Lecture 4 - Huffman Encoding

Question 1

What kind of a text compression method if huffman encoding?

Answer 1

Statistical

Question 2

What is each character replaced by?

Answer 2

A variable length code

Question 3

What are characters represented by?

Answer 3

Unique codewords of varying lengths

Question 4

What is special about frequently occuring characters?

Answer 4

They will be represented by a shorter code word than those that are less frequent.

Question 5

Can a codeword be a prefix of another codeword? WHY?

Answer 5

No
- This would give ambigous decompression

Question 6

What is this method based on?

Answer 6

The Huffman Tree.

Question 7

What is a huffman tree?

Answer 7

A binary tree where each character is represented by a leaf node and the codeword for a character is given by the path from the root to the leaf.

Question 8

Bit code for huffman tree.

Answer 8

left = 0
right = 1
- the prefix property follows from this

Question 9

Steps for building a Huffman Tree?

Answer 9

• add leaves (one per char)
• add parent to parentless nodes of smallest weight
• weight of new node is equal to sum of weights of the child nodes

Question 10

What is the weighted path length of a Huffman tree?

Answer 10

sum of (weight * distance to root) for each leaf

Question 11

What is special about a Huffman tree in relation to WPL?

Answer 11

Huffman trees have a minimum WPL over all binary trees with the given leaf nodes

Question 12

Does a Huffman tree need to be unique?

Answer 12

No, there can be many solutions that are optimal.

Question 13

Why do we care about WPL in relation to compression?

Answer 13

This is because WPL is the number of bits in the compressed file

-bits = sum over chars (frequency of char × code length of char)

Question 14

What is the complexity of building a Huffman tree?

Answer 14

O(n + mlogm) overall

• O(n) to find frequencies
• O(m logm) to construct the code
  as it takes O(m) to build tree and O(log m) to insert/remove elements to tree
• there are m-1 iterations before heap is empty

Question 15

What is the complexity of building a tree if m (number of distinct chars) is treated as a constant.

Answer 15

O(m)

Question 16

What does compression use?

Answer 16

A code table (array of codes indexed by char). It uses the built tree to get path

-> O (mlogm) to build the table as m characters so m paths of length <=log m

-> O(n) to compress. n characters in the text so n O(1) lookups

Compression is O(mlogm + n)

Question 17

What does decompression use?

Answer 17

Uses the tree directly, which means decompression is O(nlogm).

This is beacause each codeword is replaced by a char found in the Huffman Tree.

Question 18

If we assume m is a constant what is the time complexity of decompression and compression.

Answer 18

O(n)

Question 19

What is a problem with Huffman Encoding?

Answer 19

That the huffman tree must be stored with a compressed file.

Question 20

Why must a huffman tree be stored with the compressed file?

Answer 20

As otherwise decompression would be impossible.

Question 21

What are the alternatives to Huffman Encoding?

Answer 21

• use a fixed set of frequencies based on a typical values for text (will likely reduce compression ratio)
• use adaptive huffman coding: the same tree is built and adapted by the compressor and by the decompressor as characters are encoded/decoded (likely to slow down compression and decompression)