On to huffman...

[KWSW]

Thanks to the help over at http://www.thescripts.com/forum/thread686347.html on the character count program, I can now move onto getting the huffman code done. Unfortunately I am still rather new to Java and would again need help on getting started.

Not gonna ask for the whole code here cos its against the rules and its not gonna help me learn the language. I know that if I use C++, I can use the heap container which is present in the STL but for java is there something like that?

Went to try to google "java heap container" but all I got was some out of memory error...

Will still carry on searching and I hope someone will point me in the right direction.

Thanks In Advance :)

[JosAH]

Thanks to the help over at http://www.thescripts.com/forum/thread686347.html on the character count program, I can now move onto getting the huffman code done. Unfortunately I am still rather new to Java and would again need help on getting started.

Not gonna ask for the whole code here cos its against the rules and its not gonna help me learn the language. I know that if I use C++, I can use the heap container which is present in the STL but for java is there something like that?

Went to try to google "java heap container" but all I got was some out of memory error...

Will still carry on searching and I hope someone will point me in the right direction.

Thanks In Advance :)

If you think of a Huffman tree there are two types of nodes: simple character
nodes; they are the leaves of the tree, and 'internal' nodes which are the combined
nodes of the Huffman tree.

Both nodes can return their frequency: character (leaf) nodes simply return the
frequence of the character they represent and the combined nodes return the
sum of the frequencies of their child nodes.

A node X is smaller than a node Y if X's freqency is smaller than Y's frequency.
This makes those nodes ideal to store them in a SortedSet (TreeSet).
The sorted set is the equivalent of your heap structure.

Building a Huffman tree out of such a SortedSet is breeze: simply combine
the two nodes with the lowest frequency, remove them from the set and add the
newly created combined node to the set again. Keep going until there's only
one node left in the set. That'll be the root of the Huffman tree.

An abstract node class and two extending classes (character node and a
combined node) are all that is needed.

kind regards,

Jos

[KWSW]

If you think of a Huffman tree there are two types of nodes: simple character
nodes; they are the leaves of the tree, and 'internal' nodes which are the combined
nodes of the Huffman tree.

Both nodes can return their frequency: character (leaf) nodes simply return the
frequence of the character they represent and the combined nodes return the
sum of the frequencies of their child nodes.

A node X is smaller than a node Y if X's freqency is smaller than Y's frequency.
This makes those nodes ideal to store them in a SortedSet (TreeSet).
The sorted set is the equivalent of your heap structure.

Building a Huffman tree out of such a SortedSet is breeze: simply combine
the two nodes with the lowest frequency, remove them from the set and add the
newly created combined node to the set again. Keep going until there's only
one node left in the set. That'll be the root of the Huffman tree.

An abstract node class and two extending classes (character node and a
combined node) are all that is needed.

kind regards,

Jos

Thanks for the explaination... will go and try it out... think I have a rough idea how now...

Many thanks again :)

[KWSW]

question about compareTo()

so in this case, I would then need to use that to compare two nodes for sorting for the sortedSet.

would it look something like this?

public int commpareTo(Object obj)
    {        
        node aNode = (node)obj;
        return this.getFeq() - aNode.getFeq();
    }

sorry still in the planning stage thats why I dun have the full code yet...

[JosAH]

question about compareTo()

so in this case, I would then need to use that to compare two nodes for sorting for the sortedSet.

would it look something like this?

public int commpareTo(Object obj)
    {        
        node aNode = (node)obj;
        return this.getFeq() - aNode.getFeq();
    }

sorry still in the planning stage thats why I dun have the full code yet...

No need to apologize; that compareTo method is about right but beware that
you never want the return value of that method to be zero. You can get around
that by comparing the huffman code for the nodes built so far when the
frequencies compare equal. You'll bump into that later; keep my remark in mind ;-)

kind regards,

Jos

[KWSW]

No need to apologize; that compareTo method is about right but beware that
you never want the return value of that method to be zero. You can get around
that by comparing the huffman code for the nodes built so far when the
frequencies compare equal. You'll bump into that later; keep my remark in mind ;-)

kind regards,

Jos

yup just ran into that problem. WIll try out the comparing of the huffman code length.

Hmm... any advise on how I can set the code of each node?

[JosAH]

yup just ran into that problem. WIll try out the comparing of the huffman code length.

Hmm... any advise on how I can set the code of each node?

Comparing on the code length won't help: two nodes can have the same frequency
and the same code length; you have to compare on the code itself or some
other artificial criteria; as long as no two nodes compare equal in a consistent way.

The actual comparison doesn't matter much; when the nodes are taken out
of the set to be combined, a new node is entered to the set afterwards and
the two individual nodes never will occur in that set anymore by themselves.

When you combine two nodes, prepend a 0 to the code of the left node and
prepend a 1 to the code of the right node. You can append too, it depends on
how you build up those codes and in what order you want to read the bits.

In both cases you have to keep track of the length of the code (which equals
the depth of the the node in the tree).

kind regards,

Jos

[KWSW]

hmm I keep getting missing nodes... is it something to do with the compareTo() not being able to tell which is bigger?

public int compareTo(Object obj)
    {        
        Node aNode = (Node)obj;        
 
        if(this.getFeq() == aNode.getFeq())
        {
            return this.codeLen - aNode.codeLen;
        }
        else
        {
            return this.feq - aNode.feq;
        }
 
    }

my huffman algo

while (t.size() > 1)
        {
            Node left = (Node)t.first();
            t.remove(left);
            Node right = (Node)t.first();
            t.remove(right);
 
            t.add(new Branch(left,right));
 
            count++;
        }

[JosAH]

hmm I keep getting missing nodes... is it something to do with the compareTo() not being able to tell which is bigger?

    return this.codeLen - aNode.codeLen;
 

As I wrote above: don't compare the code lenghts; compare the actual codes
themselves. If the frequencies equal it is very well possible that the code lengths
compare equal as well; don't compare those lengths.

kind regards,

Jos

[KWSW]

Ok I have run out of ideas..
Here is my code

import java.util.*;
import java.io.*;
 
abstract class Node implements Comparable
{        
    // For CompareTo() to use
    protected int x = 0;
 
 
    // The code to represent the symbol
    protected String code = "";
 
    // The fequency of the symbol that this node represents
    protected int feq;    
 
    // Return fequency
    public int getFeq()
    {
        return feq;
    }
 
    public void increaseFeq()
    {
        feq++;
    }                  
 
    // Constructor that takes in the fequency
    public Node(int feq, int x)
    {
        this.feq = feq;
        this.x = x;
    }
 
    abstract protected void addCode(int b);
 
 
    public int compareTo(Object obj)
    {        
        Node aNode = (Node)obj;                
 
        if(this.getFeq() == aNode.getFeq())
        {            
            return (this.x+this.feq+this.code.length()) - (aNode.x+this.feq+this.code.length());
        }
        else
        {
            return this.feq - aNode.feq;
        }
 
    }
 
}
 
class Leaf extends Node
{
    private int ascii;
 
    public void addCode(int i)
    {
       code += i;       
    }
 
    public Leaf(int ascii, int feq)
    {
        super(feq,ascii);
        this.ascii = ascii;        
    }        
 
    public String toString() 
    { 
        return "\n["+ascii+":"+code+","+feq+","+code.length()+"]"; 
    }
}
 
class Branch extends Node
{
    private Node left;
    private Node right;
 
    public Branch(Node left, Node right) 
        {
        super(left.getFeq()+right.getFeq(),left.getFeq()+right.getFeq());
                this.left = left;
                this.right = right;
 
                left.addCode(0);        
                right.addCode(1);
    }
 
    protected void addCode(int i) 
        { 
            left.addCode(i); 
            right.addCode(i); 
        }
 
    public String toString()
        {
            return "Total: "+ (left.getFeq()+right.getFeq())+ left + right;
        }
}
 
class Tree
{
    Map<Integer,Integer> m = new HashMap<Integer,Integer>();
 
    public void add(int i)
    {
        int count;
 
        if(m.containsKey(i))
        {
            count = m.get(i);
            count++;
            m.put(i,count);
        }
        else
        {
            m.put(i,1);
        }
    }
 
    public void create()
    {
        ArrayList al = new ArrayList();                
 
        for(int i : m.keySet()) 
        {            
            Leaf l = new Leaf(i,m.get(i));
            al.add(l);            
        }                                                
 
        TreeSet t = new TreeSet();        
 
        for(int i = 0; i < al.size(); i++)
        {
            t.add(al.get(i));            
        }
 
        do
        {
            Node left = (Node)t.first(); 
            t.remove(left);
 
            Node right = (Node)t.first(); 
            t.remove(right);
 
            t.add(new Branch(left, right));
 
            System.out.println(t.size());
 
        }while(t.size() > 1);
 
        System.out.println(t.first());
 
    }
 
}

I tried adding everything I can think off to the compareTo()...

my input file just has "123456789"
but my output is just 3 nodes.... please help... been at it for about 4 hours...
TIA

[JosAH]

Your class hierarchy is (almost?) identical to mine, so I guess that's quite alright
(he said arrogantly ;-) but your compareTo() method still is incorrect, i.e. you're
still trying to compare something using the *length* of the code. That's not correct.
Here's my version (ripped straight from the code):

    public int compareTo(Object obj) {
        HuffmanNode that= (HuffmanNode)obj;
        if (this.freq == that.freq)
            return this.code-that.code;
 
        return this.freq-that.freq;
    }
 

the 'code' member represents the actual Huffman code constructed so far.
The 'freq' member is (as you can guess) the frequency count of the node.

kind regards,

Jos

[KWSW]

Your class hierarchy is (almost?) identical to mine, so I guess that's quite alright
(he said arrogantly ;-) but your comparyTo() method still is incorrect, i.e. you're
still trying to compare something using the *length* of the code. That's not correct.
Here's my version (ripped straight from the code):

    public int compareTo(Object obj) {
        HuffmanNode that= (HuffmanNode)obj;
        if (this.freq == that.freq)
            return this.code-that.code;
 
        return this.freq-that.freq;
    }
 

the 'code' member represents the actual Huffman code constructed so far.
The 'freq' member is (as you can guess) the frequency count of the node.

kind regards,

Jos

Much appreciated... will give it a try... :)

[KWSW]

Much appreciated... will give it a try... :)

hmm it seems that I cant do that as my code is a string type...

[JosAH]

Much appreciated... will give it a try... :)

Actually it's not much of a problem: it's just that the SortedSet can't store
non-unique keys: it can't deal with different nodes having just equal frequencies,
and are thus considered equal.

If the frequencies of two disjunct nodes differ, everything is fine. Just in case
those freqencies are equal you have to define, what they call, your own PTO
(Partial Total Ordering). *Any* ordering will do as long as it is consistent.

Being the lazy bastard that I am I simply picked the 'code' (which is always
unique) member variable but you can also assign a unique sequence number
to every node you create: 0, 1, 2, 3, ... and compare those:

public int compareTo(Object obj) {
   HuffmanNode that= (HuffmanNode)obj;
   if (this.freq == that.freq)
      return this.seqno-that.seqno;
   else
      return this.freq-that.freq;
}
 

This little trick (as well as my lazy little trick) makes *every* node unique so the
SortedSet can store all of them, i.e. there are no two nodes X and Y for which
X == Y holds. They're either different because of their different frequencies, and
if they're not, they're still different because of some other criteria such as sequence
numbers of because of what I did: compare the unique Huffman codes so far.

kind regards,

Jos

[JosAH]

hmm it seems that I cant do that as my code is a string type...

Use your imagination:

...
return this.code.compareTo(that.code);
...
 

assuming that you're updating your 'code' variable on the fly while you're constructing
that Huffman tree.

kind regards,

Jos

[KWSW]

thanks for all the help today... gonna give it one more shot than get some sleep can carry on tmr... :)

[KWSW]

thanks for all the help today... gonna give it one more shot than get some sleep can carry on tmr... :)

ok this is rather interesting but i added a static var(0) and another one x which is = the static var++.

somehow it works... was playing on the fact that every time a node is created the value of x would increase from the base value of the static var. than just compare the 2 node's value of x and somehow it works...

dunno how it works but thought i just share it here if anyone else is having the same problem as me. would be nice if someone explains it to me as well...

now to figure out how to encode but its time to sleep... nights and many thanks again to JosAH :)

[JosAH]

ok this is rather interesting but i added a static var(0) and another one x which is = the static var++.

somehow it works... was playing on the fact that every time a node is created the value of x would increase from the base value of the static var. than just compare the 2 node's value of x and somehow it works...

dunno how it works but thought i just share it here if anyone else is having the same problem as me. would be nice if someone explains it to me as well...

now to figure out how to encode but its time to sleep... nights and many thanks again to JosAH :)

You're welcome of course; as I wrote above: if the frequencies differ all is fine;
if they don't you need another (consistent) way to make those nodes different.
Here's a small example: A(1) B(2) B(3) B(4) C(5) C(6) D(7) E(8) etc.

The A, B, C, etc represent the frequencies; where they are equal that serial
number (your static counter) makes the difference for the nodes, so no two
nodes are considered equal and that is just what you need for that set.

kind regards,

Jos