Test Prep Question:

Step 1: Read a text file and parse it into individual characters. Count the number of characters of each type to get the frequency of each character, that is, the number of times each character appears in the text file. You can ignore the spaces and the newline returns, but count the alphabet, numbers, and punctuations. Heres a sample text file that you can use to test your program.

Txt File

POKEMON TOWER DEFENSE YOUR MISSION IN THIS FUN STRATEGY TOWER DEFENSE GAME IS TO HELP PROFESSOR OAK TO STOP ATTACKS OF WILD RATTATA. SET OUT ON YOUR OWN POKEMON JOURNEY, TO CATCH AND TRAIN ALL POKEMON AND TRY TO SOLVE THE MYSTERY BEHIND THESE ATTACKS. YOU MUST PLACE POKEMON CHARACTERS STRATEGICALLY ON THE BATTLEFIELD SO THAT THEY STOP ALL WAVES OF ENEMY ATTACKER!!! DURING THE BATTLE YOU WILL LEVEL UP AND EVOLVE YOUR POKEMON. YOU CAN ALSO CAPTURE OTHER POKEMON DURING THE BATTLE AND ADD THEM TO YOUR TEAM. USE YOUR MOUSE TO PLAY THE GAME. GOOD LUCK!

End of Txt ^^

The outcome of the first step is a list of symbols and their frequencies. For example, P 14 O 49 K 11 E 53 etc.

Step 2: Convert the frequencies into probabilities. For this, divide each frequency by total number of characters. For example, the above text file has 455 characters without spaces. You would divide each frequency by 455. The outcome of this step is a list of symbols and their probabilities. For example, Symbol Probability P 0.031 O 0.108 K 0.024 E 0.116 etc. Note that if this step is done correctly, the sum of all the probabilities is 1.0. Sometimes, you may have to round it up to the nearest decimal value, and that approximation is OK. In the above example, the probabilities have been rounded to three decimal places. You can use a different precision but make sure that they all add up to 1.0 (or very close to it). To do steps 1 and 2, you can create a class called Pair.java that defines the symbol and its probability as an object. public class Pair { private char value; private double prob; //constructor //get and set methods //toString method } You can create an arraylist of Pair objects and store the items into the arraylist as you read them. Of course, you will need other variables and methods to count the frequencies and convert them into probabilities.

Step 3: Using this set of symbols and frequencies, build the Huffman tree.

Step 3.1: Create a queue of Binary Tree nodes. Each Binary Tree node is of type Pair. The queue can be implemented as a simple arraylist, where enqueue means adding an item to the end of the arraylist and dequeue means removing the item at index 0. That is, the queue is an arraylist of type >. The queue contains these sorted according to the increasing order of their frequencies. This is your Queue S. This is done by checking the arraylist for values in increasing order, creating the binary tree nodes and enqueueing them in the queue. If you enumerate the Queue S, it should have the Pair objects in increasing order of their frequencies, something like this: (!, 0.009), (V, 0.011), etc.

Step 3.2 : Now initialize another queue T (another arraylist) of type >.

Step 3.3 : Build the Huffman tree according to the algorithm discussed in the lectures. For instance, in the above example, first (!, 0.009) and (V, 0.011) will be dequeued from S. Create a node with the combined frequency. What do you put as the character for the combined node? You can put a dummy character, say 0. So (0,0.02) will be the parent node, and (!, 0.009) and (V, 0.011) will be the left and right children. This tree will be enqueued to Queue T. You keep repeating the above procedure and building the Huffman tree according to the algorithm given below: Pick the two smallest weight trees, say A and B, from S and T, as follows: a) If T is empty, A and B are respectively the front and next to front entries of S. Dequeue them from S. b) If T is not empty, i) Find the smaller weight tree of the trees in front of S and in front of T. This is A. Dequeue it. ii) Find the smaller weight tree of the trees in front of S and in front of T. This is B. Dequeue it. 3. Construct a new tree P by creating a root and attaching A and B as the subtrees of this root. The weight of the root is the combined weights of the roots of A and B. 4. Enqueue P to T. 5. Repeat steps 2 to 4 until S is empty. 6. After step 5, if T's size is > 1, dequeue two nodes at a time, combine them and enqueue the combined tree until T's size is 1. The last node remaining in the queue T will be the final Huffman tree.

Step 4: Derive the Huffman codes. The following methods can be used for finding the encoding. They use a String array of 256 to cover all characters in the Unicode set. You can tweak them if necessary.

public static void findEncoding(BinaryTree t, String[] a, String prefix) {

if (t.getLeft() == null && t.getRight() == null) {

a[(byte) (t.getData().getValue())] = prefix;

} else {

findEncoding(t.getLeft(), a, prefix + "0");

findEncoding(t.getRight(), a, prefix + "1");

}

}

public static String[] findEncoding(BinaryTree t) {

String[] result = new String[256];

findEncoding(t, result, "");

return result;

}

Step 5: Read the sample Pokemon text file (call it Pokemon.txt) and Huffman.txt file, and encode it using the Huffman symbols. Do not encode spaces and newline characters. Leave them as they are. Write the encoded file into another text file, Encoded.txt. Step 6: Read the encoded text file and decode it. Write the decoded file into yet another text file, Decoded.txt. If you have done everything correctly, then Decoded.txt must be the same as Pokemon.txt.

Java Class Used

public class BinaryTree { private T data; private BinaryTree parent; private BinaryTree left; private BinaryTree right; public BinaryTree() { parent = left = right = null; data = null; } public void makeRoot(T data) { if (!isEmpty()) { System.out.println("Can't make root. Already exists"); } else this.data = data; } public void setData(T data) { this.data = data; } public void setLeft(BinaryTree tree) { left = tree; } public void setRight(BinaryTree tree) { right = tree; } public void setParent(BinaryTree tree) { parent = tree; } public T getData() { return data; } public BinaryTree getParent() { return parent; } public BinaryTree getLeft() { return left; } public BinaryTree getRight() { return right; } public void attachLeft(BinaryTree tree) { if (tree==null) return; else if (left!=null || tree.getParent()!=null) { System.out.println("Can't attach"); return; } else { tree.setParent(this); this.setLeft(tree); } } public void attachRight(BinaryTree tree) { if (tree==null) return; else if (right!=null || tree.getParent()!=null) { System.out.println("Can't attach"); return; } else { tree.setParent(this); this.setRight(tree); } } public BinaryTree detachLeft() { if (this.isEmpty()) return null; BinaryTree retLeft = left; left = null; if (retLeft!=null) retLeft.setParent(null); return retLeft; } public BinaryTree detachRight() { if (this.isEmpty()) return null; BinaryTree retRight = right; right =null; if (retRight!=null) retRight.setParent(null); return retRight; } public boolean isEmpty() { if (data == null) return true; else return false; } public void clear() { left = right = parent =null; data = null; } public BinaryTree root() { if (parent == null) return this; else { BinaryTree next = parent; while (next.getParent()!=null) next = next.getParent(); return next; } } public static  void preorder(BinaryTree t) { if (t!=null) { System.out.print(t.getData()+"\t"); preorder(t.getLeft()); preorder(t.getRight()); } } public static  void inorder(BinaryTree t) { if (t!=null) { inorder(t.getLeft()); System.out.print(t.getData() + "\t"); inorder(t.getRight()); } } public static  void postorder(BinaryTree t) { if (t!=null) { postorder(t.getLeft()); postorder(t.getRight()); System.out.print(t.getData() + "\t"); } } }