested.
|
||||
We will have seven arrays of size four, with one index for each method being tested.
|
||||
These arrays will be:
|
||||
\begin{itemize}
|
||||
\item \verb|long[] operations| - To count the number of primitive operations for each method.
|
||||
\item \verb|int[] decCount| - To count how many numbers that are palindromic in decimal form are found using each method.
|
||||
\item \verb|int[] binCount| - To count how many numbers that are palindromic in binary form are found using each method.
|
||||
\item \verb|int[] bothCount| - To count how many numbers that are palindromic in both decimal \& binary form are found using each method.
|
||||
\item \verb|long[] startTime| - To record the start time (in Unix epoch form) of the testing of each of the four methods.
|
||||
\item \verb|long[] totalTime| - To record the total time (in milliseconds) by the testing of each of the four methods.
|
||||
\item \verb|StringBuilder[] data| - This will be be used to create a String of CSV data for each method, which will be output to a \verb|.csv| file at the end of testing.
|
||||
Doing this saves us from having to open, write to, \& close a \verb|.csv| file every time we want to record some data, which would be very inefficient.
|
||||
\end{itemize}
|
||||
|
||||
The first thing that we'll do in the main method is initialise all the indices in the \verb|StringBuilder[] data| array to have the appropriate headings from each column, one column for the number of primitive operations
|
||||
and one for the size of the problem that used that number of primitive operations, i.e. \verb|"operations,size\n"|.
|
||||
\\\\
|
||||
We'll also want to generate a two-dimensional array of all the Strings that we'll be testing for palindromicity, with one dimension for decimal Strings and one dimension for binary Strings, both going up to
|
||||
$1,000,000_{10}$.
|
||||
Generating this once in the beginning will save us from having to re-generate the same Strings four times, which would be very inefficient.
|
||||
\\\\
|
||||
Each method that we will be testing will have its own class that implements the interface \verb|PalindromeChecker|.
|
||||
This interface will contain a single method with the signature \verb|public boolean checkPalindrome(String str)|.
|
||||
The reason for doing this will be to follow OOP principles \& the DRY principle so that we don't have unnecessary repetition of code.
|
||||
This will allow us to have one generic loop through each of the numbers from $0_{10}$ to $1,000,000_{10}$ instead of four separate ones.
|
||||
Each of the methods that we will be testing will be overridden implementations of \verb|checkPalindrome|.
|
||||
We will have four classes that implement \verb|PalindromeChecker|:
|
||||
\begin{itemize}
|
||||
\item \verb|ReverseVSOriginal| - This class will contain ``Method One'' outlined in the assignment specification, which checks if a String is palindromic by comparing the String to a reversed copy of itself, hence the name.
|
||||
\item \verb|IVersusIMinusN| - This will contain ``Method Two'' outlined in the assignment specification, which checks if a String is palindromic by looping through each character in the String using an iterator
|
||||
\verb|i| and comparing the character at index \verb|i| to the character at index \verb|n-i|, where \verb|n| is the last index in the String, i.e., comparing the first character to the last, the second character to the second-last, etc.
|
||||
\item \verb|StackVSQueue| - This will contain ``Method Three'' outlined in the assignment specification, which checks if a String is palindromic using essentially the same technique as ``Method Two''
|
||||
but instead of simply iterating through the String, each character of the String will be put onto both a Stack \& a Queue, and then items will be removed from the Stack \& Queue and compared to each other.
|
||||
The LIFO nature of the Stack and the FIFO nature of the Queue result in this comparison being first character versus last, second character versus second-last, and so on.
|
||||
\item \verb|RecursiveReverse| - This will contain ``Method Four'' outlined in the assignment specification, which checks if a String is palindromic using essentially the same technique as ``Method One''
|
||||
but using recursion to reverse the String instead of iteration.
|
||||
\end{itemize}
|
||||
|
||||
An array called \verb|PalindromeCheckers[]| will be initialised to contain an instance of each of these four classes.
|
||||
This array will be iterated over (using iterator \verb|j|) to test each method, which prevents code repetition.
|
||||
In said loop, firstly, the \verb|startTime[j]| will be recorded using \verb|System.getCurrentTimeMillis();|.
|
||||
Then, a loop will be entered that iterators over each number between $0_{10}$ to $1,000,000_{10}$.
|
||||
Both the decimal \& binary Strings at index \verb|i| of the \verb|strings[]| array will be passed to \verb|palindromeCheckers[j].checkPalindrome()| to be checked for palindromicity.
|
||||
Then, \verb|decCount[j]|, \verb|binCount[j]|, \& \verb|bothCount[j]| will be iterated or will remain unchanged, as appropriate.
|
||||
\\\\
|
||||
The count of primitive operations for each method will be iterated as they are executed.
|
||||
I won't count the \verb|return| statements at the end of methods or accessing a variable or array index as primitive operations, as these (especially the \verb|return| statements) are computationally
|
||||
insignificant, and certainly aren't on the same level as something like creating a new variable in memory.
|
||||
Unfortunately, it's not really possible to say with accuracy what is \& isn't a ``primitive operation'' in a language so high-level \& abstract as Java, but I feel that this is a reasonable approximation.
|
||||
We want to record the count of primitive operations at regular intervals of 50,000 during the process, so we will append the operations count and the current \verb|i| to \verb|data[j]| if \verb|i| is
|
||||
divisible by 50,000.
|
||||
\\\\
|
||||
Once the loop using iterator \verb|i| has ended, the total time taken will be recorded by subtracting the current time from the start time.
|
||||
The number of palindromes found in decimal, binary, and both decimal \& binary will be printed out to the screen, along with the total time taken \& the total number of primitive operations for that method.
|
||||
Finally, the data for that method will be written to a \verb|.csv| file.
|
||||
This will be repeated for all four methods in the \verb|palindromeCheckers| array.
|
||||
|
||||
\section{Code}
|
||||
\lstinputlisting[language=Java, breaklines=true, title={NewPalindrome.java}]{../code/NewPalindrome.java}
|
||||
\lstinputlisting[language=Java, breaklines=true, title={PalindromeChecker.java}]{../code/PalindromeChecker.java}
|
||||
\lstinputlisting[language=Java, breaklines=true, title={ReverseVSOriginal.java}]{../code/ReverseVSOriginal.java}
|
||||
\lstinputlisting[language=Java, breaklines=true, title={IVersusIMinusN.java}]{../code/IVersusNMinusI.java}
|
||||
\lstinputlisting[language=Java, breaklines=true, title={StackVSQueue.java}]{../code/StackVSQueue.java}
|
||||
\lstinputlisting[language=Java, breaklines=true, title={RecursiveReverse.java}]{../code/RecursiveReverse.java}
|
||||
|
||||
\newpage
|
||||
\section{Testing}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[
|
||||
title={Number of Primitive Operations per Method},
|
||||
xlabel={Size of Problem},
|
||||
ylabel={Number of Operations}
|
||||
]
|
||||
\addplot table [x=size, y=operations, col sep=comma] {../code/method0.csv};
|
||||
\addplot table [x=size, y=operations, col sep=comma] {../code/method1.csv};
|
||||
\addplot table [x=size, y=operations, col sep=comma] {../code/method2.csv};
|
||||
\addplot table [x=size, y=operations, col sep=comma] {../code/method3.csv};
|
||||
\legend{Iterative Reverse vs Original,$i$ vs $n-i$,Stack vs Queue,Recursive Reverse vs Original}
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\begin{figure}[h]
|
||||
\centering
|
||||
\includegraphics[width=0.7\textwidth]{images/output.png}
|
||||
\caption{Output of the Main Method}
|
||||
\end{figure}
|
||||
|
||||
The output from the program shows that all the methods agreed on the number of palindromes found in each category, which shows us
|
||||
that they did indeed work as intended.
|
||||
\\\\
|
||||
We can see from the graph that $i$ versus $n-i$ or \verb|IVersusNMinusI| method was the most efficient, as it used the fewest
|
||||
primitive operations out of the four methods (37,839,177) and the complexity grew relatively slowly as the size of the problem increased,
|
||||
demonstrated by the fact that its curve has quite a gentle slope.
|
||||
This is reflected in the testing times as it was by far the fastest method (referred to in the screenshot as ``Method 1'', as
|
||||
indexing was done from 0), taking only 39 milliseconds to complete.
|
||||
This makes sense, as the method consisted of just one loop and some basic operations, without using any fancy data structures.
|
||||
Furthermore, this method quickly detects non-palindromic Strings, and returns \verb|false|, saving computation.
|
||||
\\\\
|
||||
The next most efficient method was the iterative reverse versus original or \verb|ReverseVSOriginal| method.
|
||||
Despite being the next most efficient method, it took almost ten times the time that \verb|IVersusNMinusI| took to complete, taking 366
|
||||
milliseconds, which, all things considered, is still quite fast.
|
||||
The number of primitive operations and the rate of growth of this method were also accordingly higher than the previous method.
|
||||
\\\\
|
||||
The third most efficient method was the recursive reverse versus original or \verb|RecursiveReverse| method.
|
||||
This makes sense, as it uses a very similar approach to the second most efficient method, but instead did it recursively.
|
||||
Despite this solution appearing somewhat more elegant from a lines of code perspective, it was in practice less efficient than
|
||||
its iterative counterpart, both in terms of memory (as the recursive function calls had to remain in memory until all the sub-calls
|
||||
were complete) and in terms of operations, taking approximately 20 million more primitive operations to complete the same task
|
||||
as the iterative approach.
|
||||
It was also significantly slower than its iterative counterpart, taking around 200 milliseconds more to complete the task.
|
||||
We can also tell from looking at the graph that at low problem sizes that this approach is comparable \& rather similar in terms
|
||||
of efficiency to the iterative approach, but they quickly diverge, with the recursive approach having a significantly steeper
|
||||
slope.
|
||||
We can extrapolate from this that this method would be even less efficient at very large problem sizes, as its rate of growth
|
||||
is quite large.
|
||||
\\\\
|
||||
By far the least efficient method was the Stack versus Queue or \verb|StackVSQueue| method.
|
||||
It took by far the greatest number of primitive operations, and the rate of growth was ridiculously large, rapidly diverging from
|
||||
the other four techniques.
|
||||
Its rate of growth is so large that it would likely quickly become unusable for any significantly large problem.
|
||||
This is reinforced by the fact that it took 3,519 milliseconds to complete the task, being the only method that took more than one
|
||||
second to do so, and taking almost 100 times what the best-performing method took.
|
||||
|
||||
\end{document}
|
||||
@ -0,0 +1,49 @@
|
||||
\addtolength{\hoffset}{-2.25cm}
|
||||
\addtolength{\textwidth}{4.5cm}
|
||||
\addtolength{\voffset}{-3.25cm}
|
||||
\addtolength{\textheight}{5cm}
|
||||
\setlength{\parskip}{0pt}
|
||||
\setlength{\parindent}{0in}
|
||||
|
||||
%----------------------------------------------------------------------------------------
|
||||
% PACKAGES AND OTHER DOCUMENT CONFIGURATIONS
|
||||
%----------------------------------------------------------------------------------------
|
||||
|
||||
\usepackage{blindtext} % Package to generate dummy text
|
||||
\usepackage{charter} % Use the Charter font
|
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\usepackage[utf8]{inputenc} % Use UTF-8 encoding
|
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\usepackage{microtype} % Slightly tweak font spacing for aesthetics
|
||||
\usepackage[english]{babel} % Language hyphenation and typographical rules
|
||||
\usepackage{amsthm, amsmath, amssymb} % Mathematical typesetting
|
||||
\usepackage{float} % Improved interface for floating objects
|
||||
\usepackage[final, colorlinks = true,
|
||||
linkcolor = black,
|
||||
citecolor = black]{hyperref} % For hyperlinks in the PDF
|
||||
\usepackage{graphicx, multicol} % Enhanced support for graphics
|
||||
\usepackage{xcolor} % Driver-independent color extensions
|
||||
\usepackage{marvosym, wasysym} % More symbols
|
||||
\usepackage{rotating} % Rotation tools
|
||||
\usepackage{censor} % Facilities for controlling restricted text
|
||||
\usepackage{listings, style/lstlisting} % Environment for non-formatted code, !uses style file!
|
||||
\usepackage{pseudocode} % Environment for specifying algorithms in a natural way
|
||||
\usepackage{style/avm} % Environment for f-structures, !uses style file!
|
||||
\usepackage{booktabs} % Enhances quality of tables
|
||||
\usepackage{tikz-qtree} % Easy tree drawing tool
|
||||
\tikzset{every tree node/.style={align=center,anchor=north},
|
||||
level distance=2cm} % Configuration for q-trees
|
||||
\usepackage{style/btree} % Configuration for b-trees and b+-trees, !uses style file!
|
||||
\usepackage[backend=biber,style=numeric,
|
||||
sorting=nyt]{biblatex} % Complete reimplementation of bibliographic facilities
|
||||
\addbibresource{ecl.bib}
|
||||
\usepackage{csquotes} % Context sensitive quotation facilities
|
||||
\usepackage[yyyymmdd]{datetime} % Uses YEAR-MONTH-DAY format for dates
|
||||
\renewcommand{\dateseparator}{-} % Sets dateseparator to '-'
|
||||
\usepackage{fancyhdr} % Headers and footers
|
||||
\pagestyle{fancy} % All pages have headers and footers
|
||||
\fancyhead{}\renewcommand{\headrulewidth}{0pt} % Blank out the default header
|
||||
\fancyfoot[L]{} % Custom footer text
|
||||
\fancyfoot[C]{} % Custom footer text
|
||||
\fancyfoot[R]{\thepage} % Custom footer text
|
||||
\newcommand{\note}[1]{\marginpar{\scriptsize \textcolor{red}{#1}}} % Enables comments in red on margin
|
||||
|
||||
%----------------------------------------------------------------------------------------
|
||||
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@ -0,0 +1,165 @@
|
||||
% avm.sty -- for attribute-value matrices -- mar 29, 1992; rev. dec 6, 1993
|
||||
% (c) 1992 christopher manning (manning@csli.stanford.edu) -- see avm.doc.tex
|
||||
|
||||
\newif\ifavmactive\newif\ifavmsorted\newif\ifavmlabeled
|
||||
\newif\ifavmcenter\newif\ifavmbottom
|
||||
\newif\ifavmbottomright\newif\ifavmtopleft\newif\ifavmtopright
|
||||
|
||||
\newdimen\avmdimen
|
||||
\newbox\avmboxone\newbox\avmboxthree
|
||||
|
||||
\def\avmoptions#1{\avmactivefalse\avmsortedfalse\avmlabeledfalse
|
||||
\avmcenterfalse\avmbottomfalse
|
||||
\avmbottomrightfalse\avmtopleftfalse\avmtoprightfalse
|
||||
\def\more{#1}\ifx\more\empty\else\avmjoptions#1,\@nil\fi}
|
||||
\def\avmjoptions#1,#2\@nil{\def\more{#2}\csname avm#1true\endcsname
|
||||
\ifx\more\empty\else\avmjoptions#2\@nil\fi}
|
||||
|
||||
|
||||
\def\avmfont#1{\def\avmjfont{#1}}
|
||||
\def\avmjfont{}
|
||||
|
||||
\def\avmvalfont#1{\def\avmjvalfont{#1}}
|
||||
\def\avmjvalfont{}
|
||||
|
||||
\def\avmsortfont#1{\def\avmjsortfont{#1}}
|
||||
\def\avmjsortfont{}
|
||||
|
||||
\def\avmhskip#1{\def\avmjhskip{#1}}
|
||||
\def\avmjhskip{1em}
|
||||
|
||||
\def\avmbskip#1{\def\avmjbskip{#1}}
|
||||
\def\avmjbskip{0em}
|
||||
|
||||
\def\avmvskip#1{\def\avmjvskip{#1}}
|
||||
\def\avmjvskip{0.385ex}%was .3875
|
||||
|
||||
|
||||
\def\avmjprolog#1{$\mskip-\thinmuskip
|
||||
\left#1\hskip\avmjbskip\vcenter\bgroup\vskip\avmjvskip
|
||||
\ialign\bgroup\avmjfont
|
||||
\strut ##\unskip\hfil
|
||||
&&\hskip\avmjhskip\avmjvalfont ##\unskip\hfil\cr}
|
||||
\def\avmjpostlog#1{\crcr\egroup\vskip\avmjvskip\egroup
|
||||
\hskip\avmjbskip\right#1\mskip-\thinmuskip$\ignorespaces}
|
||||
|
||||
|
||||
\def\avmjcatcode{\let\lparen=(\let\rparen=)\catcode`\[=13\catcode`\]=13
|
||||
\catcode`\<=13\catcode`\@=13\catcode`\(=13\catcode`\)=13
|
||||
\catcode`\>=13\catcode`\|=13}
|
||||
|
||||
{\avmjcatcode % new group: redefine above catcodes as active
|
||||
|
||||
\gdef\specialavm{\avmjcatcode
|
||||
\def({\avmjprolog\lparen}%
|
||||
\def){\avmjpostlog\rparen}%
|
||||
\def<{\avmjprolog\langle}%
|
||||
\def>{\avmjpostlog\rangle}%
|
||||
\ifavmsorted
|
||||
\def[##1{\setbox\avmboxthree=\hbox{\avmjsortfont##1\/}\setbox2=\hbox
|
||||
\bgroup\avmjprolog\lbrack}%
|
||||
\def]{\avmjpostlog\rbrack\egroup\avmjsort}%
|
||||
\else\ifavmlabeled
|
||||
\def[##1{\def\more{##1}\setbox2=\hbox\bgroup\avmjprolog[}%
|
||||
\def]{\avmjpostlog]\egroup\node{\more}{\box2}}%
|
||||
\else
|
||||
\def[{\avmjprolog\lbrack}%
|
||||
\def]{\avmjpostlog\rbrack}%
|
||||
\fi\fi
|
||||
%
|
||||
\def\<{$\langle$}\def\>{$\rangle$}%
|
||||
\def\({\lparen}\def\){\rparen}%
|
||||
\def\[{\lbrack}\def\]{\rbrack}%
|
||||
\def|{$\,\vert\,$}%
|
||||
\def@##1{\avmbox{##1}}%
|
||||
} % end defn of \specialavm
|
||||
} % restore active catcodes
|
||||
|
||||
|
||||
\long\def\avm{\begingroup
|
||||
\ifavmactive\specialavm
|
||||
\else
|
||||
\def\({\avmjprolog(}%
|
||||
\def\){\avmjpostlog)}%
|
||||
\def\<{\avmjprolog\langle}%
|
||||
\def\>{\avmjpostlog\rangle}%
|
||||
%
|
||||
\ifavmsorted
|
||||
\def\[##1{\setbox\avmboxthree=\hbox{\avmjsortfont##1\/}\setbox
|
||||
2=\hbox\bgroup\avmjprolog[}%
|
||||
\def\]{\avmjpostlog]\egroup\avmjsort}%
|
||||
\else\ifavmlabeled
|
||||
\def\[##1{\def\more{##1}\setbox2=\hbox\bgroup\avmjprolog[}%
|
||||
\def\]{\avmjpostlog]\egroup\node{\more}{\box2}}%
|
||||
\else
|
||||
\def\[{\avmjprolog[}%
|
||||
\def\]{\avmjpostlog]}%
|
||||
\fi\fi
|
||||
%
|
||||
\def\|{$\,\vert\,$}%
|
||||
\def\@##1{\avmbox{##1}}%
|
||||
\fi % end not active
|
||||
%
|
||||
\ifx\LaTeX\undefined\def\\{\cr}% running under TeX
|
||||
\else \def\\{\@tabularcr}% Leverage off LaTeX's \\*[dimen] options
|
||||
\fi
|
||||
\def\!{\node}%
|
||||
\long\def\avmjsort{\dimen2=\ht2\advance\dimen2 by -.25\baselineskip
|
||||
\global\dimen\avmdimen=\wd\avmboxthree
|
||||
\ifavmtopleft \raise\dimen2\llap{\box\avmboxthree}\box2%
|
||||
\else\ifavmtopright \box2\raise\dimen2\box\avmboxthree%
|
||||
\else\ifavmbottomright \box2\lower\dimen2\box\avmboxthree%
|
||||
\else \lower\dimen2\llap{\box\avmboxthree}\box2%
|
||||
\fi\fi\fi}%
|
||||
\long\def\sort##1##2{\setbox2=\hbox{##2}\setbox
|
||||
\avmboxthree=\hbox{\avmjsortfont##1\/}\dimen2=\ht2
|
||||
\advance\dimen2 by -.25\baselineskip
|
||||
\ifavmtopleft \raise\dimen2\box\avmboxthree\box2%
|
||||
\else\ifavmtopright \box2\raise\dimen2\box\avmboxthree%
|
||||
\else\ifavmbottomright \box2\lower\dimen2\box\avmboxthree%
|
||||
\else \lower\dimen2\box\avmboxthree\box2%
|
||||
\fi\fi\fi}%
|
||||
\long\def\osort##1##2{\setbox2=\hbox{##2}\setbox
|
||||
\avmboxthree=\hbox{\avmjsortfont ##1\/}\avmjsort}%
|
||||
\def\avml{\avmjprolog.}%
|
||||
\def\avmr{\avmjpostlog.}%
|
||||
\def\avmb##1{\node{##1}{\lbrack\;\rbrack}}%
|
||||
\def\avmd##1{\node{##1}{---}}%
|
||||
\def\q##1{\ifx ##1\{$\lbrace$\else
|
||||
\ifx ##1\}$\rbrace$\else
|
||||
\ifx ##1<$\langle$\else
|
||||
\ifx ##1>$\rangle$\fi \fi \fi \fi}%
|
||||
\def\{{\avmjprolog\lbrace}%
|
||||
\def\}{\avmjpostlog\rbrace}%
|
||||
\def\;{\hskip\avmjhskip}%
|
||||
\def\avmspan##1{\multispan2\strut ##1\expandafter\hfil}%
|
||||
\avmjfont
|
||||
\openup\avmjvskip
|
||||
\setbox\avmboxone=\hbox\bgroup\ignorespaces
|
||||
} % end defn of \avm
|
||||
|
||||
|
||||
\def\endavm{\egroup\ifvmode\leavevmode\fi % this if is useful!
|
||||
\ifavmsorted\null\hskip\dimen\avmdimen\fi
|
||||
\ifavmcenter
|
||||
\box\avmboxone
|
||||
\else \ifavmbottom
|
||||
\lower.575\baselineskip\hbox{\vbox{\box\avmboxone\null}}%
|
||||
\else
|
||||
% the next bit is ripped off from Emma's \evnup in lingmacros.sty
|
||||
\dimen2=\ht\avmboxone\advance\dimen2 by -.725\baselineskip
|
||||
\lower\dimen2\box\avmboxone
|
||||
\fi \fi \endgroup}
|
||||
|
||||
|
||||
% based on TeXbook exercise 21.3
|
||||
\def\avmbox#1{\setbox2=\hbox{$\scriptstyle #1$}\lower.2ex\vbox{\hrule
|
||||
\hbox{\vrule\kern1.25pt
|
||||
\vbox{\kern1.25pt\box2\kern1.25pt}\kern1.25pt\vrule}\hrule}}
|
||||
|
||||
% ============ COSTOM CONFIGURATION =============
|
||||
\avmfont{\sc}
|
||||
\avmoptions{sorted,active}
|
||||
\avmvalfont{\rm}
|
||||
\avmsortfont{\scriptsize\it}
|
||||
% ===============================================
|
||||
@ -0,0 +1,131 @@
|
||||
%% Last Modified: Thu Oct 18 18:26:25 2007.
|
||||
|
||||
\NeedsTeXFormat{LaTeX2e}
|
||||
\ProvidesPackage{style/btree}
|
||||
\typeout{Document Style `weiw_BTree - Support drawing B+-Tree (ver 0.999)}
|
||||
|
||||
\RequirePackage{tikz}
|
||||
\RequirePackage{ifthen}
|
||||
|
||||
% use libraries
|
||||
\usetikzlibrary{arrows,shapes,decorations,matrix}
|
||||
|
||||
|
||||
%% global declaration
|
||||
\tikzstyle{btreeptr} = [draw, semithick, minimum height=2em]
|
||||
\tikzstyle{btreeval} = [draw, semithick, minimum size=2em]
|
||||
\tikzstyle{btreevale} = [draw,semithick, minimum size=2em]
|
||||
\tikzstyle{btlink} = [draw, semithick, ->, >=triangle 45]
|
||||
|
||||
%% macro
|
||||
%% helper macros
|
||||
\newcommand{\suppressemptystr}[1]{% leave blank for entries in leaf nodes
|
||||
\ifthenelse{\equal{#1}{}}%
|
||||
{%
|
||||
\relax%
|
||||
}%
|
||||
% Else
|
||||
{%
|
||||
#1\textsuperscript{*}%
|
||||
}%
|
||||
}%
|
||||
|
||||
\newcommand{\xyshift}[3]{% help to place the nodes
|
||||
\begin{scope}[xshift=#1, yshift=#2]
|
||||
#3
|
||||
\end{scope}%
|
||||
}
|
||||
|
||||
%% Common btree macros
|
||||
\newcommand{\btreelink}[2]{% #1: src node; #2: dest node;
|
||||
\draw[btlink] ([yshift=3pt] #1.south) -- (#2-b.north);
|
||||
}
|
||||
|
||||
\newcommand{\btreelinknorth}[2]{% #1: src node; #2: dest node;
|
||||
\draw[btlink] ([yshift=3pt] #1.south) -- (#2.north);
|
||||
}
|
||||
|
||||
\newcommand{\btreetriangle}[2]{% #1: node name; #2 text inside
|
||||
\node[anchor=north, regular polygon, regular polygon sides=3, draw] (#1) {#2};
|
||||
}
|
||||
|
||||
%%======================================================================
|
||||
%% btree with capacity = 4
|
||||
\newcommand{\btreeinodefour}[5]{%
|
||||
\matrix [ampersand replacement=\&] (#1)
|
||||
{
|
||||
\node[btreeptr] (#1-1) {\vphantom{1}}; \& \node[btreeval] (#1-a) {#2}; \&
|
||||
\node[btreeptr] (#1-2) {\vphantom{1}}; \& \node[btreeval] (#1-b) {#3}; \&
|
||||
\node[btreeptr] (#1-3) {\vphantom{1}}; \& \node[btreeval] (#1-c) {#4}; \&
|
||||
\node[btreeptr] (#1-4) {\vphantom{1}}; \& \node[btreeval] (#1-d) {#5}; \&
|
||||
\node[btreeptr] (#1-5) {\vphantom{1}}; \\
|
||||
};
|
||||
}
|
||||
\newcommand{\btreelnodefour}[5]{%
|
||||
\matrix [ampersand replacement=\&, outer sep=0pt, matrix anchor=north] (#1)
|
||||
{
|
||||
\node[btreevale] (#1-a) {\suppressemptystr{#2}}; \&
|
||||
\node[btreevale] (#1-b) {\suppressemptystr{#3}}; \&
|
||||
\node[btreevale] (#1-c) {\suppressemptystr{#4}}; \&
|
||||
\node[btreevale] (#1-d) {\suppressemptystr{#5}}; \\
|
||||
};
|
||||
}
|
||||
|
||||
%%======================================================================
|
||||
%% btree with capacity = 3
|
||||
\newcommand{\btreeinodethree}[4]{%
|
||||
\matrix [ampersand replacement=\&] (#1)
|
||||
{
|
||||
\node[btreeptr] (#1-1) {\vphantom{1}}; \& \node[btreeval] (#1-a) {#2}; \&
|
||||
\node[btreeptr] (#1-2) {\vphantom{1}}; \& \node[btreeval] (#1-b) {#3}; \&
|
||||
\node[btreeptr] (#1-3) {\vphantom{1}}; \& \node[btreeval] (#1-c) {#4}; \&
|
||||
\node[btreeptr] (#1-4) {\vphantom{1}}; \\
|
||||
};
|
||||
}
|
||||
\newcommand{\btreelnodethree}[4]{%
|
||||
\matrix [ampersand replacement=\&, outer sep=0pt, matrix anchor=north] (#1)
|
||||
{
|
||||
\node[btreevale] (#1-a) {\suppressemptystr{#2}}; \&
|
||||
\node[btreevale] (#1-b) {\suppressemptystr{#3}}; \&
|
||||
\node[btreevale] (#1-c) {\suppressemptystr{#4}}; \\
|
||||
};
|
||||
}
|
||||
|
||||
%%======================================================================
|
||||
%% btree with capacity = 2
|
||||
\newcommand{\btreeinodetwo}[4]{%
|
||||
\matrix [ampersand replacement=\&] (#1)
|
||||
{
|
||||
\node[btreeptr] (#1-1) {\vphantom{1}}; \& \node[btreeval] (#1-a) {#2}; \&
|
||||
\node[btreeptr] (#1-2) {\vphantom{1}}; \& \node[btreeval] (#1-b) {#3}; \&
|
||||
\node[btreeptr] (#1-3) {\vphantom{1}}; \\
|
||||
};
|
||||
}
|
||||
\newcommand{\btreelnodetwo}[3]{%
|
||||
\matrix [ampersand replacement=\&, outer sep=0pt, matrix anchor=north] (#1)
|
||||
{
|
||||
\node[btreevale] (#1-a) {\suppressemptystr{#2}}; \&
|
||||
\node[btreevale] (#1-b) {\suppressemptystr{#3}}; \\
|
||||
};
|
||||
}
|
||||
%%======================================================================
|
||||
|
||||
|
||||
|
||||
|
||||
%% simple example
|
||||
% \begin{center}
|
||||
% \scalebox{0.7}{
|
||||
% \begin{tikzpicture}
|
||||
% %
|
||||
% \btreeinodefour{root}{13}{17}{24}{30};
|
||||
% \xyshift{-40mm}{-20mm}{\btreelnodefour{n1}{2}{3}{5}{7}}
|
||||
% \xyshift{-0mm}{-20mm}{\btreelnodefour{n2}{14}{16}{}{}}
|
||||
% \xyshift{40mm}{-20mm}{\btreelnodefour{n3}{19}{20}{22}{}}
|
||||
% \xyshift{80mm}{-20mm}{\btreelnodefour{n4}{24}{27}{29}{}}
|
||||
% \xyshift{120mm}{-20mm}{\btreelnodefour{n5}{33}{34}{38}{39}}
|
||||
% %
|
||||
% \foreach \x in {1,2,...,5} { \btreelink{root-\x}{n\x} }
|
||||
% \end{tikzpicture}
|
||||
% }
|
||||
% \end{center}
|
||||
@ -0,0 +1,38 @@
|
||||
% Source: ss17_wissschreib (Eva)
|
||||
|
||||
\lstset{
|
||||
basicstyle=\ttfamily\scriptsize\mdseries,
|
||||
keywordstyle=\bfseries\color[rgb]{0.171875, 0.242188, 0.3125},
|
||||
identifierstyle=,
|
||||
commentstyle=\color[rgb]{0.257813, 0.15625, 0},
|
||||
stringstyle=\itshape\color[rgb]{0.0195313, 0.195313, 0.0117188},
|
||||
numbers=left,
|
||||
numberstyle=\tiny,
|
||||
stepnumber=1,
|
||||
breaklines=true,
|
||||
frame=none,
|
||||
showstringspaces=false,
|
||||
tabsize=4,
|
||||
backgroundcolor=\color[rgb]{0.98,0.98,0.98},
|
||||
captionpos=b,
|
||||
float=htbp,
|
||||
language=Python,
|
||||
xleftmargin=15pt,
|
||||
xrightmargin=15pt
|
||||
}
|
||||
|
||||
%(deutsche) Sonderzeichen
|
||||
\lstset{literate=%
|
||||
{Ä}{{\"A}}1
|
||||
{Ö}{{\"O}}1
|
||||
{Ü}{{\"U}}1
|
||||
{ä}{{\"a}}1
|
||||
{ö}{{\"o}}1
|
||||
{ü}{{\"u}}1
|
||||
{ß}{{\ss}}1
|
||||
}
|
||||
|
||||
%Verwendung im Text:
|
||||
%-> \begin{lstlisting}[language=Python,firstnumber=27] ... \end{lstlisting}
|
||||
%-> \begin{lstlisting}[language=Python,numbers=none] ... \end{lstlisting}
|
||||
%-> \lstinline[language=JAVA]{...}
|
||||
Reference in New Issue
Block a user