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- Let the following integers represent lengths of codewords: 1,2,3,4,4
- Can we produce a prefix code with the codewords of the given lengths
- Construct the code, i.e., obtain the codewords

Dr.E.Regentova

- Check Kraft-McMillan’s inequality. If holds, then an instantaneous prefix code can be designed.
- Design a binary tree with a number of levels given by the maximum length of codewords in the list.
- Each time you designate a node in the binary tree for a codeword, truncate the tree, i.e., make the node a leaf node.
- Traverse the tree from its root to the leaves assigning to the left move (branch) 0, and to the right move (1) or v.v.

Dr.E.Regentova

The optimal code for a source S has an average length lav

Dr.E.Regentova

- Symbols that occur more frequently will have shorter codewords.
- Lemma: In an optimum code, two symbols that occur least frequently will have the same length.
- Two less probable symbols of the alphabet differ in the last bit.

Dr.E.Regentova

- Consider the following alphabet
P(a1)=P(a3)=0.2;

P(a2)=0.4;P(a5)=0.1

- Sort them by descending order of probabilities.

Dr.E.Regentova

- For two less probable symbols assign codewords as
C(a4)=β1*0(1)

C(a5)=β1*1, where * is concatenation.

- Combine two least probable symbols into one (denote a45)with a probability being equal to the sum of probabilities of these two symbols. This way, we design a new alphabet.
- Resort the list.

Dr.E.Regentova

Dr.E.Regentova

- This time, two less probable symbols are assigned
c(a3) = β2*0

c(a45)= β2*1

- But, c(a45)= C(a4)+ C(a5)= β1*0+β1*1= β1, hence β1 = β2*1
- We can rewrite
c(a4)= β1*0= (β2*1) *0= β2*10

c(a5)=β1*1= (β2*1) *1= β2*11

- Combine again to obtain a new symbol a345 in the new alphabet. Resort the list.

Dr.E.Regentova

Dr.E.Regentova

- Combine again two less probable symbols, i.e, a345 and a1, assigning c(a3451)= β3
- Then
c(a345)= β3*0= β2

c(a1)=β3*1

- We can rewrite codewords as
c(a3) =β2*0 = (β3*0)*0= β3*00

c(a4)= β2*10= β3*010

c(a5)=β2*11= (β3*0) *11= β3*011

c(a1)=β3*1

Dr.E.Regentova

Since there are only two letters is the alphabet, we assign β3=0, and c(a2)=1

Dr.E.Regentova

- Assign
c(a3451)= β3 = 0

c(a2)=1

c(a1)= β3 *1=01

c(a3) = β3*00=000

c(a4)= β3*010=0010

c(a5)= β3*011=0011

Dr.E.Regentova

- Lav = 2.2b/letter
- H= 2.278 b/letter
- Redundancy :R=0.078 b/symbol
- Variance ?

Dr.E.Regentova

- When resorting probabilities, put the combined letter always before any other letter of the same probability.
- This way, the corresponding code will be of a lower variance.

Dr.E.Regentova

Instead of

Obtain this

Dr.E.Regentova

Dr.E.Regentova

Step 1: G+H = H 1 = 0.010; Step 2: E+F = F1 = 0.018

Step 3: F1 + H 1 = H 11 = 0.028;Step 4: H 11 + D = H 111 = 0.109

Step 5: B+C = C1 = 0.162; Step 6: H 111 + C1 = H 1Y =0.271

Dr.E.Regentova

Dr.E.Regentova

Dr.E.Regentova

l = 1.6 bits/symbol

H= 1.22 b/symbol

Redundancy: 0.62b/symbol

Dr.E.Regentova

Code rate for the Huffman code

For Huffman code, it is proved that if

Pmax ≥ 0.5,

the upper bound is

H(S) +Pmax;

If Pmax ≤, 0.5,

the upper bound is

H(S) +Pmax+0.086;

Dr. E.Regentova