Stonehill College
Data Structures and Discrete Mathematics Learning Community

Professors Ralph Bravaco and Shai Simonson
Text Compression and Huffman Trees

Consider a simple “message” (or file) which is stored as a string of characters using the ASCII code.  Each character in the message requires eight bits of storage.  For example character  “A,” (ASCII code 65)  would be stored as 010000001, character “B” (ASCII 66)  would bet stored as 010000010 etc.  In fact, this very paragraph  would require 3472 bits of storage (434 characters and eight bits for each character).  

Data compression
is a technique which reduces the size of a file or data collection without affecting the information contained in the file.  A compressed file, obviously, requires less storage space.  Moreover, electronic transfer of a compressed file is faster than the transfer of an uncompressed file and a smaller file is consequently less susceptible to transmission errors. 

One of the simplest but efficient methods of data compression was introduced by David Huffman in his paper ``A method for the construction of minimum redundancy codes.''  Huffman’s method is very intuitive.  Unlike the ASCII code which uses exactly eight bits to encode each character, a Huffman code uses a variable number of bits for each character. With Huffman’s scheme, a common character such as “a” or “e” might be encoded with  perhaps just two bits and less frequently used characters like “q” or “#” would be encoded with a longer bit string.  Consequently, the most frequently occurring characters are encoded with a minimal number of bits.  (Morse code uses the same technique, “E” is encodes as a single  dot, “T”  just one dash,  but “Q” is dash-dash-dot-dash). Because ASCII  encodes every character with eight bits we say that ASCII code is a fixed length code.  Huffman’s code is a variable length code.

We’ll begin with a trivial example. 

Suppose we wish to encode the string
The ASCII code would require 160 bits (20 characters each requiring 8 bits).
Notice, however, that the  characters which appear in the string have the following frequencies ( and relative frequencies)
A         1          .05
H         2          .10
-           3          .15
E          4          .20
L          4          .20
S          6          .30

A possible variable length code which uses just 2 bits to encode frequently used characters like “S” or “L “ might be  :

            Character         Codeword        Frequency of Character
A                     0000                1
H                     0001                2
 -                      001                  3
E                      10                    4
L                      11                    4
S                      01                    6
We will refer to this code as Code 1.   We will also use the term “codeword” to denote an individual character code.  Thus, 0000 and 001 are codewords.

Notice that “S” which occurs 6 times (or 30% of the time) is encoded using just 2 bits and “A” which occurs only once is encoded with 4 bits.

With this code, the string,  “SHE-SELLS-SEA-SHELLS,” can be encoded with just 49 bits:


You may wonder why we didn’t use a “more efficient” code like:
S          0  
L          1
E          00
-           11
H         01
A         10
( Code 2)

Using Code 2 we can encode the string “SHE-SELLS-SEA-SHELLS” as


which uses a mere 30 bits.  Certainly, compressing the string to 30 bits is more efficient than storing it as 49 bits.  So, why not use Code 2?  It does the job with fewer bits. 

To answer this question, consider just the simple substring, “SHE.”  Using  the ASCII coding scheme, “SHE” is encoded  as


Now, if we wish to decode (or translate this bit string back to ordinary, readable  text), we simply look at each successive group of exactly 8 bits:
01010011 -  (83)  – “S”
01001000 -  (72)  – “H”
01000101 -  (69)  – “E”
That’s all there is to it.

Now, let’s encode “SHE” using our “more efficient” Code 2:


Wow! Just 5 bits! 

Now, given this five bit string, how do we decode it?   Do we decode 00100 as  “EAS”      (00  10 0)  or  as “ SSLE”  ( 0 0  1 00 ) or perhaps as “SSLSS”  (0 0  1 0 0) ?   You really cannot tell without putting some marker between the codewords for each letter.

On the other hand, using Code 1 we encode “SHE” as


To decode this bit string , simply  scan the string from left to right  until you find a complete codeword.  You have just decoded the first character.  Continue the process until you have scanned the entire bit string.  You will see there is no ambiguity.
01    0001    10
S      H         E

Exercise 1.

            a. Decode each of the following bit strings which are encoded with Code 1:



            b. Now, using the following bit string, show that Code 2 is ambiguous.  In other words, find two possible translations of the following bit string:


            c. Explain why Code 1 is not ambiguous.  What property does Code 1 have which ensures that each bit string has a unique translation?

Prefix Codes

For arbitrary variable length codes, we may not be able determine the end of one codeword .  For example, using Code 2, do we interpret the bit string 0101 as 0.10.1 (SAL) or 01.01 (HH)  or (SLSL)?  Is the initial letter encoded as 0 (S)  or 01 (H)? Without some kind of marker between codewords, it is impossible to determine where each codeword ends.  Code 1 however, does not have this drawback.  The bit string 00010000011 can have only one interpretation:

 Let’s look at this string.  As we scan from left to right we see:
0 is not a valid code.  Continue scanning
00 is not a valid code. Continue
000 is not a valid code. Continue
0001 is the code for H. Done.

OK, we have unearthed an “H.” But how do we know that there is no code that starts with 0001 ?  How do we know that 00011 is not the code for some other letter? 

If you look closely at the codewords of Code 1, you will notice that no codeword is a prefix of another codeword.  The codeword for “H” is 0001.  No other codeword starts with 0001.  The codeword for “S” is 01.  Nothing else begins with 01.  Thus, using a left-right scan, once we read 0001 we recognize the codeword for “H.”  Further, we know that 0001 is not the start of the codeword for some other character, since no other character code begins with 0001.

            A code is a prefix code  ( or has the prefix property)  if no codeword is the prefix of another codeword.

Again, Code 1 has the prefix property but Code 2 does not.   The codes which we will develop will have the prefix property. 

A Huffman code turns out to be an “optimal” prefix code.  We will be more precise about the meaning of “optimal” a bit later.  First, let’s see how the Huffman code is constructed.
Once we look at the mechanics, we will take a closer look underneath the hood.

Codes and Binary Trees:
Let’s look at Code 1 again:
            Character         Codeword        Frequency        Relative Frequency
A                     0000                1                                  .05
H                     0001                2                                  .10
 -                     001                  3                                  .15
E                      10                    4                                  .20
L                      11                    4                                  .20
S                      01                    6                                  .30
                                                20                                1.00

We can represent this code with a binary tree.  On such a tree,
each left branch is labeled 0,
each right branch is labeled 1
the encoded characters are placed in the leaves

Notice that the concatenation of the branch labels on the path from the root to any leaf gives the code word for the symbol stored in the leaf.  For example, the path from the root to the leaf labeled “H’ is left-left-left-right and the codeword for “H” is 0001.  The path to “L” is
right-right , so the codeword for “L” is 11.

Huffman’s Algorithm constructs such a binary tree in a bottom up manner. The Huffman tree yields an optimal prefix code  for the symbols contained in the leaves.

Building the Huffman Tree.

Given the frequencies of each character in a file, the Huffman tree can be constructed as follows:

For each symbol to be encoded, make the binary tree consisting of a single root.  Each of these  root nodes should hold  the frequency (or relative frequency)  of one symbol. Ultimately, these nodes will be the leaves of  our Huffman tree.  This can be implemented as a priority queue.  The priorities are the frequencies.

2. Choose the two trees with the smallest root values and merge these  into a new tree, as in the following diagram. Here the trees labeled A and H are chosen. Priorities/frequencies are 1 and 2. .  The root of the new tree holds the value 3 , obtained as the sum of the values 1 and 2. Place this in the queue.

3.  Continue the process, always forming a new tree from the two trees with the lowest root values. Here we merge the two trees each with 3 stored in the root into a tree with root value 6.  Place the new tree in the queue.  Four trees remain  in the priority queue with root values 4,4,6 and 6.

4.  Continue this process..    Remember always merge the two subtrees which have the smallest root values.  These are the two trees with root values 4 and 4.  Merge them into a single tree with root 8 and place the new tree in the queue.

Now merge the two trees with root values 6 into a tree with root 12. Place the new tree in the queue so that  two trees remain:

Merge the two trees into one final tree. Here  is the final tree, with each branch labeled either 0 or 1.

The following is a general algorithm:      
            For each character c to be encoded
                          create a binary tree, consisting of a single root node r such that  content (r) =  f(c), the frequency of c
Place each very simple tree  into a priority queue S where the priority is the frequency.
 for i = 1 to n-1 
        From the priority queue ,S,  remove the two trees, x and y, with minimal values, f1 and f2 in the root
        create a tree with root z such that
                x  is the left child of z
                y is the right child of z
          content(z) = f1  + f2
add this new tree with root z to S.

Using the Huffman tree to determine the codewords.
Once the Huffman tree is created, it is easy to determine the codeword for each character.

Method 1: Bottom Up
As we’ve seen, each leaf represents one character.  If we begin at a leaf and proceed up ttree to the root, we can “trace out”  (in reverse) the codeword for a given symbol:
For each leaf a
       while a is not the root of the tree
            if a is a left subtree
                   output ‘0’
                   output ‘1’
            a = parent(a)

Notice that the algorithm will output codewords in reverse order. 

Let a be leaf “ -.”

a is a right subtree; output  ‘1’
a = parent(a)

a is a left subtree; output ‘0’
a = parent(a)

a is a left subtree; output ‘0’
a = parent(a)

a is the root; done

Output: 100; codeword  for “-“ is 001

Method 2 : Top Down -- this is easier

An ordinary, recursive tree traversal with an additional parameter can also be used to determine the codewords:
                                    void traverse(string s, tree root)
Each time that the function is called recursively, pass it string  s concatenated with either ‘0’ or ‘1.’ When visiting a leaf, simply output string  s –  s  is the codeword for character associated with the leaf. 

Encoding the file:
Simply translate each character using a table of codewords:

Character         Codeword       
A                     0000               
H                     0001               
 -                     001                 
E                      10                   
L                      11                   
S                      01

Translate: “SHE-SELLS.......”

S  H  E    -    S E L L S

Decoding a compressed file:
When decoding a compressed file, you will once again use the Huffman tree. Consequently, the frequency table must once again be available.  When the tree is constructed, the text characters are stored in the leaves.  Each codeword, determines a unique path from the root of the tree to to the leaf holding the translation of the code word.  Once we reach a leaf, we begin the decoding process again at the root.

Example:  Decode : 0100011.....

Begin at  the root:

100011 –  move left

0100011 – move right

A leaf—“S”  (01 is the codeword for S)
Go back to the root and continue the process.
0100011  etc.

Exercise 2:

            a. Construct a Huffman tree for the alphabet A B C D E F where the relative frequencies of each character are:

                    A         .22
                    B         .13
                    C         .33
                    D         .10
                    E          .20
                    F          .02

            b. Use the tree to determine the code word for each of the character of this short alphabet

            c. Encode the string “AECBCAF”

Program 1.  Encoding a text file.

Write a program which will encode a text file using the Huffman algorithm. 
The user interface should be very simple.  Just provide the user with two prompts:
             Input file:
             Output file:
so that the user  may supply the name of the input (text) file as well as a name for the compressed file.  Your output should actually be two files : the compressed file, and also a file with the frequency table .  This second file should have the same name as the compressed file and an extension of feq. For example if the compressed file is called then the second file should be something.feq

To encode text using a Huffman code you must
    First build a frequency chart for all characters in your text
    Build a Huffman tree
    Make a list of all codewords
    Using the list of codewords, encode your text.

Here are a few suggestions to help you with the implementation. 

Since ASCII values fall in the range  0 ...127, use a simple array (int freq[128]) to store frequencies.
For example, if 'A' has frequency 13, then freq[65] will have the value 13. Of course, not all ASCII characters will appear in your text, so many entries in the frequency array will be 0. 

When you build the Huffman tree, you must make a number of design decisions.
As you know, the algorithm builds the tree incrementally.  At each step, the sub-trees with minimum roots must be chosen.  The simplest method for determining the these two minimum valued trees would be to use a priority queue. 

If you plan to build your table of codewords in a bottom up fashion, you will need to maintain a table which holds the address of each leaf on the tree.  Also, each node must contain the address of its parent.  The top down method is easier and is really just an inorder traversal which keeps track of the codewords as you do the traversal.

Simply keep the codewords in an array indexed from 0 to 127 so that you may access the a character’s codeword directly, i.e., without a search.  Save your  frequency table to a file – simply list the 128 frequencies in the array.  

Program 2: Decoding a compressed file.

Write a program which will decode a compressed file. Your program should prompt the user for two file names:
              Compressed  file:
               Text file:
If the name of the compressed file is, your program should use that file and also the file something.feq as input. 

Exercise 3:

Use the programs in the previous exercises to compress( and decode)
a. The Declaration of Independence
b. The US Constitution
c. Shakespeare’s Hamlet
The text of these files can be downloaded.

Real Life Compression

For convenience, you encoded your text into characters (bytes), with each byte equal to either ‘0’ or ‘1’. However,
storing each bit in its own byte will result in a file 8 times larger than it needs to be.  In order to realize the true "real-life"compression, you must convert the bytes into bits.  To do this, every eight '0' or '1' characters (bytes) is packed into a single byte.  Pseudo-code to accomplish this is given below:
While there are more 0/1 characters (bytes)
    byte temp = 0;
     for i = 1 to 8 
//  This loop calculates the decimal value of 8 zeros and ones and stores it in temp.  If the number of bytes in the file is not divisible by 8, just pad the file with a few extra bytes.  
        b = getnext_0_or_1_character();  // get the next bit (stored in a byte)
        temp = temp* 2 + b;                    // note that b should be 0 or 1 (not the ascii value of b which would be 48 or 49)

          // Note that to do this correctly in Java, you will need to cast the expressions to bytes, since Java automatically up-casts bytes to int when doing arithmetic
    // store temp in a byte array

  For example, given the bits 10010011, the byte "temp" grows from 0 like this: 0, 1, 2, 4, 9, 18, 36, 73, 147. 
 These numbers represent the decimal value of the 8-bit binary number 10010011 as it grows from left to right. 
(Note that Java stores byte values as two's complement signed numbers, so if you try to print out these bytes, it will display negative numbers for any unsigned value greater than 127.  This will not affect your compression at all.)

FileOutputStream fos = new FileOutputStream("pathname");

or follow instructions  here:

Compete the following chart:

                                                            SIZE IN BYTES


Text file



Declaration of Independence












You do not need to write any programs to find these values.  You can get all this information via Windows and Word.

Huffman Codes—mathematical questions.

 Now that we have looked at the Huffman algorithm, let’s take a closer look at some of the mathematical properties of a Huffman tree.

Exercise 4:

We have stated that the Huffman’s Algorithm produces a code with the prefix property.  Explain why this is the case.  You do not have to give a rigorous mathematical proof—just think about how the tree is constructed and give a plausible explanation.  A few sentences will suffice.

Optimal codes

We have stated the the Huffman code is  “optimal” so perhaps it is now time to give a more precise definition of this term.

We first begin by defining the cost of the code or equivalently the tree.

Definition.   Suppose T is a tree representing a prefix code for a file.  The cost of T, C(T),  is the number of bits necessary to encode the file.  Thus

C(T) = Sum[ frequency(c)*length(c)] for each character c

where length(c) is the length of the path from the root of T to the leaf representing c.
Note: length(c) is also the number of bits in the codeword for character c.

Consider the following Huffman tree, T which we built to encode the message “SHE-SELLS-SEA-SHELLS”

C(T) = frequency(‘A’)length(‘A’)+ frequency(‘H’)length(‘H’) +frequency(‘-’)length(‘-’)+
            frequency(‘S’)length(‘S’)+ frequency(‘E’)length(‘E’)+ frequency(‘L’)length(‘L’) =

1*4 + 2*4+3*3 + 6*2 + 4*2 + 4*2  = 4 + 8 + 9 + 12  + 8 +8 = 49
Definition.     Suppose T is a tree representing a prefix code for a file.  The code is optimal if C(T) is minimal.

Note an optimal code is not unique – for a given file more than one optimal code may exist. (Trivially, just change all 0’s to 1’s and 1’s to 0’s in any optimal code and you have another optimal code).

With a precise definition of "optimal" in hand, we will prove (in the next several exercises) that a Huffman tree is optimal.

Exercise 5:
A full binary tree is a binary tree in which ever non-leaf node has two children.
Notice that the Huffman trees in our examples have been full trees. 

Prove: An  optimal prefix code for a file is always represented by a full binary tree
Hint:  Suppose there is a non-full tree that did represent an optimal code.  Find a contradiction by producing a better code.

Exercise 6:
Show by means of a counter example that a full binary tree may represent a prefix code which is not optimal, even if the letters with lower frequencies have the longer codewords.
Suggestion: Build another full tree for “SHE-SELLS-SEA-SHELLS” which has a higher cost than the tree of our previous examples.

Exercise 7: 
Let x and y be two characters with the lowest frequencies. (Arbitrarily, assume frequency(x) <= frequency(y)).  Prove that there is an optimal tree where x and y  are siblings at the maximum depth of the tree.
Assume that you have an optimal prefix tree, T, such that a and b are at the maximum depth.
(You can assume frequency (a)  <= frequency(b)).  Here is a partial view:

Now, because x and y have the two smallest frequencies frequency(x) <= frequency(a) and frequency(y) <= frequency(b).   Also, length(a) >= length(x) and length(b) >= length(y).
Now switch a and x, resulting in a new tree T’.  Show C(T’) <= C(T)?. Now, in T’,  switch  y and b.  Show  C(T’’) <= C(T’)? 

Exercise 8:
Prove that Huffman’s algorithm produces an optimal prefix code tree.

Outline of proof: Use mathematical induction on n, the number of characters/leaves.
For n = 1, Huffman’s algorithm produces a tree with one node, obviously optimal.
Now assume that for n leaves, Huffman’s algorithm produces an optimal tree.  Show that Huffman’s algorithm produces an optimal tree for n+1 leaves.
Let Tn+1 be an optimal tree with n+1 leaves. We may assume that in an optimal tree the two characters, say x and y, with the lowest frequencies are siblings at the lowest level of the tree. (Why may we assume that?).  Now remove x and y and replace them with a “new character z with frequency(z) = frequency(x) + frequency(y).  Call the new tree Tn.  Let  T’n be the Huffman tree constructed from these n characters   Now finish the proof.

Exercise 9:
The Huffman tree for frequencies equal to the first n Fibonacci numbers results in a very unbalanced encoding of n characters.

Suppose that the frequencies for the characters a, b, c, d, e, f, g, and h are the first 8 Fibonacci numbers: 1,1,2,3,5,8,13,and 21.

Construct the Huffman tree.

What is the cost of the tree?

You found the cost for n = 8.  Now you should determine the general cost of the Huffman tree for n characters when the frequencies are the first n Fibonacci numbers

a.   Discover a formula for the sum of the first n Fibonacci numbers, and prove it by induction..

Fill in this chart to help you get a good conjecture.   Hint: Look at Fibonacci(n + 2).

SumFib(n) = Fibonacci(1) + Fibonacci(2) + ... + Fibonacci(n)





b.  Prove that C(n) = C(n-1) + SumFib(n), where C(1) = 0.

You can prove this by looking carefully at the structure of the tree.  Draw the trees.  The next tree is built inductively from the previous tree.


Fill in this chart to help you with your proof.  Hint:  Line up the sums in the chart, and examine the difference between two rows.

C(n) = Cost of Huffman tree for the first n Fibonacci numbers
0*1   = 0
1*1 + 1*1   = 2
2*1 + 2*1 + 1*2   = 6
3*1 + 3*1 + 2*2 + 1*3 = 13





c.  Using part (a), solve C(n) = C(n-1) + SumFib(n), where C(1) = 0,. by repeated substitution, and write C(n) as a closed form in terms of F(n), for n > 1.

Movie Quiz:

1.  What was Fermat's "day job", i.e., how did he make a living?

2.  Where did Andrew Wiles live when he worked on Fermat's Last Theorem?

3.  The statement of Fermat's Last Theorem was written in the margin of Fermat's copy of Diophantus' book on Arithmetic.  Why didn't Fermat include the actual proof?

4.  How many years elapsed (to the nearest hundred) between the statement of Fermat's Last Theorem and its proof?

5.  What two mathematical subjects that "live on different planets" are bridged by the Taniyama-Shimura Conjecture?

6.  Goro Shimura, as of 2018, is 88 years old.  What happened to Taniyama?

7.  The song "One Way or Another" that plays in the movie is by what female 1980's punk vocal artist?  (singer or group name is acceptable)

8.  What did Andrew Wiles' wife say she wanted for her birthday?

9.  For how many years (within 3) did Wiles work on Fermat's Last Theorem?

10.  Which mathematician was not mentioned in the movie The Proof? John Coates, Andrew Wiles, Peter Sarnak, Ken Ribet, Noam Elkies, Sophie Germain, John Conway, Pierre de Fermat, Goro Shimura, Barry Mazur, Nick Katz.