How can I create a tree for Huffman encoding and decoding?

Question

For my assignment, I am to do a encode and decode for huffman trees. I have a problem creating my tree, and I am stuck.

Don't mind the print statements - they are just for me to test and see what the output is when my function runs.

For the first for loop, I got all the values and index from the text file I used in my main block for testing.

In the second for loop I inserted all the stuff into the priority queue.

I am so stuck about where to go next - I'm trying to make nodes, but I am confused about how to progress. Can someone tell me if I'm doing this right?

def _create_code(self, frequencies):
    '''(HuffmanCoder, sequence(int)) -> NoneType
    iterate over index into the sequence keeping it 256 elements long, '''
    #fix docstring
    p = PriorityQueue()
    print frequencies

    index = 0 
    for value in frequencies:
        if value != 0:
            print value #priority
            print index #elm
            print '-----------'       
        index = index + 1


    for i in range(len(frequencies)):
        if frequencies[i] != 0:
            p.insert(i, frequencies[i])  
            print i,frequencies[i]
            if p.is_empty():
                a = p.get_min()
                b = p.get_min()
                n1 = self.HuffmanNode(None, None, a)
                n2 = self.HuffmanNode(None, None, b)
                print a, b, n1, n2
    while not p.is_empty():
        p.get_min()

I manually inserted the first two to start my tree, is that correct?

How do I keep going? I know the idea of it, just code-wise I am very stuck.

This is using python by the way. I tried looking at Wikipedia, I know the steps, I just need help on code and how I should keep going, thanks!

The HuffmanNode comes from this nested class:

class HuffmanNode(object):

    def __init__(self, left=None, right=None, root=None):
        self.left = left
        self.right = right
        self.root = root

Grill

how so ?recursion is so confusing to me, can u sorta guide me thru it? – Grill

Actually, I think an iterative solution is pretty simple. Here are two such algorithms given on the wikipedia article Huffman Encoding. – Cassilda

Answer 1

The Huffman algorithm in Wikipedia tells you exactly how to create the node tree, so your program can be based on that algorithm, or another like it. Here is a Python program with comments showing the corresponding wikipedia algorithm step. The test data is frequencies of the letters of the alphabet in English text.

Once the node tree is created, you need to walk it down to assign Huffman codes to each symbol in your dataset. Since this is homework, that step is up to you, but a recursive algorithm is the simplest and most natural way to handle it. It's only six more lines of code.

import queue

class HuffmanNode(object):
    def __init__(self, left=None, right=None, root=None):
        self.left = left
        self.right = right
        self.root = root     # Why?  Not needed for anything.
    def children(self):
        return((self.left, self.right))

freq = [
    (8.167, 'a'), (1.492, 'b'), (2.782, 'c'), (4.253, 'd'),
    (12.702, 'e'),(2.228, 'f'), (2.015, 'g'), (6.094, 'h'),
    (6.966, 'i'), (0.153, 'j'), (0.747, 'k'), (4.025, 'l'),
    (2.406, 'm'), (6.749, 'n'), (7.507, 'o'), (1.929, 'p'), 
    (0.095, 'q'), (5.987, 'r'), (6.327, 's'), (9.056, 't'), 
    (2.758, 'u'), (1.037, 'v'), (2.365, 'w'), (0.150, 'x'),
    (1.974, 'y'), (0.074, 'z') ]

def create_tree(frequencies):
    p = queue.PriorityQueue()
    for value in frequencies:    # 1. Create a leaf node for each symbol
        p.put(value)             #    and add it to the priority queue
    while p.qsize() > 1:         # 2. While there is more than one node
        l, r = p.get(), p.get()  # 2a. remove two highest nodes
        node = HuffmanNode(l, r) # 2b. create internal node with children
        p.put((l[0]+r[0], node)) # 2c. add new node to queue      
    return p.get()               # 3. tree is complete - return root node

node = create_tree(freq)
print(node)

# Recursively walk the tree down to the leaves,
#   assigning a code value to each symbol
def walk_tree(node, prefix="", code={}):
    return(code)

code = walk_tree(node)
for i in sorted(freq, reverse=True):
    print(i[1], '{:6.2f}'.format(i[0]), code[i[1]])

When run on the alphabet data, the resulting Huffman codes are:

e  12.70 100
t   9.06 000
a   8.17 1110
o   7.51 1101
i   6.97 1011
n   6.75 1010
s   6.33 0111
h   6.09 0110
r   5.99 0101
d   4.25 11111
l   4.03 11110
c   2.78 01001
u   2.76 01000
m   2.41 00111
w   2.37 00110
f   2.23 00100
g   2.02 110011
y   1.97 110010
p   1.93 110001
b   1.49 110000
v   1.04 001010
k   0.75 0010111
j   0.15 001011011
x   0.15 001011010
q   0.10 001011001
z   0.07 001011000

Bogbean

@Dave, I test your code, but I got an error that:print(i[1], '{:6.2f}'.format(i[0]), code[i[1]]) KeyError: 'e' – Ningpo

This does not account for the lexicographical ordering of nodes with equal frequencies. – Kenwrick

Answer 2

One more solution returning a dictionary {label:code} and a recursive dictionary tree containing the resulting graph. The input vals is in form of dictionary {label:freq}:

def assign_code(nodes, label, result, prefix = ''):    
    childs = nodes[label]     
    tree = {}
    if len(childs) == 2:
        tree['0'] = assign_code(nodes, childs[0], result, prefix+'0')
        tree['1'] = assign_code(nodes, childs[1], result, prefix+'1')     
        return tree
    else:
        result[label] = prefix
        return label

def Huffman_code(_vals):    
    vals = _vals.copy()
    nodes = {}
    for n in vals.keys(): # leafs initialization
        nodes[n] = []

    while len(vals) > 1: # binary tree creation
        s_vals = sorted(vals.items(), key=lambda x:x[1]) 
        a1 = s_vals[0][0]
        a2 = s_vals[1][0]
        vals[a1+a2] = vals.pop(a1) + vals.pop(a2)
        nodes[a1+a2] = [a1, a2]        
    code = {}
    root = a1+a2
    tree = {}
    tree = assign_code(nodes, root, code)   # assignment of the code for the given binary tree      
    return code, tree

This can be used as:

freq = [
(8.167, 'a'), (1.492, 'b'), (2.782, 'c'), (4.253, 'd'),
(12.702, 'e'),(2.228, 'f'), (2.015, 'g'), (6.094, 'h'),
(6.966, 'i'), (0.153, 'j'), (0.747, 'k'), (4.025, 'l'),
(2.406, 'm'), (6.749, 'n'), (7.507, 'o'), (1.929, 'p'), 
(0.095, 'q'), (5.987, 'r'), (6.327, 's'), (9.056, 't'), 
(2.758, 'u'), (1.037, 'v'), (2.365, 'w'), (0.150, 'x'),
(1.974, 'y'), (0.074, 'z') ]    
vals = {l:v for (v,l) in freq}
code, tree = Huffman_code(vals)

text = 'hello' # text to encode
encoded = ''.join([code[t] for t in text])
print('Encoded text:',encoded)

decoded = []
i = 0
while i < len(encoded): # decoding using the binary graph
    ch = encoded[i]  
    act = tree[ch]
    while not isinstance(act, str):
        i += 1
        ch = encoded[i]  
        act = act[ch]        
    decoded.append(act)          
    i += 1

print('Decoded text:',''.join(decoded))

One can visualize the tree with Graphviz as:

The figure was generated by the following script as (Graphviz is needed):

def draw_tree(tree, prefix = ''):    
    if isinstance(tree, str):            
        descr = 'N%s [label="%s:%s", fontcolor=blue, fontsize=16, width=2, shape=box];\n'%(prefix, tree, prefix)
    else: # Node description
        descr = 'N%s [label="%s"];\n'%(prefix, prefix)
        for child in tree.keys():
            descr += draw_tree(tree[child], prefix = prefix+child)
            descr += 'N%s -> N%s;\n'%(prefix,prefix+child)
    return descr

import subprocess
with open('graph.dot','w') as f:
    f.write('digraph G {\n')
    f.write(draw_tree(tree))
    f.write('}') 
subprocess.call('dot -Tpng graph.dot -o graph.png', shell=True)

Erebus

This is awesome! @Pavel. Together with the Graphiz script, this makes a good answer – Changchun

Answer 3

@Dave walk_tree is missing tree processing code

# Recursively walk the tree down to the leaves,
# assigning a code value to each symbol
def walk_tree(node, prefix="", code={}):
    if isinstance(node[1].left[1], HuffmanNode):
        walk_tree(node[1].left,prefix+"0", code)
    else:
        code[node[1].left[1]]=prefix+"0"
    if isinstance(node[1].right[1],HuffmanNode):
        walk_tree(node[1].right,prefix+"1", code)
    else:
        code[node[1].right[1]]=prefix+"1"
    return(code)

Leland

Answer 4

@Dave class HuffmanNode(object) has a subtle bug. When the two frequencies are equal an exception is thrown: For example let

    freq = [ (200/3101, 'd'), (100/3101, 'e'), (100/3101, 'f') ]

Then you get TypeError: unorderable types: HuffmanNode() < str(). The problem has to do with the PriorityQueue implementation. I suspect when the first elements of the tuples compare equal, PriorityQueue wants to compare the second elements one of which is a python object. You add the lt method to your class and the problem is solved.

    def __lt__(self,other):
        return 0

Sergent

Answer 5

I was working out this problem today, to try and match results in above response. For most part, this solution works well but only thing i find non-intuitive is to add [0] and [1] in printing the non-node (leaf). But this answers miracles question - you can essentially print it using any traversal mechanism

import queue

class HuffmanNode(object):
    def __init__(self,left=None,right=None,root=None):
        self.left = left
        self.right = right
        self.root = root
    def children(self):
        return (self.left,self.right)
    def preorder(self,path=None):
        if path is None:
            path = []
        if self.left is not None:
            if isinstance(self.left[1], HuffmanNode):
                self.left[1].preorder(path+[0])
            else:
                print(self.left,path+[0])
        if self.right is not None:
            if isinstance(self.right[1], HuffmanNode):
                self.right[1].preorder(path+[1])
            else:
                print(self.right,path+[1])

freq = [
    (8.167, 'a'), (1.492, 'b'), (2.782, 'c'), (4.253, 'd'),
    (12.702, 'e'),(2.228, 'f'), (2.015, 'g'), (6.094, 'h'),
    (6.966, 'i'), (0.153, 'j'), (0.747, 'k'), (4.025, 'l'),
    (2.406, 'm'), (6.749, 'n'), (7.507, 'o'), (1.929, 'p'), 
    (0.095, 'q'), (5.987, 'r'), (6.327, 's'), (9.056, 't'), 
    (2.758, 'u'), (1.037, 'v'), (2.365, 'w'), (0.150, 'x'),
    (1.974, 'y'), (0.074, 'z') ]

def encode(frequencies):
    p = queue.PriorityQueue()
    for item in frequencies:
        p.put(item)

    #invariant that order is ascending in the priority queue
    #p.size() gives list of elements
    while p.qsize() > 1:
        left,right = p.get(),p.get()
        node = HuffmanNode(left,right)
        p.put((left[0]+right[0],node))
    return p.get()

node = encode(freq)
print(node[1].preorder())

Query