huffman code and lossless decomposition prof. sin-min lee department of computer science

Huffman code and Lossless Decomposition

Prof. Sin-Min LeeDepartment of Computer Science

Data Compression• Data discussed so far have used FIXED length for

representation

• For data transfer (in particular), this method is inefficient.

• For speed and storage efficiencies, data symbols should use the minimum number of bits possible for representation.

Data CompressionMethods Used For Compression:

– Encode high probability symbols with fewer bits• Shannon-Fano, Huffman, UNIX compact

– Encode sequences of symbols with location of sequence in a dictionary

• PKZIP, ARC, GIF, UNIX compress, V.42bis

– Lossy compression• JPEG and MPEG

Data CompressionAverage code length

Instead of the length of individual code symbols or words, we want to know the behavior of the complete information source

Data CompressionAverage code lengthAssume that symbols of a source alphabet {a1,a2,

…,aM} are generated with probabilities p1,p2,…,pM

P(ai) = pi (i = 1, 2, …, M)

• Assume that each symbol of the source alphabet is encoded with codes of lengths l1,l2,…,lM

Data Compression

Average code length

Then the Average code length, L, of an information source is given by:

MM plplpl 2211L

Data CompressionVariable Length Bit Codings

Rules:

1. Use minimum number of bitsAND

2. No code is the prefix of another codeAND

3. Enables left-to-right, unambiguous decoding

Data CompressionVariable Length Bit Codings

• No code is a prefix of another

– For example, can’t have ‘A’ map to 10 and ‘B’ map to 100, because 10 is a prefix (the start of) 100.

Data CompressionVariable Length Bit Codings

• Enables left-to-right, unambiguous decoding

– That is, if you see 10, you know it’s ‘A’, not the start of another character.

Data CompressionVariable Length Bit Codings

• Suppose ‘A’ appears 50 times in text, but ‘B’ appears only 10 times

• ASCII coding assigns 8 bits per character, so total bits for ‘A’ and ‘B’ is 60 * 8 = 480

• If ‘A’ gets a 4-bit code and ‘B’ gets a 12-bit code, total is 50 * 4 + 10 * 12 = 320

Data CompressionVariable Length Bit Codings

Example:

Source Symbol

P C1 C2 C3 C4 C5 C6

A 0.6 00 0 0 0 0 0B 0.25 01 10 10 01 10 10C 0.1 10 110 110 011 11 11D 0.05 11 1110 111 111 01 0

Average code length = 1.75

Data CompressionVariable Length Bit Codings

Question:

Is this the best that we can get?

Data CompressionHuffman code

– Constructed by using a code tree, but starting at the leaves

– A compact code constructed using the binary Huffman code construction method

Data CompressionHuffman code Algorithm

① Make a leaf node for each code symbolAdd the generation probability of each symbol to the leaf node

② Take the two leaf nodes with the smallest probability and connect them into a new node

Add 1 or 0 to each of the two branchesThe probability of the new node is the sum of the probabilities of the two connecting nodes

③ If there is only one node left, the code construction is completed. If not, go back to (2)

Data CompressionHuffman code Example

Character (or symbol) frequencies– A : 20% (.20) e.g., ‘A’ occurs 20 times in a

100 character document, 1000 times in a 5000 character document, etc.

– B : 9% (.09)– C : 15% (.15)– D : 11% (.11)– E : 40% (.40)– F : 5% (.05)

• Also works if you use character counts• Must know frequency of every character in the document

C .15

A.20

D.11

F.05

B.09

E.40

Huffman code Example

• Symbols and their associated frequencies.

• Now we combine the two least common symbols (those with the smallest frequencies) to make a new symbol string and corresponding frequency.

Data Compression

C .15

A.20

D.11

F.05

BF.14

B.09

E.40

Data CompressionHuffman code Example

• Here’s the result of combining symbols once.• Now repeat until you’ve combined all the symbols into a single string.

C .15

A.20

D.11

F.05

BF.14

B.09

BFD.25

AC.35

E.40

ABCDF.60

ABCDEF1.0Data Compression

Huffman code Example

• Now assign 0s/1s to each branch

• Codes (reading from top to bottom)– A: 010– B: 0000– C: 011– D: 001– E: 1– F: 0001

• Note– None are prefixes of another

ABCDEF1.0

E.40

C .15

A.20

D.11

F.05

BF.14

AC.35

BFD.25

ABCDF.60

B.09

0

0

0

0

0

1

1

11

1

Data Compression

Average Code Length = ?

Data CompressionHuffman code• There is no unique Huffman code

– Assigning 0 and 1 to the branches is arbitrary– If there are more nodes with the same probability,

it doesn’t matter how they are connected• Every Huffman code has the same average

code length!

Data CompressionHuffman code

Quiz:• Symbols A, B, C, D, E, F are being produced by the

information source with probabilities 0.3, 0.4, 0.06, 0.1, 0.1, 0.04 respectively.

What is the binary Huffman code?1) A = 00, B = 1, C = 0110, D = 0100, E = 0101, F = 01112) A = 00, B = 1, C = 01000, D = 011, E = 0101, F = 010013) A = 11, B = 0, C = 10111, D = 100, E = 1010, F = 10110

Data CompressionHuffman code

Applied extensively:• Network data transfer• MP3 audio format• Gif image format• HDTV• Modelling algorithms

Loss-less Decompositions• Definition: A decomposition of R into (R1, R2) is called

lossless if, for all legal instance of r(R):

• r = R1 (r ) R2 (r )

• In other words, projecting on R1 and R2, and joining back, results in the relation you started with

• Rule: A decomposition of R into (R1, R2) is lossless, iff:

• R1 ∩ R2 R1 or R1 ∩ R2 R2

• in F+.

Exercise

Answer

Dependency-preserving Decompositions

• Is it easy to check if the dependencies in F hold ?

• Okay as long as the dependencies can be checked in the same table.

• Consider R = (A, B, C), and F ={A B, B C}

• 1. Decompose into R1 = (A, B), and R2 = (A, C)

• Lossless ? Yes.

• But, makes it hard to check for B C

• The data is in multiple tables.

• 2. On the other hand, R1 = (A, B), and R2 = (B, C),

• is both lossless and dependency-preserving

• Really ? What about A C ?

• If we can check A B, and B C, A C is implied.

Dependency-preserving Decompositions

• Definition:

• Consider decomposition of R into R1, …, Rn.

• Let Fi be the set of dependencies F + that include

only attributes in Ri.

• The decomposition is dependency preserving, if (F1 F2 … Fn )+ = F +

Example: Decompose Lossless but not dependency preserving

Why ?

BCNF• Given a relation schema R, and a set of functional

dependencies F, if every FD, A B, is either:

• 1. Trivial• 2. A is a superkey of R

• Then, R is in BCNF (Boyce-Codd Normal Form)• Why is BCNF good ?

BCNF

• What if the schema is not in BCNF ?

• Decompose (split) the schema into two pieces.

• Careful: you want the decomposition to be lossless

Example

Achieving BCNF Schemas• For all dependencies A B in F+, check if A is a superkey

• By using attribute closure

• If not, then • Choose a dependency in F+ that breaks the BCNF rules, say A B• Create R1 = A B• Create R2 = A (R – B – A) • Note that: R1 ∩ R2 = A and A AB (= R1), so this is lossless decomposition

• Repeat for R1, and R2• By defining F1+ to be all dependencies in F that contain only attributes in R1• Similarly F2+

Example 1

B C

• R = (A, B, C)• F = {A B, B C}• Candidate keys = {A}

• BCNF = No. B C violates.

• R1 = (B, C)• F1 = {B C}

• Candidate keys = {B}• BCNF = true

• R2 = (A, B)• F2 = {A B}

• Candidate keys = {A}• BCNF = true

Example 2-1

A B

• R = (A, B, C, D, E)• F = {A B, BC D}

• Candidate keys = {ACE}• BCNF = Violated by {A B, BC D} etc…

• R1 = (A, B)• F1 = {A B}

• Candidate keys = {A}• BCNF = true

• R2 = (A, C, D, E)• F2 = {AC D}

• Candidate keys = {ACE}• BCNF = false (AC D)

• From A B and BC D by pseudo-transitivity

AC D

• R3 = (A, C, D)• F3 = {AC D}

• Candidate keys = {AC}• BCNF = true

• R4 = (A, C, E)• F4 = {} [[ only trivial ]]• Candidate keys = {ACE}

• BCNF = true

• Dependency preservation ???• We can check: • A B (R1), AC D (R3), • but we lost BC D• So this is not a dependency• -preserving decomposition

Example 2-2

BC D

• R = (A, B, C, D, E)• F = {A B, BC D}

• Candidate keys = {ACE}• BCNF = Violated by {A B, BC D} etc…

• R1 = (B, C, D)• F1 = {BC D}

• Candidate keys = {BC}• BCNF = true

• R2 = (B, C, A, E)• F2 = {A B}

• Candidate keys = {ACE}• BCNF = false (A B)

A B• R3 = (A, B)

• F3 = {A B}• Candidate keys = {A}

• BCNF = true

• R4 = (A, C, E)• F4 = {} [[ only trivial ]]• Candidate keys = {ACE}

• BCNF = true

• Dependency preservation ???• We can check: • BC D (R1), A B (R3), • Dependency-preserving• decomposition

Example 3

A BC

• R = (A, B, C, D, E, H)• F = {A BC, E HA}• Candidate keys = {DE}

• BCNF = Violated by {A BC} etc…

• R1 = (A, B, C)• F1 = {A BC}

• Candidate keys = {A}• BCNF = true

• R2 = (A, D, E, H)• F2 = {E HA}

• Candidate keys = {DE}• BCNF = false (E HA)

E HA

• R3 = (E, H, A)• F3 = {E HA}

• Candidate keys = {E}• BCNF = true

• R4 = (ED)• F4 = {} [[ only

trivial ]]• Candidate keys = {DE}

• BCNF = true

• Dependency preservation ???• We can check: • A BC (R1), E HA (R3), • Dependency-preserving• decomposition